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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 3 of 73
Marks: +1, -0
At a certain temperature and total pressure of 10510^{5} Pa, iodine vapour contains 40% by volume of I atoms. I2(g)\mathrm{I}_{2(g)} ⇌ 2I(g)2\mathrm{I}_{(g)} Calculate KpK_p for the equilibrium.
Solution:  
I2(g)1→1−xI_{\underset{1 \rightarrow 1-x}{2(g)}} ⇌ 2I(g)0−2x\underset{0 - 2x}{2\mathrm{I}_{(g)}}
Total moles at equilibrium = 1 – x + 2x = 1 + x
But 2x1+x\frac{2x}{1+x} = 0.4 (Given)
∴ x = 14\frac{1}{4}
So, mole fraction of I2\mathrm{I}_2 = 1−141+14\frac{1-\frac{1}{4}}{1+\frac{1}{4}} = 35\frac{3}{5} = 0.6, mole fraction of I = 2×141+14\frac{2 \times \frac{1}{4}}{1 + \frac{1}{4}} = 25\frac{2}{5} = 0.4
pIp_I = 0.4 × 105 Pa10^{5}\,\text{Pa}, pI2p_{\mathrm{I}_2} = 0.6 × 105 Pa10^{5}\,\text{Pa}
KpK_p = pI2PI2\frac{p_{\mathrm{I}}^2}{P_{\mathrm{I}_2}} = (0.4)2(105)2(0.6)(105)\frac{(0.4)^2 (10^{5})^2}{(0.6)(10^{5})} = 2.67 × 10410^{4}
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