Question

At \(100^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) has the value \(K_{c}=\) \(2.19 \times 10^{-10}\). Are the following mixtures of \(\mathrm{COCl}_{2}, \mathrm{CO}\), and \(\mathrm{Cl}_{2}\) at \(100^{\circ} \mathrm{C}\) at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) \(\left[\mathrm{COCl}_{2}\right]=2.00 \times 10^{-3} \mathrm{M},[\mathrm{CO}]=3.3 \times 10^{-6} \mathrm{M}\), \(\left[\mathrm{Cl}_{2}\right]=6.62 \times 10^{-6} \mathrm{M}\) (b) \(\left[\mathrm{COCl}_{2}\right]=4.50 \times 10^{-2} \mathrm{M},[\mathrm{CO}]=1.1 \times 10^{-7} \mathrm{M}\), \(\left[\mathrm{Cl}_{2}\right]=2.25 \times 10^{-6} \mathrm{M}\) (c) \(\left[\mathrm{COCl}_{2}\right]=0.0100 \mathrm{M},[\mathrm{CO}]=\left[\mathrm{Cl}_{2}\right]=1.48 \times 10^{-6} \mathrm{M}\)

Short answer

- For situation (a), the reaction will proceed to the left (towards the reactants) to achieve equilibrium. - For situation (b), the reaction will proceed to the right (towards the products) to achieve equilibrium. - In situation (c), the mixture is already at equilibrium.

Step-by-step solution

Write the equilibrium constant expression

For the given reaction, the equilibrium constant expression is: \[K_c = \frac{[\mathrm{CO}][\mathrm{Cl}_{2}]}{[\mathrm{COCl}_{2}]}\]

Calculate the reaction quotient, \(Q_c\), for situation (a)

Using the equilibrium constant expression and given concentration values, we can calculate \(Q_c\) for situation (a): \[Q_{c(a)} = \frac{[3.3 \times 10^{-6}][6.62 \times 10^{-6}]}{[2.00 \times10^{-3}]} = 1.089 \times 10^{-8}\]

Compare \(Q_c\) with \(K_c\) for situation (a)

We are given the equilibrium constant value: \[K_c = 2.19 \times 10^{-10}\] By comparing the values of \(Q_c\) and \(K_c\), \[Q_{c(a)} > K_c\] Since \(Q_c > K_c\), the reaction will proceed to the left (towards the reactants) to achieve equilibrium.

Calculate the reaction quotient, \(Q_c\), for situation (b)

Now, we can calculate \(Q_c\) for situation (b): \[Q_{c(b)} = \frac{[1.1 \times 10^{-7}][2.25 \times 10^{-6}]}{[4.50 \times10^{-2}]} = 5.50 \times 10^{-12}\]

Compare \(Q_c\) with \(K_c\) for situation (b)

By comparing the values of \(Q_c\) and \(K_c\), \[Q_{c(b)} < K_c\] Since \(Q_c < K_c\), the reaction will proceed to the right (towards the products) to achieve equilibrium.

Calculate the reaction quotient, \(Q_c\), for situation (c)

Finally, we can calculate \(Q_c\) for situation (c): \[Q_{c(c)} = \frac{[1.48 \times 10^{-6}][1.48 \times 10^{-6}]}{[0.0100]} = 2.19 \times 10^{-10}\]

Compare \(Q_c\) with \(K_c\) for situation (c)

By comparing the values of \(Q_c\) and \(K_c\), \[Q_{c(c)} = K_c\] Since \(Q_c = K_c\), the mixture in situation (c) is already at equilibrium. #Conclusion# - For situation (a), the reaction will proceed to the left (towards the reactants) to achieve equilibrium. - For situation (b), the reaction will proceed to the right (towards the products) to achieve equilibrium. - In situation (c), the mixture is already at equilibrium.

Key concepts

reaction quotient

The reaction quotient, denoted as \(Q_c\), is a vital concept in the study of chemical reactions. It allows us to determine the status of a chemical reaction at any given set of concentrations. This value is similar to the equilibrium constant \(K_c\), but unlike \(K_c\), which only considers equilibrium conditions, \(Q_c\) can be calculated at any point.To calculate \(Q_c\) for a reaction, we use the same formula as the equilibrium constant expression. For instance, given a reaction \(\mathrm{COCl}_2(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{Cl}_2(g)\), the formula is:\[Q_c = \frac{[\mathrm{CO}][\mathrm{Cl}_2]}{[\mathrm{COCl}_2]}\]By plugging the concentrations of the reactants and products into this formula, you can calculate \(Q_c\) at any stage of the reaction. Depending on whether \(Q_c\) is less than, equal to, or greater than \(K_c\), we can predict the direction in which the reaction needs to proceed to reach equilibrium.

chemical equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. This results in constant concentrations of reactants and products, meaning that the reaction has reached a state of balance.At equilibrium, the reaction quotient \(Q_c\) equals the equilibrium constant \(K_c\). This relationship indicates that the quantities of reactants and products no longer change over time. It doesn’t mean the reactions stop; they continue but at equal and opposite rates.Equilibrium is dynamic, and factors such as changes in temperature, pressure, or concentration can shift it, usually described by Le Châtelier's Principle. These changes can alter \(K_c\), but at constant temperature for a given reaction, \(K_c\) remains consistent. It is only when \(Q_c\) equals \(K_c\) that we are assured the system is at equilibrium.

reaction direction

Understanding the direction a reaction will proceed to reach equilibrium involves comparing \(Q_c\) and \(K_c\). The relationship between these two values indicates whether more reactants or products are needed to achieve equilibrium.- **If \(Q_c < K_c\):** This means there are more reactants than products in the system compared to equilibrium conditions. The reaction will proceed to the right, i.e., towards the formation of products, to reach equilibrium.- **If \(Q_c > K_c\):** In this situation, there are more products than reactants relative to equilibrium. Hence, the reaction will shift to the left, towards the formation of reactants.- **If \(Q_c = K_c\):** Here, the system is already at equilibrium, and no shift in reaction direction is required. Both forward and reverse reactions occur at equal rates.This understanding of reaction direction not only helps us predict how a reaction mixture will evolve but also assists in manipulating conditions favorably for chemical processes.