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

(i) $$2NOCl(g) ⇌ 2NO(g)+Cl_{2}(g);$$

$$K_{p}=1.8\times 10^{-2}\: \: at\: 500K$$

(ii) $$CaCO_{3}(s) ⇌ CaO(s)+CO_{2}(g);$$

$$K_{p}=167\: \: at\: 1073K$$

Asked by Pragya Singh | 1 year ago |  83

##### Solution :-

The relation between Kp and Kc is given as:

$$K_{p}=K_{c}(RT)^{\Delta n}$$

(i) Given,

R = 0.0831 bar mol–1K–1

$$\Delta n = 3-2 =1$$

T = 500 K

Kp=$$1.8\times 10^{-2}$$

Now,

Kp = Kc (RT) ∆n

$$\Rightarrow 1.8\times 10^{-2}=K_{c}(0.0831\times 500)^{1}\\ \\$$

$$\Rightarrow K_{c}=\dfrac{1.8\times 10^{-2}}{0.0831\times 500}\\ \\$$

$$=4.33\times 10^{-4}(approximately)$$

(ii) Here,

∆n =2 – 1 = 1

R = 0.0831 bar mol–1K–1

T = 1073 K

Kp= 167

Now,

Kp = K(RT) ∆n

$$\Rightarrow 167=K_{c}{(0.0831\times 1073)^{\Delta n}}\\ \\ \Rightarrow K_{c}=\dfrac{167}{0.0831\times 1073}\\ \\ =1.87(approximately)$$

Answered by Pragya Singh | 1 year ago

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