Question

One mole of N2, 2 moles of H2 and 3 moles of ammonia are taken in a one-litre flask. The equilibrium concentration of ammonia is: (Equilibrium constant for the decomposition of ammonia = 1.89 x 10-1 mol2 L-2)

Answer 1

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Explanation:

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Answer 2

One mole of N₂ , 2 moles of H₂ and 3 moles of ammonia are taken in a one-litre flask.

The equilibrium concentration of ammonia is

we know, the reaction between nitrogen and hydrogen is given by,

N₂ + 3H₂ ⇔ 2NH₃

at t = 0, 1 mol/L 2 mol/L 3mol/L

at eqlm (1 - x) 2 - 3x 3 + 2x

so equilibrium constant, Kc = [NH₃]²/[N₂][H₂]³

here given, equilibrium constant for the decomposition of ammonia = 1.89 × 10¯¹ mol²L¯²

∴ Kc = 1/(1.89 × 10¯¹) = 10/1.89 L²mol¯²

⇒10/8.9 = (3 + 2x)²/{(1 - x)(2 - 3x)³}

⇒10(1 - x)(2 - 3x)³ = 1.89(3 + 2x)²

After solving it we get, x ≈ 0.2 and 1.47 but x ≠ 1.47 because concentration of nitrogen at equilibrium (1 - x) = 1 - 1.47 = -0.47 which is impossible.

∴ x ≈ 0.2

Hence the concentration of ammonia at equilibrium = 3 + 2x = 3 + 2 × 0.2 = 3.4 Mol/L.

Therefore equilibrium concentration of ammonia is 3.4 mol/L.