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**CHEMICAL THERMODYNAMICS**

AP Chemistry Chapter 19 Notes CHEMICAL THERMODYNAMICS

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**1st Law of Thermodynamics Energy in the universe is conserved**

REVIEW 1st Law of Thermodynamics Energy in the universe is conserved

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**value depends on how process takes place (i.e. q, w)**

REVIEW Path Function value depends on how process takes place (i.e. q, w)

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heat q = mCpT

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REVIEW State Function value independent of path - is a defined reference or zero point (i.e. H)

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**Hf0 of elements in natural form at 250C**

Enthalpy - H zero point Hf0 of elements in natural form at 250C

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Enthalpy - H types: Hrxn = nHf0(prod) - nHf0(react)

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Enthalpy - H types: Hfus , Hvap

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**a measure of randomness or disorder of a system**

ENTROPY “S” a measure of randomness or disorder of a system

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**Entropy is NOT conserved. The universe seeks maximum disorder.**

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2nd Law of Thermodynamics for a spontaneous process, entropy increases

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**Entropy - state function**

zero reference: S = 0 for a perfect crystal at absolute zero

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**Ssolid < Sliquid << Sgas**

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S0 = standard entropy of elements & cmpds at P = 1 atm; T = 250C units = J/K mol [Appendix L, text]

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Calculate S using Hess’ Law S(rxn) = nS0(prod) - nS0(react)

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Example: Ca(s) + C(gr) + 3/2 O2 CaCO3(s)

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Suniv = Ssys + Ssurr if Suniv > 0 process spontaneous

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Suniv = Ssys + Ssurr if Suniv < 0 process ?

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Suniv = Ssys + Ssurr if Suniv = 0 process ?

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to relate S and H consider

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**where the water is the system & everything else is the surroundings**

H2O(s) H2O(l) where the water is the system & everything else is the surroundings entropy of water increases, therefore Ssys > 0 the vaporization of water is an endothermic process, therefore heat is flowing from the surroundings to the system & the random motion of the atoms in the surroundings decrease. thus, Ssurr < 0 so, Suniv = Ssys(+) + Ssurr(-) ; which predominates? we know that above 1000C, water spontaneously evaporates and that below 1000C, water spontaneously condenses, thus TEMPERATURE must have an effect on relative importance of Ssys and Ssurr .

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**Temperature Dependence S = J/mol (1/T) Ssurr = -H/T**

introduces central idea that the entropy changes in the surroundings are primarily determined by heat flow. exothermic process increases the entropy of the surroundings, thus is an important driving force for spontaneity. Remember that a system tends to undergo changes that lower its energy. The SIGN on Ssurr depends on the direction of the heat flow. The MAGNITUDE of Ssurr depends on the temperataure.

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**Ssys Ssurr Suniv spon**

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**Ssys Ssurr Suniv spon**

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**Ssys Ssurr Suniv spon**

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**Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - +**

It will be spontaneous if Ssys has a larger magnitude larger than Ssurr.

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**Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - + ? D**

It will be spontaneous if Ssurr has a larger magnitude larger than Ssys.

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For a phase change:

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GIBB’S FREE ENERGY G

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**most abstract of thermodynamic state functions**

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w w = reversible work w2 P V

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**G = w + PV Definition: w: reversible work PV: pressure/volume work**

isothermal, reversible path

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G = w + PV + VP at constant P P = 0 so VP = 0 G = w + PV

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G = w + PV + VP at constant P G = w + PV at constant V V = 0 so PV = 0 G = w = useful work

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**must measure G over a process**

G cannot be measured must measure G over a process

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ZERO REFERENCE G = 0 for elements in stable form under Standard Thermodynamic Conditions T = 25oC ; P = 1 atm

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**Gf0 = standard Free Energy of Formation from the elements**

Appendix L, text

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G follows Hess’ Law: G0 (rxn) = nGf0(p) - nGf0(r)

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Summary of Laws of Thermodynamics Zeroth Law: Heat Gain = Heat Loss

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Summary of Laws of Thermodynamics First Law: Law of Conservation of Energy

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Summary of Laws of Thermodynamics Second Law: Defines Entropy

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Summary of Laws of Thermodynamics Third Law: Defines Absolute Zero

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GIBB’S HELMHOLTZ EQUATION

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combine G H T

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G = -aT + H G, H are state functions, thus “a” must be a state function G = -S(T) + H

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Gibb’s Helmholtz Equation G = H - TS

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**Units on the State Functions**

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**thus, a process is spontaneous**

if and only if G is negative

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**enthalpy (minimum energy)**

Spontaneity controlled by enthalpy (minimum energy)

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**enthalpy entropy (maximum disorder)**

Sponaneity controlled by enthalpy entropy (maximum disorder)

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Sponaneity controlled by enthalpy entropy both

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Predict Spontaneity IF H(-) and S(+) G = -H - T(+S) G < 0, => spontaneous

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Predict Spontaneity IF H(+) and S(-) G = +H - T(-S) G > 0, => NOT spontan

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**Summary of Spontaneity**

H S G Spont. yes no or ? or ?

