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Question

Chemistry Question on Equilibrium

Equilibrium constants K1 K_{1} and K2K_{2} for the following equilibria

A

K2=1K1K_{2}=\frac{1}{K_{1}}

B

K2=K12K_{2}=K_{1}^{2}

C

K2=K12K_{2}=\frac{K_{1}}{2}

D

K2=1K12K_{2}=\frac{1}{K_{1}^{2}}

Answer

K2=1K12K_{2}=\frac{1}{K_{1}^{2}}

Explanation

Solution

(i) NO(g)+12O2(g)<=>[K1]NO2(g)NO (g)+\frac{1}{2} O _{2}(g) {<=>[{K_{1}}]} NO _{2}(g)
So K1=[NO2][NO][O2]12   ...(i)K_{1}=\frac{\left[ NO _{2}\right]}{[ NO ]\left[ O _{2}\right]^{\frac{1}{2}}}\,\,\,...(i)
(ii) 2NO2(g)<=>[R2]2NO+O2(g)2 NO _{2}(g) {<=>[{ R _{2}}]} 2 NO + O _{2}(g)
So K3=[NO]2[O2][NO2]2    ...(ii)K_{3}=\frac{[ NO ]^{2}\left[ O _{2}\right]}{\left[ NO _{2}\right]^{2}}\,\,\,\,...(ii)
From E (i)
K12=[NO2]2[NO]2[O2]K_{1}^{2}=\frac{\left[ NO _{2}\right]^{2}}{[ NO ]^{2}\left[ O _{2}\right]}
or 1K12=[NO]2[O2][NO2]2    ...(iii)\frac{1}{K_{1}^{2}}=\frac{[ NO ]^{2}\left[ O _{2}\right]}{\left[ NO _{2}\right]^{2}}\,\,\,\,...(iii)
From Eqs. (ii) and (iii)
K2=1K12K_{2}=\frac{1}{K_{1}^{2}}