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

(i) 2NOCl (g) 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K


(ii) CaCO3 (s) CaO(s) + CO2(g); Kp= 167 at 1073 K


We know that,

Kp = Kc[RT]Δn


Where Δn is number of gas molecule in products minus number of gas molecule in reactant in balanced chemical equation.


(i)


2NOCl (g) 2NO (g) + Cl2 (g)


Number of gaseous species on product side is 3 (2 NO and 1 Cl2 ) and on reactant side is 2 (2 NOCl )


Thus, Δn = 3 - 2 = 1


T = 500 K


R = 0.0821 L atm K-1


Given, Kp = 1.8 × 10-2atm


Kc =


=


= 4.4 × 10 �-4mol L-1


(ii)


CaCO3 (s) CaO(s) + CO2(g)


Number of gaseous species on product side is 1 (1 CO2) and on reactant side is 0.


Thus, Δn = 1 – 0 = 1


T = 1073 K


Kp = 167 atm


Kc =


=


= 1.9 mol L-1


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