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Question

A mixture containing 2 moles of N2 and 3 moles of H2 gas is brought to equilibrium in a 3 L vessel.If the molar concentration of NH3 at equilibrium is 1/3 mol/L, calculate the Kc for the formation of NH2.N2(g)+3H2(g)2NH3(g)

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Solution

N2(g)+3H2(g)2NH3(g)
At 2 3 0
t=0
At 2(1α) 3(1α) 2α
t=teq
Volume of container =3L
concentration =molvolinlitre
[NH3]=2α3=13 ( given)
α=1/2
[N2] at equilibrium =2(1α)3=13
[H2] at equilibrium =3(1α)3=1/2
KC=(13)2(12)3(13)=8/3

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