Problem 122 - Chemistry: An Atoms-Focused Approach

Question

Ammonia decomposes at high temperatures. In an experiment to explore this behavior, 2.00 mol of gascous \(\mathrm{NH}_{3}\) is sealed in a rigid \(1.00 \mathrm{L}\) vessel. The vessel is heated to \(785 \mathrm{K}\) and some of the \(\mathrm{NH}_{3}\) decomposes in the following reaction: $$2 \mathrm{NH}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$$ The system eventually reaches equilibrium and is found to contain 0.0040 mol of \(\mathrm{NH}_{3}\). What are the values of \(K_{\mathrm{p}}\) and \(K_{c}\) for the decomposition reaction at \(785 \mathrm{K} ?\)

Step-by-Step Solution

Answer

The equilibrium constants Kp and Kc for the decomposition of ammonia at 785 K are approximately 1.02 x 10^4 and 1.60 x 10^6, respectively.

Step 1: Write down the balanced chemical equation and determine the change in moles.

The balanced chemical equation for ammonia decomposition is: $$2 \ \mathrm{NH_{3}(g)} \rightleftharpoons\\ \mathrm{N_{2}(g)} + 3 \ \mathrm{H_{2}(g)}$$ Initially, we have 2.00 mol of ammonia and 0 moles of nitrogen and hydrogen gas. At equilibrium, there are 0.0040 moles of ammonia left, which means 1.996 moles of ammonia have decomposed.

Step 2: Calculate the equilibrium concentrations of all species.

Let's find the moles of other species at equilibrium: $$\mathrm{NH_{3}(g)}:~~ \Delta n = -1.996 \ \text{mol}$$ $$\mathrm{N_{2}(g)}:~~ \Delta n = +\frac{1}{2}(1.996) = +0.998\ \text{mol}$$ $$\mathrm{H_{2}(g)}:~~ \Delta n = +\frac{3}{2}(1.996) = +2.994\ \text{mol}$$ Now, we can find the equilibrium concentrations using the formula: \([\mathrm{X}]_{eq}=\frac{n_{X}}{V}\). $$[\mathrm{NH_{3}]_{eq} = \frac{0.0040 \ \text{mol}}{1.00 \ \mathrm{L}} = 0.0040 \ \mathrm{M}$$ $$[\mathrm{N_{2}]_{eq} = \frac{0.998 \ \text{mol}}{1.00 \ \mathrm{L}} = 0.998 \ \mathrm{M}$$ $$[\mathrm{H_{2}]_{eq} = \frac{2.994 \ \text{mol}}{1.00 \ \mathrm{L}} = 2.994 \ \mathrm{M}$$

Step 3: Use the equilibrium concentration to find \(K_{\mathrm{c}}\).

The expression for \(K_{\mathrm{c}}\) based on the balanced chemical equation is: $$K_{\mathrm{c}} = \frac{[\mathrm{N_{2}]_{eq}][\mathrm{H_{2}]_{eq}^{3}}}{[\mathrm{NH_{3}]_{eq}^{2}}$$ Plug in the equilibrium concentrations we found in Step 2: $$K_{\mathrm{c}} = \frac{(0.998)(2.994)^{3}}{(0.0040)^{2}}$$ Calculate the value of \(K_{\mathrm{c}}\): $$K_{\mathrm{c}} \approx 1.60 \times 10^{6}$$

Step 4: Use the Ideal Gas Law to find the pressure, then find \(K_{\mathrm{p}}\).

