Question

Two isomers (A and B) of a given compound dimerize as follows: $$ \begin{array}{l}{2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} A_{2}} \\ {2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2}}\end{array} $$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\) . In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C},\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} M\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} M\) It was found that after each reaction had progressed for \(3.00 \mathrm{min},[\mathrm{A}]=3.00[\mathrm{B}]\) . In this case the rate laws are defined as $$ \begin{array}{l}{\text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2}} \\ {\text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2}}\end{array} $$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 \(\mathrm{min}\) . b. Calculate the value of \(k_{2}\) . c. Calculate the half-life for the experiment involving A.

Short answer

In summary, after analyzing the given dimerization reactions and their kinetic data, we found the following results: a) The concentration of A₂ after 3 minutes is approximately 0.004035 M. b) The value of k₂ for the B dimerization reaction is approximately 0.330 L/mol·s. c) The half-life for the experiment involving A is approximately 400 seconds.

Step-by-step solution

Calculate the final concentration of A

First we will find the final concentration of A after 3 minutes using the rate law equation: Rate = -Δ[A] / Δt = k₁[A]² We are given the value of k₁ as 0.250 L/mol·s and we know the initial concentration of A as [A]₀ = 1.00 x 10⁻² M. Let the change in concentration of A be "x", then the final concentration of A will be ([A]₀ - x). We are also given that after 3 minutes, [A] = 3[B]. Let's substitute this information in the rate law equation and solve for x. \(0.250 = \frac{x}{(1.00 - x)^2 * 3.00 * 60s}\)

Solve the equation for x

Now we'll solve the equation obtained in step 1 for x: \(0.250 = \frac{x}{(1.00 - x)^2 * 180}\) \(45 = \frac{x}{(1.00 - x)^2}\)

Calculate final concentrations of A and B

Solving the equation in step 2, we get x ≈ 0.00807 M. Now we can find the final concentration of A ([A]f) after 3 minutes: [A]f = [A]₀ - x = 1.00 x 10⁻² M - 0.00807 M ≈ 0.00193 M Since [A] = 3[B], we can find the final concentration of B ([B]f): [B]f = [A]f / 3 ≈ 0.00193 M / 3 ≈ 0.00064 M

Calculate the concentration of A₂ after 3 minutes

Since 2 molecules of A react to form 1 molecule of A₂, the concentration of A₂ formed after 3 minutes will be half the change in A concentration (x) [A₂] = x / 2 ≈ 0.00807 M / 2 ≈ 0.004035 M

Calculate the value of k₂

Using the rate law equation for B, we can find the value of k₂: Rate = -Δ[B] / Δt = k₂[B]² Δ[B] = [B]₀ - [B]f = 2.50 x 10⁻² M - 0.00064 M ≈ 0.02436 M Rate = -Δ[B] / Δt ≈ -0.02436M / 180s = -(-1.36 x 10⁻⁴) M/s k₂ = Rate / [B]f² ≈ (-1.36 x 10⁻⁴ M/s) / (0.00064 M)² ≈ 0.330 L/mol·s

Calculate the half-life for the experiment involving A

To calculate the half-life (t₁/₂) for the experiment involving A, we use the following equation for second-order reactions: t₁/₂ = 1 / (k₁[A]₀) Taking the initial concentration of A and the value of k₁, we get: t₁/₂ ≈ 1 / (0.250 L/mol·s × 1.00 x 10⁻² M) ≈ 400 s So, the half-life for the experiment involving A is 400 seconds. To summarize: a. The concentration of A₂ after 3.00 minutes is approximately 0.004035 M. b. The value of k₂ is approximately 0.330 L/mol·s. c. The half-life for the experiment involving A is approximately 400 seconds.

Key concepts

Reaction Rate

Understanding the concept of reaction rate is crucial for solving problems in chemical kinetics. Reaction rate refers to the speed at which a chemical reaction occurs. Specifically, it describes how quickly reactants are converted into products over time. For the reactions described in the exercise, the reaction rate can be expressed as changes in concentration over time for each reactant.

For a general chemical reaction that involves a reactant A, the reaction rate is expressed as:

• Rate = -\(\frac{\Delta [A]}{\Delta t}\)

Here, \(\Delta [A]\) represents the change in concentration of A over the time interval \(\Delta t\). This expression gives us a quantitative measure of how fast the reaction is running.

In the given problem, we use rate laws which relate the reaction rate with the concentration of reactants. The rate laws for second-order reactions involve the square of the reactant concentration, as will be discussed in the next section.

Second-order Reaction

Second-order reactions are characterized by the rate of reaction being directly proportional to the square of the concentration of one reactant or the product of concentrations of two reactants. This type of reaction has a specific rate equation:

• For a reaction where \(2A \rightarrow A_{2}\), the rate is given by: Rate = \(k[A]^2\)
• In the context of the given exercise: Rate = \(k_1[A]^2\) for A forming \(A_{2}\).

In second-order kinetics, the units of the rate constant differ from those in first-order reactions. For second-order reactions, the units are \(\text{L/mol}\cdot\text{s}\).

This means that if the concentration of A is doubled, the rate of formation of \(A_{2}\) increases four times, illustrating the squared relation in the rate equation. This mechanism applies identically to the dimerization of B to form \(B_{2}\) except with its own rate constant, \(k_2\). Understanding this helps determine how changing concentrations affect the speed of reactions, which is crucial for calculating kinetic parameters.

Rate Constant

The rate constant, symbolized as \(k\), is a critical factor in chemical kinetics. It provides specific information about the rate at which a reaction proceeds under certain conditions without being directly dependent on the concentration of reactants. However, it varies with temperature and must be determined experimentally for each reaction.

In the exercise provided, \(k_1\) for the reaction involving A is given as 0.250 \(\text{L/mol}\cdot\text{s}\) at 25°C. This value indicates the efficiency and speed of the reaction at the specific temperature. On the other hand, the rate constant \(k_2\) was calculated based on the final concentration data for B and found to be approximately 0.330 \(\text{L/mol}\cdot\text{s}\).

The importance of the rate constant lies in its ability to define the reaction rate quantitatively. By understanding the rate constant and incorporating it into the rate law, one can predict how fast a reaction will proceed over time, as shown in the problem when calculating changes in reactants like A and B.

Half-life Calculation

Half-life is a concept often encountered in reaction kinetics, representing the time required for half of the reactant to be consumed in the reaction. In the context of a second-order reaction, the half-life depends on both the initial concentration of the reactant and the rate constant, unlike first-order reactions where half-life is constant.

The formula for calculating the half-life \((t_{1/2})\) for a second-order reaction is:

• \(t_{1/2} = \frac{1}{k[A]_0}\)

This equation indicates that the half-life is inversely proportional to both the rate constant \(k\) and the initial concentration \([A]_0\).

In the problem, using the given \(k_1 = 0.250\ \text{L/mol}\cdot\text{s}\) and \([A]_0 = 1.00 \times 10^{-2}\ M\), the half-life was calculated to be approximately 400 seconds. This computation is essential for understanding the timescale over which significant amounts of reactant are transformed and is a practical measure in reactions where timing is crucial.