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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 27 of 73
Marks: +1, -0
The equilibrium constant for the following reaction is 1.6 × 105 at 1024 K.
H2(g)+Br2(g)\mathrm{H}_{2(g)} + \mathrm{Br}_{2(g)} ⇌ 2HBr(g)2\mathrm{HBr}_{(g)}
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
Solution:  
H2(g)+Br2(g)\mathrm{H}_{2(g)} + \mathrm{Br}_{2(g)} ⇌ 2HBr(g)2\mathrm{HBr}_{(g)} ; KpK_{p} = 1.6 × 10510^{5}
For the reaction 2HBr(g)2\mathrm{HBr}_{(g)} ⇌ H2(g)+Br2(g)\mathrm{H}_{2(g)} + \mathrm{Br}_{2(g)}
KpK_{p} = 11.6×105\frac{1}{1.6 \times 10^{5}} = 6.25 × 10−610^{-6}
2HBr(g)2\mathrm{HBr}_{(g)} ⇌ H2(g)+Br2(g)\mathrm{H}_{2(g)} + \mathrm{Br}_{2(g)}
Initial presuure10.000At equilibrium10.0−2xxx\begin{array}{lccc} \text{Initial presuure} & 10.0 & 0 & 0 \\ \text{At equilibrium} & 10.0-2x & x & x \end{array}
∴ KpK_{p} = pH2×pBr2pHBr2\frac{p_{H_2} \times p_{Br_2}}{p_{HBr}^2}
6.25 × 10−610^{-6} = x2(10.0−2x)2\frac{x^2}{(10.0-2x)^2} or x10.0−2x\frac{x}{10.0-2x} = 2.5 × 10−510^{-5}
and x = 2.5 × 10−210^{-2} So, pH2p_{H_2} = pBr2p_{Br_2} = 2.5 × 10−210^{-2} bar
pHBrp_{HBr} = 10.0 – 5.0 × 10−210^{-2} ≈ 10.0 bar
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