Question

The reaction \(A+B \longrightarrow\) products is first order in \(A\) first order in \(\mathrm{B},\) and second order overall. Consider that the starting concentrations of the reactants are \([\mathrm{A}]_{0}\) and [ \(\mathrm{B}]_{0},\) and that \(x\) represents the decrease in these concentrations at the time \(t .\) That is, \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-x\) and \([\mathrm{B}]_{t}=[\mathrm{B}]_{0}-x .\) Show that the integrated rate law for this reaction can be expressed as shown below. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$

Short answer

The integrated rate law for the given reaction can be obtained by expressing the rate of reaction in terms of concentrations of the reacting species, substituting for concentrations at time \(t\), and integrating the resulting differential equation. The result can be manipulated to match the given integrated rate law expression.

Step-by-step solution

Understanding the Given Reaction Order and Composition of Reagents

The given reaction involves two reagents A and B, and is first order in both, which makes it an overall second order reaction. The decrease in concentrations of both reactants over time \(t\) is represented by \(x\). Therefore, at time \(t\), the concentrations of A and B will be \([A]_t = [A]_0 - x\) and \([B]_t = [B]_0 - x\), respectively.

Express the Rate of Reaction

Given that the reaction is first order with respect to both A and B, the rate of reaction can be expressed as follows: \(-\frac{d[A]}{dt} = -\frac{d[B]}{dt} = k[A][B]\) . Here, \(k\) is the rate constant.

Substitute for Concentrations at Time t

Substitute \([A]_t = [A]_0 - x\) and \([B]_t = [B]_0 - x\) into the rate equation to express the rates in terms of \(x\): \(-\frac{dx}{dt} = k([A]_0 - x)([B]_0 - x)\).

Integrate the Rate Equation

The equation obtained in Step 3 is a differential equation that can be integrated to get the time evolution of the concentrations. This is a non-linear differential equation which can be solved by separation of variables and integration. Before the integration, rearrange the equation to isolate terms involving \(x\) on one side: \(\frac{dx}{([A]_0 - x)([B]_0 - x)} = -k dt\). Integrate both sides of the equation integrating lhs from \(0\) to \(x\) and rhs from \(0\) to \(t\).

Simplify the Integrated Equation

Simplify the result obtained from the integration and rearrange into the required form to get: \(\ln \frac{[A]_0 [B]_t}{[B]_0 [A]_t} = ([B]_0 - [A]_0) k t \).

Key concepts

Reaction Order

Understanding the reaction order is essential for determining how reactant concentrations affect the rate of a chemical reaction. In the given exercise, we have a reaction where A and B are both first order.
This means that the rate of the reaction depends linearly on the concentration of each reactant:

• First order in A implies that if you double the concentration of A, the reaction rate doubles.
• First order in B means the same linear relationship holds true for B.

When combined, the reaction is overall second order. This is because you sum the orders with respect to each reactant to get the overall reaction order:

• Order with respect to A: 1
• Order with respect to B: 1
• Total reaction order: 1 + 1 = 2

By knowing the reaction order, you can understand how changes in concentration impact the rate and help in formulating the correct rate law.

Differential Equation

A differential equation describes how a quantity changes in relation to another, such as time.
For our reaction, the rate at which reactants A and B decrease is expressed as a differential equation:
-\(-\frac{d[A]}{dt} = k[A][B]\).
-\(-\frac{d[B]}{dt} = k[A][B]\).
This equation tells us the rate of change in concentration of A and B over time. Here, the rate constant \(k\) is crucial in defining the speed of the reaction.
In this exercise, since concentrations at time \(t\) are defined as \([A]_t = [A]_0 - x\) and \([B]_t = [B]_0 - x\), we substitute these into the differential equations.
This results in a modified equation specifically tailored to handle the integration in the following steps: \(-\frac{dx}{dt} = k([A]_0 - x)([B]_0 - x)\).
Solving this differential equation through integration provides us with crucial insights while developing the rate law.

Rate Constant

The rate constant, symbolized as \(k\), is a unique value that helps determine the speed of a chemical reaction. This factor is intrinsic to every reaction and remains constant under fixed conditions like temperature and pressure.
The rate constant \(k\) connects the rate of reaction to the concentrations of the reactants:

• It is a proportionality constant in the rate law.
• Higher values of \(k\) indicate a faster reaction.
• Its value varies with temperature, necessitating specification for different conditions.

In our reaction's differential equation:
-\(-\frac{dx}{dt} = k([A]_0 - x)([B]_0 - x)\),
\(k\) helps quantify exactly how quickly the concentrations of A and B decrease over time. Understanding this constant is paramount in predicting reaction kinetics and establishing how quickly equilibrium can be reached.

Separation of Variables

Separation of variables is a mathematical method used to solve differential equations. It involves rearranging the equation such that each variable and its differential are on separate sides of the equation.
This aids in solving by making it possible to integrate each side independently. For our reaction, we rearrange the equation derived from the rate expression:

• Separate variables: \(\frac{dx}{([A]_0 - x)([B]_0 - x)} = -k dt\).
• Each side now only involves terms of one variable (\(x\) or \(t\)).
• This equation can now be integrated to find a relationship between concentration changes and time.

Through integration from initial conditions to those at time \(t\), we obtain an expression for the concentrations of A and B over time, eventually leading to the integrated rate law.
By employing separation of variables, we effectively solve for how reactants diminish, giving us the broader picture of reaction kinetics.