Tro Chapter 17 - Free Energy and Thermodynamics

•  Spontaneous and non-spontaneous processes

•  Entropy and the second law of thermodynamics

•  Gibbs free energy (ΔG˚), entropy (ΔS˚) and enthalpy (ΔH˚)

•  Gibbs free energy and K

On-line HW due Thursday Nov. 11th at 11:59 PM

Thermodynamics vs. Kinetics…Equilibrium?

Perpetual motion, energy transactions and nature’s “heat tax”

Spontaneous vs. nonspontaneous change

Tro 17.2

Predicting spontaneous change

CH4 + 2 O2 CO2 + 2H2O ΔH = -802 kJ

2Fe (s) + 3/2O2 Fe2O3 ΔH = -826 kJ

H2O (l) H2O (s) ΔH = -6.02 kJ (spontaneous at T < 0˚C)

H2O (s) H2O (l) ΔH = +6.02 kJ (spontaneous at T > 0˚C)

NaCl (s) Na+ (aq) + Cl- (aq) ΔH = +3.9 kJ

Tro 17.2 Entropy and the second law of thermodynamics Tro 17.3

S = K ln W S = entropy

K =Boltzmann’s constant (1.38e-23 J/K)

W is the number of energetically equivalent ways to arrange the components of a system

Entropy and Temperature/Phase Change

Entropy and the second law of thermodynamics

The entropy of the universe is always increasing

ΔSuniv = ΔSsys + ΔSsurr > 0

Entropy and the third law of thermodynamics

A perfect crystal has zero entropy at absolute zero

Ssys = 0 at 0 K

Tro 17.3 and 17.4 Predicting relative S0 values of a system

1.  Temperature changes and S for copper metal

T (K) 273 295 298

S (J/molK): 31.0 32.9 33.2

3. Dissolution of a solid or liquid

NaCl AlCl3 CH3OH

S˚ (s or l) 72.1 (s) 167 (s) 127 (l)

S˚ (aq) 115.1 -148 132

2. Physical states and phase changes and entropy

Na H2O C (graphite)

S˚ (s or l) 51.4 (s) 69.9 (l) 5.7 (s)

S˚ (g) 153.6 188.7 158.0

Temperature and its effect on spontaneity Tro 17.4

e.g. consider the freezing of water - spontaneous at low temperatures, nonspontaneous at high temperatures

Temperature and its effect on spontaneity cont’d

• Product favored reactions for exothermic processes with net increase in entropy (+ ΔS)

• Reactant favored reactions for endothermic processes with a net decrease in entropy (- ΔS)

But, what happens with endothermic processes with a net increase in entropy

Or

What happens with exothermic processes and a decrease in entropy

Tro 17.4

Entropy, free energy and work Gibb’s free energy, G, combines a system’s enthalpy and entropy

G = H - TS

ΔSuniv = ΔSsys + ΔSsurr

ΔSsurr = -ΔHsys/T

ΔSuniv = ΔSsys -ΔHsys/T

-TΔSuniv = -TΔSsys +ΔHsys

ΔGsys = ΔHsys -TΔSsys = -TΔSuniv

Tro 17.5 Remembering the second law…

ΔSuniv>0 spontaneous

ΔSuniv = equilibrium

ΔSuniv < 0 non-spontaneous

Since absolute temperature is always positive,

-TΔSuniv < 0 for spontaneous processes

and, since ΔG = -TΔSuniv

ΔG < 0 for a spontaneous process

ΔG = 0 for a process at equilibrium

ΔG > 0 for a non-spontaneous process

Gibbs Free energy - “A decrease in G corresponds to a spontaneous process”

Calculating the change in entropy of a reaction (using table 17.2 or Appendix IIB)

ΔS˚rxn = ΣmS˚products - ΣnS˚reactants

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

ΔS˚rxn = [(2 mol NH3)(S˚ of NH3)] - [(1 mol N2)(S˚ of N2) + (3 mol H2)(S˚ of H2)]

= [2(192.8)]-[1(191.6)+3(130.7)]

= -198.76 J/K

Tro 17.6 Entropy, free energy, and work: Calculating Standard Free Energy changes (see Appendix II B)

4KClO3 (s) Δ 3KClO4 + KCl

Calculate ΔG˚ at 25˚C

1) Calculate ΔH˚ (must know ΔH˚f for each species)

