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