Problem 47 - Chemistry An Atoms First Approach

Question

The reaction $$A \longrightarrow B+C$$ is known to be zero order in 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 \([\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M}\) 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 \([\mathrm{B}]_{0}=0\)

Step-by-Step Solution

Answer

a. The integrated rate law for this zero-order reaction is \([A](t) = [A]_0 - kt\). b. The half-life for this reaction is \(t_{1/2} = 1.0 \times 10^{-4}\,\mathrm{s}\). c. The concentration of B after \(5.0 \times 10^{-3}\,\mathrm{s}\) has elapsed is \([B] = 2.5 \times 10^{-4}\,\mathrm{M}\).

Step 1: Write the integrated rate law for a zero-order reaction

According to the definition, the rate of a zero-order reaction is independent of the concentration of the reactant, which means: \(rate = k\) To obtain the integrated rate law, we first recognize that this is a differential equation for the concentration of A: \(\frac{d[A]}{dt} = -k\) We integrate both sides of the equation: \(\int_{[A]_0}^{[A]} d[A] = -\int_0^t k dt\) The result of the integration is: \([A]-[A]_0 = -kt\) Now, we can rearrange the equation to express the concentration of A as a function of time: \([A](t) = [A]_0 - kt\) This is the integrated rate law for a zero-order reaction.

Step 2: Calculate the half-life for the reaction

Half-life (t₁/₂) is the time required for the concentration of the reactant to decrease to one-half of its initial value. For a zero-order reaction, the half-life can be found using the integrated rate law: \([A]_0/2 = [A]_0 - kt_{1/2}\) Rearrange the equation to solve for t₁/₂: \(t_{1/2} = \frac{[A]_0}{2k}\) Now, substitute the values provided in the exercise: \(t_{1/2} = \frac{1.0 \times 10^{-3}\,\mathrm{M}}{2 \times 5.0 \times 10^{-2}\,\mathrm{mol / L\cdot s}}\) Calculate the half-life: \(t_{1/2} = 1.0 \times 10^{-4}\,\mathrm{s}\) The half-life for this reaction is 1.0 × 10⁻⁴ s.

Step 3: Calculate the concentration of B after a certain time has elapsed

We are asked to find the concentration of B after a given time, with the assumption that the initial concentration of B is 0. Since the stoichiometry of the reaction is 1:1, the amount of B produced is equal to the amount of A consumed. First, determine the concentration of A after 5.0 × 10⁻³ s using the integrated rate law: \([A](t) = [A]_0 - kt\) \([A](5.0 \times 10^{-3}\,\mathrm{s}) = 1.0 \times 10^{-3}\,\mathrm{M} - 5.0 \times 10^{-2}\,\mathrm{mol / L \cdot s} \cdot 5.0 \times 10^{-3}\,\mathrm{s}\) Calculate the concentration of A: \([A](5.0 \times 10^{-3}\,\mathrm{s}) = 7.5 \times 10^{-4}\,\mathrm{M}\) Now calculate the concentration of B produced, since B₀ = 0: \([B] = [A]_0 - [A](5.0 \times 10^{-3}\,\mathrm{s})\) \([B] = 1.0 \times 10^{-3}\,\mathrm{M} - 7.5 \times 10^{-4}\,\mathrm{M}\) Calculate the concentration of B: \([B] = 2.5 \times 10^{-4}\,\mathrm{M}\) The concentration of B after 5.0 × 10⁻³ s has elapsed is 2.5 × 10⁻⁴ M.

Key Concepts

Rate Constant

The rate constant is a fundamental parameter in the study of chemical kinetics, as it provides a measure of the speed at which a reaction occurs. In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. This means that the rate at which A transforms into products B and C is given by the rate constant, denoted as \( k \). In mathematical terms, the rate can be written simply as:
\[ \text{rate} = k \]
In the provided exercise, the value of \( k \) is given as \( 5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \) at \( 25^\circ \mathrm{C} \). This value remains the same throughout the reaction progress, regardless of how much reactant A is left. Understanding the rate constant is crucial because it allows us to predict how long a reaction will take to reach a certain point, such as the half-life, or to calculate concentrations of reactants or products at any given time.

Integrated Rate Law

For a zero-order reaction like the one being discussed, the integrated rate law provides a relationship between the concentration of reactant A and time. Once you've found that the differential rate equation is \( \frac{d[A]}{dt} = -k \), integrating this equation with respect to time gives us the integrated rate law for a zero-order reaction:
\[ [A](t) = [A]_0 - kt \]
This equation tells us that the concentration of A at any time \( t \) can be calculated by subtracting \( kt \) from the initial concentration \( [A]_0 \). In the context of the exercise, if you know the initial concentration of A and the rate constant \( k \), you can predict the concentration of A at any future moment. This powerful tool is vital for chemists who need to manage reaction conditions and timeframes effectively.

Reaction Half-Life

Half-life, symbolized as \( t_{1/2} \), is a term that describes the time required for the concentration of a reactant to decrease to half its initial value. The concept of half-life is widely used because it offers a straightforward way of understanding how quickly a reaction proceeds. For a zero-order reaction, the half-life can be calculated with ease from the integrated rate law by setting \( [A](t) = \frac{[A]_0}{2} \):
\[ t_{1/2} = \frac{[A]_0}{2k} \]
This equation indicates that for zero-order reactions, the half-life is directly proportional to the initial concentration of the reactant and inversely proportional to the rate constant. Hence, as seen in the exercise, if you double the initial concentration, the half-life will double as well. This contrasts with reactions of other orders, where half-life values are not always this directly connected to initial concentrations.

Concentration of Reactants

The concentration of reactants plays a pivotal role in chemical kinetics. For most reaction orders, the rate is affected by changes in the concentrations of the reactants. However, zero-order reactions are unique because their rates remain constant even as reactant concentrations decrease. This distinction is key to understanding the underlying behavior of these reactions and in predicting the outcomes. In our example, knowing that the rate of the reaction is independent of \( [A] \), one can easily calculate how much of A will be left at any given time through the equation \( [A](t) = [A]_0 - kt \). This also allows us to find the concentration of any product formed at a certain time, by knowing the stoichiometry of the reaction which in this case suggests that the production of B matches the consumption of A. Thus, knowing \( [A]_0 \), the rate constant \( k \), and the elapsed time, one can accurately deduce the concentration of reactants at any stage of the reaction as well as the amount of products formed.