Question

A 2.50 -mol sample of \(\mathrm{NOCl}\) was initially in a \(1.50-\mathrm{L}\) reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the \(\mathrm{NOCl}\) had dissociated: $$ 2 \mathrm{NOCl}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Short answer

The equilibrium constant \( K_c \) is approximately 0.283.

Step-by-step solution

Determine Initial Concentrations

First, calculate the initial concentration of \( \text{NOCl} \) using the formula: \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \). We have \( 2.50 \) moles of \( \text{NOCl} \) in a \( 1.50 \text{ L} \) container.\[[\text{NOCl}]_0 = \frac{2.50}{1.50} = 1.67 \text{ M}\]

Calculate Concentration Change

Since 28% of \( \text{NOCl} \) has dissociated, the change in concentration \([\text{NOCl}]\) is \( 0.28 \times 1.67 \text{ M} \).\[[\text{NOCl}]_{\text{diss}} = 0.28 \times 1.67 = 0.468 \text{ M}\]

Find Equilibrium Concentrations

Subtract the dissociated \([\text{NOCl}]\) from the initial concentration to find the equilibrium concentration of \([\text{NOCl}]\).\[[\text{NOCl}]_e = 1.67 - 0.468 = 1.202 \text{ M}\]The change in \( \text{NOCl} \) is converted to \( \text{NO} \) and \( \text{Cl}_2 \), thus assuming \( 2\, \text{NOCl} \rightarrow 2\, \text{NO} + \text{Cl}_2 \), the change is:\[[\text{NO}] = 2 \times 0.468 = 0.936 \text{ M}, \quad [\text{Cl}_2] = 0.468 \text{ M}\]

Write and Apply the Equilibrium Expression

Utilize the balanced equation to form the equilibrium expression:\[K_c = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2}\]Plug in the equilibrium concentrations:\[K_c = \frac{(0.936)^2 (0.468)}{(1.202)^2}\]

Calculate the Equilibrium Constant

Calculate \( K_c \) using the values obtained:\[K_c = \frac{(0.936)^2 (0.468)}{(1.202)^2} = \frac{0.8755 \times 0.468}{1.445} \approx 0.283\]

Result and Conclusion

After performing the calculations, the equilibrium constant \( K_c \) for the reaction at \( 400^{\circ} \text{C} \) is approximately 0.283.

Key concepts

NOCl dissociation

When \(\text{NOCl}\) dissociates, it breaks down into \(\text{NO}\) and \(\text{Cl}_2\) gas. This process can be described by the balanced equation: \[ 2 \, \text{NOCl}(g) \rightleftharpoons 2 \, \text{NO}(g) + \text{Cl}_2(g) \] In the context of the exercise, 28% of \(\text{NOCl}\) originally present in the container has dissociated. To understand dissociation, it's crucial to link it to changes in the amounts of reactants and products. Here, the dissociation percentage helps determine how much \(\text{NOCl}\) turns into \(\text{NO}\) and \(\text{Cl}_2\). Since dissociation changes \(\text{NOCl}\) concentrations, it inherently affects the equilibrium state of the reaction. Overall, understanding dissociation in this reaction is about seeing how initial concentrations of substances transform as the reaction progresses towards equilibrium.

Chemical equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction. In the case of \(\text{NOCl}\) dissociation, this balance is achieved when the decomposition into \(\text{NO}\) and \(\text{Cl}_2\) reaches a point where the concentrations of all gases remain constant over time.Equilibrium doesn't mean that reactants are completely used up or turned into products, but rather that there's a dynamic balance. The molecules continue to react, but their concentrations don't change because it happens at equal rates in both directions. Reaching equilibrium depends on initial reactant concentrations, temperature, and inherent reaction kinetics. Here, equilibrium for this particular reaction at 400°C has been established with specific concentrations derived through the dissociation percentage.

Reaction concentration

This term refers to the amount of a substance (reactant or product) present in a given volume. It's often expressed in molarity (M), which indicates moles of a substance per liter of solution. For the reaction \(2 \, \text{NOCl}(g) \rightleftharpoons 2 \, \text{NO}(g) + \text{Cl}_2(g)\), concentration changes play a pivotal role in determining equilibrium. Starting with 1.67 M \(\text{NOCl}\), the reaction's progress is monitored through its concentration changes. The given dissociation percentage allows us to calculate how much of \(\text{NOCl}\) converts to products, leading to equilibrium concentrations. These values are crucial for calculating the equilibrium constant, as they reflect the precise state of the system at equilibrium.

Equilibrium expression

The equilibrium expression is a mathematical representation of a chemical system at equilibrium. For \(\text{NOCl}\) dissociation, it is built on the balanced equation: \[ K_c = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2} \]This formula includes concentrations of all products over reactants, each raised to the power of their coefficients from the reaction. The equilibrium constant, \(K_c\), quantifies the ratio of concentrations at equilibrium. A given \(K_c\) value tells us about the position of equilibrium:

• If \(K_c\) is large, there are more products than reactants, favoring the forward reaction.
• If \(K_c\) is small, it indicates equilibrium favoring the reactants.

In this exercise, calculating \(K_c\) involved inserting the derived equilibrium concentrations into the expression to determine this ratio for the reaction at 400°C. The resulting \(K_c = 0.283\) offers insight into how the reaction behaves under the specified conditions.