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 $\frac{2x}{0.5-x}=7$
$2x=3.5-7x$
$\Rightarrow 9x=3.5$
$x=\frac{3.5}{9}=0.3888$
$p_{H_2}=0.5-0.3888=0.1111$