Text: Physical Chemistry, 7th Edition, Peter Atkins and J. de Paula

Text: Physical Chemistry, 7th Edition, Peter Atkins and J. de Paula
Physical Chemistry I Dr. Robert E. Barletta Text: Physical Chemistry, 7th Edition, Peter Atkins and J. de Paula

Rules of the Road Attendance: encouraged, not mandatory Disabilities
Except for exams(see below) Responsible for any supplemental material covered in lectures Cell phones/pagers off during class Students are expected to remain in class throughout period Disabilities Certify through Office of Special Student Services Help Office: Room 133 Hours: Tues. & Thurs. 11-noon Other - by appointment Homework: Problems assigned at the start of each chapter Due the day after the test on material To received credit for an assignment all work must be shown Exams - A non-programmable calculator only may be used 3 Hour Exams Exam : 1 After Chapter 4 covering Chapters 1, 24a, 2,3, and 4 Exam 2 After Chapter 8 covering Chapters 5-8 Exam 3 After Chapter 26 covering Chapters 9, 10, 24b, 25, and 26 1 Final comprehensive, Ch. 27 and portion to include ACS Thermodynamics test Make-up exams given only for documented excused absences Grading Homework - 5% Hour exams - 15% each Laboratory Grade - 25% Final Exam - 25%

Application of physics to the study of chemistry
Physical Chemistry Application of physics to the study of chemistry Develops rigorous and detailed explanations of central, unifying concepts in chemistry Contains mathematical models that provide quantitative predictions. Mathematical underpinning to concepts applied in analytical, inorganic, organic, and biochemistry Includes essential concepts for studying advanced courses in chemistry Source: American Chemical Society

Divisions of Physical Chemistry
Traditional Approach Equilibrium Thermodynamics Chapters 1-10 Chemical Kinetics Chapters 24, 25, 26, 27 Quantum Theory/ Spectroscopy Statistical Thermodynamics Special Topics Main Problems Position of Chemical Equilibrium A + B <=> C + D Rate of Chemical Reactions - Kinetics Other special topics Approaches Top down (Traditional/Analytical/Historical Approach) Begin with things we observe in the world/laboratory Examine how those observables relate to the underlying structure of matter Bottom up (Synthetic/Molecular Approach) Consider the underlying structure of matter Derive observables

Chapter 1: Properties of Gases
Homework: Exercises (a only) : 1.4,6, 9, 11, 14, 16, 17, 18, 21 Problems: 1.1, 3, 12(a & b only), 20, 32

Equations of State Boundaries between objects
Gases are the simplest state of matter Completely fills any container it occupies Pure gases (single component) or mixtures of components Equation of state - equation that relates the variables defining its physical properties Equation of state for gas: p = f (T,V,n) Gases (pure) Properties - four, however, three specifies system Pressure, p, force per unit area, N/m2 = Pa (pascal) Standard pressure = pø = 105 Pa = 1bar Measured by manometer (open or closed tube), p = pexternal + rgh g = gravitational acceleration = 9.81 m/s-2 Mechanical equilibrium - pressure on either side of movable wall will equalize Volume, V Amount of substance (number of moles), n Temperature, T, indicates direction of flow of energy (heat) between two bodies; change results in change of physical state of object Boundaries between objects Diathermic - heat flows between bodies. Change of state occurs when bodies of different temp. brought into contact Adiabatic - heat flows between bodies. No change of state occurs when bodies of different temp brought into contact

Heat Flow and Thermal Equilibrium
TA = TB High Temp. Low Temp. A B A B A B No Heat Heat No Heat Diathermic Wall Diathermic Wall Adiabatic Wall Thermal equilibrium - no change of state occurs when two objects are in contact through a diathermic boundary Zeroth Law of Thermodynamics - If A is in thermal equilibrium with B and B is in thermal equilibrium with C then A is in thermal equilibrium with C Justifies use of thermometer Temperature scales: Celsius scale, Q, · (°C) degree defined by ice point and B.P. of water Absolute scale, thermodynamic scale , (K not°K) T (K) = Q

Equation of State for Gases ( p = f(V,T,N) Ideal (Perfect) Gas Law
Approximate equation of state for any gas Product of pressure and volume is proportional to product of amount and temperature PV = nRT R, gas constant, JK-1mol -1 R same for all gases, if not gas is not behaving ideally Increasingly exact as P ® 0 a limiting law For fixed n and V, as T ® 0, P ® 0 linearly Special cases (historical precident): Boyle’s Law (1661), Charles’Law [Gay-Lussac’s Law ( )]; Avogodro’s principle (1811) Used to derive a range of relations in thermodynamics Practically important, e.g., at STP (T= , P = pø =1bar), V/n (molar volume) = L/mol For a fixed amount of gas (n, constant) plot of properties of gas give surface Isobar - pressure constant - line, V a T Isotherm - temperature constant, hyperbola, PV = constant Isochor - volume constant - line P a T

Ideal (Perfect) Gas Law - Mixtures
Dalton’s Law: Pressure exerted by a mixture of gases is sum of partial pressures of the gases Partial pressure is pressure component would exhibit if it were in a container of the same volume alone ptotal = pA + pB + pC + pD + ……. (A, B, C, D are individual gases in mixture) pJ V = nJRT This becomes : If xJ is the fraction of the molecule, J, in mixture {xJ = nJ / nTotal ), then S xJ =1 If xJ is the partial pressure of component J in the mixture, pJ = xJ p, where p is the total pressure Component J need not be ideal p = S pJ = S xJ p this is true of all gases, not just ideal gases p = pA + pB pB = xBp P pA = xAp 1 Mole Fraction B, xB

Real Gases - General Observations
Deviations from ideal gas law are particularly important at high pressures and low temperatures (rel. to condensation point of gas) Real gases differ from ideal gases in that there can be interactions between molecules in the gas state Repulsive forces important only when molecules are nearly in contact, i.e. very high pressures Gases at high pressures (spn small), gases less compressible Attractive forces operate at relatively long range (several molecular diameters) Gases at moderate pressures (spn few molecular dia.) are more compressible since attractive forces dominate At low pressures, neither repulsive or attractive forces dominate - ideal behavior

Compression Factor, Z Compression factor, Z, is ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same T & P Z = Vm/ Vm°, where Vm = V/n Using ideal gas law, p Vm = RTZ The compression factor of a gas is a measure of its deviation from ideality Depends on pressure (influence of repulsive or attractive forces) z = 1, ideal behavior z < 1 attractive forces dominate, moderate pressures z > 1 repulsive forces dominate, high pressures

Real Gases - Other Equations of State Virial Equation
Consider carbon dioxide At high temperatures (>50°C) and high molar volumes (Vm > 0.3 L/mol), isotherm looks close to ideal Suggests that behavior of real gases can be approximated using a power series (virial) expansion in n/V (1/Vm) {Kammerlingh-Onnes, 1911} Virial expansions common in physical chemistry CO2

Virial Equation (continued)
Coefficients experimentally determined (see Atkins, Table 1.3) 3rd coefficient less impt than 2nd, etc. B/Vm >> C/Vm2 For mixtures, coeff. depend on mole fractions B = x12B x1x2B12 + x22B22 x1x2B12 represents interaction between gases The compressibility factor, Z, is a function of p (see earlier figure) and T For ideal gas dZ/dp (slope of graph) = 0 Why? For real gas, dZ/dp can be determined using virial equation Substitute for Vm (Vm = Z Vm°); and Vm°=RT/p Slope = B’ + 2pC’+ …. As p ® 0, dZ/dP ® B’, not necessarily 0. Although eqn of state approaches ideal behavior as p ® 0, not all properties of gases do Since Z is also function of T there is a temperature at which Z ® 1 with zero slope - Boyle Temperature, TB At TB , B’ ® 0 and, since remaining terms in virial eqn are small, p Vm = RT for real gas

Critical Constants CO2 2 phases
Consider what happens when you compress a real gas at constant T (move to left from point A) Near A, P increases by Boyle’s Law From B to C deviate from Boyle’s Law, but p still increases At C, pressure stops increasing Liquid appears and two phases present (line CE) Gas present at any point is the vapor pressure of the liquid At E all gas has condensed and now you have liquid As you increase temperature for a real gas, the region where condensation occurs gets smaller and smaller At some temperature, Tc, only one phase exists across the entire range of compression This point corresponds to a certain temperature, Tc, pressure, Pc , and molar volume, Vc , for the system Tc, Pc , Vc are critical constants unique to gas Above critical point one phase exists (super critical fluid), much denser than typical gases CO2 2 phases

Real Gases - Other Equations of State
Virial equation is phenomenolgical, i.e., constants depend on the particular gas and must be determined experimentally Other equations of state based on models for real gases as well as cumulative data on gases Berthelot (1898) Better than van der Waals at pressures not much above 1 atm a is a constant van der Waals (1873) Dieterici (1899)

van der Waals Equation Justification for van der Waals Equation
Repulsion between molecules accounted for by assuming their impenetrable spheres Effective volume of container reduced by a number proportional to the number of molecules times a volume factor larger than the volume of one molecule Thus V becomes (V-nb) b depends on the particular gas He small, Xe large, bXe >bAr Attractive forces act to reduce the pressure Depends on both frequency and force of collisions and proportional to the square of the molar volume (n/V)2 Thus p becomes p + a (n/V)2 a depends on the particular gas He inert, CO2 less so, aCO2 >>aAr

van der Waals Equation - Reliability
Above Tc, fit is good Below Tc, deviations

van der Waal’s Loops (cont.)
CO2 Critical Temperature K (31.05°C) Below Tc, oscillations occur van der Waals loops Unrealistic suggest that increase in p can increase V Replaced with straight lines of equal areas (Maxwell construction)

van der Waals Equation - Reliability
CO2 van der

van der Waals Equation Effect of T and Vm At or below Tc
Ideal gas isotherms obtained 2nd term becomes negligible at high enough T 1st term reduces to ideal gas law at high enough Vm At or below Tc Liquids and gases co-exist Two terms come into balance in magnitude and oscillations occur 1st is repulsive term, 2nd attractive At Tc, we should have an flat inflexion point, i.e., both 1st and 2nd derivatives of equation w.r.t Vm = 0 Solving these equations for p,Vm and T gives pc,Vc and Tc in terms of a and b Hint: you must use original eqn to do this pc= a/27b3, Vc pc= 3b and Tc = 8a/27Rb Critical compression factor, Zc, can be calculated using definition for Z: pVm = RTZ This should be a constant for all gases and is (Table 1.4 )

Comparing Different Gases
Different gases have different values of p, V and T at their critical point You can compare them at any value by creating a reduced variable by dividing by the corresponding critical value preduced = pr = p / pc; Vreduced = Vr = Vm / Vc; Treduced = Tr = T/ Tc This places all gases on the same scale and they behave in a regular fashion; gases at the same reduced volume and temperature exert the same reduced pressure. Law of Corresponding States Independent of equations of state having two variables p (atm) pr

Chapter 2: The First Law: Concepts
Homework: Exercises(a only):4, 6, 9, 18, 24, 25, 26, 29, 30, 35, 40 Problems: 2.2, 2.6, 2.10, 2.24, 2.34

Basic Definitions System - volume of interest (reaction vessel, test tube, biological cell, atmosphere, etc.) Surroundings volume outside system Open system - matter can pass between system & surroundings Closed system - matter cannot pass between system & surroundings Isolated system - Neither matter nor energy can pass between system & surroundings No mechanical or thermal contact surroundings system Energy Matter Open System surroundings system Energy Closed System surroundings system Isolated System

Work, Energy and Heat Work - generalized force (intensive factor) over a generalized displacement (extensive factor) Examples Mechanical work - force x distance; -fdx Expansion work: pressure x volume; -pdV Electrical: emf x charge displacement; EdQ Magnetic: field strength x magnetization; HdM Sign Work done by a system is negative Work done on a system is positive Work is the result of organized motion of molecules Energy is the capacity to do work For an isolated system doing work reduces the energy, having work done on it increases the energy Heat is the change in energy in a system that is produced by a change in its temperature Recall adiabatic boundaries don’t allow heat flow; diathermic do In an adiabatic container endothermic and exothermic processes produce temperature change in the system In diathermic container they result in heat flow across the boundary Heat is the result of disordered (thermal) motion of molecules

First Law of Thermodynamics You can’t win!
Total energy of a system (kinetic + potential) is called the internal energy (U) The change in energy from some initial state, i, to some final state, f, DU is: DU = Uf - Ui Internal energy is a state function Value depends only on current state of the system, not how you got there - path independent Any change in state variable (e.g. pressure) results in a change in U Unit of measure for energy (U, work, heat) is the joule (J) [kg m2 s-2] Calorie (cal) = J eV = 1.6 x J (chemical processes on atomic scale are a few eV) On a molecular scale most of the internal energy of a gas is due to motion of atoms (3/2 kT {~3.7 25°C}from Boltzman distribution) so as temperature is raised U increases Internal energy is changed by either work or heat acting on system If work is done on the system (heat in), U increases If system does work (heat out), U decreases This implies that for an isolated system U is constant {1st Law of Thermodynamics} Mathematical summary of 1st Law: DU = q + w where q = heat transferred to a system and w = work done on the system No perpetual motion machines! (You can’t win!) For an isolated system doing work lowers U Isolated system DU = 0

First Law of Thermodynamics Expansion Work
In differential form, the first law is: dU = dq + dw Expansion work is work done by a change in volume Consider piston at right, change in volume is Apistondz (dV); force is pexternal If pex = 0 (e.g. vacuum),w = 0 If pex = constant (e.g. atmosphere), pex is removed from integral and w = - pex DV Indicator diagram is a plot of p vs V where area between volumes is work Note: other types of work (non-expansion work) have similar formulations (see earlier or Table 2.1). Total work is sum of individual components. pex Area = pex DV P V V1 V2

Reversible Change A reversible process is one that is carried out along a path in which all intermediate states are equilibrium states An equilibrium state is one in which an infinitesimal changes in opposite directions result in opposite changes in state Thermal equilibrium between two systems at the same temperature Reversible changes are typically slow Reversible expansions pex set equal p (pressure of gas) along the path of the expansion. Thus, If we know how p varies with V we can evaluate integral Variation of p with V is the equation of state of the gas [e.g., p =nRT/V (ideal gas)]

Isothermal Reversible Change
In an isothermal reversible expansion, temperature does not change T is not a function of V and can be removed from the integral For ideal gas, w = -nRT ln (Vf/Vi) If final volume > initial (expansion), w < 0 System has done work on surroundings internal energy(U) decreased If final volume < initial (compression), w > 0 System has work done on it, U increased Note that as T increases |w| increases

Reversible vs. Irreversible Change
Irreversible expansion: w= = -pexDV Reversible (isothermal) expansion: w=-nRTln{Vf/Vi} Which is greater? Consider indicator diagram Work(rev.) = Area under curve Work (irrev.) = Area under rectangle Work (rev.) > W (irrev.) Reversible work is the maximum which can be done True of PV work True of all work

Heat Changes - Calorimetry
Recall: dU = dq + dw If expansion work is dwexp and dwother is other work done (electrical, magnetic, etc.), then dw = dwexp + dwother dU = dq + dwexp + dwother If V is constant, dwexp is zero since no PV work can be done Assume dwother is also zero, then dU = dq = qv qv is the change in q at constant volume Thus, measuring the heat change in a system at constant volume gives a measure of the change in internal energy This process of measuring heat change is known as calorimetry One constant volume container for measuring heat change is a bomb calorimeter Experiment 1 in lab Typically, substance is burned in calorimeter and temperature rise is measured (dV is constant, but P does change in bomb) In a bomb calorimeter, DT a qv This proportionality is quantified by calibration, typically by combusting a known substance

Heat Changes - Heat Capacity
If V constant, U increases as temperature increases The rate of change of U at any temperature, (dU/dT)V is called the heat capacity, CV CV(A) is not equal to Cv(B) generally smaller if TA < TB Note volume is const. If that changes CV(T may change) CV is an extensive property (2x the amt gives twice the heat capacity) Molar heat capacity, CVm, is the Rate of change of CV with T typically small around room temperature, you can assume it’s constant This means that dU = qV= CV dT ≈ CV DT You can estimate CV by determining the amount of heat supplied to a sample Because qV ≈ CV DT, for a given amount of heat the larger CV, the smaller DT At a phase transition (boiling point) DT = 0 so CV = ∞

Heat Changes - Enthalpy
If PV work can be done by a system, at constant pressure, V changes Some of the internal energy lost through work U=q-pV or q = U+ pV dU < dq, if work is done; dU=dq, if no work is done Define new variable, H, such that H= U + pV H is called enthalpy Like U, H is a state function At constant p, dH = dU +pdV This is an important function because most experiments are carried out at constant pressure, e.g., atmospheric pressure Enthalpy is the heat supplied at constant pressure If pdV=0, dH=dq or DH=qP As with Cv, Cp is defined as dqp/dt = (dH/dT)p

Measuring Enthalpy Changes
Isobaric calorimeter measures heat changes at constant pressure Unlike bomb calorimeter in which volume constrained You can use a bomb calorimeter to measure enthalpy by converting DU to DH See Experiment 1 Assume molar enthalpy and molar internal energy are nearly the same True for most solids and liquids If process involves only solids and liquids DU ≈ DH Calculate pV term based on density differences (Example 2.2) Differential scanning calorimeter (see Atkins, p. 46) Themogram: plot of Cp vs. temperature Peaks correspond to changes in enthalpy Area in the peak = DH Typically small samples (mg) and high temperatures reached

Typical DSC Thermogram Cp(dH/dt) mJ/s Temperature (K) exothermic
Glass transition crystallization melting exothermic endothermic Thermogram Cp(dH/dt) mJ/s Temperature (K) Source:

Enthalpy Calculations
Ideal Gas Since PV term for ideal gas is nRT, DH = DU +pV = DU +nRT If a reaction changes the amount of gas, PV term becomes (nfinal -ninitial)RT or DH = DU +DngRT, where Dng= nfinal -ninitial DH - DU = DngRT Note: sign Dng of is important Enthalpy changes can be directly related to heat input Example 2.3: Calculate molar internal energy change and enthalpy changes at when g of water is vaporized by passing 12V through a resistance in contact with the water for 300 seconds

Change of Enthalpy with Temperature (Cp)
The instantaneous slope of the plot of enthalpy vs temperature is the heat capacity at constant pressure Extensive property Cp,m, the molar heat capacity is an intensive property As with U and Cv, dH = Cvdt, or for measurable intervals DH=CvDT Also, qp = CvDT, because at constant pressure DH= qp This assumes Cv is constant over the temperature range True if range is small, esp. for noble gases More generally Cv = a + bT +(c/T2) a, b,and c are constants depending on the substance If Cv is not constant with T, DH ≠ CvDT Now what happens????

