Question

Using calculus, derive the equation for (a) the concentration-time relation for a second-order reaction (see Table 11.2). (b) the concentration-time relation for a third-order reaction, \(\mathrm{A} \longrightarrow\) product.

Short answer

Question: Derive the concentration-time relations for (a) a second-order reaction and (b) a third-order reaction using calculus. Answer: (a) For a second-order reaction, the concentration-time relation is given by \([A] = \frac{[A]_0}{1+k[A]_0 t}\). (b) For a third-order reaction, the concentration-time relation is given by \([A] = \frac{1}{\sqrt{2kt + \frac{1}{[A]_0^2}}}\).

Step-by-step solution

(a) Derivation of the concentration-time relation for a second-order reaction)

To derive the equation for a second-order reaction, we first write down the rate law: Rate \(= k[A]^2\) The reaction rate can be expressed as the change in concentration of A (\([A]\)) over time, i.e., \(-\frac{d[A]}{dt}\). Therefore, the rate law equation for a second-order reaction can be written as: \(-\frac{d[A]}{dt} = k[A]^2\) Now, we need to solve this differential equation with respect to time: \(\frac{d[A]} {[A]^2} = -k\, dt\) Integrate both sides of the equation, with initial concentration \([A]_0\) and final concentration \([A]\): \(\int\limits_{[A]_0}^{[A]}\frac{d[A]}{[A]^2} = -k\int\limits_{0}^{t}dt\) After integration, we get: \(\bigg[-\frac{1}{[A]}\bigg]_{[A]_0}^{[A]} = -k\bigg[t\bigg]_{0}^{t}\) Evaluate the limits: \(-\frac{1}{[A]} + \frac{1}{[A]_0} = -kt\) Rearranging the equation for \([A]\) gives the concentration-time relation for a second-order reaction: \([A] = \frac{[A]_0}{1+k[A]_0 t}\)

(b) Derivation of the concentration-time relation for a third-order reaction

For a third-order reaction A \(\longrightarrow\) product, the rate law is given as: Rate \(= k[A]^3\) Similar to the previous case, let's express the reaction rate as the change in concentration of A (\([A]\)) over time: \(-\frac{d[A]}{dt} = k[A]^3\) To find the concentration-time relation, we need to solve this differential equation with respect to time: \(\frac{d[A]}{[A]^3} = -k\, dt\) Integrate both sides of the equation, with initial concentration \([A]_0\) and final concentration \([A]\): \(\int\limits_{[A]_0}^{[A]}\frac{d[A]}{[A]^3} = -k\int\limits_{0}^{t}dt\) After integration, we get: \(\bigg[\frac{1}{2[A]^2}\bigg]_{[A]_0}^{[A]} = -k\bigg[t\bigg]_{0}^{t}\) Evaluate the limits: \(\frac{1}{2[A]^2} - \frac{1}{2[A]_0^2} = -kt\) Rearranging the equation for \([A]\) gives the concentration-time relation for a third-order reaction: \([A] = \frac{1}{\sqrt{2kt + \frac{1}{[A]_0^2}}}\)

Key concepts

Second-Order Reaction

In chemical kinetics, a second-order reaction refers to a process where the rate of the reaction is proportional to the square of the concentration of a single reactant, or the product of the concentrations of two reactants. The mathematical expression for a second-order reaction rate law is:
\(-\frac{d[A]}{dt} = k[A]^2\),
where \(k\) is the rate constant, \([A]\) is the concentration of the reactant, and \(t\) is time. To determine how the concentration changes over time, one must solve this differential equation.

Solving the Second-Order Differential Equation

To solve this type of differential equation, separation of variables is used, leading to integration over the limits of initial concentration \([A]_0\) and final concentration \([A]\) against time. The solution provides the concentration-time relation for a second-order reaction, which is given by:
\([A] = \frac{[A]_0}{1 + k[A]_0 t}\).
This equation describes how the concentration of the reactant decreases over time in a second-order reaction. It is characterized by a plot of \(1/[A]\) versus time being a straight line, with the slope of the line related to the rate constant \(k\).

Third-Order Reaction

When dealing with a third-order reaction, the situation becomes a bit more complex. A third-order reaction rate is proportional to the cube of the concentration of a single reactant, or some combination of concentrations when multiple reactants are involved. The rate law for a homogenous third-order reaction involving a single reactant is written as:
\(-\frac{d[A]}{dt} = k[A]^3\).
The concentration-time relationship for a third-order reaction requires solving a more complex differential equation compared to the second-order reaction. Similar to the second-order, the method of separation of variables followed by integration is used here. The resulting relation is:
\([A] = \frac{1}{\sqrt{2kt + 1/[A]_0^2}}\).
This equation shows the change in the concentration of the reactant over time. Unlike second-order, third-order reactions do not yield a straight line when plotting \(1/[A]^2\) versus time, but the relationship can still be graphically determined and the rate constant \(k\) can be extracted from the curve.

Rate Law

The rate law is a fundamental concept in the study of chemical kinetics. It establishes a quantitative relationship between the rate of a chemical reaction and the concentrations of the reactants. The law is expressed by the equation:
Rate \(= k[A]^{n}[B]^{m}...\),
with \(k\) being the reaction rate constant, \([A]\), \([B]\), etc., are the molar concentrations of the reactants, and \(n\), \(m\), etc. are the orders of the reaction concerning those reactants. The overall reaction order is the sum of the orders with respect to each reactant (\(n+m+...\)).

Understanding the Reaction Order

The reaction order is determined empirically and gives insight into the mechanism of the reaction. For example, a zero-order reaction rate remains constant irrespective of the concentration of the reactants. In first-order reactions, the rate is directly proportional to the reactant concentration. As seen, second and third-order reactions have their rates proportional to the square and cube of reactant concentrations, respectively, leading to different integrated rate laws and concentration-time relations.

Differential Equations

Differential equations are powerful tools in a variety of scientific disciplines, including chemistry, physics, engineering, and economics. They are equations that involve functions and their derivatives, articulate relationships involving rates of change, and are used to describe various phenomena. In the context of chemical kinetics, differential equations are used to describe how reaction rates and concentrations change over time.

Application in Chemical Kinetics

Differential equations in chemical kinetics are solved to obtain integrated rate laws, which provide formulas for the concentration of reactants or products as a function of time. These solutions are crucial for predicting the progress of a reaction, understanding reaction mechanisms, and designing chemical processes. The complexity of solving these equations can vary from simple separable forms, solvable through direct integration, as seen in first and second-order reactions, to much more complicated forms that may require special techniques or numerical methods for third-order and other higher-order reactions.