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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 10 of 73
Marks: +1, -0
At 450 K, Kp\mathrm{K}_p = 2.0 × 1010bar\frac{10^{10}}{\text{bar}} for the given reaction at equilibrium.
2SO2(g)+O2(g)\mathrm{2SO}_{2(g)} + \mathrm{O}_{2(g)} ⇌ 2SO3(g)\mathrm{2SO}_{3(g)}
What is Kc at this temperature?
Solution:  
Kp\mathrm{K}_p = Kc(RT)Δng\mathrm{K}_c (RT)^{\Delta n_g} or Kc\mathrm{K}_c = Kp(RT)Δng\frac{\mathrm{K}_p}{(RT)^{\Delta n_g}}
Δng\Delta n_g = 2 – (2 + 1) = –1, T = 450 K, R = 0.083 bar L K−1 mol−1\mathrm{L\,K^{-1}\,mol^{-1}}
∴ Kc\mathrm{K}_c = 2.0×1010(0.083×450)−1\frac{2.0\times10^{10}}{(0.083\times450)^{-1}} = 2.0 × 101010^{10} × (0.083 × 450) = 7.47 × 1011 M−110^{11}\,\mathrm{M}^{-1}
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