Question

At 700 K, equilibrium constant for the reaction: $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is 64. If $0.5mol{L}^{-1}$ of HI(g) is present at equilibrium at 700 K, what are the concentration of $H_2\left(g\right)$ and $I_2\left(g\right)$ assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

• A. 0.0625 M
• B. 0.034 M
• C. 0.136 M
• D. 0.68 M

Answer 1

The correct option is A 0.0625 M

Let $mol{L}^{-1}$ be the equilibrium concentration of hydrogen.

Thus the equilibrium concentrations of hydrogen, iodine, and HI are x,x and 0.5M respectively.

The expression for the equilibrium constant is $K_C=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$.

Substitute values in the above expression.

$64=\frac{0.5\times 0.5}{x^2}$

$x^2=\frac{0.5\times 0.5}{64}$ or $x=0.0625M$

Thus option A is the right answer.