Problem 61 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

For the consecutive first-order reactions: \(\mathrm{A} \stackrel{K_{1}}{\longrightarrow} \mathrm{B} \stackrel{K_{2}}{\longrightarrow} \mathrm{C}\), the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are \(0.2 \mathrm{M}\) and \(0.01 \mathrm{M}\), respectively, at steady state. If \(K_{1}\) is \(2.5 \times 10^{-4} \mathrm{~min}^{-1}\), what is the value of \(K_{2} ?\) (a) \(5.0 \times 10^{-3} \mathrm{~min}^{-1}\) (b) \(2.5 \times 10^{-4} \mathrm{~min}^{-1}\) (c) \(1.25 \times 10^{-5} \mathrm{~min}^{-1}\) (d) \(5.0 \times 10^{-4} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

\(K_2 = 5.0 \times 10^{-3} \mathrm{~min}^{-1}\)

Step 1: Understand Steady State for B

At steady state, the rate at which B is formed from A is equal to the rate at which B is converted to C. This implies that the rate of appearance of B equals the rate of disappearance of B.

Step 2: Write Rate Equations

Write the rate equation for the formation of B: Rate of formation of B = \(K_1 [A]\). Write the rate equation for the disappearance of B: Rate of disappearance of B = \(K_2 [B]\).

Step 3: Set Rates Equal at Steady State

Since the system is at steady state, set the rates equal to each other: \(K_1 [A] = K_2 [B]\).

Step 4: Plug in Given Values

Substitute the given concentrations and the value of \(K_1\) into the equation to find \(K_2\): \(2.5 \times 10^{-4} \mathrm{min}^{-1} \times 0.2 \mathrm{M} = K_2 \times 0.01 \mathrm{M}\).

Step 5: Solve for \(K_2\)

Isolate \(K_2\) by dividing both sides of the equation by \(0.01 \mathrm{M}\) to find its value.

Key Concepts

Steady State Approximation

The steady state approximation is a key concept in understanding the dynamics of chemical reactions, particularly when dealing with consecutive first-order reactions. In such reactions, an intermediate species is produced and consumed simultaneously, leading to an apparent steady state concentration of that species. This approximation assumes that the rate of production of the intermediate is equal to its rate of consumption, so its concentration remains constant over time.

For instance, consider the reaction sequence \(\mathrm{A} \stackrel{K_{1}}{\longrightarrow} \mathrm{B} \stackrel{K_{2}}{\longrightarrow} \mathrm{C}\). The intermediate \(\mathrm{B}\) reaches a steady state when the formation rate from \(\mathrm{A}\) equals the disappearance rate into \(\mathrm{C}\). Mathematically, this can be expressed as \(K_1 [A] = K_2 [B]\). When dealing with homework problems or real-life processes, applying the steady state approximation simplifies the calculation of unknowns, such as the rate constants for the reactions involved.

Reaction Rate Equations

Reaction rate equations are mathematical representations of the speed at which reactants are converted into products in a chemical reaction. For first-order reactions, the rate equation is linearly dependent on the concentration of one reactant. It takes the general form \(\text{Rate} = k [\text{Reactant}]\), where \(k\) is the rate constant and \(\text{Reactant}\) denotes the concentration of the reacting species.

In a system with consecutive first-order reactions involving \(\mathrm{A}\), \(\mathrm{B}\), and \(\mathrm{C}\), two separate rate equations can be written - one for the formation of \(\mathrm{B}\) from \(\mathrm{A}\), and another for its conversion to \(\mathrm{C}\). Accurately writing and manipulating these equations is critical for solving many chemical kinetics problems, including finding the rate constants given certain concentrations at steady state, as demonstrated in the provided example.

Rate Constants

Rate constants are essential parameters in the study of chemical kinetics as they provide a quantitative measure of reaction speed. For first-order reactions, the rate constant \(k\) denotes the probability of a particular reactant undergoing reaction per unit time. Higher rate constants indicate faster reactions. Each elementary step in a reaction mechanism has its own rate constant, and these constants are influenced by factors such as temperature and the presence of catalysts.

In an exercise that involves consecutive reactions, determining the rate constants for each step is often the focus. Understanding how to use given data, like reactant concentrations and the steady state approximation, allows for the calculation of these constants. In our example, by setting up the appropriate equations under steady state conditions and substituting the known values, students can isolate and solve for the unknown rate constant \(K_2\). This application of the concepts defines the interconnectivity between steady state approximation, reaction rate equations, and rate constants in analyzing reaction mechanisms.