One mole of N2 is mixed with three moles of  H2 in a four liter vessel. If 0.25% N2 is converted into NH3 by the reaction

N2(g) +3H2 (g)             >2NH3(g)  

Calculate Kc

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