Question

A certain reaction has the following general form: $$\mathrm{aA}⟶\mathrm{bB}$$ At a particular temperature and $[\mathrm{A}{]}_0=2.00\times {10}^{-2}M,$ con- centration 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 A to decrease to $2.50\times {10}^{-3}M?$

Short answer

a. Rate law: $\text{rate}=k[\mathrm{A}]$, integrated rate law: $\ln \frac{[\mathrm{A}]}{[\mathrm{A}{]}_0}=-kt$, and rate constant: $2.97\times {10}^{-2}{\min }^{-1}$. b. Half-life: 23.4 minutes. c. Time required for concentration to decrease: 64.8 minutes.

Step-by-step solution

Identify the order of the reaction

Since a plot of $\ln [\mathrm{A}]$ versus time gives a straight line, the reaction follows first-order kinetics. A first-order reaction is characterized by having its rate dependent only on the concentration of one reactant raised to the power of 1.

Write the rate law and integrated rate law expressions

The rate law for a first-order reaction is given by: $$\text{rate}=k[\mathrm{A}]$$ where $k$ is the rate constant and $[\mathrm{A}]$ is the concentration of the reactant A. The integrated rate law for a first-order reaction is given by: $$\ln \frac{[\mathrm{A}]}{[\mathrm{A}{]}_0}=-kt$$ where $[\mathrm{A}{]}_0$ is the initial concentration and $t$ is the time.

Determine the value of the rate constant

We are given the slope of a plot of $\ln [\mathrm{A}]$ versus time, which is:$-2.97\times {10}^{-2}{\min }^{-1}$. Since the integrated rate law expression can be rewritten as: $$\ln [\mathrm{A}]=\ln [\mathrm{A}{]}_0-kt$$ The slope of the line when plotting $\ln [\mathrm{A}]$ against $t$ is equal to the negative of the rate constant: $$k=-(-2.97\times {10}^{-2}{\min }^{-1})=2.97\times {10}^{-2}{\min }^{-1}$$

Calculate the half-life for the reaction

For a first-order reaction, the half-life is given by the formula: $$t_{1/2}=\frac{\ln 2}{k}$$ Using the rate constant calculated in the previous step, we can find the half-life: $$t_{1/2}=\frac{\ln 2}{2.97\times {10}^{-2}{\min }^{-1}}=23.4\text{ min}$$

Determine the time required for the concentration to decrease to the given value

For this step, we need to find the time required for the concentration of A to decrease to $2.50\times {10}^{-3}M$. Using the integrated rate law, we can solve for the time: $$\ln \frac{2.50\times {10}^{-3}}{2.00\times {10}^{-2}}=-(2.97\times {10}^{-2}{\min }^{-1})t$$ Solve for $t$: $$t=\frac{\ln (2.50\times {10}^{-3})-\ln (2.00\times {10}^{-2})}{-2.97\times {10}^{-2}{\min }^{-1}}=64.8\text{ min}$$ In conclusion: a. The rate law is given by $\text{rate}=k[\mathrm{A}]$, the integrated rate law is given by $\ln \frac{[\mathrm{A}]}{[\mathrm{A}{]}_0}=-kt$, and the rate constant is $2.97\times {10}^{-2}{\min }^{-1}$. b. The half-life for the reaction is 23.4 minutes. c. It takes 64.8 minutes for the concentration of A to decrease to $2.50\times {10}^{-3}M$.

Key concepts

First-Order Reaction

First-order reactions are special because the rate of the reaction depends linearly on the concentration of one reactant. In the general chemical equation, $\mathrm{aA}⟶\mathrm{bB}$, the reaction is first-order with respect to A. This means that the rate at which A turns into B is directly proportional only to the concentration of A raised to the power of one.

First-order reactions are quite common in chemistry, especially in processes like radioactive decay and the decomposition of simple molecules. They are characterized by the fact that they produce a straight line when plotting the natural logarithm of the concentration of a reactant versus time in a graph. Remember, if you see a straight line in such a plot, you've likely got a first-order reaction on your hands!

• Rate is only dependent on the reactant concentration.
• Graph of $\ln [A]$ vs. time is linear for first-order reactions.

Rate Law

The rate law expresses how the rate of a reaction is related to the concentration of its reactants. For a first-order reaction, the rate law takes a very simple form:
$$extrate=k[A]$$
Individual components of the rate law are important to understand: - $k$ is the rate constant, and its value gives us an idea of the speed of the reaction. - $[A]$ represents the concentration of the reactant A.

To determine the rate constant from experimental data, you often plot $\ln [A]$ versus time. The slope of this plot for a first-order reaction is equal to $-k$. Don't forget, the units of $k$ in first-order reactions are inverse time, such as ${\min }^{-1}$.

• Rate law tells us how reaction rate changes with concentration.
• Rate constant $k$ is derived from experimental data and is crucial to understanding reaction dynamics.

Integrated Rate Law

The integrated rate law provides a relationship between concentration and time for a first-order reaction. It takes the form:
$$\ln \frac{[A]}{[A{]}_0}=-kt$$
In this equation, - $[A{]}_0$ is the initial concentration of A, - $[A]$ is the concentration at time $t$, - $k$ is the rate constant, - $t$ is time.

By rearranging this equation, you can solve for time or concentration, depending on the data available. This equation becomes super handy when trying to predict how much reactant will remain at a certain time or how long you need to wait for a given amount to react.

The neat thing about the integrated rate law for first-order reactions is its linearity when plotted as $\ln [A]$ vs. $t$, making data analysis straightforward.

Half-Life Calculation

In first-order kinetics, the half-life of a reaction is the time required for the concentration of the reactant to reach half of its original value. The half-life of first-order reactions is not dependent on the initial concentration, which makes it unique compared to other orders of reaction.

The formula used to calculate the half-life for first-order reactions is: $$t_{1/2}=\frac{\ln 2}{k}$$
Since $\ln 2$ is a constant approximately equal to 0.693, you can simply divide by the rate constant $k$ to find the half-life. This results in a consistent half-life duration regardless of how much reactant you start with. It provides a time measure that is often used to summarize the speed at which a reaction proceeds.

• Half-life in first-order reactions is a constant value, independent of concentration.
• It simply involves the rate constant $k$ for its calculation.