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 $10.0bar$ of $HBr$ is introduced into a sealed container at $1024K$.

• A. $2.8\times {10}^3$
• B. $5.2\times {10}^{-2}$
• C. $2.5\times {10}^{-2}$
• D. $3.4\times {10}^{-3}$

Answer 1

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

For the reaction $2HBr\left(g\right)\rightleftharpoons H_2\left(g\right)+B{r}_2\left(g\right)$
$K_C=\frac{1}{1.6\times {10}^5}$ , as the reaction is reverse of previous reaction.

As $\Delta n_g=0$, so
$K_P=K_C=\frac{1}{1.6\times {10}^5}$


Now decompostion,
$\begin{array}{cccccc}& 2HBr\left(g\right)& \rightleftharpoons & B{r}_2\left(g\right)+& H_2\left(g\right)& \\ initialpressure& 10& & 0& 0& \end{array}$

$\begin{array}{ccccccccc}Atequil.& 10-2p& & & & p& & & p\end{array}$

So,
$K_p=\frac{p^2}{(10-2p{)}^2}$
putting the values we get,
$\frac{1}{1.6\times {10}^5}=\frac{p^2}{(10-2p{)}^2}$
$\Rightarrow 400p=10-2p$
$\Rightarrow p=0.025bar$
so, $P_{H_2}=p=2.5\times {10}^{-2}bar$