Problem 40 - Chemical Principles: The Quest for Insight

Question

Given that \(K_{c}=61\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500 \mathrm{~K}\), calculate whether more ammonia will tend to form when a mixture of composition \(2.23 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(1.24 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(1.12 \times 10^{-4} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\) is present in a container at \(500 \mathrm{~K}\).

Step-by-Step Solution

Answer

Since the calculated reaction quotient (Q) is greater than the equilibrium constant (Kc), the reaction will shift left, towards reactants, thus more ammonia (NH3) will not form.

Step 1 - Write the Expression for the Equilibrium Constant (Kc)

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_{3}(\mathrm{~g})\), the equilibrium constant expression is \(K_{c} = \frac{[\mathrm{NH}_{3}]^{2}}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^{3}}\). This expression relates the concentrations of the reactants and products at equilibrium.

Step 2 - Calculate the Reaction Quotient (Q)

The reaction quotient (Q) is calculated using the initial concentrations of the reactants and products, in the same way as the equilibrium constant: \(Q = \frac{[\mathrm{NH}_{3}]_{\text{initial}}^{2}}{[\mathrm{N}_{2}]_{\text{initial}}[\mathrm{H}_{2}]_{\text{initial}}^{3}}\). Plugging in the given values: \(Q = \frac{(1.12 \times 10^{-4})^{2}}{(2.23 \times 10^{-3})(1.24 \times 10^{-3})^{3}}\).

Step 3 - Calculate Q Value

Calculate the value of Q using the numbers provided: \(Q = \frac{(1.12 \times 10^{-4})^{2}}{(2.23 \times 10^{-3})(1.24 \times 10^{-3})^{3}} = \frac{1.2544 \times 10^{-8}}{4.3215 \times 10^{-12}} = 2.90 \times 10^{3}\).

Step 4 - Compare Q to Kc

Now that we have the Q value, compare it with the equilibrium constant (Kc). The system will shift to reach equilibrium based on whether Q is less than, equal to, or greater than Kc.

Step 5 - Determine the Direction of the Shift

Since the given \(K_{c} = 61\) and the calculated \(Q = 2.90 \times 10^{3}\), and \(Q > K_{c}\), the reaction will shift towards the reactants to reach equilibrium. This means that more ammonia will not tend to form, instead, the concentration of ammonia will decrease.

Key Concepts

Equilibrium Constant (Kc)

Understanding chemical equilibrium is crucial for many fields such as chemistry, engineering, and environmental science. The equilibrium constant (Kc) is a pivotal concept in this regard. It is a numerical value that expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their respective coefficients in the balanced equation. In the context of our example, for the reaction \( \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \), the equilibrium constant expression is \( K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \).

When the value of \( K_c \) is known (in this case, \( K_c = 61 \) at 500 K), it serves as a vital indicator of the extent of the reaction at equilibrium. A larger \( K_c \) suggests that, at equilibrium, the reaction favors products over reactants. Conversely, a smaller \( K_c \) value indicates that the reactants are favored. This helps predict how the system behaves under a set of initial conditions and how the reaction mixture will adjust over time to achieve equilibrium.

Reaction Quotient (Q)

The reaction quotient (Q) is like a snapshot of a reaction that hasn't yet reached equilibrium. It is calculated just like \( K_c \) but uses the current or initial concentrations of the reactants and products instead of the equilibrium concentrations. The formula for the reaction quotient remains the same as that of the equilibrium constant, providing a point of comparison between the current state of the system and the equilibrium state.

For our example reaction, we calculate \( Q \) using the initial concentrations provided: \( Q = \frac{[\mathrm{NH}_3]_{\text{initial}}^{2}}{[\mathrm{N}_2]_{\text{initial}}[\mathrm{H}_2]_{\text{initial}}^{3}} \). By substituting the given concentrations into the equation, we can determine the reaction's move towards equilibrium. If \( Q < K_c \) the reaction will proceed forward, leading to the formation of more product. If \( Q > K_c \) the reaction will shift backward to produce more reactants. Lastly, if \( Q = K_c \) the system is at equilibrium, and no net change will occur.

Le Chatelier's Principle

To further understand how chemical reactions adjust to changes, we turn to Le Chatelier's principle. This principle asserts that if a dynamic equilibrium is disturbed by changing conditions, such as concentration, pressure, or temperature, the system will adjust to counteract the disturbance and restore a new equilibrium.

For instance, if more reactants are added to the system, the principle predicts an increase in the rate of the forward reaction to reduce the reactant concentrations back to equilibrium. If the temperature changes, the reaction could shift in the direction that absorbs heat (in an endothermic reaction) or releases heat (in an exothermic reaction).

In the scenario with our reaction \( \mathrm{N}_2 + 3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \) and a given \( Q > K_c \), Le Chatelier's principle helps us understand why the reaction shifts towards the reactants when \( Q \) is greater than \( K_c \)—the system is reducing the excess products to return to equilibrium. This elegant principle guides us through the prediction of a reaction's response to various disturbances, ensuring we can effectively control and manipulate chemical processes.