Question

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in \(\mathrm{A}\) and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\). An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} M\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{~s}\) has elapsed assuming \([\mathrm{B}]_{0}=0\).

Short answer

a. The integrated rate law for this zero-order reaction is: \([A] = [A]_0 - kt\) b. The half-life for this reaction is: \(t_{1/2} ≈ 0.01s\) c. The concentration of B after \(5.0 \times 10^{-3} s\) has elapsed is: \([B] = 9.0 \times 10^{-4} M\)

Step-by-step solution

Integrated rate Law

For a zero-order reaction, the rate law is given by: \( rate = k[A]^0 = k\) The relationship between the rate of reaction and the concentration of reactants is given by the integrated rate law. For a zero-order reaction, the integrated rate law is derived as: \( [A] = [A]_0 - kt \) where [A] is the concentration of A at time t, [A]₀ is the initial concentration of A, k is the rate constant, and t is the time in seconds.

Half-life calculation

Half-life (t₁/₂) is the amount of time it takes for the concentration of a reactant to decrease to one-half of its original concentration. For the half-life of a zero-order reaction, the relation is given as: \( t_{1/2} = \frac{[A]_0}{2k} \) Using the given values, we can calculate the half-life: \( [A]_0 = 1.0 \times 10^{-3} M \) \( k = 5.0 \times 10^{-2} \frac{mol}{L \cdot s} \) \( t_{1/2} = \frac{1.0 \times 10^{-3} M}{2(5.0 \times 10^{-2} \frac{mol}{L \cdot s})} \) Now we calculate the half-life: \( t_{1/2} ≈ 0.01s \)

Calculate concentration of B

To find the concentration of B after the given time, we first need to find the concentration of A at that time using the integrated rate law. \( t = 5.0 \times 10^{-3} s \) \( [A] = [A]_0 - kt \) Substituting the given values: \( [A] = 1.0 \times 10^{-3} M - (5.0 \times 10^{-2} \frac{mol}{L \cdot s})(5.0 \times 10^{-3}s) = 1.0 \times 10^{-4} M \) Since the stoichiometry of the reaction is 1:1 it means that for every mole of A consumed, 1 mole of B is produced. Initially, [B]₀ was given as 0. As a result, we can calculate the concentration of B using [B]=[A]₀-[A]. \( [B] = [A]_0 - [A] = 1.0 \times 10^{-3} M - 1.0 \times 10^{-4} M = 9.0 \times 10^{-4} M \) So, the concentration of B after \(5.0 \times 10^{-3} s\) has elapsed is \(9.0 \times 10^{-4} M\).

Key concepts

Integrated Rate Law

In chemistry, understanding the relationship between the concentration of reactants and time is pivotal, especially in zero-order reactions. For a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. This means the reaction proceeds at a constant rate. The integrated rate law for a zero-order reaction is expressed as:\[ [A] = [A]_0 - kt \]

• \([A]\) is the concentration of reactant \(A\) at time \(t\).
• \([A]_0\) is the initial concentration of reactant \(A\).
• \(k\) represents the rate constant.
• \(t\) is the elapsed time.

This equation tells us that the concentration of \(A\) decreases linearly over time. This linear relationship is a distinctive feature of zero-order kinetics. The rate does not slow down as \(A\) is consumed, because it does not depend on its concentration. For practical reasons, it's crucial to accurately determine \(k\) to predict how quickly a reaction progresses under certain conditions.

Half-Life Calculation

The concept of half-life is key to understanding how long a reactant will last in a given reaction. The half-life \(t_{1/2}\) for a zero-order reaction is unique because it decreases as the initial concentration decreases. The half-life is calculated using the formula:\[ t_{1/2} = \frac{[A]_0}{2k} \]Here, the half-life is directly proportional to the initial concentration and inversely proportional to the rate constant.

• If you begin with a higher concentration of \(A\), the half-life will be longer.
• A larger rate constant \(k\) leads to a shorter half-life, meaning the reaction is faster.

For the given example, substituting the values, we find \(t_{1/2} ≈ 0.01s\). This tells us that within a very brief period, half of the initial concentration of \(A\) would be consumed. This quick depreciation is a common technique to measure reaction speed, central to catalytic processes.

Rate Constant

The rate constant \(k\) in kinetics forms a core consideration, and for zero-order reactions, it defines the rate at which the reaction proceeds. In our example, \(k = 5.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). This constant is not influenced by the concentration of reactants in a zero-order reaction but it does vary with temperature.

• Unlike first or second-order reactions, the magnitude of \(k\) ensures that every second, a set number of moles of \(A\) are transformed into products.
• It's crucial to keep in mind that \(k\) is temperature-dependent; hence, changing the temperature alters the rate regardless of state concentrations.

A known \(k\) is essential for predicting how much reactant will remain after a specific time. The rate constant enables calculations not only for the duration of reactions but also for scaling purposes in industrial processes where control over reaction time is critical.