Question

he thermodynamic properties for a reaction are related by the equation that defines the standard free energy, ΔG∘, in kJ/mol:

ΔG∘=ΔH∘−TΔS∘

where ΔH∘ is the standard enthalpy change in kJ/mol and ΔS∘ is the standard entropy change in J/(mol⋅K). A good approximation of the free energy change at other temperatures, ΔGT, can also be obtained by utilizing this equation and assuming enthalpy (ΔH∘) and entropy (ΔS∘) change little with temperature.

Part A

For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.):

ΔH∘rxn	180.5kJ/mol
ΔS∘rxn	24.80J/(mol⋅K)

Calculate the temperature in kelvins above which this reaction is spontaneous.

Express your answer to four significant figures and include the appropriate units.

The standard free energy change, ΔG∘, and the equilibrium constant K for a reaction can be related by the following equation:

ΔG∘=−RTlnK

where T is the Kelvin temperature and R is equal to 8.314 J/(mol⋅K).

Part B

The thermodynamic values from part A will be useful as you work through part B:

ΔH∘rxn	180.5kJ/mol
ΔS∘rxn	24.80J/(mol⋅K)

Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C:

N2(g)+O2(g)→2NO(g)

Express your answer numerically to three significant figures.

Answer 1

A)
ΔHo = 180.5 KJ/mol
ΔSo = 24.8 J/mol.K
= 0.0248 KJ/mol.K

use:
ΔGo = ΔHo - T*ΔSo
for reaction to be spontaneous, ΔGo should be negative
that is ΔGo<0
since ΔGo = ΔHo - T*ΔSo
so, ΔHo - T*ΔSo < 0
180.5- T *0.0248 < 0
T *0.0248 > 180.5
T > 7278 K
Answer: 7278 K

B)
ΔHo = 180.5 KJ/mol
ΔSo = 24.8 J/mol.K
= 0.0248 KJ/mol.K
T= 25.0 oC
= (25.0+273) K
= 298 K

use:
ΔGo = ΔHo - T*ΔSo
ΔGo = 180.5 - 298.0 * 0.0248
ΔGo = 173.1096 KJ/mol

T = 298 K
ΔG = 173.1096 KJ/mol
ΔG = 173109.6 J/mol

use:
ΔG = -R*T*ln Kc
173109.6 = - 8.314*298.0* ln(Kc)
ln Kc = -69.8707
Kc = 4.524*10^-31
Answer: 4.52*10^-31