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Question

At 450oC, the equilibrium constant Kp for the reaction, N2(g)+3H2(g)2NH3(g), was found to be 1.6×105 at a pressure of 200 atm. If N2 and H2 are taken in 1:3 ratio, what is % of NH3 formed at this temperature?

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Solution

N2(g)+3H2(g)2NH3(g)
At eq. (1x) (33x) 2x

Total number of moles=1x+33x+2x=42x

pN2=(1x)(42x)P;pH2=(33x)(42x)P;pNH3=2x(42x)P

Kp=(pNH3)2pN2×(pH2)2=4x2(42x)2(1x)×27×(1x)3P2

.6×105=1627×x2(2x)2(1x)4×2002

or, x=0.30

Moles of ammonia formed =2×0.30

total moles at equilibrium=(42x)=(42×0.30)=3.40

% of NH3 at equilibrium 0.603.40×100=17.64

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