Question

Consider a reaction of the type aA \(\longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 .\) s, what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$ \text {Rate} =\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3} $$

Short answer

The time for the second half-life of the termolecular reaction can be found using the integrated rate law and the rate constant, k, which was determined from the given first half-life. After deriving the integrated rate law and solving for the time for the second half-life, we find the following expression for the second half-life: $$ t_2 = \frac{80\,\mathrm{s} \left(\frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2}\right)}{1 - \frac{1}{[0.5\mathrm{A}]_0^2}} $$ By calculating this expression, we can find the time for the second half-life of the reaction.

Step-by-step solution

Derive the integrated rate law

Start with the given differential rate law for a termolecular reaction: $$ -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^{3} $$ Separate the variables: $$ -\frac{d[\mathrm{A}]}{[\mathrm{A}]^{3}} = k \, dt $$ Integrate both sides: $$ \int_{[\mathrm{A}]_0}^{[\mathrm{A}]}\frac{-d[\mathrm{A}]}{[\mathrm{A}]^{3}} = \int_{0}^{t}k \, dt $$ Calculate the integral: $$ \frac{1}{2[\mathrm{A}]^2} - \frac{1}{2[\mathrm{A}]_0^2} = kt $$ Rearrange the equation to solve for A: $$ \frac{1}{[\mathrm{A}]^2} - \frac{1}{[\mathrm{A}]_0^2} = 2kt $$ This is the integrated rate law for a termolecular reaction.

Find the rate constant, k

For the first half-life, 50% of A would be remaining, so we have: $$ \frac{1}{[0.5\mathrm{A}]_0^2} - \frac{1}{[\mathrm{A}]_0^2} = 2k(40\,\mathrm{s}) $$ Solving for k: $$ k = \frac{1 - \frac{1}{[0.5\mathrm{A}]_0^2}}{80\,\mathrm{s}} $$

Find the second half-life

Now we need to find the time for the second half-life, when 75% of A has reacted (25% remaining): $$ \frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2} = 2k(t_2) $$ Now, plug in the k value we found earlier: $$ \frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2} = \left(\frac{1 - \frac{1}{[0.5\mathrm{A}]_0^2}}{80\,\mathrm{s}}\right)(2t_2) $$ Solve for t2, the time for the second half-life: $$ t_2 = \frac{80\,\mathrm{s} \left(\frac{1}{[0.25\mathrm{A}]_0^2} - \frac{1}{[0.5\mathrm{A}]_0^2}\right)}{1 - \frac{1}{[0.5\mathrm{A}]_0^2}} $$ After computing this expression, we will find the time for the second half-life of the reaction.

Key concepts

Integrated Rate Law

In the study of chemical kinetics, understanding how concentrations change over time is crucial. The **integrated rate law** is a mathematical expression that connects concentrations of reactants to the time variable. It is particularly useful when you know the order of the reaction and want to predict how long the reaction will take under certain conditions.
To derive the integrated rate law from the differential rate law, we start with the differential rate equation that describes how the concentration of a reactant changes with time. For a termolecular reaction like the one given, \[-\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]^{3}\].

Here's how it works:

• **Separate Variables:** This step involves moving all terms involving \([\mathrm{A}]\) to one side of the equation, resulting in, \(-\frac{d[\mathrm{A}]}{[\mathrm{A}]^{3}} = k \, dt\).
• **Integration:** The next step is to integrate both sides. This involves integrating \(\int \frac{-d[\mathrm{A}]}{[\mathrm{A}]^{3}}\) and \(\int k \, dt\), yielding the integrated rate law: \(\frac{1}{2[\mathrm{A}]^2} - \frac{1}{2[\mathrm{A}]_0^2} = kt\).

Thus, the integrated equation allows us to calculate concentrations at any given time, using the known constant \(k\) and initial conditions.

Differential Rate Law

A **differential rate law** provides an expression for the rate of a chemical reaction in terms of the concentration of the reactants. This form of the rate law is crucial for deciphering the kinetics of a reaction, including understanding how changes in concentration affect the rate.
For termolecular reactions, such as the hypothetical reaction \(aA \rightarrow \text{products}\), the differential rate law is \(\frac{-d[\mathrm{A}]}{dt} = k[\mathrm{A}]^{3}\).
Here's a breakdown of its components:

• **Dependency on Concentration:** The term \([\mathrm{A}]^{3}\) indicates that the rate depends on the cube of the concentration of \([\mathrm{A}]\). This shows a third-order dependence, which is quite rare in real-world reactions.
• **Rate Constant (k):** This constant represents the reaction rate at a unit concentration and is specific to each reaction. The value of \(k\) is influenced by factors like temperature and the presence of a catalyst.

The differential rate law helps in deducing the integrated rate law, which integrates these infinitesimal changes over time to give a broader picture of the concentration changes during the reaction.

Half-Life Calculation

The **half-life** of a reaction is the time taken for half of the reactant to be consumed in the reaction. Understanding half-life is especially useful in predicting how long a reaction will take to reach a certain point, or how much reactant will remain after a given period.
For a termolecular reaction, half-life calculations become more complex due to the third-order kinetics involved. Here's how it can be approached:

• **First Half-Life:** Utilizing the integrated rate law, the first half-life can be calculated by recognizing that the concentration of reactant reduces to 50% of its initial value.
• **Second Half-Life:** The process is repeated, but now for 25% initial concentration, as 75% would have been consumed. It shows that higher order reactions like termolecular reactions lead to a change in half-life as the reaction progresses, unlike first-order reactions where the half-life is constant.

The formula derived provides the specific time for the reactant to reduce from 50% to 25%, emphasizing the unique progression of such a reaction. By using the constants calculated from earlier steps, we can compute specific times for successive half-lives, which are not equal due to the nature of higher order kinetics.