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09/10/2021 10:55 am
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28g N2 and 6.0 g of H2 are heated over catalyst in a closed one litre flask of 450°C. The entire equilibrium mixture required 500 mL of 1.0 M H2SO4 for neutralisation. The value of Kc for the reaction
N2(g) + 3H2(g) ⇌ 2NH3(g) is
(a) 0.06 mol-2L2
(b) 0.59 mol-2L2
(c) 1.69 mol-2L2
(d) 0.03 mol-2L2
1 Answer
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09/10/2021 11:03 am
Correct answer: (b) 0.59 mol-2L2
Explanation:
Moles of N2 = \(\frac{28}{28}\) = 1
Moles of H2 = \(\frac{6}{2}\) = 3
Moles of H2SO4 required = \(\frac{500 \times 1}{1000}\) = 0.5
Moles of NH3 neutralised by H2SO4
= 1.0
(2NH3 + H2SO4 → (NH4)2SO4
Hence 1 mol of NH3 by the reaction between N2 and H2.
N2 + 3H2 ⇌ 2NH3
Initial 1 3 0
At eqn. 1-0.5 3 - 0.5 x 3 1
Kc = \(\frac{1 \times 1}{0.5 \times (1.5)^3}\)
= 0.592 mol-2L2