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Question

Find out the value of $$K_{c}$$ for each of the following equilibria from the value of $$K_{p}$$:
(a) $$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}\ ;\ K_{p} = 1.8\times 10^{-2} atm$$ at $$500\ K$$
(b) $$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}\ ;\ K_{p} = 167\ atm$$ at $$1073\ K$$


Solution

For reaction (a), change in number of moles $$(\Delta n)=$$ Number of moles of products $$-$$ Number of moles of reactants 

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

$$T=500\ K$$ (Given)

We know that, $$K_{p} = K_{c} (RT)^{\Delta n}$$

$$\therefore K_{c} = \dfrac {K_{p}}{(RT)^{\triangle n}}$$,
$$R = 0.0821\ litre\ atm\ K^{-1}\ mol^{-1}$$
$$\Rightarrow K_{c} = \dfrac {1.8\times 10^{-2}}{[(0.0821)(500)]^{1}} = 4.38\times 10^{-4}$$
For reaction (b), $$\Delta n = 2-1 =1$$

$$T=1073\ K$$ (Given)

$$\therefore K_{c} = \dfrac {167}{[(0.0821)(1073)]^{1}} = 1.90$$

Chemistry
NCERT
Standard XI

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