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

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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