Chemical EquilibriumHard
Question
When a mixture of N2 and H2 in the volume ratio of 1: 5 is allowed to react at 700 K and 103 atm pressure, 0.4 mole fraction of NH3 is formed at equilibrium. The value of KP for the below reaction N2(g) + 3H2(g) $\rightleftharpoons$2NH3(g)
Options
A.2.6 × 10–5 atm–2
B.2.6 × 10–4 atm–2
C.2.6 × 103 atm–2
D.5.1 × 10–3 atm–2
Solution
N2(g) + 3H2(g) $\rightleftharpoons$2NH3(g)
Initial moles 1 5 0
Moles at equilibrium 1 – x 5 – 3x 2x
Total moles = (1 – x) + (5 – 3x) + 2x = 6 – 2x
From question, $\frac{2x}{6 - 2x} = 0.4 \Rightarrow x = \frac{6}{7}$
$K_{P} = \frac{(2x)^{2}}{(1 - x) \times (5 - 3x)^{3}} \times \left( \frac{6 - 2x}{P} \right)^{2} = 2.6 \times 10^{- 4}\text{ at}\text{m}^{- 2}$
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