Variation of Cp with Temperature
If Cp varies with temperature, you must integrate over the temperature limits of interest. Example: Assume Cv = a + bT +(c/T2) Try Self Test 2.5

Relationship between Cv and Cp (More about this in Chapter 3)
In most cases Cp > Cv For ideal gases, Cp = Cv +nR This amounts to ~8 JK-1mol-1 difference

Adiabatic expansion of an ideal gas
Work is done, U decreases, hence T decreases To calculate the final state (Tf and Vf) assume a two step process Remember U is a state function, it doesn’t matter how you get there Step 1: Volume is changed and temperature held constant Internal energy change = 0 since it’s independent of volume the molecules occupy Step 2: Temperature changed from Ti to Tf Adiabatic so q = 0 If Cv is independent of temperature, adiabatic work = wad= CvDT DU = q + wad = 0 + CvDT= CvDT This says that for an adiabatic expansion the work done is only proportional to the initial and final temperatures The relationship between initial and final volumes can be derived using what we’ve learned about reversible adiabatic expansions and perfect gases

Adiabatic Expansion of an Ideal Gas Relationship between T and V
Given dq = 0 (adiabatic) and dw = -pdV (reversible expansion) dU = dq + dw =-pdV [1] For perfect gas dU = CvdT [2] Thus, combining [1] and [2], CvdT = -pdV [3] For an ideal gas, pV = nRT, so [3] becomes CvdT = -(nRT/V)dV [4] or, rearranging, (Cv/T)dT = -(nR/V)dV [5] To obtain the relationship between Ti, Vi and Ti, Vi [5] must be integrated Ti corresponds to Vi Cv independent of T The adiabatic work, wad= CvDT, can thus be calculated once this relationship is found

Integration of (Cv/T)dT = -(nR/V)dV

Adiabatic Work, wad, and Temperature

pV Changes During Adiabatic Expansions
This relationship means that the product, pVg doesn’t change as you go through an adiabatic expansion For ideal gas, g, the ratio of the heat capacities is >1 since Cp,m = Cv,m +R g = Cp,m /Cv,m = (Cv,m+R) /Cv,m g = 1 +(R /Cv,m) For monotomic gas, Cv,m= 3/2 R, so g = 5/2 For monotomic gas, Cv,m= 3 R, so g = 4/3 Plot of P vs V for an adiabatic change is called an adiabat p a Vg, where g>1 Recall isotherm, p a V Adiabats fall more steeply than isotherms

The First Law and Chemistry
The study of heat produced from chemical reactions is called thermochemistry We can apply what we learned from our 1st law considerations to the system of reactants and products in a chemical reaction In a reaction, heat either flows into the system (endothermic reactions) or out of the system (exothermic reactions) At constant pressure (reactions done in a beaker for example), the first law tells us for Endothermic reactions DH>0 {heat flows in} Exothermic reactions DH<0 {heat flows out} Standard enthalpy change, DHø, is the change in enthalpy for substances in their standard state Standard state is the form of the substance at 1bar (~1atm) and specified temperature DHø298 is the standard enthalypy change at 1bar and 298K (~RT)

Enthalpies of Physical Change
Changes in state (liquid to gas, solid to liquid, solid to gas, etc.) have enthalpies associated with the transition, DtransHø Liquid-to-Gas, vaporization: DvapHø Solid-to-Liquid, fusion: DfusHø Solid-to-Gas, sublimation: DsubHø Table 2.4 gives symbols for a variety of transitions Because transition temperature is an important point, transition temperatures are often listed at this point May like to know them at convenient temperature as well Recall that enthalpy is a state function. Thus, value is path independent. Can construct information if you don’t have it by making a path and summing the steps Example: You need the enthalpy of sublimation, but can’t find it Path: Step 1 - solid to liquid (DfusHø) Step 2 - liquid to gas (DvapHø) Step 3 - sum of steps1 & 2 Reverse of process involves only change of sign

Enthalpies of Chemical Change
Like state transitions reactions have enthalpies associated with them Can write a thermochemical equation Standard chemical equation plus enthalpy change Reactants and products in standard states ® enthalpy is standard enthalpy of reaction, DrHø Typically, you must calculate DrHø Again because enthapy is a state function you create a path using enthalpies you know or can find and sum them Usually a table of standard free energies of formation of components are available Given per mole of substance, DrHøm Recall that decomposition is the same as formation except sign is reversed For aA + bB ® cC + dD, DrHøm = cDrHøm(C) + dDrHøm(C) - aDrHøm(A) - bDrHøm(D) Generally, Remember state is important! More generally, the standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which it may be divided - Hess’s Law Consequence of the fact that enthalpy is a state function Component reactions could be hypothetical

Standard Enthalpies of Formation , DfHø
As above, the std enthalpy of formation refers to compound in standard state (1bar, typically 298K although other temperature could be specified) Enthalpies of reaction can be determined from enthalpies of formation as before Enthalpies of formation can be estimated by considering contributions from individual groups and summing Molecular modeling can be used to modify this to consider steric effects Not quantitatively reliable but qualitatively o.k.

Temperature Dependence of DrHø Kirkoff’s Law
To be accurate DfHø needs to be measured at temperature of interest If not available, can be estimated as follows: This equation is known as Kirkoff’s Law Assumes Cpø is independent of temperature (good approx. esp. since differences are being considered Can use functional dependence of Cp, if known (we showed this earlier) Best to use T1 as close to T2 as you can get

Chapter 3: The First Law: Machinery Relations between system properties
Homework: Exercises(a only):4, 5, 8, 9, 12, 14, 15, 18 Problems:2, 10, 28, 30a (only)

State and Path Functions
Path 1 - Adiabatic w≠0;q=0 Path 2 - Non-adiabatic w≠0;q≠0 DUpath1 = DUpath2 = Uf -Ui So far, we’ve considered several state functions, e.g., U and H Value of a state function is independent of path Depends on on initial and final state, e.g. DU = Ufinal-Uinitial Overall Change Path from initial to final state has a functional dependence For U work may be done, heat may flow or both Equations that describe how you get to final state are path functions Path functions are path specific

Thermodynamic Consequences of State Functions Internal Energy U, U(V,T)
If U changes only as function of V (T const.) U’ = U + (dU/dV)TdV [1] If U changes only as function of T (V const.) U’ = U + (dU/dT)VdT [2] If U changes only as function of V and T U’ = U + (dU/dV)TdV + (dU/dT)VdT [3] This neglects cross terms If change is infinitesimal, U’-U =dU and [3] becomes dU = (dU/dV)TdV + (dU/dT)VdT [4] (dU/dT)V is Cv (dU/dV)T is called the internal pressure, πT [4] becomes dU = πTdV + CvdT Notice that the partial derivatives have physical significance, not just the slope at a point

Internal Pressure, πT For ideal gas, πT = 0 because U independent of molecular separation for volume changes at constant T Implies ideal gas law For real gas, If dU >0 as dV increases with T constant, attractions between molecules dominate and πT > 0 If dU <0 as dV increases with T constant, repulsions between molecules dominate and πT <0

Internal Pressure, πT - Joule Experiment
Isothermal Expansion pex = 0, therefore w = 0 dT=0, therefore q = 0 DU = q +w = πT dV= 0 πT must =0 since dV>0 “No change in temperature occurs when air is allowed to expand in such a manner as not to develop mechanical power” J. P. Joule, Phil. Mag., 26, 369 (1845) “mechanical power” º no external work Joule experiment incapable of detecting small effects since the water calorimeter he used had a large heat capacity

Change in Internal Energy @ Constant P
Suppose we want to know how the internal energy, U, changes with temperature at constant pressure But the change in volume with temperature at constant pressure is the related to the thermal compressibility of the gas, a Large a means sample responds strongly to changes in T For an ideal gas, pT = 0 so

Thermodynamic Consequences of State Functions Enthalpy, H(p,T)
By analogy to a earlier, the isothermal compressibility, kT, (cont.) Like U, H is a state function For closed system with constant composition At constant volume By the chain rule, we can find (dp/dT)V

Thermodynamic Consequences of State Functions Enthalpy, H(p,T)
To evaluate (dH/dp)T apply chain rule and reciprocal identity again But, Cp= (dH/dT)p and if one defines (dT/dP)H as µ, then (dH/dp)T =- Cpµ Thus,

A Word about kT We know that if we increase the pressure the volume decreases so if dp is positive, dV is negative Since kT=-(1/V)(dV/dp)T, kT is always positive For and ideal gas V=nRT/p so (dV/dp)T= -nRT/p2 and kT=-(1/V)(-nRT/p2 ) Simplifying kT, kT= (nRT /V)(1/p2 ) kT= (p)(1/p2 ) kT= (1/p ) This means that as the pressure increases, the compressibility of a gas decreases

A Word about µ - Historical
µ was defined earlier as (dT/dP)H. It is known as the Joule-Thompson Coefficient Rate change of temperature with pressure William Thomson (Lord Kelvin) and James Prescott Joule performed a series of experiments between 1852 and 1862 Adiabatic, so q = 0 w = p1V1 - p2V2 The rate of temperature change as a function of pressure, µ,occurs at constant enthalpy, i.e., µ = (dT/dP)H adiabatic wall A B C A B C n p1, V1, T1 n p2, V2, T2 porous plug silk meerschaum

A Word about µ - Physical
100 atm 60 atm µ is a function of p and T (see Fig. for CO2) µ can be either (+) or (-) Positive µ means dT is negative when dp is negative Gas cools on expansion Negative µ means that means dT is positive when dp is negative Gas warms on expansion Transition between positive and negative µ is called the Joule-Thompson inversion temperature Gases have both high and low inversion temperatures Ideal gas µ is 0, T unchanged by change in p on expansion Modern methods measure µT not µ Isothermal J-T coefficient µT = (dH/dp)T = -Cpµ (see Atkins, p. 83) Practical applications: Gas liquefaction Isotope separation Snow

Relationship between Cv and Cp IdealGas
In Chapter 2, we said, for ideal gases, Cp = Cv +nR Proof:

Relationship between Cv and Cp Real Gases
For real gases, we can use some basic definitions and what we’ve already learned to find Cp -Cv : For ideal gas a=1/T and kT=1/p and this reduces to what we found earlier

Chapter 4: The Second Law - Concepts
Homework: Exercises(a only):4, 5, 6, 8, 10, 12, 14, 16, 18, 20 Problems:3a, 6, 7, 18, 29

2nd Law of Thermodynamics You can’t break even!
Heat engines:18th and 19th saw great economic development in due to mechanical devices to do work that was formally done by humans, animals, or simple machines (e.g. water power) Steam engine { James Watt (1796)} - converted heat from fuel to mechanical action (piston) through expanding a working fluid (steam). Cooled fluid released, piston returned to original position and expansion performed again (cycle) Heat engine withdraws heat from hot reservoir, converts some of it to heat and discards the rest to a cold reservoir In heat engine w is related to q Obviously one would like to maximize conversion Losses occur due to friction, but it was hoped that with improvements one could get better at extracting work from heat If e, the efficiency of a heat engine is defined as |w|/qh, where qh is the temperature of the hot reservoir, the ultimate goal was to get e=1 This proved unsuccessful for a more fundamental reason than poor mechanical design Hot Reservoir Heat Engine Work Cold 2nd Law of Thermodynamics (Kelvin, 19th cent.): No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion to work Result of considering heat engines from a thermodynamic viewpoint It is impossible to devise an engine which, working in a cycle, shall produce no effect other than the extraction of heat from a reservoir and the performance of an equal amount of work “Science owes more to the steam engine, than the steam engine owes to science” - L.J. Henderson

2nd Law and Spontaneous Change
Spontaneous change is change in which no work is done (just happens), e.g., gas expands to fill volume, some chemical reactions proceed in one direction rather than another Direction is determined by some aspect of the world Can reverse spontaneous change by doing work on the system (e.g.,compressing gas) Direction of spontaneous change Independent of total energy First law says that total energy is conserved in any process. Change reflects distribution of energy, not amount Direction of change is related to distribution of energy Spontaneous changes lead to a more disorderly distribution Disorder in a system is measured by entropy, S Like U (and H), entropy is a state function 2nd Law (entropy statement)The entropy of an isolated system increases during the course of a spontaneous change, i.e., DStot > 0 Stot is the entropy of system and surroundings Thermodynamically irreversible processes are spontaneous and must be accompanied by an increase in Stot

Thermodynamic Definition of Entropy
Related to heat/temperature since heat stimulates disorderly motion Work stimulates uniform motion, does not change degree of disorder - does not affect entropy Definition For measurable change Units of entropy: JK-1 Molar entropy: JK-1 mol-1 Entropy change of isothermal expansion of an ideal gas Step 1: Find heat change along a reversible path For reversible isothermal expansion of an ideal gas: DU = 0, so q = -wrev wrev = -nRTln(Vf/Vi) Step 2: Integrate expression for entropy Since isothermal, T constant

Entropy Change in Surroundings, DSsur
Suppose dqsur is transferred to the surroundings, the internal energy transferred to surroundings is dUsur dUsur is independent of how change brought about (U is state function Can assume process is reversible, dUsur= dUsur,rev Since dUsur = dqsur and dUsur= dUsur,rev, dqsur must equal dqsur,rev From the definition of entropy, dSsur= dqsur /Tsur {Note: dq can be reversible or not, see above} And for a measurable change DSsur= qsur /Tsur because Tsur is constant For adiabatic change, qsur = 0, so DSsur = 0

Entropy as a State Function
To prove entropy is a state function we must show that ∫dS is path independent Sufficient to show that the integral around a cycle is zero or Sadi Carnot (1824) devised cycle to represent idealized engine Hot Reservoir Cold Engine -w2 -w1 w3 w4 qh qc Th Tc Step 1: Isothermal reversible Th Step 2:Adiabatic expansion Th to Tc Step 3:Isothermal reversible Tc (sign of q negative) Step 4: Adiabatic compression Tc to Th

Carnot Cycle - Entropy of Ideal Gases
Total entropy around cycle Steps 2 and 4, adiabatic, q for each step = 0 Therefore change in entropy around cycle is just heat flow in adiabatic steps This value = 0 (right) For surroundings DS also is 0

Carnot Cycle - Thermodynamic Temperature Scale
Recall, the efficiency of a heat engine is the ratio of the work performed to the heat of the hot reservoir e=|w|/qh The greater the work the greater the efficiency Work is the difference between the heat supplied to the engine and the heat returned to the cold reservoir w = qh -(-qc) = qh + qc Therefore, e = |w|/qh = ( qh + qc)/qh = 1 + (qc/qh ) Wm. Thomson defined a substance-independent temperature scale based on the heat transferred between two Carnot cycles sharing an isotherm Hot Reservoir Heat Engine Work Cold qh -qc w Defined a temperature scale such that qc/-qh = Tc/Th e = 1 - (Tc/Th ) Zero point on the scale is that temperature where e = 1 Or as Tc approaches 0 e approaches 1 Efficiency can be used as a measure of temperature regardless of the working fluid Applies directly to the power required to maintain a low temperature in refrigerators (Atkins, p ) Efficiency is maximized Greater DT temperature difference between reservoirs The lower Tc, the greater the efficiency

Carnot Cycle - Entropy of Real Materials
All reversible engines have the same efficiency regardless of construction Depends only on temperature difference between hot and cold reservoir Any reversible cycle can be approximated as a collection of small Carnot cycles Total entropy is the sum of the entropy of all the individual cycles In the limit, as size of cycles becomes infinitesimal, sum becomes interval In the interior some cycles will cancel others, only those on edge will remain so the net entropy change, DSnet This implies dS is an exact differential and, hence, entropy is a state function regardless of material

Clausius Inequality & Spontaneous Change
For a system in thermal equilibrium with surroundings, dS + dSsur ≥ 0 [1] Reversible process: dS + dSsur = 0 Irreversible process: dS + dSsur > 0 [1] means dS ≥ -dSsur [2] Recall, the differential entropy change of the surroundings, dSsur = -dq/T so dS ≥ dq/T This is called the Clausius Inequality - For an isolated system, entropy cannot decrease for a spontaneous change R. Clausius introduced the function S in 1850 calling it entropy from the Greek meaning to give direction Principle of Clausius (Statement of 2nd Law): It is impossible to devise an engine which, working in a cycle, shall produce no other effect than the transfer of heat from a colder to a hotter body Entropy is a signpost of spontaneous change!!!

