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Question

At a certain temperature (T), the equilibrium constant (Kc) is 1 for the reaction N2(g)+3H2(g)2NH3(g)
If 2 moles of N2, 4 moles of H2, 6 moles of NH3 and 3 moles of inert gas are introduced into a two litre rigid vessel at constant temperature T. It has been found that equilibrium concentration of H2 and NH3 are equal then what is the equilibrium concentration of N2 (in M)?

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Solution

Qc=(62)2(22)(42)3=1.125
Qc>Kc, the reaction will proceed in backward direction.
N2(g)+3H2(g)2NH3Initial moles:246At equilibrium:2+x4+3x62x

At equilibrium [H2]=[NH3]
4+3x2=62x2x=0.4
[N2] at equilibrium =2+x2=1.2 M

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