Problem 3 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The rate constant for the forward reaction: \(\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})\) is \(1.5 \times 10^{-3} \mathrm{~s}^{-1}\) at \(300 \mathrm{~K}\). If \(10^{-5}\) moles of 'A' and 100 moles of ' \(\mathrm{B}\) ' are present in a 10 litre vessel at equilibrium, then the rate constant of the backward reaction at this temperature is (a) \(1.5 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (b) \(1.5 \times 10^{-1} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (c) \(1.5 \times 10^{-11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (d) \(1.5 \times 10^{-12} \mathrm{M}^{-1} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

The rate constant of the backward reaction at this temperature is \(1.5 \times 10^{-11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), which corresponds to option (c).

Step 1: Write the Equilibrium Expression

The equilibrium expression for the reversible reaction \(\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{B}(\mathrm{g})\) is given by the equilibrium constant \(K_c = \frac{[\mathrm{B}]^2}{[\mathrm{A}]}\), where [\mathrm{A}] and [\mathrm{B}] are the molar concentrations of A and B at equilibrium.

Step 2: Calculate the Molar Concentrations at Equilibrium

Calculate the molar concentrations using the number of moles and volume of the vessel. For A: \[ [\mathrm{A}] = \frac{10^{-5}\ \text{moles}}{10\ \text{L}} = 10^{-6}\ \text{M} \] For B: \[ [\mathrm{B}] = \frac{100\ \text{moles}}{10\ \text{L}} = 10\ \text{M} \]

Step 3: Calculate the Equilibrium Constant \(K_c\)

Using the forward reaction rate constant (\(k_f\)) and the molar concentrations: \(K_c = \frac{k_f}{k_b}\) where \(k_b\) is the backward reaction rate constant. Since \(k_f\) is given and Kc can be calculated from the equilibrium concentrations, we can find \(k_b\).

Step 4: Calculate the Equilibrium Constant \(K_c\) Using Concentrations

Substitute the calculated concentrations into the equilibrium expression: \[ K_c = \frac{[\mathrm{B}]^2}{[\mathrm{A}]} = \frac{(10)^2}{10^{-6}} = \frac{100}{10^{-6}} = 10^8 \]

Step 5: Solve for the Backward Reaction Rate Constant \(k_b\)

Rearrange the equation \(K_c = \frac{k_f}{k_b}\) to solve for \(k_b\): \[ k_b = \frac{k_f}{K_c} = \frac{1.5 \times 10^{-3}\ \text{s}^{-1}}{10^8} = 1.5 \times 10^{-11}\ \text{M}^{-1}\text{s}^{-1} \] This matches answer option (c).

Key Concepts

Understanding Chemical Equilibrium

In chemistry, chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction, meaning that the concentrations of reactants and products remain constant over time. This does not imply that the reactants and products are in equal concentrations, but rather that their ratios are stable. It's important to understand that equilibrium is dynamic; molecules continue to react, but there's no net change in concentration.

Think of it like a busy street where the number of cars entering equals the number of cars leaving, so the total number of cars on the street stays the same. In the context of the exercise, equilibrium involves the gas A transforming into gas B and vice versa. When the system has reached equilibrium, we can quantify the position of this equilibrium in terms of the equilibrium constant, \(K_c\), which compares the concentration of the products raised to the power of their coefficients to the concentration of the reactants raised to the power of their coefficients.

Rate Constant and its Role in Reactions

The rate constant, denoted as \(k\), is a crucial parameter in chemical kinetics, offering insight into the speed of a reaction. It is independent of the concentration of reactants but varies with temperature and the presence of a catalyst. In a given reaction, the rate constant for the forward reaction \(k_f\), and the backward reaction \(k_b\), are generally not equal and are reflective of how quickly products form and revert to reactants, respectively.

For instance, the exercise presents us with a scenario where the forward rate constant \(1.5 \times 10^{-3} \text{s}^{-1}\) is given, and we must infer the backward rate constant using the equilibrium concentrations. It's similar to knowing how fast a pendulum swings in one direction and using that information, along with the equilibrium point, to determine how fast it swings back.

Reaction Kinetics Fundamentals

Reaction kinetics is the field that studies the rates of chemical reactions and the factors that affect these rates. It tells us not just how long a reaction will take to reach equilibrium, but also how the presence of different substances or changes in conditions like temperature can speed up or slow down this process. Kinetics is central to understanding reaction mechanisms—the steps that take place during a reaction.

Effects of Concentration and Temperature

For example, increasing the concentration of reactants typically increases the rate of reaction because there are more particles to collide and react. Temperature also plays a key role; as it increases, particles move faster, leading to more frequent and energetic collisions, thus increasing the rate of reaction. In the given exercise, the reaction kinetics principles allow us to determine the backward reaction rate constant by understanding the relationship between the equilibrium constant and the forward reaction rate constant.