Question

The equilibrium constant for the following reactions is $1.6\times {10}^5$ at $1024K$.

$H_{2\left(g\right)}+B{r}_{2\left(g\right)}\rightleftharpoons 2HB{r}_{\left(g\right)}$

Find the equilibrium pressure of all gases if $10.0$ bar of $HBr$ is introduced into a sealed container at $1024K$.

Answer 1

For the equilibrium reaction $2HBr\left(g\right)\rightleftharpoons H_2\left(g\right)+B{r}_2\left(g\right)$ , the equilibrium constant is $K=\frac{1}{1.6\times {10}^5}$.

Let the decease in the equilibrium pressure of HBr be p bar.
The initial pressures of HBr, $H_2,B{r}_2$ are 10 bar, 0 bar and 0 bar respectively.
The equilibrium pressures are 10-p, p/2 and p/2 respectively.
The equilibrium constant expression is
$K_p=\frac{P_{H_2}{P}_{B{r}_2}}{(P_{HBr}{)}^2}=\frac{\left(p/2\right)\times \left(p/2\right)}{(10-p{)}^2}=\frac{1}{1.6\times {10}^5}$

$\frac{p^2}{4(10-p{)}^2}=\frac{1}{1.6\times {10}^5}$

$400p=20-2p$

$p=0.0498$ bar

The equilibium pressures are
$P_{H_2}=P_{B{r}_2}=\frac{0.0498}{2}=0.0249$ bar
$P_{HBr}=10-0.0498=10$ bar