Question

The equilibrium constants of the reactions,

$S{O}_2+\frac{1}{2}{O}_2\rightleftharpoons S{O}_2$

$2S{O}_2+O_2\rightleftharpoons 2S{O}_3$

and $K_1$ and $K_2$ respectively. The relationship between $K_1$ and $K_2$ is :

• A. $K_1=K_2$
• B. $K_2=K_1^2$
• C. $K_1=\sqrt{K_2}$
• D. $K_2=\sqrt{K_1}$

Answer 1

Answer: (B), (C)

$K_2=K_1^2$
$K_1=\sqrt{K_2}$

The equilibrium constants of the reactions

$S{O}_2+\frac{1}{2}{O}_2\rightleftharpoons S{O}_2$ and

$2S{O}_2+O_2\rightleftharpoons 2S{O}_3$

are $K_1$ and $K_2$ respectively.

The relationship between $K_1$ and $K_2$ is $K_2=K_1^2$ and $K_1=\sqrt{K_2}$

When first reaction is multiplied with 2, second reaction is obtained. Hence, the equilibrium constant for the first reaction is equal to the square root of the equilibrium constant for second reaction. In other words, the equilibrium constant for the second reaction is equal to the square of the equilibrium constant for the first reaction.

$K_1=\frac{\left[S{O}_3\right]}{\left[S{O}_2\right][O_2{]}^{\frac{1}{2}}}$

$K_2=\frac{[S{O}_3{]}^2}{\left[S{O}_2{]}^2\right[O_2]}$