The following equilibrium constants were determined at 1120 K: 2CO(g)⇌C(s)+CO2(g)Kp1=10−14atm−1 CO(g)+Cl2(g)⇌COCl2Kp2=6×10−3atm−1
What is the equilibrium constant Kp for the following reaction at 1120 K. C(s)+CO2(g)+2Cl2(g)⇌2COCl2(g)
A
36×104atm−1
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B
3.6×109atm−1
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C
3.6×105atm−1
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D
3.6×1010atm−1
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Solution
The correct option is B3.6×109atm−1 2CO(g)⇌C(s)+CO2(g);Kp1=10−14atm−1......(1) CO(g)+Cl2(g)⇌COCl2(g);Kp2=6×10−3atm−1.....(2)
Multiply eqn (2) by (2), we get 2CO(g)+2Cl2(g)⇌2COCl2(g)K′p2=(6×10−3)2atm−1 ...(3)
reverse eqn (1) we get C(O)+CO2(g)⇌2COKp11=110−14atm−1 ...(4)
add eqn (3) and (4) C(s)+CO2(g)+2Cl2(g)⇌2COCl2(g) K=Kp12×Kp11 K=(6×10−3)2×110−14 ,K=36×10−610−14 K=36×108 K=3.6×109atm−1