Thermochemistry Chapter 6

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Thermochemistry Chapter 6
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Energy is the capacity to do work
Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position Kinetic energy is energy associated with the motion of an object. 6.1

Energy cannot be created or destroyed.
Energy can be converted from one form of energy to another. Potential Energy Kinetic Energy Energy cannot be created or destroyed. => Law of conservation of energy = total quantity of energy in the universe is constant

Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. Temperature = Thermal Energy °C K °F 1 joule = 1 kg m /sec 1 calorie = joule (exactly) 400C 900C greater thermal energy

Study the interconversion of energy from one form to another.
Thermochemistry is the study of heat change in chemical & physical processes. => chemical reactions => phase changes Study the interconversion of energy from one form to another.

The system is the specific part of the universe that is of interest in the study.
The surroundings is the rest of the universe. SURROUNDINGS SYSTEM System Types: open closed isolated Exchange: mass & energy energy nothing 6.2

State – specific characteristics of a system (physical or chemical)
Examples: Pressure Temperature Composition State Function – a property of a system that characterizes a state of a system and is independent of how the system got to that state. Examples: Pressure Volume Composition

Energy it took each hiker to get to the top is different !!
Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature DE = Efinal - Einitial DP = Pfinal - Pinitial DV = Vfinal - Vinitial DT = Tfinal - Tinitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. Energy it took each hiker to get to the top is different !!

2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) Hg (l) + O2 (g) 6.2

Exothermic Endothermic 6.2

First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. DEsystem + DEsurroundings = 0 or DEsystem = -DEsurroundings E is a state function. C3H8 + 5O CO2 + 4H2O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings 6.3

Another form of the first law for DEsystem
DE = q + w Energy is expressed in units of joules of kJ. DE is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -PDV when a gas expands against a constant external pressure Note the conditions that define each process !!

Work is done by the surroundings on the system
Work Done On The System w = Fd Work is done by the surroundings on the system DV < 0 -PDV > 0 wsys > 0 w = -P DV P x V = x d3 = Fd = w F d2 initial final Work is not a state function! Dw = wfinal - winitial

Work is done by the system on the surroundings
Work Done By The System w = Fd Work is done by the system on the surroundings DV > 0 -PDV < 0 wsys < 0 w = -P DV P x V = x d3 = Fd = w F d2 Work is not a state function! Dw = wfinal - winitial initial final 6.3

W = -0 atm x 3.8 L = 0 L•atm = 0 joules
A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P DV (a) DV = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L•atm = 0 joules (b) DV = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = L•atm Energy needs to be converted to joules. Use 1 L . atm = J w = L•atm x 101.3 J 1L•atm = J The change in volume was the same in both cases, but the work done was not the same => work is not a state function.

Chemistry in Action: Making Snow
DE = q + w Inside machine = air + water vapor @ 20 atm Outside machine = 1 atm Air w/ water vapor rapidly expands q = adiabatic process DE = w w < 0 DE < 0 Kinetic Energy (gas) a T (K) DE = CDT DT < 0, SNOW !!

Note the conditions that define each process !!
Enthalpy and the First Law of Thermodynamics DE = q + w Note the conditions that define each process !! q = DH and w = -PDV At constant pressure: DE = DH - PDV DH = DE + PDV 6.4

Enthalpy is a state function.
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) Enthalpy is a state function. DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.4

Thermochemical Equations
Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ 6.4

Thermochemical Equations
Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = kJ 6.4

Enthalpy is a state function.
We only care about where we start and where we finish, not the route. DHnet = DH1 + DH2 + DH3 + DH4 DH2 DH3 Energy DH1 end DHnet DH4 start => We can construct a convenient path.

Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) H2O (l) DH = 2 x 6.01 = 12.0 kJ You must pay attention to the physical state and the quantity of all reactants and products !!!

Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = kJ 1 mol P4 123.9 g P4 x 3013 kJ 1 mol P4 x 266 g P4 = 6470 kJ 6.4

Flameless Ration Heaters
“Feeding the Army” Flameless Ration Heaters Chemical Equation: Mg (s) + 2H2O (l) Mg(OH)2 (s) + H2 (g) + 353kJ Solid magnesium and water yields solid magnesium hydroxide, hydrogen gas & kilojoules of heat energy Source- Tocci, Viehland Holt Chemistry Visualizing Matter. Austin, Texas: Holt, Rinehart and Winston, Inc. pg. 271, 303 and

It takes about 353kJ of heat energy is needed to heat the MRE to 60°C
It takes about 353kJ of heat energy is needed to heat the MRE to 60°C . Calculate the amount of magnesium needed. 353 kJ = 1 mol Mg = g Mg Calculate the volume of water needed. 353 kJ = 2 mol H g / mol H20 H20 density = 1.00 g/mL (2 mol H20) x (18.01 g / mol H20) x (1 mL H20 / 1.00 g H20) = mL H20 Calculate the mass of magnesium hydroxide Mg(OH)2 produced. 24.0g Mg = 1 mol Mg = 1 mol Mg(OH) g / mol Mg(OH)2 58.0 g Mg(OH)2 Calculate the number of moles of hydrogen gas H2 produced. 353 kJ = 1 mol Mg = 1.00 mol H2 Calculate the volume of hydrogen gas H2 produced at 1.00 atm and 25°C. PV = nRT => V = nRT/P V = (1.00 mol)( L atm/K mol)(298 K)/1.00 atm = L

What would happen if 25 mL of water is added?
What would happen is the soldier were smoking while heating dinner? Is the material in a used flameless ration heater safe to touch? Can the heater be reused?

