Problem 30 - Chemistry An Atoms First Approach

Question

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

Step-by-Step Solution

Answer

The equilibrium constant \(K_\mathrm{p}\) for the given reaction at the given temperature is approximately 54.28. When given new pressure values (\(P_{N_2} = 0.525 \mathrm{atm}\), \(P_{NH_3} = 0.0167 \mathrm{atm}\), and \(P_{H_2} = 0.00761 \mathrm{atm}\)), we find that the reaction quotient, \(Q_\mathrm{p} \approx 58.25\). Since \(Q_\mathrm{p} > K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the left.

Step 1: Write the expression for the equilibrium constant Kp

For the given reaction: \[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\] The expression for the equilibrium constant \(K_\mathrm{p}\) can be written as: \[K_\mathrm{p} = \frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3}\]

Step 2: Substitute the given pressures into the Kp expression

We are given: \(P_{NH_3} = 3.1 \times 10^{-2} \mathrm{atm}\), \(P_{N_2} = 8.5 \times 10^{-1} \mathrm{atm}\), and \(P_{H_2} = 3.1 \times 10^{-3} \mathrm{atm}\). Substitute these pressure values into the expression for Kp: \[K_\mathrm{p} = \frac{(3.1 \times 10^{-2})^2}{(8.5 \times 10^{-1}) \times (3.1 \times 10^{-3})^3}\]

Step 3: Calculate Kp

Now, calculate the value of Kp: \[K_\mathrm{p} = \frac{(3.1 \times 10^{-2})^2}{(8.5 \times 10^{-1}) \times (3.1 \times 10^{-3})^3} \approx 54.28\] So, the equilibrium constant \(K_\mathrm{p}\) at the given temperature is approximately 54.28.

Step 4: Analyze the new pressures

We are given new pressure values: \(P_{N_2} = 0.525 \mathrm{atm}\), \(P_{NH_3} = 0.0167 \mathrm{atm}\), and \(P_{H_2} = 0.00761 \mathrm{atm}\). We need to check if these pressures represent a system at equilibrium. To do this, we can calculate \(Q_\mathrm{p}\) using the same expression as for \(K_\mathrm{p}\), and compare the two values. \[Q_\mathrm{p} = \frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3}\]

Step 5: Calculate Qp

Substitute the new pressure values into the expression for Qp: \[Q_\mathrm{p} = \frac{(0.0167)^2}{(0.525) \times (0.00761)^3} = 58.25\] Now compare \(Q_\mathrm{p}\) to \(K_\mathrm{p}\): - If \(Q_\mathrm{p} < K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the right. - If \(Q_\mathrm{p} = K_\mathrm{p}\), the reaction is at equilibrium. - If \(Q_\mathrm{p} > K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the left.

Step 6: Compare Qp to Kp

Comparing our calculated \(Q_\mathrm{p}\) and \(K_\mathrm{p}\) values: \(Q_\mathrm{p} = 58.25\), \(K_\mathrm{p} \approx 54.28\). Since \(Q_\mathrm{p} > K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the left.

Key Concepts

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a powerful tool used to determine if a chemical reaction is at equilibrium. It essentially measures the relative amounts of products and reactants present at any point during a reaction.

• The expression for \( Q \) is similar to that for the equilibrium constant \( K \), but it uses the current concentrations or pressures, rather than those at equilibrium.
• If the value of \( Q \) is less than \( K \), the reaction will proceed in the forward direction to form more products.
• If \( Q \) is greater than \( K \), it means there are too many products, so the reaction will proceed in the reverse direction to form more reactants.
• If \( Q \) equals \( K \), the reaction is at equilibrium, and the concentrations or pressures of the reactants and products remain constant.

In our exercise, the calculated \( Q_\mathrm{p} \) was 58.25, compared to \( K_\mathrm{p} \) which was found to be approximately 54.28. Since \( Q_\mathrm{p} > K_\mathrm{p} \), the reaction is not at equilibrium, indicating it will shift to the left to reach equilibrium.

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the forward and reverse reactions occur at the same rate. As a result, the concentrations or pressures of the reactants and products remain constant over time.

• At equilibrium, it may seem like the reaction has stopped, but in reality, both forward and reverse reactions continue to take place.
• The equilibrium constant \( K \) is unique for a given reaction at a specific temperature and provides insight into the ratio of products to reactants at equilibrium.
• An understanding of equilibrium helps chemists in predicting the extent of reactions and how different conditions will affect the position of equilibrium.

The equilibrium constant \( K_\mathrm{p} \) calculated in the exercise was 54.28. This value signifies the ratio of pressures of products to reactants when the system has reached equilibrium. By comparing \( Q \) with \( K \), we can determine the direction in which a reaction needs to shift to achieve equilibrium.

Pressure Calculations

In the context of gas-phase reactions, pressure plays a crucial role in determining equilibrium and reaction direction.

• The partial pressures of gases can be used in equilibrium expressions to calculate \( K_\mathrm{p} \), which reflects the degree to which a reaction has proceeded.
• For a reaction like \( N_2 + 3H_2 \leftrightarrows 2NH_3 \), the expression for \( K_\mathrm{p} \) involves the pressures of ammonia and hydrogen and nitrogen gases:

\[K_\mathrm{p} = \frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3} \]

• The pressures are measured in atmospheres (atm), but consistent units must be used throughout the calculation.

In the exercise, specific pressure values were plugged into this equation to find \( K_\mathrm{p} \). By substituting new pressure values, \( Q_\mathrm{p} \) was also calculated to assess the reaction’s current state. Consistent care in measuring, converting, and calculating pressures ensures accurate results in evaluating chemical equilibria.