Question
In the reaction N2(g) + 3H2(g) $\rightleftharpoons$2NH3(g) + Heat. One mole of N2 reacts with three moles of H2. If at equilibrium ‘x’ moles of N2 combined, then the value of ‘x’ in terms of KP and the total pressure of gases at equilibrium P is (x << 1)
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Solution
N2(g) + 3H2(g) $\rightleftharpoons$2NH3(g)
Initial moles 1 3 0
Moles at equilibrium 1 – x 3 – 3 x 2x
Total moles of gases = (1 – x) + (3 – 3x) + 2x = 4 – 2x
Equilibrium partial pressure
$\frac{1 - x}{4 - 2x} \times P\frac{3 - 2x}{4 - 2x} \times P\frac{2x}{4 - 2x} \times P$
Now, $K_{P} = \frac{\left( \frac{2x}{4 - 2x} \times P \right)^{2}}{\left( \frac{1 - x}{4 - 2x} \times P \right)\left( \frac{3 - 3x}{4 - 2x} \times P \right)^{3}}$
$= \frac{4x^{2}}{27(1 - x)^{4}} \times \frac{(4 - 2x)^{2}}{P^{2}} \approx \frac{4x^{2} \times 16}{27 \times P^{2}}$
$\therefore x = \sqrt{\frac{27K_{p}.P^{2}}{64}} = \frac{3P.\sqrt{3K_{p}}}{8}$
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