2 moles of N2 is mixed with 6 moles of H2 ina closed vessel of 1 litre capacity .If 50%N2 is converted into NH3 at equilibrium,the value of kc for the reaction:-N2(g)+3H2(g)------>2NH3(g)

N2 (g) + 3H2 (g) → 2 NH3(g)

Initial 1mole (given) 3 mole (given) 0

At equilibrium 1-(0.25 * 1)/100 3--(0.25 * 1)/100 2(0.25 * 1)/100

0.9975 moles 2.9925 moles 0.005 moles

Given that 0.25% of nitrogen is converted to ammonia (given) therefore 1-0.25% is done

Kc = [NH3]2/[N2][H2]3 = (0.005/4)2/(2.9925/4)(0.9975/4)3 = 1.8 * 10-8

Regards

Arun (askIITians forum expert)