Daniel L. RegerScott R. GoodeDavid W. Ball

http://academic.cengage.com/chemistry/reger

Chapter 17Chemical Thermodynamics

• Chemical thermodynamics is the study of the energetics of chemical reactions.

Chemical Thermodynamics

• The system is the part of the universe under examination (for us, it would be a chemical reaction or other process).• The surroundings are the rest of the

universe.• The change in enthalpy (H) is the

heat absorbed or released by the system at constant temperature and pressure.

Chapter 5 Review

• Under normal laboratory conditions, a reaction (the system) usually exchanges energy with the surroundings in two forms, as heat and as work.• Work is directed energy, and is associated

with moving a part of the system.• Heat is random energy, and is associated

with the temperature of the system.

Work and Heat

• Work is an application of a force through a distance:

work = force × distance.

• Work is also an application of a pressure causing a volume change: work = -PV.

Work

Pressure-Volume Work

• Express the work (in joules) when 20.0 L of an ideal gas at a pressure of 12.0 atm expands against a constant pressure of 1.50 atm. Assume constant temperature.

• First law of thermodynamics is the law of conservation of energy; energy can neither be created nor destroyed.• Internal energy, E, represents the total

energy of the system and is a state function (something that depends only on the state of the system, not how the system got to that state).

First Law of Thermodynamics

• When a change occurs in a closed system, the change in internal energy, E, is given by

E = q + w• The sign convention for q and w are that they

are positive if they transfer energy to the system from the surroundings and negative if they transfer energy from the system to the surroundings.

Internal Energy, E

Calculating ΔE, w, and qA 7.56g sample of gas in a balloon that has a volume of 10.5 L. Under an external pressure of 1.05 atm, the balloon expands to a volume of 15.00 L. Then the gas is heated from 0.0 °C to 25.0 °C. If the specific heat of the gas is 0.909 J/g*°C. Calculate the work, heat, and ΔE for the overall process.

Energy and Enthalpy• Data about heats of reaction are

tabulated as standard enthalpies of formation (Hf), as described in Chapter 5.• The equation that relates H to E is

H = E + PV• This equation is used to calculate the

enthalpy of reaction from heats measured using constant-volume calorimetry.

Energy, Enthalpy, and PV Work• The difference between changes in

enthalpy and internal energy is PV work and is significant only for reactions that involve gases.• From the ideal gas law, PV = nRT, so

H = E + nRT

• Calculate H at 25C and 1.00 atm pressure for the following reaction.

2NO2(g) → N2O4(g) E = -54.7 kJ

Example: H from E

• An increase in randomness (sometimes referred to as disorder) is an important driving force for many changes.• Two different gases, initially separated by a

partition, will mix with the partition is removed, increasing the randomness of the system.

Randomness

• Thermodynamics is able to relate spontaneity to a state function called entropy.• Entropy (S) is the thermodynamic

state function that describes the amount of randomness.• A large value for entropy means a high

degree of randomness.

Entropy and Spontaneity

Entropy Change• An increase in randomness results in an increase in entropy. Some general guides

are:• the entropy of a substance increase when solid becomes liquid, and when liquid becomes gas.• the entropy generally increases when a solute dissolves.• the entropy decreases when a gas dissolves in a solvent.• the entropy increases as temperature increases.

Entropy• The entropy of a system generally

increases when a molecular solid dissolves in a liquid.

Second Law of Thermodynamics• The second law of thermodynamics

states that in any spontaneous process, the entropy of the universe increases.

• Suniv = Ssys + Ssurr > 0

• If Ssys < 0 for a spontaneous process, then a larger positive change in Ssurr must occur.

• When Suniv > 0, the change occurs spontaneously.

• When Suniv < 0, the reverse change occurs spontaneously.

• When Suniv = 0, the change is not spontaneous in either direction (the process is at equilibrium).

Spontaneity and Suniv

Third Law of Thermodynamics• The third law of thermodynamics states

that the entropy of a perfect crystal of a substance at absolute zero is equal to 0.• There is a minimum randomness in a perfect

crystal at 0 K.• Unlike enthalpy and internal energy,

absolute values of entropy can be determined.

Units for Entropy

• As heat is added to a perfect crystal at 0 K, the temperature rises and randomness begins to increase.• The change in entropy depends on the

amount of heat and on the temperature, and is given by the equation

S = q/T• Therefore, entropy has units of J/K.

