Question

\(\begin{array}{lll}\text { For the } & \text { sequential } & \text { reactions: }\end{array}\) \(\mathrm{A} \stackrel{K_{1}=0.02 \mathrm{~min}^{-1}}{\longrightarrow} \mathrm{B} \stackrel{K_{2}=0.02 \mathrm{~min}^{-1}}{\longrightarrow} \mathrm{C}\), the initial concentration of 'A' was \(0.2 \mathrm{M}\) and initially 'B' and 'C' were absent. The time at which the concentration of ' \(\mathrm{B}\) ' becomes maximum and the maximum concentration of ' B' are, respectively, (a) \(50 \mathrm{~min},\left(\frac{0.2}{e}\right) \mathrm{M}\) (b) \(50 \mathrm{~min}, 0.2 \mathrm{M}\) (c) infinite, \(0.2 \mathrm{M}\) (d) \(25 \mathrm{~min},\left(\frac{0.2}{e}\right) \mathrm{M}\)

Short answer

None of the options provided correctly match the calculated results. However, option (a) is closely related but slightly inaccurate with the concentration given as \(\left(\frac{0.2}{e}\right) \mathrm{M}\).

Step-by-step solution

Understanding the reaction sequence

In the given sequential reactions, species A converts to B at first-order kinetics with rate constant \(K_1\), and subsequently, B converts to C also at first-order kinetics with rate constant \(K_2\). Note that both rate constants are equal (0.02 min^-1). The concentration of A starts at 0.2 M, while B and C are initially absent.

Determining the rate equations

The rate expressions for the reactions can be written as: \(\frac{d[A]}{dt} = -K_1[A]\) and \(\frac{d[B]}{dt} = K_1[A] - K_2[B]\). Since \(K_1 = K_2\), we will reach a point where the formation rate of B equals its consumption rate, which is when [B] is at its maximum.

Finding the time when concentration of B is maximum

When \(\frac{d[B]}{dt} = 0\), which implies \(K_1[A] = K_2[B]\), the concentration of B is at a maximum. Since \(K_1 = K_2\), this simplifies to \([A] = [B]\) at the maximum of B. Using the integrated rate law for a first-order reaction \([A] = [A]_0e^{-K_1t}\), we set \([A] = [A]_0/2\) to solve for time (t), giving \( t = \frac{\ln 2}{0.02} = 50 \min\).

Calculating maximum concentration of B

Since the reaction is first-order and the rate constant is the same for both steps, the maximum concentration of B will be when \([A] = [B]\). At the time when we found B to be at its maximum, half of A has been converted to B. Therefore, the maximum concentration of B will be \([B]_\max = [A]_0/2 = 0.1 \text{M}\). However, when comparing with the options, the choices suggest using \([B]_\max = \frac{[A]_0}{e}\), which indicates a misinterpretation. Since we're considering B at its maximum for first-order kinetics with equal rate constants, the exact calculation should be \(0.1 \text{M}\) and not \(\frac{0.2}{e}\).

Comparing with given options

Comparing our calculated values with the given options, none of the options correctly matches the calculated time and maximum concentration of B. However, option (a) closely relates to the approach but uses \(\frac{0.2}{e}\) for the concentration, which is not obtained from our calculation.

Key concepts

First-Order Kinetics

Understanding first-order kinetics is crucial when studying how chemical reaction rates depend on the concentration of the reactants. In first-order reactions, the rate at which the reactant is consumed is directly proportional to its concentration. This means that as the reactant's concentration decreases, so does the rate at which it is used up.

The rate equation for a first-order reaction can be written as \( \frac{d[A]}{dt} = -k[A] \) where \( [A] \) represents the concentration of the reactant A, \( t \) is time, and \( k \) is the first-order rate constant, which has units of inverse time (e.g., min^-1 or s^-1).

A key characteristic of first-order kinetics is that the time required for the concentration of the reactant to drop to half its initial value, known as the half-life (\( t_{1/2} \) ), is independent of the initial concentration. The half-life can be calculated using the formula \( t_{1/2} = \frac{\ln 2}{k} \) which reflects that consistent ratio.

Reaction Rate Equations

Reaction rate equations are mathematical expressions that describe the rate of change in concentration of reactants or products over time during a chemical reaction. These equations provide a quantitative way of representing the dynamics of a chemical reaction.

For sequential first-order reactions like the one described in the exercise, we have two separate rate equations for the two steps: \( \frac{d[A]}{dt} = -K_1[A] \) for the conversion of A to B and \( \frac{d[B]}{dt} = K_1[A] - K_2[B] \) for the conversion of B to C. Notice that the formation rate of B includes the term \( K_1[A] \) and the consumption rate of B includes \( -K_2[B] \) reflecting the processes that are respectively adding to and subtracting from the concentration of B.

Equating the two rates gives the point in time at which the concentration of B reaches its maximum value, as the rate at which B is produced is equal to the rate at which it is consumed. It is at this exact point that we can utilize integrated rate laws to find the concentrations of the reactants and products at different times.

Integrated Rate Law

The integrated rate law is a powerful tool for determining the concentrations of reactants or products at any given time. It extrapolates from the rate of the reaction to provide an equation that relates concentration to time directly.

For first-order reactions, the integrated rate law is expressed as \( [A] = [A]_0e^{-kt} \) where \( [A]_0 \) is the initial concentration and \( e \) is the base of the natural logarithm. This equation implies that the plot of \( \ln[A] \) versus time \( t \) will be a straight line with a slope of \( -k \) for a first-order reaction.

In the context of the exercise, we use the integrated rate law to find the time when the concentration of B is maximum by recognizing the point at which \( [A] \) equals \( [B] \) since \( K_1 = K_2 \) . However, there seems to be a discrepancy in the solution with respect to the given options, as it suggests an error in the interpretation of the integrated rate law. The correct approach would involve solving the integrated rate equation with respect to the known rate constants and initial concentrations. This highlights the significance of mastering the integrated rate law as part of understanding chemical kinetics.