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Question

The following equilibrium constants were determined at 1120 K:
2CO(g)C(s)+CO2(g); Kp1=1014atm1
CO(g)+Cl2(g)COCl2(g);Kp2=6×103atm1

What is the equilibrium constant KC for the following reaction at 1120 K;

C(s)+CO2(g)+2Cl2(g)2COCl2(g)

A
3.31×1011M1
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B
5.5×1010M1
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C
5.51×106M1
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D
None of these
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Solution

The correct option is C 3.31×1011M1
2CO(g)C(s)+CO2(g), KP1=1014atm1
Reverse this reaction will get, C(s)+CO2(g)2CO(g), Kp1=11014atm1=1014atm1...........(1)
CO(g)+Cl2(g)COCl2(g),Kp2=6×103atm1
Multiply this equation by 2, we get
2CO(g)+2Cl2(g)2COCl2(g),K2p2=(6×103)2=36×106atm1................(2)
Adding (1) and (2) equation, we get
C(s)+CO2)(g)+2Cl2(g)2COCl2(g)
Kp=Kp1.K2p2
= 1014×36×106=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.K1mol1 , T = 1120 K
therefore Kc=Kp(RT)2
Kc=36×108×(0.0821)2(1120)2=3.31×1011M1


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