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Question

Consider following equilibrium at 400 K and 24 atm.
3A2B+2C+D
If the equilibrium pressure of A is 4 atm then Kc (in mol2 L2) will be:
(Take R=0.08 L atm K1 mol1)


A
0.5
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B
0.25
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C
2.5
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D
5.0
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Solution

The correct option is B 0.25
3A2B+2C+DInitialP00+0EquilibriumP2x2x2xx
Equilibrium pressure =P2x+2x+2x+x24=P+3x   .........(1)
From question 4=P2x   .........(2)
From equation (1) and (2)
20=5xx=4Kp=P2BP2CPDP3A=(8)2(8)2443=256 atm2Kp=Kc(RT)ΔnKc=Kp(RT)Δn=256 atm2(0.08 L atm K1 mol1×400 K)2=0.25 mol2 L2

Chemistry

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