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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 16 of 73
Marks: +1, -0
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl(g)2\mathrm{ICl}_{(g)} ⇌ I2(g)+Cl2(g)\mathrm{I}_{2(g)} + \mathrm{Cl}_{2(g)} ; KcK_c = 0.14
Solution:  
2ICl(g)2\mathrm{ICl}_{(g)} ⇌ I2(g)+Cl2(g)\mathrm{I}_{2(g)} + \mathrm{Cl}_{2(g)} ; KcK_c = 0.14
Initial molar conc.0.7800Eqm. molar conc.0.78−2xxx\begin{array}{lccc} \text{Initial molar conc.} & 0.78 & 0 & 0 \\ \text{Eqm. molar conc.} & 0.78 - 2x & x & x \end{array}
Applying law of chemical equilibrium,
KcK_c = [I2][Cl2][ICl]2\frac{[\mathrm{I}_2][\mathrm{Cl}_2]}{[\mathrm{ICl}]^2} ⇒ 0.14 = x.x(0.78−2x)2\frac{x.x}{(0.78-2x)^2}
x2x^2 = 0.14 (0.78−2x)2(0.78-2x)^2
or x0.78−2x\frac{x}{0.78-2x} = 0.14\sqrt{0.14} = 0.374 or x = 0.292 - 0.748 x
or 1.748x = 0.292 or x = 0.167
Hence at equilibrium, [I2][\mathrm{I}_2] = [Cl2][\mathrm{Cl}_2] = 0.167 M
[ICl] = 0.78 – 2 × 0.167 = 0.446 M
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