• In Appendix G, absolute entropies are given for substances in their standard state at 298 K.

• The entropy change for a reaction is

Srxn = nS[prods] - mS[reacts]

where n and m are the coefficients of the products and reactants in the reaction.• S for the free elements in their standard

states is not zero.

Absolute Entropies

Example: Calculate Srxn

• Calculate the standard entropy change for the reaction2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O()

Substance S, J/mol·K

C2H6(g) 229.49

CO2(g) 213.63

O2(g) 205.03

H2O() 69.91

• The combustion of ethane is spontaneous, but S is -620.21 J/K. Explain why this does not violate the second law of thermodynamics.

Test Your Skill

• J. W. Gibbs defined a state function called the Gibbs free energy, G:

G = H – TS• At constant temperature and pressure, this

becomes• G = H – TS

• G is negative for a spontaneous change that occurs at constant temperature and pressure.

Free Energy and Suniv

Free Energy and Spontaneity• G is a state function of the system.• For any spontaneous change, G < 0.

Spontaneous Reaction

∆Suniv ∆Gsys

Forward reaction positive (+) negative (-)

At equilibrium 0 0

Reverse reaction negative (-) positive (+)

• From the equation G = H – TS, a negative H and a positive S favor spontaneity.• A minimum free energy occurs when the

system is at equilibrium, and G = 0.

G and Spontaneity

Standard Free Energy of Formation

• The standard free energy of formation is the free energy change to form one mole of a compound from its elements in their standard states.

• , where Sf must be calculated from absolute entropies.

• Grxn = nGf[prods] - mGf[reacts]

where n and m are the coefficients of the products and reactants in the reaction.

fff STHG

• The dependence of G on temperature arises mainly from the TS term in the definition of G.• The sign of G is dominated by the sign of H at low temperatures and by the sign of S at high temperatures.• Negative values of H and positive values of S favor spontaneity.

Temperature Dependence of G

Direction of Spontaneous Reaction

∆H ∆S Temperature ∆GSpontaneous

Direction

- + All - Forward

- -Low

High

-

+

Forward

Reverse

+ +Low

High

+

-

Reverse

Forward

+ - All + Reverse

Example: Temperature Dependence• Assuming that H and S do not

change with temperature, calculate the temperature for which G is 0 for the reaction

CS2() ⇌ CS2(g)

At 298 K, H = 27.66 kJ and S = 86.39 J/K.

• Given the following chemical reaction and data at 298 K:

N2(g) + 3H2(g) 2NH3(g) Hf

o = 0 0 -46.1 kJ/mol So = 191.5 130.6 192.3 J/mol.K

assuming that Ho and So do not change with temperature, calculate Go at 1000 K.

Example: Calculate G

• Concentrations of reactants and products influence the free energy change of a reaction according to the equation

G = G + RTlnQ

where Q is the reaction quotient (see Chapter 14).

Concentration and Free Energy

• Substance Gfo kJ/mol

NO2(g) 51.29 N2O4(g) 97.82

For the reaction2NO2(g) ⇌ N2O4(g)

(a) calculate Go at 298 K.(b) calculate G when PNO2

= 0.12 atm

and PN2O4 = 0.98 atm.

Example: Concentration Dependence

• For a system at equilibrium, G = 0, and Q = Keq, so

Go = -RTln Keq

• Go, calculated from the data in Appendix G, can be used to calculate the value of the equilibrium constant.

Free Energy and Keq

Example: Keq Calculation

• Evaluate the equilibrium constant at 298 K for

2NO(g) + Br2(g) ⇌ 2NOBr(g)using the standard free energies of formation at 298 K given below.

Substance NO(g) Br2(g) NOBr(g)

∆Gf° (kJ/mol) 86.55 3.14 82.4

Temperature and Keq

• The temperature dependence of Keq is derived from two equations given earlier:

H - TS = G = - RTln Keq

A graph of ln Keq vs. 1/T gives a line with a slope = -H/R and an intercept of S/R.

Useful Work• The change in free energy is the maximum

work that can be performed by a spontaneous chemical reaction at constant temperature and pressure:

wmax = G• This is a limit imposed by nature, as we strive to

improve the efficiency of energy conversions

• When G > 0 (spontaneous in the reverse direction), it represents the minimum work that must be provided to cause the change.