Manipulate equilibrium constant expressions.

Question

The equilibrium constant for the following reaction is 2.54×10⁸ at 523 °C.

1. 2 H₂(g) + S₂(g)2 H₂S(g)   K = 2.54×10⁸ at 523 °C

Calculate the equilibrium constant for the following reactions at 523 °C.

(a)   2 H₂S(g)2 H₂(g) + S₂(g)   K = 

(b)   H₂S(g) H₂(g) + 1/2 S₂(g)   K =

Answer 1

When you reverse a reaction, the equilibrium constant for the reversed reaction is the reciprocal of the original equilibrium constant.

When you multiply a reaction by some factor, the equilibrium constant for the resulting reaction is the original equilibrium constant raised to the same factor.

(a) You have reversed the original reaction, so K = 1 / 2.54x10⁸

K = 3.94x10^-9

(b) You have reversed the original reaction AND have divided it by 2 (or multiplied it by 1/2), so K = 3.94x10^-9

(calculated above for reversal of rxn) raised to the 1/2 power, or square root of 3.94x10^-9

K = 6.28x10^-5