Question

The reaction: 2 SO2(g) + O2(g) --> 2 SO3(g) has an equilibrium constant of K1. What is the K value for the reaction: SO3(g) --> SO2(g) + ½ O2(g)?

K1^½
1/K1
½ K1
(1/K1)^½

Answer 1

Answer: The value of equilibrium constant for reverse reaction is $(\frac{1}{K_1})^{1/2}$

Explanation:

The given chemical equation follows:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equilibrium constant for the above equation is $K_1$

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$SO_3(g)\rightarrow SO_2(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '$\frac{1}{2}$', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(\frac{1}{K_1})^{1/2}$

Hence, the value of equilibrium constant for reverse reaction is $(\frac{1}{K_1})^{1/2}$