Problem 28 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Reaction: \(\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons \mathrm{C}(\mathrm{g})\) \(+\mathrm{D}(\mathrm{g})\), occurs in a single step. The rate constant of forward reaction is \(2.0 \times 10^{-3} \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1} .\) When the reaction is started with equimolar amounts of \(\mathrm{A}\) and \(\mathrm{B}\), it is found that the concentration of \(\mathrm{A}\) is twice that of \(\mathrm{C}\) at equilibrium. The rate constant of the backward reaction is (a) \(5.0 \times 10^{-4} \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\) (b) \(8.0 \times 10^{-3} \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\) (c) \(1.25 \times 10^{2} \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\) (d) \(2.0 \times 10^{3} \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

The rate constant of the backward reaction is (b) \(8.0 \times 10^{-3} \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\).

Step 1: Understanding the Rate Law for the Forward Reaction

The rate of the forward reaction can be expressed as rate = k_forward[A][B], where k_forward is the rate constant for the forward reaction, and [A] and [B] are the molar concentrations of reactants A and B, respectively.

Step 2: Expression of Equilibrium Constant (K)

At equilibrium, the rate of forward reaction equals the rate of the backward reaction. The equilibrium constant, K, is defined by the concentrations of the products divided by the concentrations of the reactants, raised to the power of their stoichiometric coefficients. For the given reaction, this is expressed as K = [C][D]/[A][B].

Step 3: Relating K to the Rate Constants

The equilibrium constant K can also be expressed in terms of the rate constants of the forward and backward reactions as K = k_forward/k_backward.

Step 4: Applying the Given Relationship at Equilibrium

It's given that at equilibrium, concentration of A is twice that of C. If we call the concentration of C at equilibrium x, then the concentration of A will be 2x and B will also be 2x, since the reaction started with equimolar amounts of A and B and they react in a 1:1 ratio.

Step 5: Determining the Equilibrium Constant, K

Using the relationship [C] = x, [A] = 2x, [B] = 2x and [D] = x (since C and D are produced in 1:1 ratio), we can plug these into the expression for K: K = [C][D]/[A][B] = (x)(x)/(2x)(2x) = 1/4.

Step 6: Calculating the Backward Rate Constant, k_backward

With K = 1/4 and k_forward given as 2.0 x 10^{-3} mol^{-1} L s^{-1}, we can solve for k_backward: k_backward = k_forward / K = (2.0 x 10^{-3}) / (1/4) = 8.0 x 10^{-3} mol^{-1} L s^{-1}.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for interpreting reaction dynamics and predicting the concentrations of substances in a closed system.
When a reaction reaches a state where the rate of the forward reaction equals that of the reverse reaction, it is said to be in chemical equilibrium. At this point, the concentrations of reactants and products remain constant over time, though they are not necessarily equal.

For the reaction \(A(g) + B(g) \rightleftharpoons C(g) + D(g)\), assuming the reaction occurs in a closed vessel, we can infer that, at equilibrium, there will be no net change in the concentrations of A, B, C, and D even though both the forward and reverse reactions are still occurring.

In our exercise, we considered the relationship between the reactants and products at equilibrium to calculate the equilibrium constant, which is a quantitative measure of the ratio of concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation. This concept is pivotal for students to understand how to predict and manipulate reaction conditions in the lab or industry.

Rate Constant

The concept of the rate constant is fundamental in kinetics and is used to understand how rapidly a reaction proceeds.
The rate constant, denoted as \(k\), is a proportionality factor that relates the rate of a chemical reaction to the concentrations of the reactants. For a given reaction at a specific temperature, the rate constant is fixed and provides an intrinsic measure of the reaction's speed.

In our exercise, we were given the rate constant for the forward reaction, \(k_{\text{forward}} = 2.0 \times 10^{-3} \mathrm{mol}^{-1} \mathrm{L} \mathrm{s}^{-1}\). This constant enables us to establish a rate law expressing the rate of reaction in terms of the concentration of the reactants, \(\text{rate} = k_{\text{forward}}[A][B]\), where \([A]\) and \([B]\) are the molar concentrations of reactants A and B.

Understanding the rate constant helps students foresee how changes in concentration or conditions could affect the speed of the reaction, impacting the time to reach equilibrium and the amounts of products formed.

Reaction Rates

Reaction rates bring kinetic theory to life by quantifying how fast a reaction proceeds and are central to the study of chemical processes.
Reaction rates can be defined as the change in concentration of a reactant or product per unit time. The rate can be expressed in terms of either an increase in the concentration of the products or a decrease in the concentration of the reactants.

In the context of our problem, the rates of the forward and reverse reactions determine when the system will reach equilibrium. Initially, the forward reaction rate is higher than the reverse, as reactants are being converted to products. Over time, as product concentrations rise, the reverse reaction rates increase until they are equal to the forward rates.

To provide a more comprehensive understanding, students should realize that factors such as temperature, catalysts, and concentration can greatly influence reaction rates and thereby the position and time it takes to achieve equilibrium. An increased understanding of reaction rates aids students in predicting and controlling the outcomes of laboratory experiments and industrial processes.