Clausius Inequality - Examples
1. Irreversible isothermal expansion of a gas: dU = 0 so dq = -dw If pex = 0 (gas expands into vacuum), dw = 0 so dq = 0 hence dS ≥ 0 For surroundings, since change is isothermal, no heat flows so dSsur = 0 For total system: dStot = dS + dSsur ≥ {Clausius Inequality} 2. Spontaneous cooling: Heat flows from hot source to cold sink Let amount of heat = dq Entropy from hot source:dSsource= -|dq|/Th Entropy decreases in hot source since heat is leaving Entropy from cold sink: dSsink= +|dq|/Tc Entropy increases in cold sink since heat is entering Overall entropy change: dS = dSsource + dSsink = +|dq|/Tc -|dq|/Th dS = |dq| x (1/Tc - 1/Th ) since Th > Tc (1/Tc - 1/Th )>0 so dS > 0 Cooling spontaneous {Clausius Inequality} If Th = Tc, (1/Tc - 1/Th ) = 0 so dS = 0 Thermal equilibrium {Clausius Inequality}

Entropy Changes for Phase Transition at Ttrns
At a phase transition (freezing, boiling, etc.) there is a change in molecular order so there should be a change in entropy Consider a situation where system and surroundings are at temperature in which two phases exist at equilibrium at 1 atm (normal transition temperature), Ice point Ttrns = K (0°C); Water boiling point Ttrns = K Any heat transfer between phases is reversible since they are in equilibrium p is constant so q = qrev = DtrnsH Thus, DtrnsS = qrev /T = DtrnsH/T {like DtrnsH, DtrnsS is also a molar quantity} If transition is exothermic (DtrnsH<0) {freezing, condensation, etc.} DtrnsS<0 System becomes more ordered If transition is endothermic (DtrnsH>0) {melting, boiling, etc.} DtrnsS>0 System becomes less ordered Trouton’s Rule (empirical): Liquids have approximately the same DvapS Not true if molecules in liquid are ordered (H2O) Not true if entropy of gas is low (methane)

Entropy of Expansion - Ideal Gas
Previously showed that for an ideal gas, DS = nRln(Vf/Vi) DS path independent, so for system it doesn’t matter if change is reversible or not Total change in entropy does depend on path, however, because entropy of surroundings must be considered {DStotal = DSsystem + DSsurroundings} Reversible: DStotal = 0 so DSsurroundings = - DSsystem If isothermal, qsur = 0 so DSsurroundings = 0 and DStotal = DSsystem As you increase volume disorder increases

Temperature Variation of Entropy
Cv,m = 4R 3R 2R Above true if Cp (Cv)not a function of temperature This means there is a logarithmic dependence of entropy on temperature To evaluate S(T2) need to know S(T1) Look at Example Note how choosing path impt! R

Calculating and Measuring Entropy
Determining entropy over a large range of temperatures requires accounting for both entropy changes at transitions and entropy changes between transitions Fortunately entropy can be summed For a substance with 3 states (solid, liquid and gas) Many of the above terms can be measured or calculated Calorimetric measurements Integrals evaluated Graphically Area under plot of Cp vs, T gives ∫CpdT Numerically Polynomial fit followed by explicit integration Measurement of Cp at very low temperatures difficult Debye Extrapolation Measure Cp as low as possible Assume below that point Cp proportional to T3

Debye Extrapolation Example 4.4
Molar Cpof a solid at 10K is 0.43 JK-1mol-1. What is entropy at that temperature? Question: How do we evaluate S(0)???

Third Law of Thermodynamics
Observation: At T = 0 K, all the energy of thermal motion is quenched In a perfect crystal, all atoms (ions) are in a regular uniform array The absence of motion and disorder suggest Kthe entropy is zero Nernst Heat Theorum: The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero (1906) S ® 0 as T ® 0 provided all substances involved are perfectly ordered If S = T = 0 for all perfect crystalline elements, then, from the Nernst Theorem, S of all perfect crystalline T = 0 is also zero Compound formation is a chemical transformation This corollary of the Nernst Heat Theorem is the 3rd Law of Thermodynamics 3rd Law of Thermodynamics The Entropy of all perfect crystalline substances at T = 0K is zero Choice of zero value for entropy may seem to be arbitrary, but it is in fact consistent with molecular observations (above)

Nernst Heat Theorum Example
Elemental sulfur has many forms, two of them are orthorhombic (a) and monoclinic (b) Transition between a and monoclinic b occurs at 369K with an enthalpy (DH) = -402 Jmol-1 DtrsS = Sm (a) - Sm (b) = (DH/T) = -402 Jmol-1/369K = Jmol-1K-1 Using this information and a measure of Sm for orthorhombic and monoclinic one can estimate S(0) sulfur At 369K, Sm, trs(a) = 37 Jmol-1K-1 and Sm, trs(b) = 38 Jmol-1K-1 Sm (a) = Sm (a,0) + Sm, trs(a) and Sm (b) = Sm (b,0) + Sm, trs(b) Sm (a) = Sm (a,0) + 37 Jmol-1K-1 and Sm (b) = Sm (b,0) + 38 Jmol-1K-1 DtrsS = Sm (a) - Sm (b) = Sm (a) - Sm (b) DtrsS = Sm (a,0) + 37 Jmol-1K- - Sm (b,0) - 38 Jmol-1K-1 DtrsS = {Sm (a,0) - Sm (b,0) }+ {37 Jmol-1K Jmol-1K-1} = 1 Jmol-1K-1 This is essentially what was calculated from DH considerations Implies Sm (a,0) - Sm (b,0) = 0 Consistent with Nernst Theorem

3rd Law of Thermodynamics You can’t get there from here!
Successive isothermal magnetization and adiabatic demagnetization {The engine that almost could} Principle: Unpaired e- of a parmagnetic substances are randomly oriented, can be aligned with external magnetic field Process: 1. Turn on magnetic field in isothermal environment. e- oriented. S decreases. 2. Turn off magnetic field in adiabatic environment. e- randomize, and energy decreases. T decreases. Repeat Temperature can be successively reduced Each temperature increment gets smaller and smaller as temperature gets lower and lower In limit as number of steps ® ∞, T ® 0K Unfortunately, you can never reach the limit! Adiabatic (S decreases) Isothermal (T decreases) S T Field Off H = 0 Field On H = H1 3rd Law (alternate statement): It is impossible by any procedure, no matter how idealized to reduce the temperature of any system to absolute zero in a finite number of operations

3rd Law Entropies Entropies reported on the basis that S(0) = 0 are 3rd Law entropies If substance is in standard state these are expressed Sø(T) Standard reaction entropy (DrSø) is analogous to standard reaction enthalpy (DrHø) Sum of std. entropies of products -Sum of std. entropies of reactacts Appropriate stoichiometric factors used

Other State Functions -1
Recall: Clausius Inequality: dS - (dq/T) ≥ 0 At constant volume: dS - (dU /T) ≥ 0 or TdS ≥ dU With no additional work (non PV work), spontanety is expressed solely in terms of state functions (q not a state function) At constant energy this becomes TdSU,V ≥ 0 @ const. energy and volume S increases for spontaneous change Re-statement of 2nd Law At constant entropy this becomes dUS,V ≤ 0 (note reversed inequality) In a spontaneous change, heat flows out of the system if entropy and volume of the system is unchanged. This means entropy of the surroundings must increase! Recall, at constant pressure, with no additional work, dq = dH, so dS - (dq/T) ≥ 0 becomes dS - (dH/T) ≥ 0 or TdS ≥ dH At constant enthalpy this becomes TdSH,p ≥ 0 @ const. enthalpy and pressure S increases for spontaneous change At constant entropy this becomes dHS,p ≤ 0 (note reversed inequality) In a spontaneous change, if entropy and pressure of the system is constant entropy of the surroundings must increase!

Other State Functions - 2
Summary At constant volume: TdS ≥ dU, dU-TdS ≤ 0 At constant pressure: TdS ≥ dH , dH-TdS ≤ 0 Define two new state functions: A and G A = U - TS and G = H - TS At const. T dA = U - TdS and dG = H - TdS Clausius inequality becomes @ Vconst: dAT,V ≤ 0 pconst dGT,V ≤ 0 These are two statements of spontaneous change e.g., if dG ≤ 0, change is spontaneous A is called the Helmholtz energy G is called the Gibbs energy Both are important in chemical thermodynamics, esp. G

Helmholz Energy Change in Helmholz energy is an expression of the maximum work of a process, i.e., dA = dwmax Proof ® For macroscopic change, wmax = DA Recall DA = DU - TDS Depending on sign of TDS, not all the energy change may be available for doing work TDS<0 (entropy decreases/-TDS = +), and wmax < DU In a spontaneous change, some of the energy must flow out of the system TDS>0 (entropy increases/ -TDS = -), and wmax > DU In a spontaneous change, some of the energy must flow into the system

Gibbs Energy/Gibbs Free Energy
More common than A since usually changes is at constant pressure rather than volume Statement of Clausius inequality for Gibbs energy [dGT,V ≤ 0] is a familiar observation Spontaneous reactions occur in direction of decreasing Gibbs free energy p=const. and T=const.) Spontaneous endothermic reactions In edothermic reaction, DH>0 For DG<0 (i.e, for the reaction to be spontaneous), TDS must be > DH Entropy changes drive the reaction rather than energy changes Standard molar Gibbs energy of reaction Obtained from standard enthalpies and entropies: DrGø = DrHø - TDrSø As with DfHø , the standard Gibbs energy of formation, DfGø, is the Gibbs energy for the formation of the compound from elements in their standard states DfGø (elements) = 0 As with DrHø and DrSø , for a complex reaction Calorimetry one way of determining Gibbs energies (DH and DS) Electrochemistry Spectroscopy

Chapter 5: The Second Law - Machinery
Homework: Exercises(a only):4, 5, 6, 8, 9, 10, 12, 13, 14 Problems:1, 4a, 6, 9, 14, 19

Combining 1st and 2nd Laws The Fundamental Equation
First Law of Thermodynamics: dU = dq +dw For reversible change with only pV work: dwrev = -pV and dqrev = TdS (2nd Law) 1st Law becomes: dU = TdS - pdV [1] Because dU is a state function this applies to any path reversible or not Above is a combination of the 1st and 2nd Laws Fundamental Equation of Thermodynamics Properties of internal energy (dU) Since U =f(S,V) dU =(dU/dS)vdS + (dU/dV)sdV [2] From [1] above, we conclude T =(dU/dS)v [3] and p =-(dU/dV)p [4] Recall Kelvin’s (Thomson’s) earlier definition of a purely thermodynamic temperature scale [3] is a demonstration that this is legitimate More generally deriving these relationships between T, U & S and p, U &S demonstrates how thermodynamics can be used to develop relationships

Maxwell Relationships
Other Maxwell Relations Mathematical statement: Suppose you can write an differential, df in the form df = gdx + hdy where, g & h are functions of x and y If df is an exact differential, then (dg/dy)x = (dh/dx)y For the fundamental equation (dU = TdS - pdV) this means (dT/dV)S = -(dp/dS)V This is known as a Maxwell relation Other Maxwell relations can be easily derived for entalpy (dH), Helmholtz energy (dA), Gibbs energy (dG) Enthalpy Helmholtz Energy Gibbs Energy

Using Maxwell Relationships
Proof In Ch. 3, when looking at the relationship between Cp and Cv for real gases, it was asserted that, for the internal pressure of a gas, pT, pT = T(dp/dT)V - p This can now be proved using a Maxwell relationship pT How do you know what relationship to use? Look at variables: Constant variable and partial variable in denominator (T,V) give infinitesimals in the state function (dT and dV) This gives you the state function for which you can either look up or derive Max. rel. This takes practice!

Deriving Themodynamic Relationships
A thermodynamic proof is one in which only equations of state and general thermodynamic relationships are used No molecular arguments allowed Use definitions, Maxwell relations, etc. and limiting assumptions (e.g., isothermal process, ideal gas, etc. Example: Compute pT for van der Waal’s gas (part b example 5.1) Additional assignment: Self Test 5.1

Gibbs Energy Properties General Considerations
Recall, G =H-TS or dG=dH-TdS-SdT H = U+pV so dG = (dU + pdV +Vdp)-TdS-SdT But for a closed system dU = TdS - pdV (Fund. Thrm. of Thermo.) Thus, dG = {(TdS -pdV )+ pdV + Vdp } - TdS -SdT Rearranging and cancelling: dG = Vdp - SdT [1] This means G=f(p, T) or dG = (dG/dp)Tdp + (dG/dT)pdT [2] Combining [1] and [2] we find, (dG/dp)T = V [3a] and (dG/dT)p = -S [3b] At const. p and composition, [3b] implies that as T increases G decreases (S=+) G decreases most sharply with temp when S is large, so G for gases are more sensitive to temperature change than it is for liquids or solids (dG/dT)p = -S (dG/dp)T = V At const. T and composition, [3a] implies that as p increases G increases (V=+) Because molar volumes of gases are larger than liquids, G for gases are more sensitive to pressure change than it is for liquids or solids

Gibbs-Helmholtz Equation Variation of Gibbs Energy with Temperature
Definition of Gibbs energy is G = H -TS so -S = (G-H)/T [1] and earlier we saw that -S = (dG/dT)p [2] From [1] and [2], (dG/dT)p = (G-H)/T Because the equilibrium constant (K) is related to G/T, we wish to know (d(G/T)/dT)p rather than (dG/dT)p This is the Gibbs -Helmholtz equation Alternate form of it is (d(G/T)/d(1/T)p = H (see Justification 5.2) In terms of changes of state (physical and chemical) Gibbs Helmholz equation: If we know change in enthalpy (DH) we know how Gibbs energy varies with temperature

Variation of Gibbs Energy with Pressure Liquids and Solids
For liquids and solids, volume doesn’t change much with pressure so Usually VDp is small and can be neglected, so for solids and liquids G is constant with changes in temperature Exceptions: extreme conditions If V=f(p), must integrate full expression

Variation of Gibbs Energy with Pressure Gases
For gases, volume varies strongly with pressure For ideal gas, V = nRT/p so p ® 0, Gm ® ∞

Variation of Gibbs Energy with Pressure Real Gases ~ Fugacity
Sometimes convenient to keep the same form of an equation for real system that was derived for ideal system One would expect pressure dependence of G for a real gas to be similar to that for ideal gas Note that differences in equation of state are small Define a quantity known as fugacity which is related to the pressure of a real gas such that Gm = Gmø +ln(f/pø) Gilbert Lewis (1908) - Am. physical chemist best known for work on valence bond theory (1902). Also coined term photon (1926). Developed Lewis dot structures. Home schooled until age of 9, at 14 went to U. Netraska Often we define equilibrium constants in terms of pressure, but they are strictly true only for fugacities Example

f = f p where is a dimensionless constant
Fugacity and Pressure Fugacity can be related to pressure by assuming a simple functional dependence f = f p where is a dimensionless constant f is called the fugacity coefficient Relationship between f and compressibility

Fugacity and Pressure Consequences
At low to moderate pressures, Z<1 Integral is negative, ø<1, i.e., attractive forces dominate between molecules p > f and Greal<Gideal At high pressures, Z>1 Integral is positive, ø>1, i.e., repulsive forces dominate between molecules f > p and Greal>Gideal

Estimating Fugacities
High pr Tr = 5 Low pr Tr = 3.0 Using an equation of state for real gas, fugacities can be plotted in terms of reduced variables Remember reduced variables allow you to compare differnent gases on the same curve Reduced variables are determined by dividing the actual variable by the variable at the critical point, e.g., Tr =T/Tc, pr =p/pc, etc.

