Problem 99 - Chemistry The Science in Context

Question

A 100 -mL reaction vessel initially contains \(2.60 \times 10^{-2} \mathrm{mol}\) of \(\mathrm{NO}\) and \(1.30 \times 10^{-2} \mathrm{mol}\) of \(\mathrm{H}_{2} . \mathrm{At}\) equilibrium, the concentration of NO in the vessel is \(0.161 M .\) The vessel also contains \(\mathrm{N}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{H}_{2}\) at equilibrium. What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for the following reaction? $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$

Step-by-Step Solution

Answer

Answer: The equilibrium constant (Kc) is approximately 0.218.

Step 1: Calculate Initial Concentrations

First, we'll calculate the initial concentrations of \(\mathrm{NO}\) and \(\mathrm{H}_{2}\) in the 100-mL reaction vessel: $$ [\mathrm{NO}]_0 = \frac{2.60 \times 10^{-2} \mathrm{mol}}{0.100 \mathrm{L}} = 0.260 M $$ $$ [\mathrm{H}_2]_0 = \frac{1.30 \times 10^{-2} \mathrm{mol}}{0.100 \mathrm{L}} = 0.130 M $$

Step 2: Find the Change in Moles at Equilibrium

Since the equilibrium concentration of NO is given as \(0.161 M\), we can determine the change in moles of NO at equilibrium: $$ \Delta[\mathrm{NO}] = [\mathrm{NO}]_{eq} - [\mathrm{NO}]_0 = 0.161 M - 0.260 M = - 0.099 M $$

Step 3: Determine Equilibrium Concentrations for All Species

Now, using the stoichiometry from the balanced reaction, we can determine the equilibrium concentrations for the other species: $$ [\mathrm{H}_2]_{eq} = [\mathrm{H}_2]_0 - 2 \times \Delta[\mathrm{NO}] = 0.130 M - 2 \times (-0.099 M) = 0.328 M $$ According to the reaction, when 2 moles of NO react, 1 mole of N2 is produced. Therefore: $$ [\mathrm{N}_2]_{eq} = - \frac{1}{2} \times \Delta[\mathrm{NO}] = - \frac{1}{2} \times (-0.099 M) = 0.0495 M $$ Similarly, when 2 moles of NO react, 2 moles of H2O are produced, so: $$ [\mathrm{H}_2 \mathrm{O}]_{eq} = - 2 \times \Delta[\mathrm{NO}] = - 2 \times (-0.099 M) = 0.198 M $$

Step 4: Express the Equilibrium Constant in Terms of Concentrations

The expression for the equilibrium constant \(K_{\mathrm{c}}\) is: $$ K_{\mathrm{c}} = \frac{[\mathrm{H}_2 \mathrm{O}]_{eq}^2 [\mathrm{N}_2]_{eq}}{[\mathrm{H}_2]_{eq}^2 [\mathrm{NO}]_{eq}^2} $$

Step 5: Calculate the Equilibrium Constant

Substitute the equilibrium concentrations we found in Step 3 into the expression for \(K_{\mathrm{c}}\): $$ K_{\mathrm{c}} = \frac{(0.198 M)^2 \times (0.0495 M)}{(0.328 M)^2 \times (0.161 M)^2} \approx 0.218 $$ The equilibrium constant \(K_{\mathrm{c}}\) is approximately 0.218.

Key Concepts

Chemical Equilibrium

In the realm of chemistry, chemical equilibrium represents a state in which the rate of the forward reaction is equal to the rate of the reverse reaction. This balance means that the concentrations of the reactants and products remain constant over time, though they are not necessarily equal. To visualize it, imagine a scale perfectly balanced with weights on each side.

For a reaction like
\[2 \text{H}_2(g) + 2 \text{NO}(g) \rightleftharpoons 2 \text{H}_2\text{O}(g) + \text{N}_2(g),\]
equilibrium is reached when the rates at which the reactants are converting into products and the products are reverting to reactants are the same. This concept is crucial as it also allows us to calculate the equilibrium constant, which is a measure of the extent at which a reaction will proceed.

Reaction Quotient

The reaction quotient, often denoted as \( Q \), plays a pivotal role in understanding the direction in which a reaction will proceed before reaching equilibrium. It is calculated using the same formula as the equilibrium constant, \( K_{\text{c}} \), but with the concentrations or partial pressures of the reactants and products at any point in time before equilibrium.

Consider the formula for the reaction quotient of the given reaction:
\[Q = \frac{[\text{H}_2\text{O}]^2 [\text{N}_2]}{[\text{H}_2]^2 [\text{NO}]^2}.\]
Comparing \( Q \) to \( K_{\text{c}} \) tells us that if \( Q < K_{\text{c}} \), the reaction will proceed in the forward direction to produce more products. Conversely, if \( Q > K_{\text{c}} \), the reaction will shift to form more reactants. When \( Q = K_{\text{c}} \), the system is at equilibrium.

Le Chatelier's Principle

Moving to Le Chatelier's Principle, it provides an intuitive way to predict how a system at equilibrium will respond to changes. Essentially, it states that if an external change is imposed on a system at equilibrium, the system will adjust to counteract that change and restore a new balance.

For instance, an increase in the concentration of reactants or decrease in the concentration of products will shift the reaction forward, favoring the formation of more products. Similarly, changes in temperature or pressure can also cause the position of equilibrium to shift. This principle helps chemists control the outcome of reactions in industrial processes by manipulating conditions to favor the production of desired products.

Equilibrium Concentrations

The term equilibrium concentrations refers to the amounts of reactants and products present in a reaction mixture when equilibrium has been achieved. These concentrations are constant as long as the system is undisturbed. In the context of our calculation example, the equilibrium concentration of each compound—\( \text{NO} \), \( \text{H}_2 \), \( \text{N}_2 \), and \( \text{H}_2\text{O} \)—is a critical piece of data required to compute the equilibrium constant, \( K_{\text{c}} \).

To calculate these concentrations, it's necessary to know the initial concentrations and the changes in concentrations that occur as the system moves towards equilibrium, considering the stoichiometry of the balanced chemical equation. By comparing the equilibrium concentrations to the initial concentrations, we can understand how far the reaction has proceeded.