Thermodynamics: Entropy,

Free Energy

Direction of Chemical Reactions

Chapter 20 Entropy and Free Energy - Spontaneity of Reaction

1. The Second Law of Thermodynamics: Predicting Spontaneous Change

First Law and enthalpy, Second Law and entropy, Standard Entropy and Third Law

2. Calculating the Change in Entropy of a ReactionThe standard entropy of reaction, entropy change and equilibrium

3. Entropy, Free Energy, and WorkFree energy change and spontaneity, Standard free energy changes

4. Free Energy, Equilibrium, and Reaction Direction

Important criterions

Spontaneous reaction, in chemical reaction terms, is one which occurs

with the system releasing releasing free energyfree energy in some form (often, but not

always, heat) and moving to a lower energylower energy, hence more

thermodynamically stablethermodynamically stable, state.

Enthalpy, H, is the sum of the sum of the internal energyinternal energy and the product of product of

the pressure and volumethe pressure and volume of a thermodynamic system.

Entropy, S, is a state function which measures the degree of disorderdegree of disorder

(randomness)(randomness) of a system. The larger the disorder, the greater the

value of S in units of J/K.

Free Energy is the measure of the spontaneityspontaneity of a process and of the

useful energy available from it.

The First Law of Thermodynamics indicates that to conserve energyconserve energy,

the change in energy must equal the heat transferred plus the work

performed, E = q + wE = q + w.

The Second Law of Thermodynamics states that for a spontaneousspontaneous

process, ∆∆SSuniverseuniverse = ∆Ssystem + ∆Ssurroundings > 0 > 0 ((∆∆SSuniverseuniverse positive))

Figure 1 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

Spontaneous reaction and exothermic /endothermic

Common mistake – exothermic is spontaneous.

The Concept of Entropy (S)

Entropy refers to the state of order, measures the degree of disorder of a system. .

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

Figure 2

1 atm evacuated

Spontaneous expansion of a gas

stopcock closed

stopcock opened

0.5 atm 0.5 atm

In thermodynamic terms, a change in order is the number of ways of arranging the particles.

The change in order is likely the key factor to determine the direction of a spontaneous reaction.

The number of possible arrangement is increased.

The number is of ways of arranging a system is related to the entropy.

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.

Figure 4 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

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.

Thermodynamics:

Entropy, Free Energy, and the Direction of Chemical Reactions

1 Three Laws of Thermodynamics: First Law : ∆Universe = 0, Second Law: ∆S>0, G<0, Third law :

Ssolid=0

2 Calculating the Change in Entropy of a Reaction

∆S0 = mS0products - nS0

reactants

Combine standard molar entropies to find the standard entropy for a reaction

3 Entropy, Free Energy, and WorkG = H-TS, ∆ G= ∆H- ∆TS, ∆G= ∆H- T∆S, ∆G0= ∆H0 - T∆S0,

4 Free Energy, Equilibrium, and Reaction Direction

Lecture 2

Summary

Spontaneous reaction happens at a specific condition without continuous input of energy.

Neither first law of thermodynamic or the sign of ∆H predicts the direction of reaction.

The spontaneous reaction involves a change in the universe from more to less order.

Entropy is the measure of disorder and directly related to the number of ways of arrangement for components in the system.

The second law of thermodynamic states that the entropy of universe increases in a spontaneous process.

Entropy value are absolute since the pure single crystal has a zero entropy at absolute zero degree.

Standard molar entropy is dependant on the temperature, phase change, dissolution, atomic size, and molecular complexity.

Sample Problem

SOLUTION:

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)

PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature.

(a) 1mol of SO3(g) - more atoms

(b) 1mol of CO2(g) - gas > solid

(c) 3mol of O2(g) - larger #mols

(d) 1mol of KBr(aq) - solution > solid

(e) 230C - higher temperature

(f) CCl4 - larger mass

Important criterions

Spontaneous reaction, in chemical reaction terms, is one

which occurs with the system releasing free energy in some

form (often, but not always, heat) and moving to a lower

energy, hence more thermodynamically stable, state.

Entropy, S, measures the degree of disorder of a system.

The larger the disorder, the greater the value of S in units of

J/K.

Free Energy, G, is the measure of the spontaneity of a

process and of the useful energy available from it.

Sample Problem Calculating the Standard Entropy of Reaction, S0rxn

PROBLEM: Calculate S0rxn for the combustion of 1mol of propane at 25 0C.

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

PLAN: Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas.

SOLUTION: Find standard entropy values in the Appendix or other table.

S = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)]

S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)]

S = - 374 J/K

When a reaction occurs, chemists expect to learn how to predict and calculate the change in entropy using the following equation:

∆S0 = mS0products - nS0

reactants

Lecture 2

In many spontaneous reaction, the system become more ordered at a given condition, given condition, ∆∆SS00

rxnrxn<0<0.

The Second Law of Thermodynamics indicated that the decrease in entropy of a system only can occur when the increase the entropy in surrounding overweigh them.

The entropy change of the surrounding is related to an opposite change in the heat of the system. ∆S0

surr - qsys

The entropy change of the surrounding is also reversely proportional to the temperature of the heat transferred. ∆S0

surr 1/T

Combining these changes, we obtain the equation:

∆S0surr = -

∆S0surr = -

Determining Reaction Spontaneity

qsys

T When a pressure is constant, the heat is ∆H. ( ∆H = q + ∆ PV)∆ Hsys

T

Lecture 2

Sample Problem

SOLUTION:

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.

