Daniel L. Reger Scott R. Goode David W. Ball

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

Chapter 17 Chemical Thermodynamics

Chemical Thermodynamics

•

Chemical thermodynamics

is the study of the energetics of chemical reactions.

Chapter 5 Review

• 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 (

D

H)

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

•

Work and Heat

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

• • 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 = -

P

D

V

.

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

• •

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).

Internal Energy,

E

• When a change occurs in a closed system, the change in internal energy, D

E

, is given by D

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.

Calculating ΔE, w, and q

A 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 ( D

H

f  ), as described in Chapter 5.

The equation that relates D

H

= D

E

+

P

D

V

D

H

to D

E

is • 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

D

H

= D

E

+ D

nRT

=

nRT

, so

Example:

D

H

from

D

E

• Calculate D

H

at 25  C and 1.00 atm pressure for the following reaction.

2NO 2 (g) → N 2 O 4 (g) D

E

= -54.7 kJ

Randomness

• 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.

Entropy and Spontaneity

• • 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 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.

D

S

univ • = D

S

sys + D

S

surr > 0 If D

S

sys < 0 for a spontaneous process, then a larger positive change in D

S

surr must occur.

Spontaneity and

D

S

univ • When D

S

univ > 0, the change occurs spontaneously.

• When D

S

univ < 0, the reverse change occurs spontaneously.

• When D

S

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

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 D

S

=

q

/

T

• Therefore, entropy has units of J/K.

Absolute Entropies

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

The entropy change for a reaction is D

S

rxn = S

nS

 [prods] S

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.

Example: Calculate

D

S

rxn

• Calculate the standard entropy change for the reaction 2C 2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6H 2 O( l )

Substance

C 2 H 6 (g) CO 2 (g) O 2 (g) H 2 O( l )

S



, J/mol ·K

229.49

213.63

205.03

69.91

Test Your Skill

• The combustion of ethane is spontaneous, but D

S

 is -620.21 J/K. Explain why this does not violate the second law of thermodynamics.

Free Energy and

D

S

univ • • • J. W. Gibbs defined a state function called the

Gibbs free energy

,

G

:

G = H – TS

At constant temperature and pressure, this becomes • D

G

= D

H

D

G

–

T

D

S

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

Free Energy and Spontaneity

• • D

G

is a state function of the system.

For any spontaneous change, D

G

< 0.

Spontaneous Reaction

Forward reaction

∆S

univ

positive (+) At equilibrium 0 Reverse reaction negative (-)

∆G sys

negative (-) 0 positive (+)

• • D

G

and Spontaneity

From the equation negative D

H

D

G

= D

H

– and a positive D

S T

D

S

, a favor spontaneity.

A minimum free energy occurs when the system is at equilibrium, and D

G

= 0.

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.

D

G f

  D

H f

 

T

D

S f

 , where D

S

f  must be calculated from absolute entropies.

D

G

rxn = S

nG

f  [prods] S

mG

f  [reacts] where

n

and

m

are the coefficients of the products and reactants in the reaction.

•

Temperature Dependence of

D

G

The dependence of D

G

on temperature arises mainly from the

T

D

S

term in the definition of D

G

.

• • The sign of D

G

D

H

is dominated by the sign of at low temperatures and by the sign of D

S

at high temperatures.

Negative values of D

H

D

S

favor spontaneity.

and positive values of

Direction of Spontaneous Reaction

∆

H

+ + ∆

S

+ + Temperature All Low High Low High All ∆

G

+ + + Spontaneous Direction Forward Forward Reverse Reverse Forward Reverse

Example: Temperature Dependence

• Assuming that D

H

 and D

S

 do not change with temperature, calculate the temperature for which D

G

 is 0 for the reaction CS 2 ( l ) ⇌ At 298 K, D

H

 CS 2 (g) = 27.66 kJ and D

S

 86.39 J/K.

=

•

Example: Calculate

D

G

 Given the following chemical reaction and data at 298 K: N 2 (g) + 3H 2 (g)  D

H

f o = 0 0 2NH 3 (g) -46.1 kJ/mol

S

o = 191.5 130.6 192.3 J/mol .

K assuming that D

H

o and D

S

o do not change with temperature, calculate D

G o

at 1000 K.

Concentration and Free Energy

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

G

= D

G

 +

RT

ln

Q

where

Q

is the reaction quotient (see Chapter 14).

Example: Concentration Dependence

• Substance NO 2 (g) N 2 O 4 (g) D

G

f o kJ/mol 51.29

97.82

For the reaction 2NO 2 (g) ⇌ N 2 O 4 (g) (a) calculate D

G

o at 298 K.

(b) calculate and

P

N2O4 D

G

when

P

NO2 = 0.98 atm.

= 0.12 atm

Free Energy and

K

eq • For a system at equilibrium, D

G

• = 0, and

Q

=

K

eq , so D

G

o = -

RT

ln

K

eq D

G

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

Example: K

eq

•

Calculation

Evaluate the equilibrium constant at 298 K for 2NO(g) + Br 2 (g) ⇌ 2NOBr(g) using the standard free energies of formation at 298 K given below.

Substance ∆G f ° (kJ/mol)

NO(g) 86.55

Br 2 (g) 3.14

NOBr(g) 82.4

Temperature and

K

eq • The temperature dependence of

K

eq derived from two equations given is earlier: D

H

 -

T

D

S

 = D

G

 = -

RT

ln

K

eq A graph of ln

K

eq

vs

. 1/

T

with a slope = D

H

 /

R

gives a line and an intercept of D

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 :

w

max = D

G

• This is a limit imposed by nature, as we strive to improve the efficiency of energy conversions When D

G

> 0 (spontaneous in the reverse direction), it represents the

minimum work

that must be provided to cause the change.