The following equilibrium constants were determined at 1120 K:
2CO(g) ⇌ C(s)+CO2(g); Kp1=10−14atm−1
CO(g)+Cl2(g) ⇌ COCl2(g);Kp2=6×10−3atm−1
What is the equilibrium constant KC for the following reaction at 1120 K;
C(s)+CO2(g)+2Cl2(g)⇌2COCl2(g)
The correct option is C 3.31×10−11M−1
2CO(g) ⇌C(s)+CO2(g), KP1=10−14atm−1
Reverse this reaction will get, C(s)+CO2(g) ⇌ 2CO(g), Kp1=110−14atm−1=1014atm−1...........(1)
CO(g)+Cl2(g) ⇌COCl2(g),Kp2=6×10−3atm−1
Multiply this equation by 2, we get
2CO(g)+2Cl2(g) ⇌2COCl2(g),K2p2=(6×10−3)2=36×10−6atm−1................(2)
Adding (1) and (2) equation, we get
C(s)+CO2)(g)+2Cl2(g) ⇌ 2COCl2(g)
Kp=Kp1.K2p2
= 1014×36×10−6=36×108
We know that Kp=Kc×(RT)ΔngKc=Kp×(RT)−Δng
For given reaction Δng = -2 (number of gaseous molecules of products(2) - number of gaseous molecules of reactants(4) ), R = 0.0821 L.atm.K−1mol−1 , T = 1120 K
therefore Kc=Kp(RT)2
Kc=36×108×(0.0821)2(1120)2=3.31×10−11M−1