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

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml