Question

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\), equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

Short answer

The original partial pressure of ammonia before any decomposition occurred was approximately 34.38 atm.

Step-by-step solution

Write the balanced chemical equation

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

Write the equilibrium constant expression for the reaction

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.

Set up the initial conditions and the changes at equilibrium

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: \(P_{\mathrm{N}_2} = 0.5 \cdot \dfrac{1}{2}P_0 = 0.25P_0\) \(P_{\mathrm{H}_2} = 0.5 \cdot \dfrac{3}{2}P_0 = 0.75P_0\)

Substitute the equilibrium partial pressures into the constant expression and solve for P0

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.