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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 5 of 73
Marks: +1, -0
Find out the value of Kc of each of the following equilibria from the value of KpK_p :
(i) 2NOCl(g)2\mathrm{NOCl}_{(g)} ⇌ 2NO(g)+Cl2(g)2\mathrm{NO}_{(g)} + \mathrm{Cl}_{2(g)} ; KpK_p = 1.8 × 10−210^{-2} at 500 K
(ii) CaCO3(s)\mathrm{CaCO}_{3(s)} ⇌ CaO(s)+CO2(g)\mathrm{CaO}_{(s)} + \mathrm{CO}_{2(g)} ; KpK_p = 167 at 1073 K
Solution:  
(i) KpK_p = Kc(RT)ΔngK_c(RT)^{\Delta n_g}, Δng\Delta n_g = (2 + 1) – 2 = 1
1.8 × 10−210^{-2} = KcK_c (0.0821 × 500), KcK_c = 4.38 × 10−410^{-4}
(ii) Δng\Delta n_g = 1 , KcK_c = KpRT\frac{K_p}{RT} = 1670.0821×1073\frac{167}{0.0821 \times 1073} = 1.896
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