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Question

Chemistry Question on Equilibrium

Consider the following gaseous equilibria with equilibrium constants K1K_1 and K2K_2 respectively. SO2(g)+12O2(g)SO3(g)SO_{2(g)} + \frac{1}{2} O_{2(g)} \leftrightharpoons SO_{3(g)} 2SO3(g)2SO2(g)+O2(g)2SO_{3(g)} \leftrightharpoons 2SO_{2(g)} + O_{2(g)} The equilibrium constants are related as

A

2K1=K222 \, K_1 = K^2_2

B

K12=1K2K^2_1 = \frac{1}{K_2}

C

K22=1K1K^2_2 = \frac{1}{K_1}

D

K2=2K12K_2 = \frac{2}{K^2_1}

Answer

K12=1K2K^2_1 = \frac{1}{K_2}

Explanation

Solution

SO2(g)+12O2(g)SO3(g)SO _{2}(g)+\frac{1}{2} O _{2}(g) \rightleftharpoons SO _{3}(g) ...(i) K1K_{1}

Equation (i) is reversed and multiplied by 2

2SO3(g)2SO2(g)+O2(g)2 SO _{3}(g) \rightleftharpoons 2 SO _{2}(g)+ O _{2}(g)...(i) 1K12\frac{1}{K_{1}^{2}}
2SO3(g)2SO2(g)+O2(g)2 SO _{3}(g) \rightleftharpoons 2 SO _{2}(g)+ O _{2}(g)(ii) K2K_{2}
K2=1K12\therefore K_{2} =\frac{1}{K_{1}^{2}}
K12=1K2K_{1}^{2} =\frac{1}{K_{2}}