Solveeit Logo
Login

Question

Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

For the following equilibrium, Kc=6.3×1014K_c = 6.3 × 10^{14} at 1000 K1000\ K
NO(g)+O3(g)⇋NO2(g)+O2(g)NO (g) + O_3 (g) ⇋ NO_2 (g) + O_2 (g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc′K'_c for the reverse reaction?

Answer

It is given that KcK_c for the forward reaction is 6.3×10146.3 × 10^{14}.
Then, KcK_c for the reverse reaction will be,
Kc′=1KcK'_c = \frac {1}{K_c }

Kc′=16.3×1014K'_c= \frac {1}{6.3 × 10^{14 }}

Kc′=1.59×10−15K'_c= 1.59 × 10^{-15}