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  • Find out the value of K_c for each of the following equilibria from the value of K_ p: (i) 2NOCl (g) 2NO (g)+ Cl_2 (g) K_p=1.8 X 10^-2 at 500 K

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

       (i)     2NOCl _{(g)}\rightleftharpoons 2NO_{(g)}+Cl_{2}_{(g)};K_{p}=1.8 \times 10^{-2} at 500K  

        (ii)     CaCO_{3}_{(s)}\rightleftharpoons CaO_{(s)}+CO_{2}_{(g)};K_{p}=167 at 1073K

Answers (1)

best_answer

We know that the relation between K_p and K_c is expressed as;

K_p=K_c(RT)^{\Delta n}............................(i)
here {\Delta n} = (no. of moles of product) - (no. of moles of reactants)

R = 0.0831 bar L /mol/K, and

For (i) 
K_p = 1.8\times 10^{-2} and Temp (T) = 500K

{\Delta n} = 3 - 2 = 1
By putting the all values in eq (i) we get

K_c = \frac{1.8\times 10^{-2}}{0.0831\times 500} = 4.33\times 10^{-4}

For (ii)
K_p = 167 and temp(T) = 1073 K
{\Delta n} = 2 - 1 = 1
Now, by putting all values in eq (i) we get,

K_c = \frac{167}{0.0831\times 1073} = 1.87

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manish

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