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

PLAN: S0universe must be > 0 in order for this reaction to be spontaneous, so

S0surroundings must be > 197 J/K. To find S0

surr, first find Hsys; Hsys = Hrxn which can be calculated using H0

f values from tables.

S0universe = S0

surr + S0sys.

H0rnx = [(2 mol)(H0

fNH3)] - [(1 mol)(H0fN2) + (3 mol)(H0

fH2)]

H0rnx = -91.8 kJ

S0surr = -H0

sys/T = -(-91.8x103J/298K) = 308 J/K

S0universe = S0

surr + S0sys = 308 J/K + (-197 J/K) = 111 J/K

S0universe > 0 so the reaction is spontaneous.

Lecture 2

G0system = H0

system - TS0system

G0rxn = mG0

products - nG0reactants

Gibbs Free Energy (G)

G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.

G < 0 for a spontaneous process

G > 0 for a nonspontaneous process

G = 0 for a process at equilibrium

Lecture 2

Sample Problem

SOLUTION:

Calculating G0 from Enthalpy and Entropy Values

PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated.

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 (G0rxn) at 250C for this reaction.

PLAN: Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve.

H0rxn

= mH0products - nH0

reactants

H0rxn

= (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) -

(4 mol)(-397.7 kJ/mol)

H0rxn

= -144 kJ

Lecture 2

Sample Problem Calculating G0 from Enthalpy and Entropy Values

continued

S0rxn

= mS0products - nS0

reactants

S0rxn

= (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) -

(4 mol)(143.1 J/mol*K)

S0rxn

= -36.8 J/K

G0rxn

= H0rxn - T S0

rxn

G0rxn

= -144 kJ - (298K)(-36.8 J/K)(kJ/103 J)

G0rxn

= -133 kJ

Lecture 2

Entropy, Free Energy, Work

Free Gibbs energy is the measure of the spontaneity of a process. • ∆G0

sys = ∆H0

sys - T∆S0sys

• T∆S0univ = - (∆H0

sys - T∆S0sys)

• The sign for ∆G0sys

and ∆S0univ can tell us the spontaneity of a reaction.

• ∆G0sys < 0 for a spontaneous reaction;

• ∆G0sys = 0 for a reaction at equilibrium;

• ∆G0sys > 0 for a nonspontaneous reaction;

• ∆S0univ > 0 for a spontaneous reaction;

• ∆S0univ = 0 for a reaction at equilibrium;

• ∆S0univ < 0 for a nonspontaneous reaction.

Lecture 2

The spontaneity of reaction and the sign of ∆H, ∆G, ∆S.

∆G0 = ∆H0 - T∆S0

∆H0 ∆S0 -T∆S0 ∆G0 Description

- + - - Spontaneous at all T

+ - + + Non-spontaneous at all T

+ + - + or - Spontaneous at high TNon-spontaneous at low T

- - + + or - Spontaneous at low TNon-spontaneous at high T

PROBLEM:

Determining the effect of temperature on G

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

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

At 298K, ∆G0 = -141.6 kJ/mol, ∆H0 = -198.4 kJ/mol, ∆S0 = -187.9kJ/mol,

(a). Use the data to predict if this reaction is spontaneous reaction at 25 C.

(b). Predict how ∆G0 changes when temperature increases.

(c). Assume ∆H0 and ∆S0 are constant with increasing T, is the reaction spontaneous at 900 C?

PLAN: Note the sign of ∆G0, ∆H0, and T∆S0

Use the equations ∆G0 = ∆H0 - T∆S0

SOLUTION:

(a) Noting G = -141.6 kJ, is negative, so the reaction at room temperature is spontaneous.

Determining the effect of temperature on G

(b) Noting G = H – TS

= -198.4 - (- 0.1879T) = -198.4 + 0.1879T

If T increases, the G is less negative. Therefore, the reaction is less spontaneous.

(c) Calculating G = H – TS at 900 C

= -198.4 - (- 0.1879T)

= -198.4 + 0.1879(900+273.15)

= 22.0 kJ

At 900 C, G is positive. Therefore, the reaction is non- spontaneous.

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 ln Q = 0 so

G0 = - RT lnK

FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

Table 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

PROBLEM:

Calculating G at Nonstandard Conditions

The oxidation of SO2, 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.

PLAN: Use the equations and conditions. G0 = -RTlnK

SOLUTION:

Calculating G at Nonstandard Conditions

(a) Calculating K at the two temperatures:

G0 = -RTlnK so

K e (G 0 /RT)

At 298, the exponent is -G0/RT = -(-141.6kJ/mol)(103J/kJ)

(8.314J/mol*K)(298K)= 57.2

K e (G 0 /RT) = e57.2 = 7x1024

At 973, the exponent is -G0/RT(-12.12kJ/mol)(103J/kJ)

(8.314J/mol*K)(973K)= 1.50

K e (G 0 /RT) = e1.50 = 4.5

Calculating G at Nonstandard Conditions

(b) The value of Q =pSO3

2

(pSO2)2(pO2)=

(0.100)2

(0.500)2(0.0100)= 4.00

Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight.

(c) The nonstandard G is calculated using G = G0 + RTLnQ

G298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(Ln4.00)

G298 = -138.2kJ/mol

G973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(Ln4.00)

G298 = -0.9kJ/mol

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

Important equations

• ∆G0sys

= ∆H0sys - T∆S0

sys

• ∆S0univ = - qsys/T

• ∆G0 = ∆H0 - T∆S0

• ∆S0 = mS0products - nS0

reactants

• ∆G0 = mG0products - nG0

reactants

• ∆G0 = - RT lnK

• ∆G = ∆G0 - RT lnQ

Thermodynamic