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 0.000413 atm.

Step-by-step solution

Set up the ICE table

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.

Apply the given condition to the ICE table

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

Write the expression for Kp and plug in the values from the ICE table

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

Solve for P

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.

Key concepts

ICE Table

An ICE Table is an essential tool in solving equilibrium problems in chemistry. It's named for its columns: Initial, Change, and Equilibrium. This table helps us keep track of the concentrations or pressures of species involved in a reaction as it reaches equilibrium.

• **Initial**: It refers to the starting conditions of the reactants and products. In this case, we started with an initial partial pressure of ammonia (\[ NH_3 \]) labelled as \( P \). For nitrogen (\[ N_2 \]) and hydrogen (\[ H_2 \]), we start at zero as they are not present initially.
• **Change**: This represents how much of the reactants or products shifts to form a balance. If \( x \) is the amount of ammonia that decomposes, the change is \(-2x\) because 2 moles of ammonia form 1 mole of nitrogen and 3 moles of hydrogen. Therefore, the changes for nitrogen and hydrogen are \( +x \) and \( +3x \), respectively.
• **Equilibrium**: Finally, we calculate the pressures at equilibrium. As provided in the problem, 50% of ammonia decomposes, allowing us to deduce that \( x = 0.5P \). Thus, the equilibrium pressures are \( 0.5P \) for each of \([ NH_3 ]\) and \( [ N_2 ] \), and \( 1.5P \) for \( [ H_2 ] \).

The ICE Table is structured to help us clearly visualize the progress of the reaction and use the given conditions effectively to solve the problem.

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. It's a crucial concept in understanding chemical equilibrium, especially in gaseous reactions.

• In a mixture of gases, each gas behaves independently and exerts pressure as if it were alone in the container. This is the essence of Dalton's Law of Partial Pressures.
• The initial partial pressure of a gas influences how the equilibrium is established. Here, ammonia's initial partial pressure is vital because it dictates the decomposition that leads to equilibrium.
• The concept helps in determining how much of each gas is present at equilibrium, which in turn influences the equilibrium constant.

In this exercise, identifying the partial pressure of ammonia before decomposition is key. It allowed us to calculate how the system evolved towards equilibrium. Understanding partial pressures gives insight into how gases interact and affect the reaction kinetics and equilibria.

Equilibrium Constant

The Equilibrium Constant, often represented as \( K_p \) for gases, is significant for assessing the position of equilibrium in the reaction. It is calculated using the partial pressures of the products and reactants at equilibrium.

• For the reaction \( N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g) \), \( K_p = \frac{[NH_3]^2}{[N_2] [H_2]^3} \).
• The value of \( K_p \) tells us about the extent of the reaction. A large \( K_p \) (like \( 5.3 \times 10^5 \) in our problem) suggests that at equilibrium, the reaction heavily favors product formation, meaning more ammonia is made.
• Equilibrium constants are vital for understanding how changes in conditions like pressure or concentration can shift the equilibrium. They provide a quantitative measure to predict the direction of reaction shifts.

By solving the \( K_p \) expression with the equilibrium pressures, we found the initial partial pressure of ammonia needed to achieve the reported equilibrium position. This illustrates how powerful \( K_p \) can be in predicting the behavior of gaseous reactions.