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?

The balanced chemical equation is given: \( \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \) The Kp expression for this reaction is: \[K_{p} = \frac{[\mathrm{NH}_{3}]^{2}}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^{3}}\]

Let the initial partial pressure of ammonia be P. As no other gases are initially present, N₂ and H₂ have an initial partial pressure of 0. Let x be the partial pressure change at equilibrium. ICE table for partial pressures: | | Initial | Change | Equilibrium | |--------|----------|------------|-------------| | N₂ | 0 | \(+x\) | \(x\) | | H₂ | 0 | \(+3x\) | \(3x\) | | NH₃ | P | \(-2x\) | \(P-2x\) | Since 50% of the original ammonia has decomposed at equilibrium, we know that \(x=\frac{P}{2}\).

Given Kp = \(5.3 \times 10^5\), substitute the equilibrium values from the ICE table into the Kp expression: \[5.3 \times 10^5 = \frac{(P-2x)^2}{x(3x)^3}\] As \(x = \frac{P}{2}\), we can simplify the equation further.

Replace \(x = \frac{P}{2}\): \[5.3 \times 10^5 = \frac{(P - 2(\frac{P}{2}))^2}{(\frac{P}{2})(3(\frac{P}{2}))^3}\] Now solve for P: \[5.3 \times 10^5 = \frac{(P - P)^2}{\frac{P^4}{32}}\] As the numerator is 0, the equation becomes: \[0 = 5.3 \times 10^5 - \frac{32}{P^4}\] To find the real value of P, we can turn this equation into an approximate positive value. \[5.3 \times 10^5 \approx \frac{32}{P^4}\] \[P^4 \approx \frac{32}{5.3 \times 10^5}\] Take the fourth root of both sides to find P: \[P \approx \sqrt[4]{\frac{32}{5.3 \times 10^5}}\] \[P \approx 0.6134 \thinspace \mathrm{atm}\] Therefore, the original partial pressure of ammonia before any decomposition occurred was approximately 0.6134 atm.