Question

The equilibrium constant of the reaction, $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is $50$. If the volume of the container is reduced to one half of its original value, the equilibrium constant will be:

• A. $25$
• B. $50$
• C. $75$
• D. $100$

Answer 1

Answer: (B)

$50$

The equilibrium constant of the reaction, $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is $50$. If the volume of the container is reduced to one half of its original value, the equilibrium constant will be 50.
Note: The value of the equilibrium constant will not be affected by the change in volume as the total number of moles of reactants is equal to total number of moles of products.
$K=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$
$K=\frac{(\frac{n_{HI}}{V}{)}^2}{\left(\frac{n_{H_2}}{V}\right)\times \left(\frac{n_{I_2}}{V}\right)}$
$K=\frac{[n_{HI}{]}^2}{n_{\left[H_2\right]}{n}_{\left[I_2\right]}}$