The equilibrium constant for the following reaction is $1.6 \times 10^{5}$ at $1024 \, K$:
$H_{2(g)} + Br_{2(g)} \longleftrightarrow 2HBr_{(g)}$
Find the equilibrium pressure of all gases if $10.0 \, bar$ of $HBr$ is introduced into a sealed container at $1024 \, K$.

(N/A) Given,$K_{p}$ for the reaction $H_{2(g)} + Br_{2(g)} \longleftrightarrow 2HBr_{(g)}$ is $1.6 \times 10^{5}$.
For the reverse reaction $2HBr_{(g)} \longleftrightarrow H_{2(g)} + Br_{2(g)}$,the equilibrium constant $K'_{p}$ is:
$K'_{p} = \frac{1}{K_{p}} = \frac{1}{1.6 \times 10^{5}} = 6.25 \times 10^{-6}$.
Let $p$ be the pressure of both $H_{2}$ and $Br_{2}$ at equilibrium.
Reaction: $2HBr_{(g)} \longleftrightarrow H_{2(g)} + Br_{2(g)}$
Initial pressure: $10, 0, 0$
Equilibrium pressure: $10 - 2p, p, p$
Expression: $\frac{p_{H_{2}} \times p_{Br_{2}}}{p_{HBr}^{2}} = K'_{p}$
$\frac{p \times p}{(10 - 2p)^{2}} = 6.25 \times 10^{-6}$
Taking the square root of both sides:
$\frac{p}{10 - 2p} = 2.5 \times 10^{-3}$
$p = 0.025 - 0.005p$
$1.005p = 0.025$
$p \approx 0.0249 \, bar = 2.49 \times 10^{-2} \, bar$.
At equilibrium:
$p_{H_{2}} = p_{Br_{2}} = 2.49 \times 10^{-2} \, bar$
$p_{HBr} = 10 - 2(0.0249) = 9.95 \, bar$.