Chapter 6: Physical Transformations of Pure Substances
Homework: Exercises(a only):4, 5, 8, 12, 15 Problems:4, 6,10, 15, 21

Phase Diagrams Phase - a form of matter that is uniform throughout in chemical composition and physical state (J. Willard Gibbs) Homogeneous - one phase present, e.g., glass of cracked ice (not including the glass) Heterogeneous - more than one phase, e.g., glass of cracked ice with water Phase transition - spontaneous conversion of one phase to a another Occurs at characteristic temperature for a given pressure, transition temperature, Ttrs At Ttrs, phases are in equilibrium and Gibbs energy minimized Example ice and water at 0°C This says nothing about the rate the transition occurs Graphite is the stable phase of carbon at R.T., but diamonds exist because rate of thermodynamically “spontaneous” to low Such kinetically persistent unstable phases are called metastable phases Phase Diagram is a P,T plot showing regions of thermodynamically stable phases Lines separating phases are called phase boundaries Vapor-liquid boundary shows variation of vapor pressure with temperature Solid-liquid boundary shows variation of sublimation pressure with temperature

Phase Boundaries (cont.)
Boiling Points Temperature at which vapor pressure equals external pressure Normal boiling point (Tb) external pressure = 1 atm (H2O: 100.0°C) Standard boiling point external pressure = 1 bar (0.987 atm) (H2O: 99.6°C) Critical Point In closed vessel boiling does not occur. Pressure increases and fluid level drops. rliquid decreases and rvapor increase as T increases At some point rliquid = rvapor and boundary disappears: critical temperature (Tc),vapor pressure is critical pressure, pc Above a single uniform phase exists, supercritical fluid Highest temperature liquid can exist Melting Points Melting temperature both solid and liquid phases exist. Equivalent to freezing temperature Normal freezing(melting) point (Tf) external pressure = 1 atm (H2O: 0°C) Standard freezing point external pressure = 1 bar (0.987 atm) Triple Point (T3) - place at which 3 phase boundaries meet All 3 phases exist simultaneously Typically solid, liquid and gas Lowest pressure and temperature liquid can exist Invariant, property of substance (H2O: K, 6.11 mbar) pc Tb 1 atm = pex supercritical fluid

Phase Diagrams - Water Ice VI,VII, liquid 22 kbar, 81.6 °C Ice VI,VII, VIII 21 kbar, ~5 °C Slope of solid-liquid boundary means large Dp necessary to significantly change melting temp. Negative slope means Tmelt decreases as p increase Due to structure of water Liquid has lower volume than solid (rliquid >rsolid) High pressure phases Different crystal structure and density Ice I (hexagonal); Ice III (Tetragonal) At -175°C and 1 atm: Ice I (r = 0.94 g/cc); Ice VI (r = 1.31 g/cc); Can melt at higher temperatures than Ice I Ice VII melts at 100°C but only exists at p>25 kbar 8 triple points (6 plotted) Only one between solid (Ice I), liquid and gas

Phase Diagrams - Carbon Dioxide
Melting temperature increases a p increases Unlike water& more typical T3 > 1 atm Liquid CO2 doesn’t exist at 1 atm High pressure CO2 tanks contain liquid At 25°C ( K) and 67 atm, gas and liquid co-exist

Phase Diagrams - Helium
Note temperature scale (< 6K) Solid and gas never in equilibrium Two liquid phases Phase boundary l-line (type of phase transition) Higher temperature liquid (He-I) is regular liquid Low temperature liquid (He-II) superfluid (zero viscosity) It rather than solid exists close to 0K 4He Phase diagram depends on nuclear spin 3H - non-zero spin, 4H -zero spin 3H diagram different than 43H, esp. low temperature 3H Sliquid < Ssolid, melting exothermic

Gibbs Energies and Phase Diagrams
Chemical potential (µ) - for one component system µ =Gm Measure of the potential of a substance to bring about change More detail in Chapters 7 & 9 At equilibrium, µ is the same throughout the system Regardless of number of phases If µ1 is chem. potential of phase 1 and µ2 is chem. potential of phase 2, at equilibrium µ1 = µ2 Consequence of 2nd Law of Thermodynamics If dn is transferred from one location (phase) to another, -µ1dn, is the change in Gibbs energy in that phase and µ2dn is change in free energy in the second phase. dG = -µ1dn + µ2dn = (µ2 - µ1)dn If µ1 > µ2 dG < 0 and change spontaneous If µ1 = µ2 dG = 0 and system at equilibrium Transition temperature is that temperature where µ1 = µ2 Chemical potentials of phases change with temperature At low T and reasonable p, solid has lowest µ and, hence, most stable As T raised, µ of other phase may become less than solid(at that temp.) so it becomes stable phase

T and p Dependence on Melting
Recall, (∂G/∂T)p = -S so (∂µ/∂T)p = -Sm Since Sm> 0 for all pure substances, (∂µ/∂T)p < 0 or a plot of µ vs. T will have a negative slope Because Sm(gas)> Sm(liquid)> Sm(solid)> slopes going from gas ® solid increasingly negative Transitions (melting, vaporization) occur when µ of one phase becomes greater than another so substances melt when µ(liquid)>µ(solid) Phase change means modifying relative values of µ for each phase Given differences in slopes, change in T easiest way to do it Vm(l)> Vm(s) Similarly, (∂G/∂p)T = V so (∂µ/∂T)p = Vm Because Vm increases with p, graph of µ vs. T translates upward as p increases For most substances Vm(l)> Vm(s), µ(l) increases more than µ(s), Tf increases as p increases (a) Consistent with observations and physical sense higher p retards movement to lower density Water exception since Vm(s)> Vm(l), as p increases µ(s) increases more than µ(l), so Tf decreases as p increases (b) Vm(s)> Vm(l)

Assessing Effect of Pressure on Melting
Example 6.1 Calculate Dµ for each state over pressure range, Dµ (water) D p = 1 bar; Dµ (ice) D p = 1 bar Dµ (water) = 1.80 J mol-1 and Dµ (ice) = 1.97 J mol-1 J mol-1 are units of µ just like G Dµ (water) < Dµ (ice), so tendency for ice to melt Look at Self Test 6.2 (opposite of water)

Effect of Pressure on Vapor Pressure
When pressure applied to a condensed phase (solid or liquid), vapor pressure increases, i.e., molecules move to gas phase Increase in p can be mechanical or with inert gas Ignore dissolution of pressurizing gas in liquid Ignore gas solvation, attachment of liquid molecules to gas-phase species Vapor pressure in equilibrium with condensed phase is the partial vapor pressure of the substance, p* p = p*eVmDP/RT [1] Proof ® Math Moment: ex = 1+ x + 1/2x2 +…. If x<<1, .e ≈ 1 + x [1] becomes p = p*(1 + VmDP/RT) If VmDP/RT<<1 or

Location of Phase Boundaries
Phase boundaries occur when chemical potentials are equal and phases are in equilibrium, or for phases a and b: µa(p,T) = µb(p,T) Need to solve this equation for p, p= f(T) On plot of p vs. T, f(T) is a gives the phase boundary Slopes of phase boundaries Slope is dp/dT If p and T are changed such that two phases, a and b , are in equilibrium, dµa = dµb But, dµ = -SmdT + Vdp So, -Sa,m, dT + Va,mdp = -Sb,m, dT + Vb,mdp Or (Sb,m, -Sa,m, )dT + = (Vb,m - Va,m )dp But (Sb,m, - Sa,m, ) = DtrsS and (Va,m - Vb,m ) = DtrsV This rearranges to: dp/dT = DtrsS / DtrsV where and are the entropy and volume of transition This is called the Clapeyron Equation Exact expression for the phase boundary at equilibrium Can be used to predict appearance of phase diagrams and form of boundaries µa = µb

Solid-Liquid Boundary
For solid-liquid boundary, Clapeyron Equation becomes dp/dT = DfusS / DfusV where DfusV is the change in molar volume on melting DfusS is always positive (except for 3He) and DfusV is usually small so dp/dT is large (steep slope) and positive Formula for phase boundary comes from integrating Clapeyron equation This is straight line of slope [DfusH / (T*DfusV)] Math Moment: ln(1+x) = x - 1/2x2 +…. If x<<1, ln(1 + x) = x

Liquid Vapor Boundary Again, Clapeyron equation can be used
DvapV is large and positive so dp/dT is positive, but smaller than for solid-liquid transition DvapH/T is Trouton’s constant Because Vm(gas) >> Vm(liquid), DvapV ≈ Vm(gas) For ideal gas, Vm(gas) = RT/p so DvapV ≈ RT/p Clapeyron equation becomes Integrating Clausius-Clapeyron equation gives variation of vapor pressure with temperature Assumes DvapH is independent of T and p* is vapor pressure at T* and p the vapor pressure at T This is a curve, not a line Does not extend beyond Tc

Solid-Vapor Boundary Solid-vapor boundary same as liquid vapor boundary except use DsubH instead of DvapH Since DsubH > DvapH, slope of curve is steeper Curves coincide at triple point along with solid-liquid boundary

Classification of Phase Transitions Ehrenfest Clasification
We’ve been talking a lot about the slopes of phase transitions, (∂µ/∂T)p or (∂µ/∂V)T Transitions are accompanied by changes in entropy and volume At transition from a phase, a, to another phase, b (∂µb /∂T)p - (∂µa /∂T)p = -Sb,m + Sa,m = - Strs = -DtrsH/T (∂µb /∂p)T - (∂µa /∂p)T = Vb,m - Va,m = -DtrsV 1st Order Transitions (e.g., melting, vaporization) Since DfusH, DvapH and DfusV, DvapV are non-zero, the changes in µ {(∂µ/∂T)p or (∂µ/∂p)T} as you approach the transition are different. There is a discontinuity at the transition A transition in which the slope of µ, (∂µ/∂T)p , is discontinuous is called a 1st order transition Cp is slope of plot of H vs. T (∂H/∂T)p at 1st order transition is infinite Infinitesimal change in T produces finite change in H Ttrs V, H. S Cp

Classification of Phase Transitions Ehrenfest Classification (continued)
2nd Order Transition (glass transition, superconducting to conducting transition) (∂µ/∂T)p , is continuous Volume and entropy don’t change at transition (∂2µ/∂T2)p is discontinuous Heat capacity is discontinuous, but not infinite Ttrs Cp V, S, H l -Transition (4He super-fluid to liquid transition, order-disorder transition in b-brass) Not 1st order, (∂µ/∂T)p , is continuous Heat capacity is discontinuous, and infinite at transition temperature

Liquid Surfaces Surface effects can be expressed in terms of thermodynamic functions since work is required to change the surface area of a liquid If s is the surface area of a liquid, ds is its infinitesimal change when an amount of work, dw is done dw is proportional to the surface area of the liquid, i.e., dw = gds g the proportionality constant is defined as the surface tension Dimensions of g : energy/area (J/m2 or N/m) To calculate the work needed to create or change a surface by a particular area increment you only need to calculate the area since is a constant Look at Self Test 6.4 At constant volume and temperature the work is the Helmholtz energy (A) dA = gds As ds decreases the Helmholtz energy minimizes This is the direction of spontaneous change so surfaces of liquids tend to contract Often minimizing of dA results in curved surfaces

Curved Liquid Surfaces
Bubble - a region in which vapor (+ air) is trapped by thin film Bubbles have 2 surfaces one on each side of the film Cavity is a vapor-filled hole in liquid (commonly called bubbles) but 1/2 the surface area Droplet - small volume of liquid surrounded by vapor (+air) Pressure inside a concave surface (pin) is always greater than pressure outside (pout) Difference depends on surface area and surface tension pin = pout + 2g/r Laplace eqn As r ® ∞ (flat surface), pin = pout As r ® small(small bubbles), 2g/r important Earlier we saw, the vapor pressure in the presence of external pressure Dp is p = p*eVmDP/RT So, for bubble since Dp = 2g/r p = p*e2gVm/rRT Kelvin Equation (bubble) In a cavity, pout < pin, so sign of exponential term is reversed p = p*e-2gVm/rRT Kelvin Equation (cavity)

Nucleation, Superheating &Supercooling
For water droplets r p/p* = 1.001 Small effect but mat have important consequences r p/p* = 3 Clouds form when water droplets condense. Warm moist air rises, condenses at colder altitude Initial droplets small so Kelvin equation tells us vapor pressure of droplet increases Small droplets tend to evaporate Unless large numbers of molecules congregate (spontaneous nucleation) Air becomes supersaturated and thermodynamically unstable Nucleation centers (dust particles, sea salt) allow clouds to form by allowing condensation to occur on larger surfaces Superheating - liquids persist above boiling point Vapor pressure inside a small cavity in liquid is low so cavities tend to collapse Spontaneous nucleation causes larger more stable cavities (and bumping!) Nucleation centers allow for stabilization of cavity Basis of bubble chamber Supercooling, persistence of liquids past freezing point, is analogous

Capillaries Capillary action - tendency of liquids to rise up/fall in narrow bore (capillary) tubes Capillary rise/fall If liquid has tendency to adhere to the tube walls (e.g. water), energy lowest when most surface is covered Liquid creeps up wall (concave meniscus) Pressure beneath curve of meniscus is lower than atmosphere by 2g/r, where r is radius (Kelvin equation, cylindrical tube) Pressure at flat surface = atmospheric pressure (r ® ∞) Liquid rises in capillary until hydrostatic equilibrium is reached As liquid rises p increases by rgh (r = density; h= height) At equilibrium, pcapillary = pexternal or 2g/r = rgh Height of capillary rise h = 2g/rgr As tube gets smaller, h gets higher Can be used to measure surface tension of liquids g is temperature dependent If liquid has a tendency not to adhere (e.g. Hg) liquid will fall in capillary because pressure less under meniscus Treatment the same, except sign reversed

Contact Angle Contact angle is the angle between edge of meniscus and wall If qc ≠ 0, then the equation for capillary rise becomes h = 2g(Cosqc)/rgr Arises from balance of forces at the point of contact between liquid and solid The surface tension is essentially the energy needed to create a unit area of each of the interfaces gsg = Energy to create unit area at gas-solid interface glg = Energy to create unit area at gas-liquid interface gsl = Energy to create unit area at solid-liquid interface At equilibrium, the vertical forces in capillary are in balance so gsg = gsl + Cos(qc) glg Or Cos(qc) =( gsg -gsl )/ glg See diagram® Work of adhesion (wad) of liquid to solid is gsg + glg - gsl So Cos(qc) =( gsg -gsl )/ glg =(wad -glg )/ glg =(wad / glg ) - 1 If liquid wets surface 0° < qc < 90° , 1> Cos(qc) > 0 so 1< (wad / glg ) < 2 If liquid doesn’t wet surface 90° < qc < 180° , 0> Cos(qc) > -1 so 1> (wad / glg ) >0 Takes more work to overcome cohesive forces in liquid “wets” “doesn’t wet”

Chapter 7: Simple Mixtures
Homework: Exercises(a only):7.4, 5,10, 11, 12, 17, 21 Problems: 1, 8

Chapter 7 - Simple Mixtures
Restrictions Binary Mixtures xA + xB = 1, where xA = fraction of A Non-Electrolyte Solutions Solute not present as ions

