Question

Question 7.6 For the following equilibrium, Kc = 6.3 x 10^14 at 1000K

NO(g) + O3(g) <---> NO2(g) + O2(g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

Class XI Equilibrium Page 224

Answer 1

Kc'=1/Kc
Kc = 6.3*10^14
1/Kc = 1/6.3*10^14
kc'= 1.59 * 10^-15

Answer 2

for the reaction , NO(g) + O3(g) <=> NO2(g) + O2(g)

Given,
equilibrium constant ( Kc) = 6.3 × 10¹⁴ at 1000K
but we know, Kc for above reaction ,
Kc = [NO2][O2]/[NO] [O3]

now, for reverse reaction ,
NO2(g) + O2(g) <=> NO(g) + O3(g)
K'c = [NO][O3]/[NO2][O2]
= 1/Kc = 1/6.3 × 10¹⁴ = 1.587 × 10^-15