Problem 6 - Chemistry and Chemical Reactivity

Question

The equilibrium constant, \(K_{c}\), for the reaction $$ 2 \mathrm{NOCl}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ is \(3.9 \times 10^{-3}\) at \(300^{\circ} \mathrm{C} .\) A mixture contains the gases at the following concentrations: \([\mathrm{NOCl}]=\) \(5.0 \times 10^{-3} \mathrm{mol} / \mathrm{L},[\mathrm{NO}]=2.5 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{Cl}_{2}\right]=2.0 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Is the reaction at equilib- rium at \(300^{\circ} \mathrm{C}\) ? If not, in which direction does the reaction proceed to come to equilibrium?

Step-by-Step Solution

Answer

The reaction is not at equilibrium; it will shift right to reach equilibrium.

Step 1: Write the expression for the reaction quotient, Qc

The reaction quotient, \(Q_c\), for the given reaction \(2 \ \mathrm{NOCl}(\mathrm{g}) \rightleftarrows 2 \ \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is calculated using the expression:\[ Q_c = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2} \]This expression is similar to the equilibrium constant expression, \(K_c\), but uses initial concentrations.

Step 2: Substitute the concentrations into the Qc expression

Plug in the given concentrations into the \(Q_c\) expression:\[ Q_c = \frac{(2.5 \times 10^{-3})^2 \times (2.0 \times 10^{-3})}{(5.0 \times 10^{-3})^2} \]

Step 3: Calculate the numerical value of Qc

Calculate \(Q_c\) with the substituted values:\[ Q_c = \frac{(6.25 \times 10^{-6}) \times (2.0 \times 10^{-3})}{(25.0 \times 10^{-6})} \]\[ Q_c = \frac{1.25 \times 10^{-8}}{25.0 \times 10^{-6}} \]\[ Q_c = 5.0 \times 10^{-4} \]

Step 4: Compare Qc with Kc

The equilibrium constant \(K_c\) is given as \(3.9 \times 10^{-3}\). Compare \(Q_c = 5.0 \times 10^{-4}\) with \(K_c\):- Since \(Q_c < K_c\), the reaction will shift to the right to reach equilibrium. This means more NO and \(\mathrm{Cl}_2\) will be produced.

Key Concepts

Reaction Quotient

The reaction quotient, represented as \(Q_c\), is an important concept when analyzing chemical equilibrium. It is a number that describes the relative amounts of products and reactants at any point during a reaction. Unlike the equilibrium constant \(K_c\), which is calculated using equilibrium concentrations, \(Q_c\) uses the initial concentrations of the substances.
To calculate \(Q_c\), you use the formula specific to the balanced chemical equation. For a reaction such as \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\), the expression is:

• \( Q_c = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2} \)

This expression involves raising the concentrations of the products and reactants to their coefficients from the balanced equation. Substituting concentrations from any given state into this expression allows us to calculate \(Q_c\) and compare it with \(K_c\) to understand the reaction's progression.

Equilibrium Constant

The equilibrium constant, \(K_c\), is a fundamental concept in chemical equilibrium that helps predict the direction in which a reaction will proceed to reach equilibrium. It is a ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced chemical equation, at equilibrium. For the reaction \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\), the expression is the same as for \(Q_c\), but using the concentrations at equilibrium:

• \( K_c = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2} \)

In the given problem, \(K_c\) is provided as \(3.9 \times 10^{-3}\) at a temperature of \(300^{\circ} \mathrm{C}\). This constant value signifies whether the reactants or products are favored when the system is at equilibrium under these conditions.
Studying \(K_c\) helps chemists understand how a reaction might respond to changes in conditions like temperature, pressure, or the presence of catalysts.

Reaction Direction

Understanding the direction in which a reaction proceeds is crucial for assessing whether a system is at equilibrium. This direction can be predicted by comparing \(Q_c\) and \(K_c\). If \(Q_c < K_c\), the reaction shifts to the right, toward the formation of more products to achieve equilibrium. On the other hand, if \(Q_c > K_c\), the reaction shifts to the left, favoring the reactants.
In the given exercise, \(Q_c\) was calculated to be \(5.0 \times 10^{-4}\), which is less than \(K_c = 3.9 \times 10^{-3}\). Therefore, the reaction shifts to the right; more \(\mathrm{NO}\) and \(\mathrm{Cl}_2\) will be formed as it approaches equilibrium. Recognizing the reaction direction allows us to predict changes in concentrations over time and helps control industrial chemical processes.

Equilibrium Concentration

Equilibrium concentrations are the concentrations of all reactants and products when the reaction has reached a state of balance, where no further changes occur in the concentrations over time. At this point, the rate of the forward reaction equals the rate of the reverse reaction.
Equilibrium concentrations are crucial for chemists to understand the dynamic nature of chemical systems and to design experiments or industrial processes that maximize yield or efficiency.
By determining \(K_c\) and using \(Q_c\) to predict the direction of shifts, one can estimate the concentrations of reactants and products once equilibrium is established. This information helps in deciding the conditions (temperature, pressure) needed for achieving the desired levels of reactants and products in a chemical reaction setup.