Partial Molar Quantities -Volume
Partial molar volume of a substance slope of the variation of the total volume plotted against the composition of the substance Vary with composition due to changing molecular environment VJ = (V/ nJ) p,T,n’ pressure, Temperature and amount of other component constant Partial Molar Volumes Water & Ethanol

Partial Molar Quantities & Volume
If the composition of a mixture is changed by addition of dnA and dnB dV = (V/ nA) p,T,nA dnA + (V/ nB) p,T,nB dnB dV =VAdnA + VBdnB At a given compositon and temperature, the total volume, V, is V = nAVA + nBVB

Measuring Partial Molar Volumes
Measure dependence of volume on composition Fit observed volume/composition curve Differentiate Example - Problem 7.2 For NaCl the volume of solution from 1 kg of water is: V= b b b2 What are the partial molar volumes? VNaCl = (∂V/∂nNaCl) = (∂V/∂nb) = (1.77 x 1.5)b0.5 + (0.12 x 2) b1 At b =0.1, nNaCl = 0.1 VNaCl = b b = cm3 /mol V = cm3 nwater = 1000g/(18 g/mol) = 55.6 mol V = nNaClVNaCl + nwaterVwater Vwater = (V - VNaCl nNaClr )/ nwater = ( )/55.6= cm3 /mol

Partial Molar Quantities - General
Any extensive state function can have a partial molar quantity Extensive property depends on the amount of a substance State function depends only on the initial and final states not on history Partial molar quantity of any function is just the slope (derivative) of the function with respect to the amount of substance at a particular composition For Gibbs energy this slope is called the chemical potential, µ

Partial Molar Free Energies
Chemical potential, µJ, is defined as the partial molar Gibbs constant P, T and other components µJ = (G/ nJ) p,T,n’ For a system of two components: G = nAµA + n B µB G is a function of p,T and composition For an open system constant composition, dG =Vdp - SdT + µA dnA + µB dn B Fundamental Equation of Thermodynamics @ constant P and T this becomes, dG = µA dnA + µB dn B dG is the the non expansion work, dwmax FET implies changing composition can result in work, e.g. an electrochemical cell

Chemical Potential Gibbs energy, G, is related to the internal energy, U U = G - pV + TS (G = U + pV - TS) For an infinitesimal change in energy, dU dU = -pdV - Vdp + TdS + SdT + dG but dG =Vdp - SdT + µA dnA + µB dn B so dU = -pdV - Vdp +TdS +SdT + Vdp - SdT + µA dnA + µB dn B dU = -pdV + TdS + µA dnA + µB dn B at constant V and S, dU = µA dnA + µB dn B or µJ = (U/ nJ)S,V,n’

µ and Other Thermodynamic Properties
Enthalpy, H (G = H - TS) dH = dG + TdS + SdT dH= (Vdp - SdT + µA dnA + µB dn B) - TdS SdT dH = VdP - TdS + µA dnA + µB dn B at const. p & T : dH = µA dnA + µB dn B or µJ = (H/ nJ)p,T,n’ Helmholz Energy, A (A = U-TS) dA = dU - TdS - SdT dA = (-pdV + TdS + µA dnA + µB dn B ) - TdS - SdT dA = -pdV - SdT + µA dnA + µB dn B at const. V & T : dA = µA dnA + µB dn B or µJ = (A/ nJ)V,T,n’

Gibbs-Duhem Equation Recall, for a system of two components:
G = nAµA + n B µB If compositions change infinitesimally dG = µA dnA + µB dn B + nAdµA + n Bd µB But at constant p & T, dG = µA dnA + µB dn B so µA dnA + µB dn B = µA dnA + µB dn B + nAdµA + n Bd µB or nAdµA + n Bd µB = 0 For J components, nidµi = 0 (i=1,J) {Gibbs-Duhem Equation}

Significance of Gibbs-Duhem
Chemical potentials of multi-component systems cannot change independently Two components, G-D says, nAdµA + n Bd µB = 0 means that d µB = (nA/ n B)dµA Applies to all partial molar quantities Partial molar volume dVB = (nA/ n B)dVA Can use this to determine on partial molar volume from another You do this in Experiment 2

Example Self Test 7.2 VA = 6.218 + 5.146b - 7.147b2
dVA = *7.147b = b db dVA/db = b b If *MB is in kg/mol dVB = -nA/nB (dVA); b=nA/nB*MB or b *MB = nA/nB dVB = -nA/nB (dVA ) = nA/nB dVA = b *MB dVA dVB = -b* MB ( b) db =- MB(2.573b-4.765b2) VB =VB* + MB (4.765b b) from data VB* = cm3mol-1 and MB = kg/mol so VB = cm3mol b b

Thermodynamics of Mixing
Gibbs Energy of Mixing Of Two Ideal Gases For 2 Gases (A &B) in two containers, the Gibbs energy, Gi Gi = nAµA + nBµB But µ = µ° + RTln(p/p°) so Gi = nA(µA° + RTln(p/p°) )+ nB(µB° + RTln(p/p°)) If p is redefined as the pressure relative to p° Gi = nA(µA° + RTln(p) )+ nB(µB° + RTln(p) ) After mixing, p = pA + pB and Gf = nA(µA° + RTln(pA) )+ nB(µB° + RTln(pB) ) So Gmix = Gf - Gi = nA (RTln(pA/p) )+ nB(RTln(pB /p) Replacing nJ by xJn and pJ/p=xJ (from Dalton’s Law) Gmix = nRT(xA ln (xA ) + xBln(xB )) This equation tells you change in Gibbs energy is negative since mole fractions are always <1

Example :Self-Test 7.3 2.0 mol atm) + 4 mol N2 atm) mixed at const. V. What is Gmix? Initial: pH2= 2 atm;VH2= 24.5 L; pN2= 3 atm;VN2= 32.8 L{Ideal Gas} Final: VN2= VH2= 57.3 L; therefore pN2= atm; pH2= atm;{Ideal Gas} Gmix = RT(nA ln (pA /p) + nBln(pB /p)) Gmix = (8.315 J/mol K)x(298 K)[2mol x ( ln(0.855/2) + 4 mol x (ln(1.717/3)] Gmix = -9.7 J What is Gmix under conditions of identical initial pressures? xH2 = 0.333; xN2 = 0.667; n = 6 mol Gmix = nRT(xA ln (xA ) + xBln(xB )) Gmix = 6mol x( 8.315J/molK)x K{0.333ln ln0.667) Gmix = -9.5 J

Entropy and Enthalpy of Mixing
For Smix, recall G = H - TS Therefore Smix = -Gmix / T Smix = - [ nRT(xA ln (xA ) + xBln(xB ))] / T Smix = - nR(xA ln (xA ) + xBln(xB ) It follows that Smix is always (+) since xJln(xJ ) is always (-) For Hmix H = G + TS ={nRT(xA ln (xA ) + xBln(xB )} +T{- nR(xA ln (xA ) + xBln(xB )} H ={nRT(xA ln (xA ) + xBln(xB )} - {nRT(xA ln (xA ) + xBln(xB )} H = 0 Thus driving force for mixing comes from entropy change Entropy of Mixing Two Ideal Gases

Chemical Potentials of Liquids Ideal Solutions
At equilibrium chem. pot. of liquid = chem. pot. of vapor, µA(l) = µA(g,p) For pure liquid, µ*A(l) = µ°A + RT ln(p *A) [1] For A in solution, µA(l) = µ°A + RT ln(p A) [2] Subtracing [1] from [2] : µA(l) - µ*A(l) = RT ln(pA) + RT ln(p *A) µA(l) - µ*A(l) = RT{ln(pA) - ln(p *A)} = RT{ln(pA/p *A)} µA(l) = µ*A(l) + RT{ln(pA/p *A)} [3] Raoult’s Law - ratio of the partial pressure of a component of a mixture to its vapor pressure as a pure substance (pA/p*A) approximately equals the mole fraction, xA pA = xA p*A Combining Raoult’s law with [3] gives µA(l) = µ*A(l) + RT{ln(xA)}

Ideal Solutions/Raoult’s Law
Mixtures which obey Raoult’s Law throughout the composition range are Ideal Solutions Phenomenology of Raoult’s Law: 2nd component inhibits the rate of molecules leaving a solution, but not returning rate of vaporization  XA rate of condensation  pA at equilibrium rates equal implies pA = XA p*A

Deviations from Raoult’s Law
Raoult’s Law works well when components of a mixture are structurally similar Wide deviations possible for dissimilar mixtures Ideal-Dilute Solutions Henry’s Law (William Henry) For dilute solutions, v.p. of solute is proportional to the mole fraction (Raoult’s Law) but v.p. of the pure substance is not the constant of proportionality Empirical constant, K, has dimensions of pressure pB = xBKB (Raoult’s Law says pB = xBpB) Mixtures in which the solute obeys Henry’s Law and solvent obeys Raoult’s Law are called Ideal Dilute Solutions Differences arise because, in dilute soln, solute is in a very different molecular environment than when it is pure

Applying Henry’s Law & Raoult’s Law
Henry’s law applies to the solute in ideal dilute solutions Raoult’s law applies to solvent in ideal dilute solutions and solute & solvent in ideal solutions Real systems can (and do ) deviate from both

Applying Henry’s Law What is the mole fraction of dissolved hydrogen dissolved in water if the over-pressure is 100 atmospheres? Henry’s constant for hydrogen is x 107 PH2= xH2K; xH2 = PH2 /K= 100 atm x 760 Torr/atm/ 5.34 x 107 xH2 = 1.42 x 10-3 In fact hydrogen is very soluble in water compared to other gases, while there is little difference between solubility in non-polar solvents. If the solubility depends on the attraction between solute and solvent, what does this say about H2 -water interactions?

Properties of Solutions
For Ideal Liquid Mixtures As for gases the ideal Gibbs energy of mixing is Gmix = nRT(xA ln (xA ) + xBRTln(xB )) Similarly, the entropy of mixing is Smix = - nR(xA ln (xA ) + xBln(xB ) and Hmix is zero Ideality in a liquid (unlike gas) means that interactions are the same between molecules regardless of whether they are solvent or solute In ideal gases, the interactions are zero

Real Solutions In real solutions, interactions between different molecules are different May be an enthalpy change May be an additional contribution to entropy (+ or - ) due to arrangement of molecules Therefore Gibbs energy of mixing could be + Liquids would separate spontaneously (immiscible) Could be temperature dependent (partially miscible) Thermodynamic properties of real solns expressed in terms of ideal solutions using excess functions Entropy: SE = Smix - Smixideal Enthalpy: HE = Smix(because Hmixideal = 0) Assume HE = nbRTxAxB where is const. b =w/RT w is related to the energy of AB interactions relative to AA and BB interactions b > 0, mixing endothermic; b < 0, mixing exothermic solvent-solute interactions more favorable than solvent-solvent or solute-solute interactions Regular solution is one in which HE  0 but SE  0 Random distribution of molecules but different energies of interactions GE = HE Gmix = nRT{(xA ln (xA ) + xBRTln(xB )) + bRTxAxB (Ideal Portion + Excess)

Activities of Regular Solutions
Recall the activity of a compound, a, is defined a = gx where g = activity coefficient For binary mixture, A and B, consideration of excess Gibbs energy leads to the following relationships (Margules’ eqns) ln gA = bxB2 and ln gB = bxA [1] As xB approaches 0, gA approaches 1 Since, aA = gAxA, from [1] If b = 0, this is Raoult’s Law If b < 0 (endothermic mixing), gives vapor pressures lower than ideal If b > 0 (exothermic mixing), gives vapor pressures higher than ideal If xA<<1, becomes pA = xA eb pA* Henry’s law with K = eb pA*

Colligative Properties of Dilute Solutions

Colligative Properties
Properties of solutions which depend upon the number rater than the kind of solute particles Arise from entropy considerations Pure liquid entropy is higher in the gas than for the liquid Presence of solute increases entropy in the liquid (disorder increases) Lowers the difference in entropy between gas and liquid hence the vapor pressure of the liquid Result is a lowering chemical potential of the solvent Types of colligative properties Boiling Point Elevation Freezing Point Depression Osmotic pressure

Colligative Properties - General
Assume Solute not volatile Pure solute separates when frozen When you add solute the chemical potential, µA becomes µA = µA * + RT ln(xA) where µA * = Chemical Potential of Pure Substance x A = mole fraction of the solvent Since ln(xA) in negative µA > µA*

Boiling Point Elevation
At equilibrium µ(gas) = µ(liquid) or µA(g) = µA *(l) + RTln(xA) Rearranging,(µA(g) - µA *(l))/RT = ln(xA) = ln(1- xB) But , (µA(g) - µA *(l)) = G vaporization so ln(1- xB) = G vap. /RT Substituting for G vap. (H vap. -T S vap. ) {Ingnore T dependence of H & S) ln(1- xB) = (H vap. -T S vap.)/RT = (H vap. /RT) - S vap./R When xB = 0 (pure liquid A), ln(1) = (H vap. /RTb) - S vap./R = 0 or H vap. /RTb = S vap./R where Tb= boiling point Thus ln(1- xB) = (H vap. /RT) - H vap. /RTb = (H vap. /R)(1/T- 1/Tb) If 1>> xB, (H vap. /R)(1/T- 1/Tb)  - xB and if T  Tb and T= T  Tb Then (1/T- 1/Tb) = T/Tb2 and (H vap./R) T/Tb2 = - xB so T= - xB Tb2 /(H vap./R) or T= - xB Kb where Kb = Tb2 /(H vap./R)

Boiling Point Elevation
Kb is the ebullioscopic constant Depends on solvent not solute Largest values are for solvents with high boiling points Water (Tb = 100°C) Kb = 0.51 K/mol kg -1 Acetic Acid (Tb = 118.1°C) Kb = 2.93 K/mol kg -1 Benzene (Tb = 80.2°C) Kb = 2.53 K/mol kg-1 Phenol (Tb = 182°C) Kb = 3.04 K/mol kg -1

Freezing Point Depression
Derivation the same as for boiling point elevation except At equilibrium µ(solid) = µ(solid) or µA(g) = µA *(l) + RTln(xA) Instead of the heat of vaporization, we have heat of fusion Thus, T= - xB Kf where Kf = Tf2 /(H fus./R) Kf is the cryoscopic constant Water (Tf = 0°C) Kf = 1.86 K/mol kg -1 Acetic Acid (Tf = 17°C) Kf = 3.9 K/mol kg -1 Benzene (Tf = 5.4°C) Kf = 5.12 K/mol kg-1 Phenol (Tf = 43°C) Kf = 7.27 K/mol kg -1 Again property depends on solvent not solute

Temperature Dependence of Solubility
Not strictly speaking colligative property but can be estimated assuming it is Starting point the same - equilibrium µ is equal for two states First state is solid solute, µB(s) Second state is dissolved solute, µB(l) µB(l) = µB*(l) + RT ln xB At equilibrium, µB(s) = µB(l) µB(s) = µB*(l) + RT ln xB Same as expression for freezing point except that use xB instead of xA

Temperature Dependence of Solubility
To calculate functional form of temperature dependence you solve for mole fraction ln xB = [µB(s) - µB*(l)]/ RT = -G fusion/RT = -[H fus-T S fus]/RT ln xB = -[H fus-T S fus]/RT = -[H fus /RT] + [S fus/R] {1} At the melting point of the solute, Tm, G fusion/RTm = 0 because G fusion = 0 So [H fus-Tm S fus]/RTm = 0 or [H fus /RTm] -[S fus/R] = 0 Substituting into {1}, ln xB = -[H fus /RT] + [S fus/R]+ [H fus /RTm] -[S fus/R] This becomes ln xB = -[H fus /RT] + [H fus /RTm] Or ln xB = -[H fus /R] [1/T - 1/Tm] Factoring Tm, ln xB = [H fus /R Tm] [1 - (Tm /T)] Or xB = exp[H fus /R Tm] [1 - (Tm /T)-1] The details of the equation are not as important as functional form Solubility is lowered as temperature is lowered from melting point Solutes with high melting points and large enthalpies of fusion have low solubility Note does not account for differences in solvent - serious omission

ln xB = -[H fus /R] [1/T - 1/Tm] or xB = exp[H fus /R Tm] [1 - (Tm /T)-1]

Osmotic pressure J. A. Nollet (1748) - “wine spirits” in tube with animal bladder immersed in pure water Semi-permiable membrane - water passes through into the tube Tube swells , sometimes bladder bursts Increased pressure called osmotic pressure from Greek word meaning impulse W. Pfeffer (1887) -quantitative study of osmotic pressure Membranes consisted of colloidal cupric ferrocyanide Later work performed by applying external pressure to balance the osmotic pressure Osmotic pressure , , is the pressure which must be applied to solution to stop the influx of solvent

