Question

The equilibrium constant for the following reaction is $\text{1} \text{.6} \times \text{10}^{5}$ at $1024 \, \text{K}$
$\left(\text{H}\right)_{2} \text{(g)} + \left(\text{Br}\right)_{2} \text{(g)} \rightleftharpoons \text{2HBr(g)}$
​Find the equilibrium pressure of $\text{HBr} \left(\text{g}\right)$ if $10$ bar of $\text{HBr}$ is introduced into a sealed container at $\text{1024} \, \text{K} \text{.}$

• A. 9.95
• B. $10.1$
• C. $9.8$
• D. $9.9$

Answer 1

Answer: (A)

$\left(\text{H}\right)_{2} \text{(g)} + \left(\text{Br}\right)_{2} \text{(g)} \rightleftharpoons \text{2HBr(g)} \text{;} \left(\text{K}\right)_{\text{p}} = \text{1.6} \times \left(\text{10}\right)^{5} \text{at 1024 K}$

$ \, \, \, \text{2HBr(g)} \, \overset{}{\rightleftharpoons}\left(\text{H}\right)_{\text{2}}\text{(g)} \, \text{+} \, \left(\text{Br}\right)_{\text{2}} \, \text{(g)} \, \text{;} \, \left(\text{k}\right)_{\text{p}}\text{=}\frac{\text{1}}{\text{1} \left(\text{.6\times 10}\right)^{\text{5}}}\text{at} \, \text{1024} \, \text{K} \\ \text{Initial} \, \text{pressure} \, \, \, \text{10}\text{.0} \, \, \text{bar} \, \, \text{0} \, \, \, \text{0} \\ \text{Equilibrium} \, \text{pressure} \, \, \, \text{(10-x)} \, \, \frac{\text{x}}{\text{2}} \, \, \frac{\text{x}}{\text{2}}$
$\text{K}_{\text{p}} = \frac{\text{p}_{\text{H}_{2}} \cdot \text{p}_{\text{Br}_{2}}}{\text{p}_{\text{HBr}}^{2}} = \text{0.625} \times \text{10}^{- 5} = \frac{\frac{\text{x}}{2} \cdot \frac{\text{x}}{2}}{\text{10} \times \text{10}}$
$\frac{1}{\text{1.6} \times \text{10}^{5}} = \text{0.625} \times \text{10}^{- 5} = \frac{\frac{\text{x}}{2} \cdot \frac{\text{x}}{2}}{\text{10} \times \text{10}}$
[(10 - x) ≈ 10 because magnitude of K_p is small.)]
2 × 2 × 10 × 10 × 0.625 × 10^-5 = x² or x = 0.050
$\text{p}_{\text{H}_{2}} = \text{p}_{\text{Br}_{2}} = \frac{\text{x}}{2} = \frac{\text{0.050}}{2} = \text{0.025 bar = 2.5} \times \text{10}^{- 2} \text{bar}$
p_HBr = 10 - 0.050 = 9.95 .