Question

$\text{15}$ moles of $H_{2}$ and $ \, 5.2$ moles of $I_{2}$ are mixed and allowed to attain equilibrium at $500^\circ C$ . At equilibrium, the number of moles of $HI$ is found to be $10$ mole. The equilibrium constant for the formation of $HI$ is

• A. $50$
• B. $15$
• C. $100$
• D. $25$

Answer 1

Answer: (A)

$H_{2} \, \, \, + \, I_{2} \, \rightleftharpoons \, \, 2HI$
$15$ $5.2$ $0$
$\left(15 - 5\right)$ $\left(5.2 - 5\right)$ $10$
Equilibrium constant
$\left(\left(\text{K}\right)_{\text{c}}\right)=\frac{\left(\left[H I\right]\right)^{2}}{\left[H_{2}\right] \left[\right. I_{2} \left]\right.}=\frac{10 \times 10}{10 \times 0.2}=50$