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Question

Find out the value of $$K_C$$ for each of the following equilibrium respectively from the value of $$K_P$$.
(i) $$2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g); K_P=1.8\times 10^{-2}$$ at $$500$$ K
(ii) $$CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g); K_P=167$$ at $$1073$$ K


A
4.4×104 & 1.90
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B
8.8×104 & 3.8
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C
4.4×104 & 1.90
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D
8.8×194 & 3.8
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Solution

The correct option is A $$4.4\times 10^{-4}$$ & $$1.90$$
The relationship between $$K_p$$ and $$K_c$$ is $$K_p=K_c(RT)^{\Delta n}$$.
For the reaction $$2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)$$, the value of $$\Delta n$$ is $$1+2-2=1$$.
Substituting this value in the above expression, we get
$$K_p=K_c(RT)^{\Delta n}=K_c(RT)^{1}$$.
But $$ K_P=1.8\times 10^{-2}, R=0.082 \text {L atm /mol. K}, T=500K$$.
Substituting values in the above expression, we get
$$ 1.8\times 10^{-2}=K_c(0.082 \times 500)^{1}$$.
$$K_c=4.4\times 10^{-4}$$.

For the reaction $$CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$$, the value of $$\Delta n$$ is $$=1$$.
Substituting this value in the above expression, we get
$$K_p=K_c(RT)^{\Delta n}=K_c(RT)^{1}.$$
But $$K_p=167, R=0.082 \text {L atm /mol. K}, T=1073K$$.
Substituting values in the above expression, we get
$$167=K_c(0.082 \times 1073)^{1}.$$
$$K_c=1.90$$.

Chemistry

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