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 \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium? (c) What additional information would you need in order to decide whether the reaction as written is endothermic or exothermic?

Short answer

a) The equilibrium constant (K_c) at 25°C is \(1.5 \times 10^{-39}\) b) Reactants are more plentiful at equilibrium c) Additional information needed to determine whether the reaction is endothermic or exothermic is the enthalpy change (ΔH) of the reaction.

Step-by-step solution

Determine the equilibrium constant (K_c)

Using the rate constants for the forward and reverse reactions, we can calculate the equilibrium constant (K_c) using the following equation: \(K_c = \cfrac{k_f}{k_r}\) In this equation, \(k_f\) is the rate constant for the forward reaction, and \(k_r\) is the rate constant for the reverse reaction. We are given these values as \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (forward) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (reverse). Plug in the values for \(k_f\) and \(k_r\) and calculate K_c: \(K_c = \cfrac{(1.4\times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1})}{(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1})}\)

Calculate the value of K_c

Now we can calculate the value of K_c: \(K_c = 1.5 \times 10^{-39}\)

Determine the relative amounts of reactants and products at equilibrium

Based on the value of K_c, we can determine whether reactants or products are more plentiful at equilibrium. If K_c > 1, products are more plentiful. If K_c < 1, reactants are more plentiful. Since \(K_c = 1.5 \times 10^{-39}\), which is much less than 1, reactants are more plentiful at equilibrium.

Identify additional information needed to decide if the reaction is endothermic or exothermic

To determine whether a reaction is endothermic or exothermic, we need information about the heat of the reaction (enthalpy change, ΔH). This information is not provided in the problem statement. Hence, we would need information about the enthalpy change of the reaction to determine whether it's endothermic or exothermic. In summary: a) The equilibrium constant (K_c) at 25°C is \(1.5 \times 10^{-39}\) b) Reactants are more plentiful at equilibrium c) Additional information needed to determine whether the reaction is endothermic or exothermic is the enthalpy change (ΔH) of the reaction.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant, often expressed as K_c, is crucial for students studying chemical reactions that can occur in both forward and reverse directions until they reach a state of balance. This delicate point of balance is what we call chemical equilibrium, where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time.

The equilibrium constant is a numerical value that gives us insight into the relative concentrations of reactants and products at equilibrium. It is determined by dividing the rate constant of the forward reaction (k_f) by the rate constant of the reverse reaction (k_r). This can be represented by the formula:
\( K_c = \frac{k_f}{k_r} \).

In our exercise, the incredibly small value of K_c indicates a large excess of reactants over products at equilibrium. This is a valuable piece of information that can inform us about the system's behavior under different conditions and is foundational to many calculations in chemical kinetics and thermodynamics.

Rate Constants

Rate constants, symbolized as k, play a pivotal role in the kinetics of a chemical reaction. They are values that quantify the speed of a given reaction—the larger the rate constant, the faster the reaction proceeds. Comprehending rate constants involves recognizing that they are determined by various factors including temperature, catalysts, and the molecular structure of the reactants.

The rate constants for both forward and backward reactions provide essential data to compute the equilibrium constant, as shown in the exercise. They are unique for each direction of the reaction and are influenced by different activation energies. For example, in our exercise, the extremely low rate constant for the forward reaction relative to the reverse signifies that the formation of products occurs very slowly.

Elementary Steps

Chemical reactions often proceed through a series of steps known as elementary steps, each of which represents a single molecularity event. These are the smallest divisions into which we can break down a complex reaction mechanism. Understanding them is crucial because they reveal important details such as order and molecularity, which explain how particles must collide to yield a chemical change.

In the context of our exercise, recognizing that both the forward and reverse reactions are elementary helps simplify the calculation of the equilibrium constant. For elementary reactions, the stoichiometric coefficients can be used directly in rate law expressions, thus simplifying the kinetics analysis.

Le Chatelier's Principle

Le Chatelier's Principle is a guiding concept in chemistry that deals with the response of a system in equilibrium to external changes. Put simply, it states that if a dynamic equilibrium is disturbed by altering the conditions, the system will respond or shift its position to counteract the change and restore a new equilibrium.

This principle allows us to predict how changes in concentration, pressure, volume, or temperature will affect the position of equilibrium. For example, if more reactants are added to the system, the principle suggests that the equilibrium will shift to the right, favoring the production of more products. Conversely, an increase in temperature for an endotheric reaction shifts the equilibrium towards the product side, as it absorbs the added heat. The adaptability of chemical equilibria as described by Le Chatelier's principle is a fundamental concept for anyone studying chemistry to grasp.