Question

Calculate Kc for the reaction: 2 HI(g) ⇄ H2(g) + I2(g) given that the concentrations of each species at equilibrium are as follows: [HI] = 0.85 mol/L, [I2] = 0.60 mol/L, [H2] = 0.27 mol/L.

Answer 1

Answer: The value of $K_c$ for the given reaction is 0.224

Explanation:

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

The expression of $K_c$ for given equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$[HI]_{eq}=0.85M$

$[H_2]_{eq}=0.27M$

$[I_2]_{eq}=0.60M$

Putting values in above expression, we get:

$K_c=\frac{0.27\times 0.60}{(0.85)^2}=0.224$

Hence, the value of $K_c$ for the given reaction is 0.224