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.

Step by step solution

01

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. \(0.250 = \frac{x}{(1.00 - x)^2 * 3.00 * 60s}\)

02

Solve the equation for x

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

03

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

04

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

05

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

06

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.

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