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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 33 of 73
Marks: +1, -0
The value of Kc for the reaction 3O2(g)3\mathrm{O}_{2(g)} ⇌ 2O3(g)2\mathrm{O}_{3(g)} is 2.0 × 10−5010^{-50} at 25°C. If the equilibrium concentration of O2\mathrm{O}_2 in air at 25°C is 1.6 × 10−210^{-2}, what is the concentration of O3\mathrm{O}_3?
Solution:  
3O2(g)3\mathrm{O}_{2(g)} ⇌ 2O3(g)2\mathrm{O}_{3(g)}
KcK_c = [O3]2[O2]3\frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3} ⇒ 2.0 × 10−5010^{-50} = [O3]2(1.6×10−2)3\frac{[\mathrm{O}_3]^2}{(1.6 \times 10^{-2})^3}
[O3]2[\mathrm{O}_3]^2 = 2.0 × 10−5010^{-50} × (1.6)2×10−6(1.6)^2 \times 10^{-6}
[O3][\mathrm{O}_3] = 2.0×(1.6)3×10−28\sqrt{2.0 \times (1.6)^3} \times 10^{-28} = 3.2×(1.6)2×10−.28\sqrt{3.2 \times (1.6)^2} \times 10^{-.28}
= 1.6 × 10−283.210^{-28} \sqrt{3.2} = 1.6 × 1.79 × 10−28mol L−110^{-28} \text{mol L}^{-1} = 2.864 × 10–28molL–110^{–28} mol L^{–1}
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