Chemical EquilibriumHard
Question
An amount of 16 moles H2 and 4 moles of N2 is confined in a vessel of volume one litre. The vessel is heated to a constant temperature until the equilibrium is established. At equilibrium, the pressure was found to be 9/10th of the initial pressure. The value of KC for the reaction N2(g) + 3H2(g) $\rightleftharpoons$2NH3(g) is
Options
A.8100
B.6.07 × 10–4
C.1647.75
D.8.99 × 10–5
Solution
N2(g) + 3H2(g) $\rightleftharpoons$2NH3(g)
Initial moles 4 16 0
Moles at equilibrium 4 – x 16 – 3x 2x
Total moles = (4 – x) + (16 – 3x) + 2x = 20 – 2x
From question, $20 \times \frac{9}{10} = 20 - 2x \Rightarrow x = 1$
Now, $K_{C} = \frac{(2x)^{2}}{(4 - x)(16 - 3x)^{3}}V^{2} = 6.07 \times 10^{- 4}\text{ }\text{M}^{- 2}$
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