Question

A certain reaction has the following general form: $\mathrm{aA}⟶\mathrm{bB}$ At a particular temperature and $[\mathrm{A}{]}_0=2.00\times {10}^{-2}M$, concentration versus time data were collected for this reaction, and a plot of $\ln [\mathrm{A}]$ versus time resulted in a straight line with a slope value of $-2.97\times {10}^{-2}{\min }^{-1}$. a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of $\mathrm{A}$ to decrease to $2.50\times {10}^{-3}M?$

Short answer

a. The rate law is $Rate=k[A]$, the integrated rate law is $ln[A]=-kt+ln[A{]}_0$, and the rate constant $k=2.97\times {10}^{-2}{\min }^{-1}$. b. The half-life of the reaction is approximately 23.37 minutes. c. Approximately 85.94 minutes are required for the concentration of A to decrease to $2.50\times {10}^{-3}M$.

Step-by-step solution

Identifying the rate law and the integrated rate law for the reaction based on the graph

Since there is a straight line in a plot of ln[A] versus time, the reaction rate can be described by a first-order rate law: Rate = k[A] where k is the rate constant and [A] is the concentration of A at any time. The integrated rate law for a first-order reaction is: ln[A] = -kt + ln[A]₀ where [A]₀ is the initial concentration of A and t is the time elapsed.

Calculate the rate constant k using the given slope

Since the slope of the plot of ln[A] vs time is -2.97 × 10⁻² min⁻¹, we can equate the slope to -k and solve for k: k = 2.97 × 10⁻² min⁻¹

Calculate the half-life of the reaction

The half-life t₁/₂ of a first-order reaction is given by the formula: t₁/₂ = ln(2) / k Using the calculated k value, we can find the half-life: t₁/₂ = ln(2) / (2.97 × 10⁻² min⁻¹) t₁/₂ ≈ 23.37 minutes

Calculate the time required for [A] to decrease to 2.50 × 10⁻³ M

We can use the integrated rate law to find the time required for the concentration of A to decrease to a specific value: ln([A] / [A]₀) = -kt Using the given [A]₀, k, and target [A] value: ln((2.50 × 10⁻³ M) / (2.00 × 10⁻² M)) = -(2.97 × 10⁻² min⁻¹) * t Solving for t: t ≈ 85.94 minutes In summary: a. The rate law is Rate = k[A], the integrated rate law is ln[A] = -kt + ln[A]₀, and the rate constant k is 2.97 × 10⁻² min⁻¹. b. The half-life of the reaction is approximately 23.37 minutes. c. Approximately 85.94 minutes are required for the concentration of A to decrease to 2.50 × 10⁻³ M.

Key concepts

Rate law

Understanding the rate law is crucial when studying reaction kinetics. For a first-order reaction, the rate law is given by the expression $extRate=k[extA]$. This equation emphasizes how the reaction rate depends on the concentration of reactant A. Here, $k$ is the rate constant, which is specific to the reaction and conditions like temperature. It's important to note that in first-order reactions, the rate directly depends on the concentration of A. So, a change in concentration results in a proportional change in the reaction rate. Recognizing the order of the reaction, which is first-order in this case, helps in predicting how the concentration will vary over time during the reaction.

Integrated rate law

The integrated rate law for a first-order reaction provides a way to quantify how concentrations change over time. It is captured by the equation $extln[A]=-kt+extln[A{]}_0$. This form indicates the logarithmic relationship between the concentration of the reactant A and time. The parameters here include:

• $[A{]}_0$: the initial concentration of A
• $t$: the elapsed time
• $k$: the rate constant

Using the integrated rate law, chemists can predict the future concentrations of reactants. It also highlights the characteristic that a plot of $extln[A]$ versus time yields a straight line when the reaction is first-order. The slope of this line is equal to $-k$, and the intercept is $extln[A{]}_0$.

Rate constant

The rate constant $k$ is a critical parameter in the rate law that influences how quickly a reaction proceeds. For a first-order reaction, it can be directly related to the slope of a $extln[A]$ versus time plot. From the problem statement, we know this slope is $-2.97\times {10}^{-2}{\text{min}}^{-1}$, implying $k=2.97\times {10}^{-2}{\text{min}}^{-1}$. The value of $k$ gives insights into the reaction's sensitivity to changes in concentration. Higher values of $k$ suggest a faster reaction under the given conditions. In practical terms, $k$ helps to calculate other kinetic parameters, such as half-life and time for changes in concentration.

Half-life calculation

The half-life of a reaction, denoted as $t_{1/2}$, refers to the time required for the concentration of a reactant to decrease to half of its initial concentration. For first-order reactions, the half-life is given by the formula:$$t_{1/2}=\frac{\ln (2)}{k}$$The distinct feature of first-order reactions is that the half-life remains constant regardless of the initial concentration. This makes it particularly useful for predicting reaction progress over time. For instance, using the calculated rate constant $k=2.97\times {10}^{-2}{\text{min}}^{-1}$, the half-life is approximately $23.37\text{minutes}$. Understanding half-life helps in fields like pharmacology, where it is essential to understand how quickly drugs are metabolized.