Question

1 mole of N, and 2 moles of H, are allowed to react in a 1 dm vessel. At equilibrium, 0.8 mole o
formed. The concentration of H, in the vessel is :
(1) 0.6 mol/L
(2) 0.8 mol/L
(3) 0.2 mol/L
(4) 0.4 mol/L​

Answer 1

N2​​(g) +3H2(g) ----> 2NH3(g)

Given Initially , amount of N2 = 2 moles

amount of H2 = 2 moles

When equilibrium is reached,  

Amount of N2 used = 0.5 moles

Thus according to the balanced chemical reaction

1 mole of N2 on reaction will produce = 2 moles of NH3

Thus 0.5 mole ​of N2 on reaction will produce = 2 × 0.5 moles of NH3 = 1 mole of NH3

Thus at equilibrium 1 mol of NH3 will be present.