Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. At $500\,K$, the equilibrium constant for the reaction $H_{2(g)} +I_{2(g)}\rightleftharpoons 2HI_{(g)}$ is $24.8$. If $\frac{1}{2}\text{mol/L}$ of $HI$ is present at equilibrium, what are the concentrations of $H_2$ and $I_2$, assuming that we started by taking $HI$ and reached the equilibrium at $500\,K$?

Equilibrium

Solution:

image
$K_{c}=\frac{\left[H_{2}\right]\left[I_{2}\right]}{\left[HI\right]^{2}}$ or $\frac{1}{24.8}=\frac{xx}{0.5\times0.5}$
or $x^{2}=\frac{0.25}{24.8}=0.010$
$x=0.10\,\text{mol L}^{-1}$