At $T(K)$, the equilibrium constant for the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g})…

At $T(K)$, the equilibrium constant for the reaction $\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})$ is $1.6 \times 10^5$. If 10 bar ' of HBr is introduced into a sealed vessel at $\mathrm{T}(\mathrm{K})$, the equilibrium pressure of HBr (in bar) is approximately
  1. $10.20$
  2. $10.95$
  3. $9.95$
  4. $11.95$

Solution

Given $\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{~g}) \cdot \mathrm{K}_{\mathrm{p}}=1.6 \times 10^5$
Reverse reaction, $2 \mathrm{HBr}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}), \mathrm{K}_{\mathrm{p}}^{\prime}=\frac{1}{\mathrm{~K}_{\mathrm{p}}}$ $\mathrm{HBr}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Br}(\mathrm{~g}), \mathrm{K}_{\mathrm{p}}^{\prime \prime}=\left(\frac{1}{\mathrm{~K}_{\mathrm{p}}}\right)^{1 / 2}$ $\text {At }t=0 \quad 10 \text { bar } 00$ $\text {At eqm, } 10-\mathrm{x} \quad \frac{\mathrm{x}}{2} \quad \frac{\mathrm{x}}{2}$ $\left(\frac{1}{\mathrm{~K}_{\mathrm{p}}}\right)^{\frac{1}{2}}=\frac{\mathrm{p}_{\mathrm{H}_2}^{1 / 2} \times \mathrm{p}_{\mathrm{Br}}^{1 / 2}}{\mathrm{P}_{\mathrm{HBr}}}$ $\left(\frac{1}{1.6 \times 10^5}\right)^{1 / 2}=\frac{\left(\frac{x}{2}\right)^{1 / 2}\left(\frac{x}{2}\right)^{1 / 2}}{(10-x)}$ $\therefore 10 \gt \gt x$ $2.5 \times 10^{-3}=\frac{x}{20}$ $\Rightarrow \mathrm{x}=0.05 \Rightarrow \mathrm{p}_{\mathrm{HBr}}=10-0.05=9.95 \text { bar. }$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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