Question

HI decomposes to H2 and I2 by the following equation: 2HI(g) → H2(g) + I2(g);Kc = 1.6 × 10−3 at 25∘C If 1.0 M HI is placed into a closed container and the reaction is allowed to reach equilibrium at 25∘C, what is the equilibrium concentration of H2 (g)?

Answer 1

Answer: The concentration of hydrogen gas at equilibrium is 0.037 M

Explanation:

We are given:

Initial concentration of HI = 1.0 M

The given chemical equation follows:

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

Initial:               1.0

At eqllm:        1.0-2x          x           x

The expression of $K_c$ for above equation follows:

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

We are given:

$Kc=1.6\times 10^{-3}$

Putting values in above expression, we get:

$1.6\times 10^{-3}=\frac{x\times x}{(1.0-2x)^2}\\\\x=-0.043,0.037$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.037 M

Hence, the concentration of hydrogen gas at equilibrium is 0.037 M