Question

The $ K_{p} $ value for the reaction, $ H_{2} + I_{2} {\rightleftharpoons} 2HI $ at $ 460^{\circ}C $ is $ 49 $ . If the initial pressure of $ 2 $ and $ 2 $ is $ 0.5 $ atm, respectively, what will be the partial pressure of $ H_{2} $ at equilibrium?

• A. 0.111 atm
• B. 0.123 atm
• C. 0.113 atm
• D. 0.222 atm

Answer 1

Answer: (A)

$K_{p}=\frac{\left(P_{HI}\right)^{2}}{P_{H_2}\cdot P_{I_2}}$
$\Rightarrow \frac{\left(2x\right)^{2}}{\left(0.5-x\right)\left(0.5-x\right)}=49$
or $\frac{\left(2x\right)^{2}}{\left(0.5-x\right)^{2}}=49$ or $ \frac{2x}{0.5-x}=7$
$2x=3.5-7x$
$\Rightarrow 9x=3.5$
$x=\frac{3.5}{9}=0.3888$
$p_{H_2}=0.5-0.3888=0.1111$