Question

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

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)

Answer 1

Answer:

a. 7278 K

b. 4.542 × 10⁻³¹

Explanation:

a.

Let´s consider the following reaction.

N₂(g) + O₂(g) ⇄ 2 NO(g)

The reaction is spontaneous when:

ΔG° < 0  [1]

Let's consider a second relation:

ΔG° = ΔH° - T × ΔS° [2]

Combining [1] and [2],

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)

T >  7278 K

b.

First, we will calculate ΔG° at 25°C + 273.15 = 298 K

ΔG° = ΔH° - T × ΔS°

ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K

ΔG° = 173.1 kJ/mol

We can calculate the equilibrium constant using the following expression.

ΔG° = - R × T × lnK

lnK = - ΔG° / R × T

lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K

K = 4.542 × 10⁻³¹