For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data : ΔH∘rxn 180.5kJ/mol
ΔS∘rxn 24.80J/(mol⋅K)

Required:
a. Calculate the temperature in kelvins above which this reaction is spontaneous
b. Calculate the equilibrium constant for the following reaction at room temperature, 25°C

a. 7278 K

b. Kc = 4.542 × 10⁻³¹

Explanation:

a.

The reaction is spontaneous when ΔG° < 0. We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T × ΔS°

Then, the reaction will be spontaneous when,

ΔH° - T × ΔS° < 0

T > ΔH°/ΔS

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

T > 7278 K

b.

First, we will calculate ΔG° at 25 °C (298 K)

ΔG° = ΔH° - T × ΔS°

ΔG° = (180.5 × 10³ J/mol) - 298 K × (24.80J/mol⋅K) = 1.731 × 10⁵ J/mol

Then, we will calculate the equilibrium constant (Kc) using the following expression.

ΔG° = - R × T × ln Kc

-ΔG°/R × T =  ln Kc

-(1.731 × 10⁵ J/mol)/(8.314 J/mol.K) × 298 K =  ln Kc

Kc = 4.542 × 10⁻³¹

c8h10 is your answerr

the answer is c.the valence of the atoms combine .