Osmotic Pressure van’t Hoff Equation
J. H. van’t Hoff (1885) - In dilute solutions the osmotic pressure obeys the relationship, V=nBRT nB/V = [B] {molar concentration of B, so =[B] RT Derivation- at equilibrium µ solvent is the same on both sides of membrane: µA *(p) = µA (x A,p +) {1} µA (x A,p +) = µ*A (x A,p +) + RTln(x A) {2} µ*A (x A,p +) = µA *(p) + ∫p p +Vm dp {3} [Vm = molar volume of the pure solvent] Combining {1} and {2} : µA *(p) = µA *(p) + ∫p p +Vm dp + RTln(x A) For dilute solutions, ln(x A) = ln(x B) ≈ - x B Also if the pressure range of integration is small, ∫p p +V m dp = Vm∫p p +dp = Vm  So 0 = V m + RT - x B or Vm= RT x B Now nA V m = V and, if solution dilute x B ≈ - n B /nA so =[B] RT Non-ideality use a virial expansion =[B] RT{1 + B[B] + ...} where B I s the osmotic virial coef. (like pressure)

Application of Osmotic Pressure
Determine molar mass of macromolecules =[B] RT{1 + B[B] + ...} but [B] = c/M where c is the concentration and M the molar mass so = c/M RT{1 + Bc/M + ...} g h = c/M RT{1 + Bc/M + ...} h/c = RT/(Mg) {1 + Bc/M + ...} h/c = RT/(Mg) + RTB/(M2g) + ...} Plot of h/c vs. c has intercept of RT/(Mg) so Intercept= RT/(Mg) or M= RT/(intercept xg) Units (SI) are kg/mol typical Dalton (Da) {1Da = 1g/ mol

Non-Ideality & Activities Solvents
Recall for ideal solution µA =µA * + RTln(xA) µA* is pure liquid at 1 bar when xA =1 If solution does not ideal xA can be replaced with activity aA activtiy is an effective mole fraction aA = pA/pA* {ratio of vapor pressures} Because for all solns µA =µA * + RTln(pA/ p*A) As xA-> 1, aA -> xA so define activity coefficient,, such that aA =  A xA As xA-> 1, A -> 1 Thus µA =µA * + RTln(xA) + RTln(A) {substiuting for a A}

Non-Ideality & Activities Solutes (Ideal & Non-Ideal)
For ideal dilute solutions Henry’s Law ( pB = KBxB) applies Chemical potential µB = µ B * + RTln(pB /pB*) µB = µ B * + RTln(KBxB /pB*) = µ B * + RTln(KB/pB*) + RTln(xB ) KB and pB* are characteristics of the solute so you can combine them with µ B * µB† = µ B * + RTln(KBxB /pB*) Thus µB = µ B† + RTln(xB ) Non-ideal solutes As with solvents introduce acitvity and activity coefficient aB = pB/KB*; aB =  B xB As xA-> 0, aA -> xA and A -> 1

Activities in Molalities
For dilute solutions x B ≈ n B /nA , and x B =  b/b° kappa,  , is a dimensionless constant For ideal-dilute solution, µB = µ B† + RTln(xB ) so µB = µ B† + RTln( b/b°) = µ B†+ RTln() + RTln( b/b°) Dropping b° and combing 1st 2 terms, µB = µ Bø+ RTln( b) µ Bø = µ B†+ RTln() µB has the standard value (µBø ) when b=b° As b ->0, µB ->infinity so dilution stabilizes system Difficult to remove last traces of solute from a soln Deviations from ideality can be accounted for by defining an activity aB and activity, B, aB =  B bB/b° where  B ->1, bB -> 0 The chemical potential then becomes µ = µø + RTln( a) Table 7.3 in book summarizes the relationships

Chapter 8: Phase Diagrams
Homework Exercises (a-only): 8.3,4, 6, 13, 15 Problems 8.2 & 8.4 1

Definitions Phase(P) -“State matter which is uniform throughout not only in chemical composition but also in physical state” J. Willard Gibbs Solid Various phases [e.g. crystal structures (diamond; graphite) or compositions (UC;UC2)] Alloys (sometimes its difficult to tell this - microscopic examination may be necessary {dispersions uniform on macroscopic scale}) Miscible one phase (P=1) Immisible multiple phases (P>1) Liquid Miscible liquids are one phase Immiscible liquids are multiple phases (P>1) Gas Systems consisting of gases can have only one phase Shape or degree of subdivision irrelevant 2

Definitions Heterogeneous and homogeneous systems
Systems with one phase are homogeneous Systems with more than one phase are heterogeneous Constituent- a chemical species (ion or molecule which is present Component (C) - chemically independent constituents of a system C = #of independent chemical constituents - # of distinct chemical reactions #of independent chemical constituents = total # of constituents minus the number of any restrictive conditions (charge neutrality, material balance etc.) 3

Counting Components Components: Example: CaCO3(s) ® CaO(s) + CO2(g)
Phases: P = 2 solid + 1 gas = 3 Component: C = 3 consitiuents - 1 reaction = 2 Example: Consider the system NaCl, KBr and H2O. Suppose you can also isolate the following from it KCl(s), NaBr(s), NaBr. H2O(s), KBr. H2O(s), and NaCl. H2O(s) Phases: P = 5 solid Components: There are 8 constituents You can write the following reactions four: NaCl + KBr ® KCl + NaBr; NaCl + H2O ® NaCl. H2O KBr + H2O ® KBr. H2O; NaBr + H2O ® NaBr. H2O Conditions: 1 -Material balance moles of KCl = moles of NaBr + NaBr. H2O Components: C = (8 constituents -1 condition) - 4 reactions = 3 4

Counting Components (continued)
Remember you must count reactions which actually occur not those which could occur Consider a system in which we has the following : O2(g), H2(g), H2O(g) If conditions are such O2(g) + H2(g) ® H2O(g) does not occur, C=3 If conditions are such (T, catalyst) O2(g), H2(g) ® H2O(g) occurs, then C=3components -1reaction =2 If conditions are such (T, catalyst) O2(g), H2(g) ® H2O(g) occurs and you impose the condition that all the hydrogen and oxygen come from dissociation of water, then C=3 constituents - 1 condition -1 reaction = 1 Identity of components is a matter of some choice, number isn’t Chose components that cannot be converted into one another by reactions E.g. CaCO3(s) ® CaO(s) + CO2(g) 5

Phase Rule In discussing phase equilbria, you need only consider intensity factors (temperature, pressure, and concentration) Only certain of these can be varied independently Some are fixed by the values chosen for the independent variables and by the requirements of thermodynamic equilibrium e.g. if you chose  and T for a system P is fixed Number of variables which can be varied independently without changing the number of phases is called the degrees of freedom of the system The number of degrees of freedom or variance of a system, F, is related to the number of components(C) and number of phases(P) F = C - P + 2 6

Phase Rule Proof Only C-1 mole fractions are needed since the xJ = 1
In any system the number of intensive variables are: pressure, temperature plus the mole fractions of each component of each phase. Only C-1 mole fractions are needed since the xJ = 1 Thus for P phase, the number of intensive variables = P(C-1) + 2 At equilibrium the chemical potential of phase must be equal, i.e. µJP1= µJP2= µJP3= µJP4= µJP5….{there are P-1 such equations} Since there are C components, equilibrium requires that there are C(P-1) equations linking the chemical potentials in all the phases of all the components F = total required variables - total restraining conditions F = P(C-1) C(P-1) = PC - P + 2 -CP + C = C- P + 2 7

One Component Systems Phase rule says that you can have at most 3 phases F = C- P +2; C=1 so F=3-P If P=3, F=0 system is invariant Specified by temperature and pressure and occurs at 1 point (called the triple point) If one phase is present, F = 2 that is P and T can be varied independently This defines an area in a P,T diagram which only one phase is present If two phases are present, F = 1 so only P or T can be varied independently. This defines a line in a P, T diagram 8

Single Component Phase Diagram
Point a: only vapor present Point b: Liquid boils - 1 atm= Tb but vapor and liquid co-exist along line(vapor pressure curve) Point C : Liquid freezes Point D: Only solid Triple point: 3 phases in equilibrium Line Below triple point: Vapor pressure above solid Note: You don’t necessarily have to have 3 phases and they don’t have to be solid liquid and gas 9

Cooling Curve You can generate a cooling constant pressure (isobar) from previous phase diagram Halts occur during 1st order phase transitions (e.g. freezing) 10

Experimental Measurements
Phase changes can be measured by performing DTA (differential thermal analysis) on samples In DTA sample is heated vs. a reference 1st order transitions can be measured even when they can’t be observed They will occur as peaks in DTA High Pressures can be achieved with diamond anvil cells See text descriptions 11

Two Component Systems For two component systems, F = 2-P+2 = 4-P
If P or T is held constant, F’ = 3-P (’ indicates something is constant) Maximum value for F’ is 2 If T is constant one degree of freedom is pressure and the other is mole fraction Phase diagram (Constant T) is map of pressure and compositions at which each phase is stable If P is constant one degree of freedom is temperature and the other is mole fraction Phase diagram (Constant P) is map of temperature and compositions at which each phase is stable Both Useful 12

Vapor Pressure Diagrams
By Raoult’s law (pA = xApA * & pB = xBpB * ), the total pressure p is p = pA + pB = xApA * + xBpB * But xB = (1-xA)so xApA * + xBpB * = xApA * + (1-xA) pB * = pB * + xA(pA * -pB *) @ Constant T, total vapor pressure is proportional to xA (or xB ) The composition of the vapor is given by Raoult’s law so the mole fraction in the gas phase, yA and yB is yA = pA/p and yB = pB/p {also yA = 1-yA) From above yA = xApA * /[pB * + xA(pA * -pA *)] If pA * /pB *= pA/B then yA = (xA * pA/B) /(1+ (xA * pA/B) - xA ) Or yA = (xA * pA/B) /(1+ (xA *( pA/B - 1) ) 13

Effect of Ratio of Vapor Pressure on Mole Fraction in Vapor
This shows the vapor is richer in the more volatile component If B is non volatile then yB = 0 14

Pressure Composition Diagrams
Assume the composition on the x axis is the overall composition, zA (as mole fraction) In Liquid region zA = xA In vapor Region region zA = yA In between two phases present F’=1 so at given pressure compositions are fixed by tie lines 15

Isopleth Compositions in Each Phase
A vertical line represents a line of constant composition or isopleth Until pressure = p1sample is liquid vapor phase composition is a1 At p1, vapor composition is given by tie line to vapor curve At p2 vapor composition is a’2, liquid composition is a2 and overall composition is a At p3 vitually all the liquid is vapor and trace of liquid has composition given by tie line to liquid 16

Determining proportions of Phases (lever rule)
The composition of each phase is given by the each end of the tie line The relative proportion of each phase is given by the length of the tie line nl = nl or n= nl / l 17

Temperature-Compositions
Assume A more volatile than B Region between two curves is 2-phase region F’=1 (pressure is fixed) At given temperature compositions are fixed by tie lines Region outside lines composition & temperature variable Heat liquid with composition a1 Hits boiling curve, vapor has composition a2’, liquid a2 (=a1) vapor is richer in more volatile component Distillation Vapor condensed (a2’-a3) New concentration a3’ (richer still) New etc until nearly pure liquid obtained

Distillation/Theoretical Plates
A theoretical plate is a vaporization-condensation step Previous example has 3 theortetical plates If the two curves move closer together, more theoretical plates are required to achieve same degree of separation Curves more together if components have similar vapor pressures

Non-Ideal T-C Diagrams - High Boiling Azeotropes
Maximum in phase diagram occurs when interactions in liquid between A & B stabilize the liquid GE is more negative If such a liquid is boiled, as vapor is removed, composition of liquid is richer in B (less A) As vapor is removed you move to right up the curve until you reach point b At b liquid boils with constant composition Called an azeotrope (unchanging Gr.) Example HCl-water boils @80 wt % water at 108.6°C

Non-Ideal T-C Diagrams - Low Boiling Azeotropes
Minimum in phase diagram occurs when interactions in liquid between A & B destabilize the liquid GE is more positive If such a liquid is boiled, & vapor is condensed , composition of vapor is richer in B (less A) As vapor is removed you move to right down the curve until you reach point b At b liquid boils with constant composition Example ethanol-water boils at constant water content of 4 78°C

Non-Ideal T-C Diagrams - Immisicble Liquids
If two liquids immiscible and in intimate contact then p is nearly the sum of vapor pressures of pure components (p = pA* +pB*) mixture will boil when p = atmospheric pressure intimate contact (& trace level saturation maintained) If two liquids immiscible and not in intimate contact then p for each is the vapor pressures of pure components (p = pA* and p = pB*) Each will boil separately when respective pA* = atmospheric pressure and/or pB* = atmospheric pressure

Liquid-Solid Phase Diagrams
Liquids miscible & solids immiscible Consider Cooling along isopleth from a1 At a2 pure B starts to come out of solution At a3 solution is mixture of B + Liquid with composition b3 (ratio by lever rule) At a4 liquid has composition “e” and freezes In solid region there are two phases pure A and pure B Composition given by tie line, ratio by lever rule “e” is called a eutectic

Examples of Simple Eutectic Systems

Eutectics In previous diagram, the eutectic (easily melted, Gr.) point is a temperature at which a mixture freezes without first depositing pure A or B Like a melting point in that it it is a definite temperature That’s because, since C=2 and P=3, by Phase rule, F’=0 A cooling (or heating) curve will have a halt at the eutectic temperature If pure A and pure B are in contact a liquid will form at the eutectic temperature Examples solder lead/tin (67/33) melting point 183°C NaCl and water (23/77) melting point -21.1°C

Liquid-Solid Phase Diagrams - Reacting Systems
Some Binary systems react to produce one (or more) compounds Definite composition Unique melting point Congruent melting point, I.e. melts to a liquid of identical composition Maximum in phase diagram Phase diagram interpreted as before except now there are additional regions

PARTICLE BED REACTORS (PBRs)
High Power Density Nuclear Sources for Space Power & Propulsion Performance superior to chemical rockets (H2/O2) Enabling technology for Mars mission Multiple coolants possible He for power applications H2, NH3 for propulsion applications High power densities (10’s MW/liter) Superior performance to NERVA system (70’s era nuclear propulsion system) Typical operating temperatures >2500 K for propulsion 2

PBR - Schematic

Materials Requirements for PBR Hot Components
Hot frit, nozzle, etc. Withstand H2 Environment ~2800 K 70 atmospheres Large Temperature Gradients 12K to Tmax over a few cm Multiple Thermal Cycles Long Exposure Times 10’s minutes Launch Stresses Withstand Radiation Fields Not Affect Reactor Criticality Fuel Same as general components Provide for adequate reactivity Optimize coating thickness and type to maintain criticality Maintain (Keff) No HfC coatings Maintain coolable geometry No large gaps between layers No particle clumping No reaction with other components, e.g. hot frit Minimum F.P. and U release Criticality and safety criterion 3

Hot Component Material Selection
Reactor Components (Hot Frit) Rhenium (monolithic & coatings) Large neutronic penalty for monolithic Re High radiation heating for monolithic Re Extensive alloying of Re with fuel T > 2760 K Pyrolytic BN High cost 11BN required Unacceptable thermal decomposition (3 wt % in K Reaction with baseline ZrC fuel coating Carbide-Coated Carbon (graphite & carbon-carbon) Potential for CTE mismatch between coatings and substrate Fuels Baseline fuel HTGR-type with ZrC coating Conclusion: Materials development program focused on carbide coatings of carbon materials 4

“HTGR” Type Fuel UC Outer carbide shell Pyrocarbon layer(s)
HTGR - SiC PBR - ZrC Pyrocarbon layer(s) “Spongy” layer Dense layer Inner kernel HTGR - UO2 PBR - UC UC melting point 2525°C UC °C UC

Carbide Phase Diagrams
Tantalum-Carbon Zirconium-Carbon

Liquid-Solid Phase Diagrams - Reacting Systems (Incongruent Melting)
If the compound is not stable as a liquid incongruent melting occurs Compound melts into components Called the peritectic melting point One solid phase “melts around” the other Isopleths “a” a1-> a2 liquid phase with A + B a2 solid B precipitates a3->a4 Solid B + compound “b” b2-> b3 liquid phase with A + B b3 B reacts to form compound b3->b4 Solidcompound + liquid b5 solid A precipitates with compound