ΔH˚f = -397.7 kJ/mol ΔH˚f =-432.8 kJ/mol ΔH˚f = -436.5 kJ/mol

S˚ = 143.1 J/molK S˚ = 151.0 J/molK S˚ = 82.6 J/molK

2) Calculate ΔS˚ (must know S˚ for each species)

3) Calculate ΔG using ΔG˚ = ΔH˚ - TΔS˚

Tro 17.7

Calculating ΔG˚rxn from ΔG˚f values

4KClO3 (s) Δ 3KClO4 + KCl ΔG˚f = -296.3 kJ/mol ΔG˚f = -303.1 kJ/mol ΔG˚f = -408.5 kJ/mol

ΔG˚rxn = [3(ΔG˚f KClO4) + ΔG˚f KCl] - [4 ΔG˚f KClO3]

Calculating ΔG˚rxn at elevated temperatures

2SO2+ O2 2SO3 At 298 K, ΔG˚ = -141.6 kJ, ΔH˚ = -198.4 kJ and ΔS˚ = -187.9 J/K

Is the reaction spontaneous at 25˚C?

Is the reaction spontaneous at 900˚C (assume ΔH˚ and ΔS˚ don’t change much with temperature)?

ΔG˚rxn = ΣmG˚products - ΣnG˚reactants

Free energy for Non-standard states: ΔG˚rxn and ΔGrxn

ΔG˚ = -RT ln K (by choosing standard conditions for Q)

ΔG = ΔG˚ + RT ln Q

ΔG = RT ln Q/K = RT ln Q - RT ln K

Sign of ΔG˚ allows us to predict spontaneity and thus reaction direction…

If Q<K, ln Q/K < 0; ΔG < 0

If Q>K, ln Q/K > 0; ΔG > 0

If Q=K, ln Q/K = 0; ΔG = 0

Tro 17.8 and 17.9

Tro 17.9 Free Energy and Equilibrium

ΔG˚ = -RT ln K

Free Energy and Equilibrium Calculating ΔGrxn at non-standard conditions

2SO2+ O2 2 SO3 At 298 K, ΔG˚ = -141.6 kJ

a) Calculate K at 298 K and at 973 K (ΔG˚298 = -141.6 kJ/mol and ΔG˚973 = -12.12 kJ/mol)

Useless expressions:

ΔG˚ = -RTlnK

ΔG = ΔG˚ + RTlnQ

b) Which direction would the above reaction proceed if we had 0.500 atm of SO2, 0.0100 atm of O2 and 0.100 atm of SO3 at 25˚C? At 700˚C?

ΔG = -138.16 kJ mol-1 @ 298 K ΔG = -.906 kJ mol-1 @ 973 K

1) For each of the following pairs, choose (circle) the substance with the higher entropy per mole at a given temperature:

a) Ar (l) or Ar (g)

b) He (g) at 3 atm pressure or He (g) at 1.5 atm pressure

c) 1 mol of Ne (g) in 15.0 L or 1 mol of Ne (g) in 1.50 L

d) CO2 (g) or CO2 (s)

2) Using S˚ values provided above, calculate ΔS˚ values for the following reactions. In each case, account for the sign of ΔS˚.

a) C2H4 (g) + H2 (g) C2H6 (g)

b) N2O4 (g) 2NO2 (g)

S˚ C2H4 = 219.4 Jmol-1K-1 S˚ N2O4 = 304.3 Jmol-1K-1 S˚ C2H6 = 229.5 Jmol-1K-1 S˚ NO2 = 240.45 Jmol-1K-1 S˚ H2 = 130.58 Jmol-1K-1

3). For a certain chemical reaction, ΔH˚ = -35.4 kJ and ΔS˚ = -85.5 J/K.

Is the reaction exothermic or endothermic

Does the reaction lead to an increase or decrease in the disorder of the system?

Calculate ΔG˚ for the reaction at 298 K

Is the reaction spontaneous at 298 K

The value of Ka for nitrous acid (HNO2) at 25˚C is 4.5x10-4

By using the value of Ka, calculate ΔG˚ for the dissociation of nitrous acid in aqueous solution.

What is the value of ΔG at equilibrium

What is the value of ΔG when [H+] = 5.0 x 10-2 M, [NO2-] = 6.0 x 10-4 M, and

[HNO2] = 0.20 M?