91Ó°ÊÓ

We have the balanced chemical equation for the reaction: \[ \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g) \]

For the given reaction, the equilibrium constant expression in terms of partial pressures (\(K_p\)) is: \[ K_p = \frac{P_{\mathrm{NH}_3}^2}{P_{\mathrm{N}_2} \cdot P_{\mathrm{H}_2}^3} \] where \(P_{\mathrm{NH}_3}\), \(P_{\mathrm{N}_2}\), and \(P_{\mathrm{H}_2}\) represent the partial pressures of ammonia, nitrogen, and hydrogen gases at equilibrium, respectively.

Let the initial partial pressure of ammonia be \(P_0\). Since the vessel is initially empty of N2 and H2 gas, their partial pressures are 0. At equilibrium, 50% of the original ammonia has decomposed, so the partial pressures at equilibrium can be expressed as: \(P_{\mathrm{NH}_3} = 0.5P_0\) For every 2 moles of NH3 that decompose, 1 mole of N2 and 3 moles of H2 form. Therefore, at equilibrium: \(P_{\mathrm{N}_2} = 0.5 \cdot \dfrac{1}{2}P_0 = 0.25P_0\) \(P_{\mathrm{H}_2} = 0.5 \cdot \dfrac{3}{2}P_0 = 0.75P_0\)

We can substitute the equilibrium partial pressures into the equilibrium constant expression: \[ 5.3 \times 10^5 = \frac{(0.5P_0)^2}{(0.25P_0) \cdot (0.75P_0)^3} \] Now, we need to solve this equation for \(P_0\). First, we can simplify the equation: \[ 5.3 \times 10^5 = \frac{0.25P_0^2}{0.25P_0 \cdot 0.421875P_0^3} \] \[ 5.3 \times 10^5 = \frac{P_0^2}{1.7578125P_0^4} \] Then, multiply both sides by \(1.7578125P_0^4\), and we get: \[ 5.3 \times 10^5 \cdot 1.7578125P_0^4 = P_0^2 \] Simplify the equation further by dividing both sides by \(P_0^2\): \[ 5.3 \times 10^5 \cdot 1.7578125 = P_0^2 \] Now, find the square root of both sides: \[ P_0 = \sqrt{5.3 \times 10^5 \cdot 1.7578125} \approx 34.38 \mathrm{atm} \] So, the original partial pressure of ammonia before any decomposition occurred was approximately 34.38 atm.