First, we need to use the Ideal Gas Law to find the pressure at equilibrium: $$PV=n_{total}RT$$ $$P=\frac{n_{total}RT}{V}=\frac{(0.004 + 0.998 + 2.994) \times 8.314 \times 785}{1.00} \ \mathrm{Pa}$$ Calculate the pressure P: $$P \approx 2.58 \times 10^{5} \ \mathrm{Pa}$$ Now, we can relate \(K_{\mathrm{p}}\) and \(K_{\mathrm{c}}\) using the formula: \(K_{\mathrm{p}} = K_{\mathrm{c}}(RT)^{\Delta n_{gas})\) Plug in the values of \(K_{\mathrm{c}}\), R, T, and \(\Delta n_{gas}\): $$K_{\mathrm{p}} = (1.60 \times 10^{6})[(8.314 \times 785) ^{1}]$$ Calculate the value of \(K_{\mathrm{p}}\): $$K_{\mathrm{p}} \approx 1.02 \times 10^{4}$$ The values of \(K_{\mathrm{p}}\) and \(K_{\mathrm{c}}\) for the decomposition reaction at \(785 \ \mathrm{K}\) are approximately \(1.02 \times 10^{4}\) and \(1.60 \times 10^{6}\), respectively.

Key Concepts

Ammonia Decomposition

Ammonia decomposition is a reaction where ammonia (\(\text{NH}_{3}\)) breaks down into nitrogen (\(\text{N}_{2}\)) and hydrogen (\(\text{H}_{2}\)). This process typically occurs at high temperatures. This is because heat provides the energy necessary for breaking the strong nitrogen-hydrogen bonds within ammonia molecules.
In this process, two molecules of \(\text{NH}_{3}\) decompose to form one molecule of \(\text{N}_{2}\) and three molecules of \(\text{H}_{2}\):

\[2 \text{NH}_{3}(g) \rightleftharpoons \text{N}_{2}(g) + 3 \text{H}_{2}(g)\]
This reaction showcases dynamic equilibrium, where the forward and reverse reactions occur at the same rate. Thus, the concentrations of reactants and products remain constant over time.

Equilibrium Constant

The equilibrium constant (\(K\)) provides an understanding of how far a reaction proceeds before reaching equilibrium. It is a numerical representation of the ratio of product concentrations to reactant concentrations, each raised to their stoichiometric coefficients.
For the ammonia decomposition described, the equilibrium expression is:

\[K_{c} = \frac{[\text{N}_{2}][\text{H}_{2}]^{3}}{[\text{NH}_{3}]^{2}}\]
The value of \(K_{c}\) indicates the extent of decomposition. If \(K_{c}\) is large, products are favored at equilibrium, meaning the reaction proceeds significantly toward the right. Conversely, a small \(K_{c}\) value would indicate that the decomposition of ammonia is less extensive. In our example, the calculated \(K_{c}\) is approximately \(1.60 \times 10^{6}\), showing a strong tendency for ammonia to decompose under the given conditions.

Ideal Gas Law

The Ideal Gas Law (\(PV=nRT\)) is essential for understanding the behavior of gases, particularly under equilibrium conditions. It connects four fundamental properties of gases: pressure (\(P\)), volume (\(V\)), the number of moles (\(n\)), and temperature (\(T\)), along with the gas constant (\(R\)).
In our exercise involving the decomposition of ammonia, the Ideal Gas Law helps calculate the total pressure at equilibrium. Given the constant volume and temperature, combined with the total number of moles of gas, the pressure is determined. The calculated pressure is then used to convert between \(K_{c}\) and \(K_{p}\), which accounts for pressure-based equilibrium expressions.

Reaction Quotient

The reaction quotient (\(Q\)) is a valuable tool for predicting the direction in which a chemical reaction will proceed to reach equilibrium. Similar to the equilibrium constant, \(Q\) is calculated using the same formula as \(K\).
However, \(Q\) is calculated using the initial concentrations or pressures of reactants and products at any specified moment. By comparing the values of \(Q\) and \(K\), we can predict:

• If \(Q < K\), the reaction will proceed forward to produce more products.
• If \(Q > K\), the reaction will shift backward, increasing the concentration of reactants.
• If \(Q = K\), the reaction is at equilibrium.

This concept is critical in scenarios where the initial state of the system is not known to be at equilibrium, allowing for control over the reaction's direction to achieve desired results.