Question

The equilibrium constant for the following reaction is $1.6\times {10}^5$ at 1024K.
$H_2\left(g\right)+B{r}_2\left(g\right)\rightleftharpoons 2HBr\left(g\right)$
Find the equilibrium pressure of $H_2$ gas if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

• A. $2.8\times {10}^3$
• B. $5.2\times {10}^{-3}$
• C. $2.5\times {10}^{-2}$
• D. $2\times {10}^{-3}$

Answer 1

The correct option is C $2.5\times {10}^{-2}$

The equilibrium reaction is $H_2\left(g\right)+B{r}_2\left(g\right)\rightleftharpoons 2HBr\left(g\right)$.

Initially, 10 bar is the pressure of HBr.

At equilibrium, the partial pressures of hydrogen, bromine, and HBr are x bar, x bar and (10-2x) bar respectively.

The expression for the equilibrium constant is $K_p=\frac{P_{HBr}^2}{P_{H_2}{P}_{B{r}_2}}=16\times {10}^4=\frac{(10-2x{)}^2}{x^2}$.

$4\times {10}^2=\frac{10-2x}{x}\simeq \frac{10}{x}$

$x=\frac{10}{4\times {10}^2}=2.5\times {10}^{-2}$