Zone Refining Ultra high purity can be obtained by moving a small molten zone across a sample. Impurities more soluble in liquid than solid so they continually move down the liquid front One end becomes purer while other end is dirtier Multiple passes can be used to achieve high purity

Ternary Phase Diagrams
Each composition must be defined by two compositions or mole fractions Composition diagrams are therefore two dimensional Triangle with each edge one line of binary phase diagram Pressure or temperature add third dimension Usually temperature Phase diagrams are usually given as a succession of surfaces at constant temperature To examine temperature variation you hold composition constant

Ternary Phase Diagrams

Chapter 9: Chemical Equilibrium
Homework: Exercises (a only): 9.5, 6, 7, 9, 11, 12 Problems: 9.1, 8

Reaction Gibbs Energy Consider the simple reaction A ˛ B
Examples d-alanine to l- alanine Suppose an infintesimal amount of nA converts to nB, then dnA =-dnB If we define x (ksi)as the extent of reaction then d x = -dnA =dnB Dimensions of x are moles For finite reaction the change is Dx and nA goes to nA - Dx and nB goes to nB +Dx Reaction Gibbs Energy is the slope of the Gibbs energy vs. x or DrG = (dG/d x) p,T Remember, dG = µA dnA + µB dnB so dG = -µA d x + µB d x or DrG = (dG/d x ) p,T = µB -µA Reaction Gibbs energy is the difference of the product chemical potential and the reactant chemical potential

Gibbs Energy and Equilibrium
Initially µB <µA so DrG <0 and the reaction is spontaneous Called an exergonic reaction (work producing) If µB >µA so DrG >0 and the reverse reaction is spontaneous Called an endergonic reaction (work consuming) When µB =µA so DrG >0 or system is at equilibrium Thus a minimum in the reaction Gibbs energy plot represents equilibrium At a minimum, the slope = 0

Ideal Gas Equilibria Consider homogenous gas phase reactions
Reactants and products are gases Gases to 1st approximation behave as ideal gases Since DrG = (dG/d x ) p,T = µB -µA and µ = µ° + RT ln(p), DrG = µB° + RT ln(p B) - (µA°+ RT ln(p A)) = µB°- µA° + RT ln(pB /pA) DrG = DrG ° + RT ln(Q) where DrG ° = µB° - µA°, the standard reaction Gibbs energy Like std reaction enthalpy it is the difference in standard free energies of formation of the products - that of reactants Q = pB /pA, the reaction quotient which ranges from 0 (pure A) to infinity (pure B) At equilibrium, DrG = 0, so DrG ° = - RT ln(K) or K = exp(-DrG °/RT) K is the equilibrium value of Q or K = (pB /pA )equilibrium

Comments on the Above The minimum in Gibbs energy arises from the mixing of reactant and products If there were no mixing, G would change linearly in proportion to the amount of B formed Slope of G vs. x would be DrG ° In chapter 7, we learned DmixG ° = nRT(x A ln(x A) + x B ln(x B) whi9ch is a U shaped function (minimum at 50% B) If DrG ° <0 then ln(K) >0 so product is favored If DrG ° >0 then ln(K) <0 so reactant is favored

Generalization of Above
If you have a number of products and reactants (all ideal gases), Dalton’s Law tells you PiV = vi RT where vi is the number of moles of the ith component (stoichiometric factor in the equation) For each component, the Gibbs energy difference Gi -Gi° = RT ln(Pi) At constant temperature, dG =VdP =nRTd lnP so DRG -DRG° = RTS vi ln(Pi) equilibrium DRG =0 Remember DRGi° = DRGi° (products ) - DRGi° (reactants) = RT(S vi ln(Pi(prod) - S vi ln(Pi(reactants) ) Thus -DRG° = RTS vi ln(Pi)equilibrium or -DRG° /RT = S vi ln(Pi) equilibrium DRG° is a constant at t=constant so exp(-DRG° /RT ) = K Further, S vi ln(Pi) = ln(P Pi vi )equilibrium So K = P Pi vi vi is positive for products and negative for reactants In terms of activities (aA = pA/p*A), K =( P ai vi )equilibrium

Notes on Generalization
For pure solids and liquds a=1 so they don’t contribute to the reaction quotient (or K) For ideal gases (aA = pA/p*A), but for real gases (aA = fA/p*A), where fA is the fugacity K is a function of temperature and DRG° is also a function of temperature K is independent of total pressure and variation in partial pressures Pressures varied by changing proportions of reactants and products consistent with K =( P ai vi )equilibrium K is called the thermodynamic equilibrium constant Like activities it is dimensionless You can approximate K by replacing fugacities by partial pressures or molalities or molar concentrations Poor approximation in electrolyte solutions or in conc. solutions

Application to a Gas Phase Equation - Formation of NO @ 25°C
Reaction: N2 (g) + O2 (g) ˛ 2NO(g) Data: DGf°(N2 ) = 0; DGf°(O2 ) = 0; DGf°(NO ) = kJ/mol DRG° = 2(86.6 kJ/mol) = kJ/mol K = exp(-DRG° /RT ) = exp ( kJ/mol/(2.48 kJ/mol)) K = 4.67 x 10-31 4.67 x = a2NO/ aN2 aO2 = (fNO p °)2/ (fN2) (fO2) At low pressures fA =pA ,so K= 4.67 x = (pNO p °)2/ (pN2) (pO2) At atmospheric pressure pNO2 =(4.67 x )( x0.2094) = 2.86 x atm; pNO = 1.69 x atm

Degree of Dissociation (Self Test 9.2)
For the equation, H2O(g), ˛ H2 (g) + 1/2O2 (g), what is the mole fraction of oxygen resulting from passing steam through a tube at 2000 K if the Gibbs energy is kJ/mol and P = 200kPa? K = exp(-DRG°/RT ) = exp (( kJ/mol)/(8.3145*2000 J/mol)) =exp(-8.13)=2.945 x 10-4 If x moles of water dissociate then fraction left is 1-x, fraction of hydrogen = x and fraction of oxygen=x/2 Total pressure = {(1-x) + x +x/2}p = {1 + 1/2x} [p=200 kPa] Partial pressures :water = (1-x)p/(1+1/2x); H2 = xp/(1+1/2x); O2 = 1/2xp/(1+1/2x) K = {(xp/(1+1/2x)( 1/2xp/(1+1/2x))0.5)/[ (1-x)p/(1+1/2x)]; if 1>>x this simplifies to K = (1/2) 0.5(xp) 1.5 /p = (p/2) 0.5(x) 1.5 = x 10-4 Substituting for p (200 kPa/105 Pa)and solving for x , (x) 1.5 = x 10-4 ;so x = Mole fraction of oxygen = x/2=

Equilibrium Constant & Activity Coefficients
Recall that the activity, a is the product of the activity coefficient, g, and mole fraction, x or the activity coefficient, g, and concentration, b Thus for A + B˛ C + D K = (gCaC)(gDaD )/(gAaA)(gBaB) or K = (gCgD)/ (gAgB) ( aCaD )/ (aAaB)= Kg KB For dilute solutions, Kg = 1 and K is approximately KB

A Statistical View of Equilibrium
All molecules have energy levels corresponding to electronic vibrational and rotational energies The total internal energy is the sum of the vibrational, rotational and electronic energies, eJ = ev + er + ee and the partition functions are also additive, zJ = zv + zr + ze Partition function is the denominator in the Boltzman distribution Boltzman distribution is the population distribution between states. For a simple reaction A-> B, the equilibrium constant is the sum of the probabilities that the probabilities that the system be found at in one of the energy levels of B divided by the sum of the probabilities that it will be found in one of the states of A or K = z°B / z°B exp (-DeO/kT), where DeO is the difference between the lowest states of A and B

Statistical View (cont.)
K = z°B / z°B exp (-DeO/kT), If the spacing of energy levels the same, then the dominate species has the lower energy levels K dominated by DeO Enthalpy dominates If the spacing of the levels is higher for one species than the other (a greater higher density of states), the partition function will be greater and it will dominate K Entropy donimates

Example Consider an endothermic reaction Na2˛ 2Na
DeO = 0.73 eV and exp (-DeO/kT) = 2.09 x 10-4 Energetically equilibrium would be toward reactants But K =2.4248 Ground state of Na2 (A)is a singlet whereas ground state of atomic Na(B) is a doublet Atomic Na has two nearly superimposed states and a much higher statistical weight

Le Chatelier-Braun Principle
Henry Le Chatelier (1888) and F. Braun (1887) “Any system in chemical equilibrium, as a result of the variation in one of the factors determining the equilibrium, undergoes a change such that, if this change had occurred by itself, it would have introduced a variation in the opposite direction” H. Le Chatelier or If a system in equilibrium is perturbed by subjecting it to a small variation in one of the variables that define the equilibrium, it will tend to return to an equilibrium state , which is usually somewhat different than the initial state

Proof of Le Chatelier-Braun
Recall, dG = -SdT + Vdp + (S viµi) d x Define the affinity, A , as (dG/d x ) p,T At equilibrium, -A = 0 = S viµI (recall T and p are constant) Now -d A = d (dG/d x ) p,T or from above -d A = - (dS/d x ) p,T dT + (dV/d x p,T )dp + (d2G/d2x ) p,T d x For all equilibrium states -d A = d (dG/d x ) p,T = 0 or - (dS/d x ) p,T dT + (dV/d x p,T )dp + - (d2G/d2x ) p,T d x = 0

Proof of Le Chatelier-Braun (cont.)
At constant T, (d xe / dP)T = -[(dV/d x ) p,T ]/[(d2G/d2x ) p,T ], xe is the equilibrium extent of reaction (d2G/d2x ) p,T >0 at stable equilibrium (definition of stable equilibrium) so if P is increased at constant T, the extent of reaction increases in the direction that the volume of the system is decreased at constant T and P At constant p, (d xe / dT) p = [(dS/d x ) p,T ]/[(d2G/d2x ) p,T ]= [T(dq/dx ) p,T]/[(d2G/d2x ) p,T ] dq is the reversible work and xe is the equilibrium extent of reaction (d2G/d2x ) p,T >0 at stable equilibrium (definition of stable equilibrium) so if T is increased at constant p, the extent of reaction increases in the direction that heat is absorbed by the system at constant T and P

Response of Equilibria to Pressure
Recall, K depends on DrG° DrG° is defined at a standard presure, hence is independent of pressure so (dK/dp)T = 0 Just because the constant is independent of pressure doesn’t mean the composition of gases is Effect of additional inert gas - none because partial pressures unchanged (assumes ideality Compression (changing volume): Le Chatelier’s Principle says reaction will adjust to minimize pressure increase

Response of Equilibria to Pressure
Example: A -> 2B; K = pB2/pAp° On compression [A] will increase to compensate for pressure change If extent of reaction is a then we start with n moles the A = (1- a)n and b=2 a n pb Mole fraction of A, xA = (1- a )n/((1- a)n + 2 a n) = (1- a )/(1+ a) Mole fraction of B, xA = 2a /(1+ a) K = [(2a /(1+ a))p]2/(1- a )/(1+ a)p = p[(4a 2/(1+ a)(1- a )] or K = 4a 2 p/(1- a 2 ) or 4a 2 p + K a 2 - K = (4p + K) a 2 - K = 0 Note : p = p/p° Solving for a 2 = K/ (4p + K) = 1/(1 + 4p/K) or a =1/(1 + 4p/K)1/2 This says as p increases then the extent of reaction decreases

Example: 3H2 + N2 -> 2NH2 K = pNH32/p2H2pN2
K = xNH32p2/x3H2p3xN2 p where pA =xAp K = [xNH32 /x3H2 xN2 ] [p2/p 4 ] = [xNH32 /x3H2 xN2 ] [p2/p 4 ] K = KX 1/ p2 where, KX = [xNH32 /x3H2 xN2 ] or KX a p2 Since K is constant, if p increases must increase as the square Self test 9.3: If there is a ten-fold increase in pressure, what will be the effect on the constituents KX a p2 so a ten fold increase in pressure will increase KX by 100 This means that the products must increase or the reactants must decrease Le Chatleier’s principle tells you that by inspection since reactants have the greatest contribution to the total pressure

Effect of Temperature on Equilibrium
-DG/RT = lnK so dlnK /dT =- (1/R)(d(DG /T)/dT) Recall Gibbs-Helmhotz equation (Ch. 5): d(DG /T)/dT = [d(DG /T)/d(1/T)][d(1/T)/dT] d(DG /T)/dT = [d(DG /T)/d(1/T)][1/T2] Now, DG = DH - TDS or DG/T = DH/T - DS d(DG /T)/dT = [d(DH /T -DS)/d(1/T)][1/T2] = DH /T2 (G-H eqn) Thus, d(lnK) /dT = DH /RT2 van’t Hoff Equation Because d(1/T)/dT = -1/T2 , dT = -T2 d(1/T) d(lnK) /d(1/T) = -DH /R van’t Hoff Equation (alt. Form) Says that plot of K vs. 1/T is a straight line with a slope = -DH /R Can be used to estimate reaction enthalpy (see example 9.3)

d(lnK) /d(1/T) = -DH /R If reaction is exothermic DH <0 slope of ln(K) vs. 1/T is positive (>0) As T increases 1/T decreases so as T increases thus K gets smaller Increase in temperature favors reactants (le Chatelier) If reaction is endothermic DH >0 slope of ln(K) vs. 1/T is negative (>0) As T increases 1/T decreases so as T increases thus K gets larger Increase in temperature favors products(le Chatelier)

Statistical Mechanical Explanation
As T increases population of higher energy states is increased at the expense of lower states For an endothermic reaction product states higher than reactants to it shifts to products For an exothermic reaction, reactant states higher than products so reaction shifts to them

Estimating K at Different Temperatures
dlnK /d(1/T) = -DH /R Integrating between 1/T1 and 1/T2 lnK2 - lnK1 = -DH /R (1/T2 - 1/T1 ) This assumes H does not vary much over the temperature range Self test 9.6 For N2O4(g) ˛ 2NO2 298 K, K= 0.15 What is value at 100°C (398K)? ln K1 = , 1/T1 = , 1/T2 = , (1/T2 - 1/T1 )= DH = *33.18 = 57.2 kJ; DH/R = -DH /R (1/T2 - 1/T1 ) = * ( ) = 5.8 Ln K2 = = 3.903; K2 = 49.6

Chapter 10: Equilibrium Electrochemistry
Homework: Exercises(a only):5, 6,7,12, 18, 20, 25, 26 Self Test: 8 and 9

Thermodynamic Properties of Ions in Solution Enthalpy and Gibbs Energy
Enthalpy and Gibbs energy of formationValues of DfHø and DfGø refer to formation of ions from reference state of parent ions Individual enthalpies and Gibbs energies for ions not directly measurable Ag(s) ® Ag+(aq) + 1e- DfHø = ?, DfGø = ? Only measure overall reactions Ag(s) + 1/2Cl2(g)® Ag+(aq) + 2Cl- (aq) DfHø = kJ mol-1 Since H and G state functions, the overall reaction is the sum of component reactions DrHø = DfHø(Ag+, aq) + DfHø (Cl-,aq) Could measure a number of reactions with similar components and by difference get DfHø Need to know one DfHø however Define reference ion and assign value of zero for DfHø and DfGø for it Ion is H+: DfHø(H+(aq)) = 0 and DfGø(H+(aq)) = 0 definition Measure other ions relative to H+ Chlorine: DfHø(Cl-, aq) = kJ mol-1 + DfGø(Ag+, aq) = kJ mol-1 Determine ions by difference from reaction enthalpy /Gibbs energy Silver: DrHø = DfHø(Ag+, aq) + DfHø (Cl-,aq) or =DfHø(Ag+, aq) = DrHø - DfHø (Cl-,aq) DfHø(Ag+, aq) = kJ mol-1 - ( kJ mol-1 ) = kJ mol-1

Contribution to DfGø 2) Remove e- from H(g) 3) Dissociate Cl2(g) 4) Add e- from Cl(g) 5) Solvate Cl- 6) Solvate H+ 6) Form H2(g) and Cl2(g) from solvated ions (-DrG) 1) Dissociate H2(g) Contributions can be seen by constructing a thermodynamic cycle Sum around cycle is zero (G is state function) DfGø of an ion includes contribution from dissociation, ionization and solvation of hydrogen All Gibbs energies except DG of solvation estimated from standard tables DfGø of two ions is related to DsolvGø DfGø(Cl-, aq) = 1272 kJ mol-1 + DsolvGø(Cl-) + DsolvGø(H+) Can be estimated from Born equation

