Question

Two isomers (A and B) of a given compound dimerize as follows:

2A→A₂

2B→B₂

Both processes are known to be second order in the reactant, and k₁ is known to be 0.250 L mol^-1s^-1 at 25˚ C. In a particular experiment A, and B were placed in separate containers at 25˚C, where [A]˳=1.00×10^-2 M and [B]˳=2.50×10^-2 M. After each reaction had progressed for 3.00 min, [A]=3.00[B]. In this case the rate laws are defined as follows:

a. Calculate the concentration of A₂ after 3.00 min.

b. Calculate the value of k₂.

c. Calculate the half-life for the experiment involving A.

Short answer

Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. In general, a rate law (or differential rate law, as it is sometimes called) takes this form:

rate=k[A]m[B]n[C]p…

Step-by-step solution

 Calculation of concentration of A2

$$\begin{gathered} rate=-\frac{\Delta A}{\Delta T}=K_1{\left[A\right]}^2 \\ =0.270L/mo{l}^{-s}\times {\left(1\times {10}^{-2}\right)}^2 \\ =2.7\times {10}^{-5}mol/L^{-s} \end{gathered}$$

For time

role="math" localid="1663772588203" $$\begin{gathered} 1\min ute=60seconds \\ 3\min utes= ? \\ 3\min utes\times 60second/\min ute=180seacond \\ \Delta \left[A\right]=180second\times 2.7\times {10}^{-5}mol/L^{-s} \\ =4.86\times {10}^{-3}M \end{gathered}$$

The concentration of A₂ is $2.7\times {10}^{-5}mol/L^{-s}$

Calculate the value of rate constant, k2 as follows:

$2B\to B_2$

The integrated rate equation for second order for the above reaction is as follows:

$\frac{1}{\left[B\right]}=K_{2}t+\frac{1}{\left[B_0\right]}$

Calculate the final concentration of B as follows:

$$\begin{gathered} Since\left[A\right]=3\left[B\right] \\ Calculate{k}_2 \\ \frac{1}{\left[B\right]}=k_{2}t+\frac{1}{\left[B_{\circ }\right]} \\ \frac{1}{0.002298M} =k_2\times 180s+\frac{1}{2.5\times {10}^{-2}} \\ 435M{}^{-1}=K_2\times 180s+45M^{-1} \\ {K}_2\times 180s=390M^{-1} \\ {K}_2=2.166M^{-1}{s}^{-1} \\ {K}_2=2.166L/mol/s \end{gathered}$$

Hence, rate constant, k₂ is 2.1666 L/mol/s.

Calculate half-life of second order reaction

The half-life for second-order reaction is given as follows:

$$\begin{gathered} t_{1/2}=\frac{1}{k_1\left[A_{\circ }\right]} \\ =\frac{1}{0.250Lmo{l}^{-1}{s}^{-1}\times 1.00\times {10}^{-2}M} \\ =\frac{1}{0.250Lmo{l}^{-1}{s}^{-1}\times 1.00\times {10}^{-2}mol{L}^{-1}} \\ {t}_{1/2}=400seaconds \end{gathered}$$

Hence, the half-life for an experiment involving A is 400 seconds.