chapter 20 thermodynamics: entropy, free energy and the direction of chemical reactions

Chapter 20

Thermodynamics:

Entropy, Free Energy and the Direction of Chemical Reactions

Limitations of the First Law of Thermodynamics

E = q + w

Euniverse = Esystem + Esurroundings

Esystem = -Esurroundings

The total energy-mass of the universe is constant.

However, this does not tell us anything about the direction of change in the universe.

A spontaneous endothermic chemical reaction

water

Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l).

H0rxn = +62.3 kJ

The Concept of Entropy (S)

Entropy refers to the state of order.

A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.

solid liquid gasmore order less order

crystal + liquid ions in solution

more order less order

more order less order

crystal + crystal gases + ions in solution

The number of ways to arrange a deck of playing cards

1 atm evacuated

Spontaneous expansion of a gas

stopcock closed

stopcock opened

0.5 atm 0.5 atm

1877 Ludwig Boltzman S = k ln W

where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/NA (R = universal gas constant, NA = Avogadro’s number.

•A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy.

•A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy.

Suniverse = Ssystem + Ssurroundings > 0

This is the second law of thermodynamics.

Random motion in a crystal

The third law of thermodynamics.

A perfect crystal has zero entropy at a temperature of absolute zero.

Ssystem = 0 at 0 K

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Predicting Relative S0 Values of a System

1. Temperature changes

2. Physical states and phase changes

3. Dissolution of a solid or liquid

5. Atomic size or molecular complexity

4. Dissolution of a gas

S0 increases as the temperature rises.

S0 increases as a more ordered phase changes to a less ordered phase.

S0 of a dissolved solid or liquid is usually greater than the S0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.

A gas becomes more ordered when it dissolves in a liquid or solid.

In similar substances, increases in mass relate directly to entropy.

In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

The increase in entropy from solid to liquid to gas


The entropy change accompanying the dissolution of a salt

pure solid

pure liquid

solution

MIX


Ethanol Water Solution of ethanol

and water

The small increase in entropy when ethanol dissolves in water


The large decrease in entropy when a gas dissolves in a liquid

O2 gas


NO

NO2

N2O4

Entropy and vibrational motion

Sample 1 Predicting Relative Entropy Values

PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]:

(a) 1mol of SO2(g) or 1mol of SO3(g)(b) 1mol of CO2(s) or 1mol of CO2(g)(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)(d) 1mol of KBr(s) or 1mol of KBr(aq)(e) Seawater in midwinter at 20C or in midsummer at 230C(f) 1mol of CF4(g) or 1mol of CCl4(g)

Sample 2 Calculating the Standard Entropy of Reaction, S0rxn

PROBLEM: Calculate S0rxn for the combustion of 1mol of propane at 250C.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Components of S0universe for spontaneous reactions

exothermic

system becomes more disordered

exothermic

system becomes more ordered

endothermic

system becomes more disordered

Sample 3 Determining Reaction Spontaneity

PROBLEM: At 298K, the formation of ammonia has a negative S0sys;

Calculate S0rxn, and state whether the reaction occurs

spontaneously at this temperature. H0fNH3 = -45.9 kJ/mol

N2(g) + 3H2(g) 2NH3(g) S0sys = -197J/K

G0system = H0

system - TS0system

G0rxn = mG0

products - nG0reactants

Sample 4 Calculating G0 from Enthalpy and Entropy Values

PROBLEM: Potassium chlorate, one of the common oxidizing agents in explosives, fireworks, and matchheads, undergoes a solid-state redox reaction when heated. In this reaction, note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation):

4KClO3(s) 3KClO4(s) + KCl(s)

+7 -1+5

Use H0f and S0 values to calculate G0

sys at 250C for this reaction.

Sample 5

PROBLEM:

Calculating G0rxn from G0

f Values

Use G0f values to calculate Grxn for the reaction:

4KClO3(s) 3KClO4(s) + KCl(s)

Sample 6

PROBLEM:

Determining the Effect of Temperature on G0

An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g):

2SO2(g) + O2(g) 2SO3(g)

At 298K, G0 = -141.6kJ; H0 = -198.4kJ; and S0 = -187.9J/K

(a) Use the data to decide if this reaction is spontaneous at 250C, and predict how G0 will change with increasing T.

(b) Assuming H0 and S0 are constant with increasing T, is the reaction spontaneous at 900.0C?

Sample 7

PROBLEM:

Determining the Effect of Temperature on G0

A reaction is nonspontaneous at room temperature but is spontaneous at -40 0C. What can you say about the signs and relative magnitudes of H0 S0 and -TS0

Reaction Spontaneity and the Signs of H0, S0, and G0

H0 S0 -TS0 G0 Description

- + - -

+ - + +

+ + - + or -

- - + + or -

Spontaneous at all T

Nonspontaneous at all T

Spontaneous at higher T;nonspontaneous at lower T

Spontaneous at lower T;nonspontaneous at higher T

The effect of temperature on reaction spontaneity

G and the Work a System Can Do

For a spontaneous process, G is the maximum work obtainable from the system as the process takes place: G = workmax

For a nonspontaneous process, G is the maximum work that must be done to the system as the process takes place: G = workmax

An example

The coupling of a nonspontaneous reaction to the hydrolysis of ATP.

The cycling of metabolic free enery through ATP

Free Energy, Equilibrium and Reaction Direction

•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0)

•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0)

•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (G = 0)

G = RT ln Q/K = RT lnQ - RT lnK

Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and lnQ = 0 so

G0 = - RT lnK

FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

The Relationship Between G0 and K at 250C

G0(kJ) K Significance

200

100

50

10

1

0

-1

-10

-50

-100

-200

9x10-36

3x10-18

2x10-9

2x10-2

7x10-1

1

1.5

5x101

6x108

3x1017

1x1035

Essentially no forward reaction; reverse reaction goes to completion

Forward and reverse reactions proceed to same extent

Forward reaction goes to completion; essentially no reverse reaction

Sample Problem 6

PROBLEM:

Calculating G at Nonstandard Conditions

The oxidation of SO2, which we considered in Sample Problem 6

2SO2(g) + O2(g) 2SO3(g)

is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature.

(a) Calculate K at 298K and at 973K. (G0298 = -141.6kJ/mol of reaction as

written using H0 and S0 values at 973K. G0973 = -12.12kJ/mol of reaction

as written.)(b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO2, 0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature?(c) Calculate G for the system in part (b) at each temperature.

Sample Problem 7

PROBLEM:

Calculating G at Nonstandard Conditions

At 298 K hypobromous acid (HBrO) dissociates in water with aKa of 2.3 x 10-9.

(a) Calculate G0298

(b) Calculate G if [H3O+] = 6.0 x 10-4M, [BrO-] = 0.1 M and [HBrO]=0.20 M

The relation between free energy and the extent of reaction

G0 < 0 K >1 G0 > 0

K <1

chapter 20 thermodynamics: entropy, free energy and the direction of chemical reactions

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