Born Equation - Solvation Gibbs Energies
Solvation Gibbs energy estimated from the electrical work required to transfer an ion to a solvent - Born Equation Solvent treated as a dielectric with permittivity, eG Good example of how work need not be PV work to calculate Gibbs energies {look at Justification 10.1, p.255} For water, Born equation becomes DsolGø = 6.86 x 104(zi2/ri) kJ mol-1 ri is radius in pm Values turn out to be in reasonable agreement with experimental

Standard Entropies of Ions in Solution
Like G and H can’t measure entropies of isolated ions in solution Assume DfSø for H+ in solution is 0 Derive values for other ions relative to it Some ions have positive DfSø and some negative relative to H+ DfSø (Cl-,aq) = 57 J K-1 mol-1 DfSø (Mg2+,aq) = -128 J K-1 mol-1 Means that relative to water Mg2+ induces more order and Cl- less Can be rationalized in terms of charge on ion affecting local order around ion Small highly charged ions (Mg2+) induce more order than bigger less highly charged ions Depends a bit on your model for the liquid state how you think about it Various models include distorted ice-like structure, flickering clusters (Frank-Wen clusters), glassy water (J. Gibbs) For structure breaking ions Zone B can encroach on Zone A Estimate of DfSø (H+) on absolute (3rd Law scale) is -21 J K-1 mol-1 H+ induces order Zone B: Structure somewhat broken Zone C: Structure unaffected + Zone A: Structure Highly Affected

Ion Activities Definition
The activity of a solution, a, is related to the chemical potential, µ µ = µø + RTln(a) Tend to associate activity with concentration (molality) Its related but not equivalent Replacement valid in very dilute solution (<10-3 mol/kg total ion concentration) Given a solution whose ions behave ideally with a molality, bø of 1 mol/kg a = g(b/bø) g is the activity coefficient Depends on composition, concentration (molality) and temperature g ® 1 and a ® b/bø as b ® 0 From [1] µ = µø + RTln(b) + RTln(g) where b = b/bø µ = µideal + RTln(g) µideal (= µø + RTln(b) ) is the chemical potential of an ideal dilute solution of molality bhere b = b/bø

Mean Activity coefficients
Consider a solution of two monovalenent cations (M+) and anions (X-) Deviation from ideality contained in term RTln(g+ g-) Define (g+ -) = (g+ g-)0.5 (geometric mean) Reflects fact you can’t really separate deviation from non-ideality (g+ -) is the mean activity coefficient for monovalent ions Then, µ+ = µ+ideal + RTln(g+-) and µ- = µ-ideal + RTln(g+-) Generally for compound MpXq that dissolves into p cations and q anions, by same process define mean activity coefficient as (g+ -) = (g+ g-)1/s where s = p + q The chemical potential, µi, becomes µi = µi ideal + RTln(g+-) And G becomes G = p µ+ + q µ- Again non-ideality is shared

Estimating (g+ -) - Debye-Hückel Theory
Coulomb interactions imply oppositely charged ions attract each other In solutions, near an ion counter ions are found (ionic atmosphere) Coulomb potential(f) drops as 1/r: fi = Zi/r Zi a ionic charge G (& µ) of ion lowered by electrostatic interactions Since µi = µi ideal + RTln(g+-), lowering is associated with RTln(g+-) ln(g+-) can be calculated by modeling these interactions Debye-Hückel Limiting Law (proof Justification 10.2): zi is charged number on ions Must sum all ions in solution Sign of charge included, e.g, zNa+ = +1; zSO42- = -2 You’ll be using this in lab (Expt. 7, 9) Works well at dilute solutions (b < 1 mmol/kg) Extended Debye-Hückel Limiting Law (1 mmol/kg <b < 0.1 mol/kg): B dimensionless const., adjustable empirical parameter b>0.1 mol/kg (e.g. sea water): Model dependence of g of solvent on solute and use Gibbs-Duhem equation (SnJdµJ) = 0 to estimate g of solute

Estimating (g+ -) - Debye-Hückel Theory
Limiting Law vs. Ionic Strength

Electrochemical Cells
Electrochemical cell - two electrodes in contact with an electrolyte Electrolyte is an ionic conductor (solution, liquid, or solid) Electrode compartment = electrode + electrolyte If electrolytes are different compartments may be connect with salt bridge Electrolyte solution in agar Galvanic cell - an electrochemical cell that produces electricity Electrolytic cell - an electrochemical cell in which a non-spontaneous reaction is driven by an external source of current

Types of Electrodes Gas (b) Metal/insoluble salt (c) Redox (d)
Metal/metal ion (a) Designation:M(s)|M+(aq) Redox couple: M+ /M Half reaction: M+(aq) + 1e- ® M(s) Gas (b) Designation*: Pt(s)|X2(g)|X+(aq) or Pt(s)|X2(g)|X-(aq) Redox couple: X+ /X2 or X2 / X- Half reaction: X+(aq) + 1e- ® 1/2X2(g) or 1/2X2(g) + 1e- ® X-(aq) Metal/insoluble salt (c) Designation:M(s)|MX(s)|X-(aq) Redox couple: MX /M,X- Half reaction: MX(s) + 1e- ® M(s) + X-(aq) Redox (d) Designation*: Pt(s)| M+(aq), M2+(aq) Redox couple: M+/M2+ Half reaction: M2+(aq) + 1e- ® M+(aq) *Inert metal (Pt) source or sink of e-

Half-Reactions Recall definition of redox:
Redox reaction is one involving transfer of electrons OILRIG Oxidation is loss of electrons Reduction is gain of electrons Reducing agent (reductant) is electron donor Oxidizing agent (oxidant) is electron acceptor Any redox reaction can be expressed as the difference of two reduction half reactions (sum of oxidation and reduction half reaction) Cu2+(aq) + 2e- ® Cu (s) Zn2+(aq) + 2e- ® Zn (s) (copper - zinc): Cu2+(aq) + Zn (s) ® Cu (s) + Zn2+(aq) Redox couples are the reduced and oxidizing species in a redox reaction Written Ox/Red for half reaction Ox + ne- ® Red Example above: Cu2+/Cu; Zn2+/Zn

Half Reactions Reaction quotient for half-reaction (Q)
Like reaction quotient for overall reaction (activity of product over activity reactant raised to appropriate power for stoichiometry) except electrons omitted Cu2+(aq) + 2e- ® Cu (s) Q = 1/aCu {Metal in standard state aM = 1} O2(g) + 4H+(aq) + 4e- ® 2H2O (l) O2 assumed to be ideal gas Redox couples in an electrochemical cell separated in space Oxidation in one compartment, reduction in another Oxidation compartment Red1 ® Ox1 + ne- Electrode at which this occurs is the cathode Reduction compartment Ox2 + ne- ® Red2 Electrode at which this occurs is the anode

Half Reactions Galvanic cell (produces electricity)
Cathode at higher potential than anode Species being reduced withdraws electrons from cathode giving it a relative (+) charge Species being oxidized deposits electrons in anode giving it a relative (-) charge Electrolytic cell(electricity supplied) Oxidation still occurs at anode Oxidation doesn’t occur spontaneously Electrons come from the species in that compartment Anode relatively positive to cathode Cathode Supply of electrons drives reduction Electrolytic Cell

Types of Cells Daniel Cell Commonest cell has single electrolyte in contact with both electrodes Daniel cell - electrode compartments separated Different electrolyte in each compartment Electrolyte concentration cell - same electrolyte, different concentration Electrode concentration cell - electrodes have different concentration Gas cells at different pressures Amalgams at different cocentrations Additional potential difference across interface of two electrolytes - liquid junction potential Present in electrolyte concentration cells Due to differing mobility of ions of different sizes across interface Can be reduced with salt bridge Potential is then independent of concentration of electrolyte solution

Cell Reactions Notation - overall cell reaction denoted by cell diagram Phase boundary denoted by single vertical line (|) Liquid junction denoted by single vertical dotted line (:) Daniel cell Interface with no junction potential double verticla line (||) Salt bridge Cell Reactions: reaction occurring in the cell with the right-hand as the cathode (spontaneous reduction) To write cell reaction: 1. Write r.h.s. as reduction 2. Write l.h.s. as reduction 3. Subtract 2 from 1 Cell Potential - electrical work that can be done through the transfer of electrons in a cell Depends on the potential difference between electrodes If overall reaction at equilibrium cell potential is zero If reaction spontaneous, w is negative. At constant p and T we,max = DG Work is maximum only if cell is operating reversibly G is related to composition, work is constant composition, i.e., no current Under these conditions the potential difference is the electromotive force (emf) of the cell, E

Cell Reactions: Relation between E and DrG
For a reversible cell, the Gibbs energy of reaction, DrG, is proportional to the cell potential, DrG = -nFE where n is the number of electrons transferred and F is Faraday’s constant  Proof When potential is high, DrG is negative and cell reaction is spontaneous The emf (driving power of cell) is related to the extent of reaction

Nernst Equation: Relationship of emf to Activity
Recall that the Gibbs energy of reaction is related to composition: DrG = DrGø + RTlnQ where Q is reaction quotient(anJproducts/anJreactants) Since DrG = -nFE E = -(DrGø/nF) - (RT/nF)lnQ Define -(DrGø/nF) as Eø, the standard emf of the cell So, E = Eø - (RT/nF)lnQ (Nernst Equation) Eø is the emf when all reactants and products are in their standard states aproducts = 1 and areactants = 1, so Q=1 and ln(Q) = 0 Nernst Equation indicates a plot of E vs. ln Q will have Slope = -(RT/nF) Intercept = Eø You’ll see this in Lab #6 At 25°C, RT/F = mV, Nernst equation becomes E = Eø - ( V/nF)lnQ

Nernst Equation: Concentration Cells
M|M+(aq, L)|| M+(aq, R)|M In above cell the only difference is the concentration of electrolyte in each cell Left cell - molality is L; right cell - molality is R Cell reaction: M+(aq, R) ® M+(aq,L) n is 1 Eø is 0 because when R=L the two compartments are identical and no driving force Nernst Equation: E = Eø - (RT/nF)lnQ In this case: E = - (RT/F)ln(aL/aR) = - (RT/F)ln(bL/bR) If R>L, ln(bL/bR)<0, E>0, concentration will be lowered by reduction in right compartment If L>R, ln(bL/bR)>0, E<0, concentration will be lowered by reduction in left compartment This has biological application - nerve firing involves change in permeability of cell membrane to Na+ This changes nerve cell potential. (see text)

Nernst Equation: Equilibrium Cells
At equilibrium, by definition no work can be done E = 0 Concentrations are fixed by the equilibrium constant(K) K=Q Nernst Equation: E = Eø - (RT/nF)lnQ In this case: E = 0 so Eø = (RT/nF)lnQ = (RT/nF)lnK Rearranging, lnK = Eø/ (RT/nF) = nFEø/ RT or K = exp(nFEø/ RT) This means cell potentials can be used to determine equilibrium constants

Standard Potentials Since you can’t measure the potential of a single electrode, one pair has been assigned, by convention a potential of 0 Standard hydrogen electrode (SHE): Other potentials determined by constructing cells in which SHE is left hand electrode: Silver Chloride|Silver Pt(s)|H2(g)|H+(aq)||Cl-(aq)|AgCl(s)|Ag Eø(AgCl, Ag, Cl-)=+0.22V Reaction: AgCl(s) + 1/2H2(g) ® Ag(s) + H+(aq) + Cl-(aq) Because all potentials are relative to the hydrogen electrode, the reaction is listed without the contribution of the SHE, AgCl(s) ® Ag(s) + Cl-(aq) Numerical factors If std emf reaction is multiplied by numerical factor, DrG increases by that factor Standard potential does not increase! Recall Eø = -(DrGø/nF) If DrGø(new) = n x DrGø, n(new) = n x n Eø (new) = -(DrGø(new) / n(new) F) =-(n x DrGø /n n(new) F)= -(DrGø/nF) = Eø

Cell emf & Standard Potentials
Cell emf of any cell can be calculated from table of standard potentials emf of cell is just difference in standard potential Procedure: 1) Write new cell diagram 2) Eø = Eø(right) - Eø(left) Sc(s)|Sc3+(aq)||Al3+(aq)|Al(s) Al3+(aq) + 3e- ® Al(s) Eø = 1.69 V Sc3+(aq) + 3e- ® Sc(s) Eø = V Al3+(aq) + Sc(s) ® Sc3+(aq) +Al Eø = 1.69 V - (-2.09 V) = 3.78V Recall: Eø = -(DrGø/nF) If Eø > 0, DrGø < 0 and K>1 Example above, at 25°C, K = exp(nFEø/ RT) = exp (3 x 3.78V/0.0257V) = exp(441) = 4.27 x 10191 K = [Sc3+]/[Al3+]

Measuring Standard Potentials Harned Cell
From Nernst Equation E = E°(AgCl/Ag,Cl-) - (RT/F)ln Q E = E°- (RT/F)ln ((aH+ aCl-)/(fH2/pØ)0.5 = E°- (RT/F){ln (aH+ aCl-) -ln(fH2/pØ)0.5 } Let fH2= pØ. E = E°- (RT/F)ln (aH+ aCl-) But a =bg± and bH+= bCl-) so E = E°- (RT/F)ln (b2g± ) = E°- (RT/F)ln (b2) - (RT/F)ln (g± 2) = E°- (2RT/F)ln(b) - (2RT/F)ln(g±) Debye-Hückel Limiting Law log(g±) a -I0.5 so log(g±) a -b0.5 or ln(g±) a -b0.5 E = E°- (2RT/F)ln(b) + Cb0.5 or E + (2RT/F)ln(b) = E° + Cb0.5 This means a plot of {E + (2RT/F)ln(b)} vs. b0.5 has Eø as intercept Measuring activity coefficients Since E = E°- (2RT/F)ln(b) - (2RT/F)ln(g±) , ln(g±) = {(Eø - E)/(2RT/F)}- ln(b) Knowing Eø and measuring E at known molality allows you to calculate activity coefficient

Applications of Standard Potential
More Reducing More Oxidzing Electrochemical series Because Eø = -(DrGø/nF), if Eø > 0, DrGø < 0 Since Eø = Eø2 - Eø1 , the reaction is spontaneous as written Red1 has tendency to reduce Ox2 , if Eø1 <Eø2 More directly in the electromotive series elements arranged such that low on the chart reduces high Calcium reduces platinum (Eø = 4.05 V) Platinum reduces gold (Eø = 0.51) Tin reduced lead (Eø = V) Sodium reduces magnesium (Eø = 0.34 V)

pH and pKa For hydrogen electrode (1/2 reaction above), Eø = 0
Glass Electrode For hydrogen electrode (1/2 reaction above), Eø = 0 If fH2= pø, Q = 1/aH+ and E = (RT/F) ln(aH+) E = Eø - (RT/nF) lnQ Converting ln to log (ln =2.303log), E = (RT/F) 2.303log(aH+) Define pH=-logaH so E = (RT/F)pH At 25°C, E= mVpH Measurement Direct method: hydrogen electrode + saturated calomel reference electrode (Hg2Cl2) At 25°C, pH = (E + E(calomel))/ (-59.16mV ) Indirect method: Replace hydrogen electrode with glass electrode sensitive to hydrogen activity (but not permeable to H+ E(glass) a pH, E(glass) = 0 when pH = 7 pKA Since we learned pH = pKa when concentration of acid and conjugate base are equal pKa can be measured directly from pH measurement Ion-Selective electrodes - related to glass electrode except potentials sensitive to other species (see Box 10.2, p 278)

Electrochemical Cells and Thermodynamic Functions
Since the standard emf of a cell is related to the Gibbs energy, electrochemical measurements can be used to obtain other thermodynamic functions Complementary to calorimetric measurements Esp. useful for ions in solutions (aqueous, molten salts, etc.) Starting point: DrGø = -nFEø and thermodynamic relationships we saw earlier DrGø can be used to calculate Eø directly (or reverse) Look at Example important caution about the relationship between numerical factors and Eø Entropy (S) Thermodynamic relationship: (∂G/∂T)p =-S DrGø = -nFEø At constant p, (dGø/dT) =-nF(dEø/dT) -DS =-nF(dEø/dT) or (dEø/dT) =DS/nF Enthalpy (S) Thermodynamic relationship: DrHø = DrGø + T DrSø DrHø = -nFEø + T(nF(dEø/dT)) DrHø =-nF(Eø - TdEø/dT)

Text: Physical Chemistry, 7th Edition, Peter Atkins and J. de Paula

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Text: Physical Chemistry, 7th Edition, Peter Atkins and J. de Paula

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