Problem 57 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) The value of the equilibrium constant at 25°C is \(15.05 \times 10^{-39}\). (b) Reactants (CO and Cl2) are more plentiful at equilibrium.

Step 1: Calculate the equilibrium constant

To calculate the equilibrium constant, we use the relation: Equilibrium constant (K) = Forward rate constant (kf) / Reverse rate constant (kr) Given, kf = \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{s}^{-1}\) kr = \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1}\) Now, we can divide the forward rate constant by the reverse rate constant to find the equilibrium constant: K = \(\frac{1.4 \times 10^{-28}}{9.3 \times 10^{10}}\)

Step 2: Analyze the equilibrium constant

Now we will calculate K: K = \(15.05 \times 10^{-39}\) Since K is a very small number, it means the reaction favors the formation of reactants (CO and Cl2) over products (COCl and Cl). Therefore, reactants are more plentiful at equilibrium.

Step 3: Final answers

(a) The value of the equilibrium constant at 25°C is \(15.05 \times 10^{-39}\). (b) Reactants (CO and Cl2) are more plentiful at equilibrium.

Key Concepts

Equilibrium Constant

The equilibrium constant, symbolized as K, is a fundamental concept in chemical equilibrium that represents the ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients, when a reaction is at equilibrium. The magnitude of K indicates the extent to which a reaction proceeds before reaching equilibrium.

When K is significantly greater than 1, the equilibrium mixture is rich in products. Conversely, when K is much less than 1, as in the given exercise where K is calculated to be a very small value (\(15.05 \times 10^{-39}\)), there is a higher concentration of reactants compared to products at equilibrium. This small value of K signifies that the reaction favors the reactants, meaning they are more plentiful when the system has reached a state of balance.

Rate Constants

Rate constants are pivotal in understanding the dynamics of a reaction. They indicate the speed at which products form from reactants (forward rate constant, k_f) or the reverse (reverse rate constant, k_r). In the given exercise, the forward rate constant (k_f) is much smaller than the reverse rate constant (k_r), with values of \(1.4 \times 10^{-28} \text{M}^{-1} \text{s}^{-1}\) and \(9.3 \times 10^{10} \text{M}^{-1} \text{s}^{-1}\), respectively.

These rate constants are crucial ingredients in the calculation of the equilibrium constant. A much larger reverse rate constant suggests that the reverse reaction happens at a significantly faster pace relative to the forward reaction, which also helps explain why in this exercise the reactants are predominant at equilibrium.

Elementary Steps

Elementary steps are the basic single-stage processes in a complex reaction mechanism. Each elementary step can be classified as unimolecular, bimolecular, or termolecular, depending on the number of reacting species involved. In our exercise, we assume that the forward and reverse reactions encompass single elementary steps involving two molecules reacting together, indicating a bimolecular process.

Understanding the nature of these steps helps chemists deduce reaction mechanisms and predict the rate law directly from the coefficients of the elementary step. It's essential to recognize that the overall reaction order can only be inferred in this simple manner for elementary steps, not for the overall reaction if it involves multiple steps.

Reaction Favorability

Reaction favorability can be assessed through the equilibrium constant, but it's also crucial to consider thermodynamic factors such as Gibbs free energy. A reaction is said to be favorable if it proceeds forward under given conditions. This does not necessarily mean that the products will dominate in the equilibrium state, as the reaction may reach equilibrium with a higher amount of reactants, as is the case in the exercise.

Furthermore, environmental conditions like temperature and pressure also play significant roles in determining reaction favorability. In this context, Le Chatelier’s Principle explains how a system at equilibrium responds to changes in its conditions. If a reaction is exothermic, increasing temperature would typically shift the equilibrium toward the reactants, which complements the idea that reactant-favored equilibria can indeed be the product of highly favorable reactions under certain conditions.