A Comparison of DH and DE
2Na (s) + 2H2O (l) NaOH (aq) + H2 (g) DH = kJ/mol DE = DH - PDV At 25 0C, 1 mole H2 = 24.5 L at 1.0 atm PDV = 1.0 atm x 24.5 L = 2.5 kJ DE = kJ/mol – 2.5 kJ/mol = kJ/mol 6.4

DHnet = DH1 + DH2 + DH3 + DH4 + DH5

Heat (q) absorbed or released:
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial 6.5

Dt = tfinal – tinitial = 50C – 940C = -890C
How much heat is given off when an 869 g iron bar cools from 940C to 50C? s of Fe = J/g • 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = msDt = 869 g x J/g • 0C x –890C = -34,000 J 6.5

DHnet = DH1 + DH2 + DH3 + DH4 + DH5
How much heat is required to take exactly one mole of water from a solid at -25.0°C to steam at 125.0°C? DHnet = DH1 + DH2 + DH3 + DH4 + DH5 DH1 = q1 = msDt1 m = (1 mol ) (18.01 g/mol) = g s = J / g °C Dt1 = (0°C - -25°C) = 25°C DH1 = (18.01 g) (2.030 J / g °C) (25°C) = 914 J = kJ DH2 = DHfus = (6.01 kJ/mol ) (1 mol) = 6.01 kJ H2O (s) H2O (l) DH = 6.01 kJ DH3 = q3 = msDt s = J / g °C Dt3 = (100.°C -0°C) = 100.0°C DH3 = (18.01 g)(4.184 J / g °C)(100.0°C) = 7540 J = 7.54 kJ DH4 = DHvap = (44.0 kJ/mol ) (1 mol) = 44.0 kJ H2O (l) H2O (g) DH = 44.0 kJ DH5 = q5 = msDt s = J / g °C Dt5 = (125.0°C – 100.0°C) = 25.0°C DH5 = (18.01 g) (2.043 J / g °C) (25.0°C) = 920 J = kJ DHnet = DH1 + DH2 + DH3 + DH4 + DH5 = kJ kJ kJ kJ kJ = 56.1 kJ

Constant-Volume Calorimetry
qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = msDt qbomb = CbombDt Reaction at Constant V DH = qrxn DH ~ qrxn No heat enters or leaves! 6.5

Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves! 6.5

What happens is we mix two liquids at different temperatures?
Example: Mix 100 g water at 25°C with 100 g water at 75°C What is the final temperature? Since both liquids are the same (water) and both have the same weight (100 g) => take the average 75°C Loss of thermal energy 50°C Energy Gain of thermal energy 25°C What happens if the substances are different or the amounts are different?

| Heat lost by copper | = | Heat gained by water |
If g of Copper at °C is added to g water at 25.0°C What is the final temperature? | Heat lost by copper | = | Heat gained by water | qsys = qwater + qcal + qCu = 0 assume qcal is very very small - qCu = qH2O -qCu = - mCusCuDtCu = mH2OsH2ODtH2O Dt = (tfinal – tinitial) sCu = J / g °C sH2O = J / g °C -(60.56 g)(0.385 J/g°C)(tfinal °C) = (150. g)(4.184 J/g°C)(tinitial -25.0°C) (-23.3 J/g°C) tfinal J = (628 J/g°C) tfinal – J 15700 J J = (( ) J/g°C) tfinal tfinal = (18000/651.3)°C tfinal = 27.7°C

6.5

Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol 1 cal = J 1 Cal = 1000 cal = 4184 J

(24.76 kJ / g Mg)(24.31 g Mg / mol Mg) = 601.9 kJ / mol Mg
A g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity of 3027 J/°C. The temperature increases by °C. Calculate the heat change by the burning Mg in kJ/g and in kJ/mol. | Heat given off by reaction | = | Heat gained by calorimeter | qCal = CpDt = (3024 J/°C)(1.126 °C) = x103 J qRxn = (3.405 x103 J)(1 kJ / 1000 J) = kJ / g Mg g Mg (24.76 kJ / g Mg)(24.31 g Mg / mol Mg) = kJ / mol Mg Is the reaction exothermic or endothermic? What would happen to the temperature of the water if it were the opposite type reaction?

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (O3) = 142 kJ/mol f DH0 (C, diamond) = 1.90 kJ/mol f

6.6

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = kJ 2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = (2x-296.1) = 86.3 kJ rxn

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol. 2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x– x–187.6 ] – [ 2x49.04 ] = kJ -5946 kJ 2 mol = kJ/mol C6H6 6.6

DHsoln = Hsoln - Hcomponents
The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 6.7

The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7

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