Ncert Exercises & Solutions

7.5 Find out the value of Kc for each of the following equilibria from the value of Kp:

(i) 2NOCl (g) 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K

(ii) CaCO3 (s) CaO(s) + CO2(g); Kp= 167 at 1073 K

The relation between $K_{p} a n d K_{c}$ is given as:

$K_{p} = K_{c} {(RT)}^{∆ n} (a) Here, n = 3 - 2 = 1 R = 0.0831 {barLmol}^{- 1} K^{- 1} T = 500 K K_{p} = 1.8 \times 10^{- 2} Now, K_{p} = K_{c} {(RT)}^{∆ n} \Rightarrow 1.8 \times 10^{- 2} = K_{c} {(0.0831 \times 500)}^{1} \Rightarrow K_{c} = \frac{1.8 \times 10^{- 2}}{0.0831 \times 500} = 4.33 \times 10^{- 4} (approximately) (b) Here, ∆ n = 2 - 1 = 1 R = 0.0831 {barLmol}^{- 1} K^{- 1} T = 1073 K K_{p} = K_{c} {(RT)}^{∆ n} \Rightarrow 167 = K_{c} {(0.0831 \times 1073)}^{1} \Rightarrow K_{c} = \frac{167}{0.0831 \times 1073} = 1.87 (approximately)$

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