Question

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{s}^{-1}\) and \(9.3 \times 10^{10} M^{-1} \mathrm{s}^{-1}\) respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium?

Short answer

The equilibrium constant at \(25^{\circ} \mathrm{C}\) is approximately \(K_c \approx 1.505 \times 10^{-39}\), and at equilibrium, the reactants are more plentiful than the products.

Step-by-step solution

Write down the given information

We are given the following information: The elementary equilibrium reaction: \(\mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g) + \mathrm{Cl}(g)\) Rate constant for the forward reaction, \(k_f = 1.4 \times 10^{-28} \, M^{-1} s^{-1} \) Rate constant for the reverse reaction, \(k_r = 9.3 \times 10^{10} \, M^{-1} s^{-1} \) Temperature, \(T = 25^{\circ} \mathrm{C}\)

Calculate the equilibrium constant

For a simple elementary equilibrium reaction, the relationship between the equilibrium constant, \(K_c\), and the rate constants for the forward and reverse reactions, \(k_f\) and \(k_r\), is as follows: \[K_c = \frac{k_f}{k_r}\] Substitute the given rate constants into the equation: \[K_c = \frac{1.4 \times 10^{-28} \, M^{-1} s^{-1}}{9.3 \times 10^{10} \, M^{-1} s^{-1}}\] Solve for the equilibrium constant: \[K_c \approx 1.505 \times 10^{-39}\]

Reactants or products more plentiful at equilibrium

To determine whether the reactants or products are more plentiful at equilibrium, we need to analyze the equilibrium constant, \(K_c\). If \(K_c > 1\), the equilibrium is product-favored and products are more plentiful. If \(K_c < 1\), the equilibrium is reactant-favored and reactants are more plentiful. Since \(K_c \approx 1.505 \times 10^{-39} < 1\), it is reactant-favored, and the reactants are more plentiful at equilibrium.

Write down the final answers

(a) The equilibrium constant at \(25^{\circ} \mathrm{C}\) is approximately: \(K_c \approx 1.505 \times 10^{-39}\) (b) At equilibrium, the reactants are more plentiful than the products.

Key concepts

Elementary Equilibrium Reactions

When we discuss chemical reactions, we often talk about complex processes that can involve multiple steps. However, elementary reactions are reactions that occur in a single step and involve a direct interaction between molecules to form products. In an equilibrium scenario, an elementary reaction happens in both the forward and reverse directions.

Understanding elementary equilibrium reactions requires familiarity with the concept of dynamic equilibrium. This is where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactors or products over time, even though both reactions are still proceeding. To visualize this, imagine a room with two doors where people are entering through one door and exiting through the other at the same rate; the number of people in the room stays the same, reflecting a state of balance.

Getting to grips with this concept helps students to better visualize chemical processes and relate them to real-world systems. As we explore how reactions can 'settle' into balance, we gain a deeper insight into the behavior of reactions under different conditions.

Rate Constants

The rate constants for a chemical reaction measure the speed at which the reactants are converted into products. In the language of kinetics, they reflect the 'rate' or 'propensity' of a reaction under specific conditions, typically temperature and pressure. The rate constant for the forward reaction is denoted as k_f, while the reverse reaction is represented as k_r.

Mathematically, the value of the rate constant depends on several factors such as the nature of the reactants, the presence of a catalyst, temperature, and the reaction environment. When breaking down problems related to kinetics and equilibrium, students need to handle these constants with care, as they are keys to unlocking the behavior of the reaction.

To enhance understanding, we might compare the rate constants to the speed of cars: k_f and k_r could represent cars traveling in opposite directions with different speeds. The major takeaway for students is that these 'speeds' have profound implications for the position of equilibrium and the amounts of reactants and products at that point.

Reaction Favorability

The term reaction favorability signals which side of a chemical equilibrium is favored: the reactants or the products. This can be quantitatively determined by calculating the equilibrium constant (\(K_c\)). If the value of the equilibrium constant is greater than 1 (\(K_c > 1\)), the products are favored and the reaction proceeds mostly towards the product side under equilibrium conditions. Conversely, if the equilibrium constant is less than 1 (\(K_c < 1\)), the reactants are favored.

The size of the equilibrium constant provides a clue as to the extent of the reaction—how far it 'pushes' towards completion before settling into equilibrium. Clearly articulating this idea helps learners forecast the outcome of reactions and interpret scientific data more effectively. In the case of our CO and Cl₂ reaction, the extremely low value of the equilibrium constant indicates that the reactants are significantly more favored, akin to a pendulum that barely swings towards the product side before swinging back.