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?

We'll start by writing the balanced chemical equation and setting up the Initial, Change, and Equilibrium (ICE) table for the reaction. \(N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g)\) | | N₂ | H₂ | NH₃ | |---------|----|----|-----| | Initial | 0 | 0 | P | |Change | x | 3x | -2x | |Equilibrium| x | 3x | P-2x Where P is the initial partial pressure of ammonia and x is the amount of ammonia that decomposes.

We are given that 50% of the ammonia has decomposed at equilibrium. Therefore, x = 0.5P (since 50% of the original amount decomposed). We can update the equilibrium row of the ICE table. | | N₂ | H₂ | NH₃ | |---------|----|----|-----| | Initial | 0 | 0 | P | | Change | x | 3x | -2x | |Equilibrium| 0.5P | 1.5P | 0.5P

The expression for Kp for the reaction is: \[K_p = \frac{[\text{NH}_3]^2}{[\text{N}_2] [\text{H}_2]^3}\] At equilibrium: \[5.3 \times 10^5 = \frac{(0.5P)^2}{(0.5P)(1.5P)^3}\]

To find the initial partial pressure of ammonia, we need to solve for P in the above equation. \[5.3 \times 10^5 = \frac{(0.5P)^2}{(0.5P)(1.5P)^3} \Rightarrow 5.3 \times 10^5 = \frac{(0.5P)^2}{(0.5P)(3.375P^3)}\] Now cancel out 0.5P from the numerator and denominator: \(5.3 \times 10^5 = \frac{P}{(3.375P^3)}\) Multiply both sides by \(3.375P^3\): \((5.3 \times 10^5)(3.375P^3) = P\) Divide both sides by \(P\) \((5.3 \times 10^5)(3.375P^2) = 1\) Now, solving for P^2: \(P^2 = \frac{1}{(5.3 \times 10^5)(3.375)}\) Finally, to get the initial partial pressure of ammonia (P), take the square root of above expression: \(P = \sqrt{\frac{1}{(5.3 \times 10^5)(3.375)}} \approx 0.000413\, \text{atm}\) So, the original partial pressure of ammonia before any decomposition occurred was approximately 0.000413 atm.