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Question

If a mixture of N2 and H2 in the ratio 1 : 3 at 50 atmosphere and 650oC is allowed to react till equilibrium is reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate the equilibrium constant (KP) for the reaction.
N2(g)+3H2(g)2NH3(g)

A
1.677×103 atm2
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B
2.677×103 atm2
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C
0.677×103 atm2
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D
3.677×103 atm2
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Solution

The correct option is A 1.677×103 atm2
Equilibrium of formation of NH3 may be given as,
N2+3H22NH3At t=0132At equilibrium(1x)3(1x)2x
Mole fraction of NH3=2x(42x)
Partial pressure of NH3=2x(42x)×p=2x(42x)×50
25=2x×50(42x) or =0.666
Kp=16x2(2x)227p2(1x)4=16×(0.666)2(20.666)227×(50)2×(0.334)2=1.677×103 atm2

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