Question

At a certain temperature, 0.88 mol N2N2 and 2.654 mol H2H2 are placed in a container.

N2(g)+3H2(g)−⇀↽−2NH3(g)

At equilibrium, there is 0.82 mol NH3NH3 present. Determine the number of moles of N2N2 and H2H2 that are present when the reaction is at equilibrium.

Answer 1

Given data: Initial moles of 12 = 0.68 mole Initial moles of H2 = 2.654 mot The number of moles of NH3 at equilibrium = 0.82 The given chemical eam is N2 + 3H22NH₃ Bitial o.bb 2.654 det a moles dissociate after time t. 211H3 Ng + 3Hq - After 0.86 x 2.654-3x 2x No. of moles of NH3 at equilibrium = 0.62 : 21 = 0.62 se = 0.82 20:41 : Number of moles of Ng at equilibrium = 0.86-2 = 0.bf-0.41 - 0.47 moles. similarly the number of motes of Ma at equilibrium - 2.654-3x = 2.654-3x0.41 = 1.424 moles