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Uses of the Gibb’s Helmholtz Equation

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1. Find the molar entropy of formation for ammonia.

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**2. Elemental boron, in thin fibers, can be made from a boron halide:**

BCl3(g) + 3/2 H2(g) ->B(s) + 3HCl(g)

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Calculate: H0, S0 and G0. Spontaneous? Driving force?

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3. Using thermodynamic information, determine the boiling point of bromine.

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Thermodynamic Definition of Equilibrium Geq = 0

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by definition G = H - TS & H = E + PV

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**thus, G = E + PV - TS take derivative of both sides**

dG = dE + PdV + VdP - TdS - SdT

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**for a reversible process**

TdS = q

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**derivative used for state function while **

partial derivative used for path function

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if the only work is PV work of expansion PdV = w

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First Law of Thermodynamics E = q - w dE = q - w = 0

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thus q = w or TdS = PdV

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by substitution dG = 0 + PdV + VdP - PdV - SdT or dG = VdP - SdT

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let’s assume we have a gaseous system at equilibrium, therefore, examine Kp at that constant temperature

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at constant T SdT = 0 thus dG = VdP

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but

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if “condition 2” is at standard thermodynamic conditions, then G2 = G0 and P2 = 1 atm

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thus

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determine G for aA + bB cC + dD where all are gases

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G = Gprod - Greact = cGC + dGD - aGA - bGB

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but and likewise for the others

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and

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but

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Thus, DG = DG0 + (RT) ln Q

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But aA + bB cC + dD G = 0

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thus or in general

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**THERMODYNAMICS & EQUILIBRIUM**

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**information, determine**

3. Using thermodynamic information, determine the boiling point of bromine.

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**Thermodynamics and Keq**

FACT: Product-favored systems have Keq > 1.

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**Thermodynamics and Keq**

Therefore, both ∆G˚rxn and Keq are related to reaction favorability.

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**Thermodynamics and Keq**

Keq is related to reaction favorability and thus to ∆Gorxn. The larger the value of K the more negative the value of ∆Gorxn

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**Thermodynamics and Keq**

∆Gorxn= - RT lnK where R = J/K•mol

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**Thermodynamics and Keq**

∆Gorxn = - RT lnK Calculate K for the reaction N2O4 2 NO2 ∆Gorxn = +4.8 kJ K = 0.14 When ∆G0rxn > 0, then K < 1

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∆G, ∆G˚, and Keq ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient

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**∆G, ∆G˚, and Keq When Q < K or Q > K, reaction is spontaneous.**

When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

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**Product favored reaction**

∆G, ∆G˚, and Keq Product favored reaction –∆Go and K > 1 In this case ∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion.

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**∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2 N2O4**

∆Gorxn = – 4.8 kJ Here ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.

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**Thermodynamics and Keq Overview**

∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.

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**Keq is related to reaction favorability.**

When ∆Gorxn < 0, reaction moves energetically “downhill”

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**4. For the following reaction, calculate the temperature at which the reactants are favored.**

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**THERMODYNAMICS OF CHEMICAL REACTIONS**

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**5. How much useful work can be obtained from an engine fueled with 75**

5. How much useful work can be obtained from an engine fueled with 75.0 L of hydrogen at 10 C at 25 atm?

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**reacting silver with water.**

6. The reaction to split water into hydrogen and oxygen can be promoted by first reacting silver with water. 2 Ag(s) + H2O(g) Ag2O(s) + H2(g) Ag2O(s) 2 Ag(s) + 1/2 O2(g)

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**Calculate H0, S0 and G0 for each reaction.**

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**Combine the reactions and calculate H0 and G0 for the combination.**

Is the combination spontaneous?

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**At what temperature does the second reaction become spontaneous?**

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**7. The conversion of coal into hydrogen for fuel is:**

C(s) + H2O(g) CO(g) + H2(g)

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Calculate G0 and Kp at 250C. Is the reaction spontaneous?

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**At what temperature does the reaction become spontaneous?**

Calculate the temperature at which K = 1.0 x 10-4.

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8. The generation of nitric acid in the upper atmosphere might destroy the ozone layer by: NO(g) + O3(g) NO2(g) + O2(g)

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**Calculate G0 (reaction)**

and K at 250C.

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Copyright 1999, PRENTICE HALLChapter 191 Chemical Thermodynamics Chapter 19 David P. White University of North Carolina, Wilmington.

Copyright 1999, PRENTICE HALLChapter 191 Chemical Thermodynamics Chapter 19 David P. White University of North Carolina, Wilmington.

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