Question

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

• A. 8
• B. 2
• C. 3
• D. 5

Answer 1

The correct option is C 3

$H_2\left(g\right)+B{r}_2\left(g\right)\rightleftharpoons 2HBr\left(g\right)t=00010bar\frac{+x+x-2x}{t=t_{eq}xx10-2x}{K}_p=\frac{P_{HBr}^2}{P_{B{r}_2}{P}_{H_2}}$
$K_p=16\times {10}^{-6}=\frac{(10-2x{)}^2}{x^2}\Rightarrow 4\times {10}^{-3}=\frac{10-2x}{x}=\frac{10}{x}-2$
$\Rightarrow 4\times {10}^{-3}=\frac{10}{x}-2$
$\Rightarrow 2.004=\frac{10}{x}$
$x=4.989$
Hence, equilibrium pressure of $H_2$ gas is 5 bar.