Problem 71 - Chemistry: The Central Science in SI Units, Global Edition

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 reactions are \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?(\mathbf{b})\) Are reactants or products more plentiful at equilibrium?

Step-by-Step Solution

Answer

The equilibrium constant (K) at \(25^{\circ}C\) can be calculated as the ratio of the forward and reverse rate constants (k_f and k_r, respectively): \(K = \frac{k_f}{k_r} = \frac{1.4 \times 10^{-28}}{9.3 \times 10^{10}} \approx 1.5 \times 10^{-39}\). Since K is a very small value (K << 1), this indicates that the reaction favors the reactants, and thus, reactants are more plentiful at equilibrium than products.

Step 1: Write down the equilibrium constant expression

According to the equilibrium constant expression, the equilibrium constant (K) can be calculated as the ratio of the forward and reverse rate constants (k_f and k_r, respectively): \[K = \frac{k_f}{k_r}\]

Step 2: Calculate the equilibrium constant

Now we can plug in the given values of forward and reverse rate constants: \(k_f = 1.4 \times 10^{-28} M^{-1} s^{-1}\) \(k_r = 9.3 \times 10^{10} M^{-1} s^{-1}\) \[K = \frac{1.4 \times 10^{-28}}{9.3 \times 10^{10}}\] Calculate the value of K: \[K \approx 1.5 \times 10^{-39}\]

Step 3: Determine whether reactants or products are more plentiful at equilibrium

Since K is a very small value (K << 1), this indicates that the reaction favors the reactants and not the products. Therefore, reactants are more plentiful at equilibrium than products. In conclusion, the value of the equilibrium constant at \(25^{\circ}C\) is approximately \(1.5 \times 10^{-39}\), and reactants are more plentiful at equilibrium.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a central concept in the study of chemical reactions and equilibrium. It provides a numerical value that describes the ratio of concentrations of products to reactants at equilibrium. For the reaction \( \mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g) + \mathrm{Cl}(g) \), the equilibrium constant expression can be represented as \( K = \frac{[\mathrm{COCl}][\mathrm{Cl}]}{[\mathrm{CO}][\mathrm{Cl}_2]} \). However, when reactions are elementary, as in this case, \( K \) can also be derived from the ratio of the forward and reverse rate constants, given by the equation \( K = \frac{k_f}{k_r} \).

In the given exercise, the forward reaction rate constant \( k_f \) is \( 1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{s}^{-1} \), and the reverse reaction rate constant \( k_r \) is \( 9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1} \). Using these values, \( K \) is calculated to be approximately \( 1.5 \times 10^{-39} \). This very small value indicates that at equilibrium, the concentration of reactants is much higher than that of products, meaning the reaction predominantly remains in the reactant state.

Reaction Rate

The reaction rate is the speed at which reactants are converted into products in a chemical reaction. This rate can vary significantly and is influenced by factors such as concentration, temperature, and the presence of catalysts. For elementary steps, which this reaction involves, the reaction rate is directly proportional to the concentration of the reactants.

In the context of the given exercise, we have separate rate constants for the forward and reverse reactions: \( k_f \) and \( k_r \). The forward rate constant \( k_f = 1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{s}^{-1} \), and the reverse rate constant \( k_r = 9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1} \).
The extraordinarily small value of \( k_f \) compared to \( k_r \) suggests a slow conversion of reactants to products, but a rapid reformation of reactants from products. This imbalance contributes to the low equilibrium constant \( K \) observed, reinforcing that the reactants are much more prevalent than products at equilibrium.

Elementary Steps

In chemical kinetics, a reaction mechanism can involve multiple steps, with each individual step classified as an elementary step. An elementary step refers to a single reaction event that cannot be broken down into simpler ones. It involves direct interactions between reacting molecules at the most fundamental level.

The given chemical reaction \( \mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g) + \mathrm{Cl}(g) \) is described as consisting of elementary steps. This implies that both the forward and reverse reactions proceed in a single step involving one molecule of carbon monoxide and one molecule of chlorine to form phosgene and a chlorine atom, and vice versa.

• Direct Collisions: Elementary steps rely on direct molecular collisions and involve the simplest possible pathway from reactants to products or vice versa.
• Rate Expressions: The rate law for an elementary step can be directly written from the stoichiometry of the step since each reactant is involved as it appears in the chemical equation.

Understanding these concepts helps explain why the rate constants differ so significantly in this exercise, impacting the equilibrium constant and distribution of reactants and products.