Question

For the reactions,

$$\begin{gathered} {\text{I+I+M→I}}_2+\text{M} \\ {\text{Br+Br+M→Br}}_2+\text{M} \end{gathered}$$

The rate laws are

$$\begin{gathered} -\frac{d\left[\text{I}\right]}{dt}=k_I{\left[\text{I}\right]}^2\left[\text{M}\right] \\ -\frac{d\left[Br\right]}{dt}=k_{Br}{\left[\text{Br}\right]}^2\left[\text{M}\right] \end{gathered}$$

The ratio k_I>k_Br at 500 °C is 3.0 when M is an Ar molecule. Initially,${\left[\text{I}\right]}_0=2{\left[\text{B}r\right]}_0$ while [M] is the same for both reactions, it is much greater than [I]₀. Calculate the ratio of [I]₀the time required for [I] to decrease to half its initial value and the same time for [Br] at 500 °C.

Short answer

The ratio of half-lives are 1:6

Step-by-step solution

Rate Laws:

For the given reaction in the question

$$\begin{gathered} {\text{I+I+M→I}}_2+\text{M.}..........\left(1\right) \\ {\text{Br+Br+M→Br}}_2+M..........\left(2\right) \end{gathered}$$

The rate laws for the above reaction

$$\begin{gathered} \text{-}\frac{d\left[\text{I}\right]}{dt}=k_I{\left[\text{I}\right]}^2\left[\text{M}\right].............\left(3\right) \\ -\frac{d\left[\text{Br}\right]}{dt}=k_{Br}{\left[Br\right]}^2\left[\text{M}\right]..........\left(4\right) \end{gathered}$$

The concentration of M is greater so its concentration is neglected in the equations.

So the new rate law are

$$\begin{gathered} \text{-}\frac{d\left[\text{I}\right]}{dt}=k_I{\left[\text{I}\right]}^2.............\left(4\right) \\ -\frac{d\left[\text{Br}\right]}{dt}=k_{Br}{\left[\text{Br}\right]}^2..........\left(5\right) \end{gathered}$$

Half-Life of the Reaction

Half life of the reaction is defined as the time required for the reaction to decrease to half of its concentration.For the second order reaction, the half- life is given as

$$\begin{gathered} t_{1/2}=\frac{1}{k{\left[\text{reactent}\right]}_0} \\ \text{Wherekisrateconstanthfill} \\ {\left[reactent\right]}_0=\text{Initialconcendartionofreactents} \end{gathered}$$

Therfore for the first reaction

${\text{t}}_{1/2}=\frac{1}{{\text{k}}_0{\left[\text{I}\right]}_0}.....\left(6\right)$

For seacond reaction

${\text{t}}_{1/2}=\frac{1}{{\text{k}}_0{\left[\text{Br}\right]}_0}.....\left(7\right)$

Hence the ratio of half-lives are as follows

$\frac{{\text{t}}_{1/2\left(I\right)}}{{\text{t}}_{1/2\left(Br\right)}}=\frac{\frac{1}{k_0{\left[I\right]}_0}}{\frac{1}{k_0{\left[Br\right]}_0}}=\frac{{\text{k}}_0{\left[\text{Br}\right]}_0}{k_0{\left[\text{I}\right]}_0}.....\left(8\right)$

Ratio of half lives:

According to given condition in the above reaction.

$$\begin{gathered} {\left[\text{I}\right]}_0=2{\left[\text{Br}\right]}_0 \\ {\text{k}}_1/k_2=3.0 \\ \text{IfMisAromaticmolecule} \\ \frac{{\text{k}}_I}{{\text{k}}_{Br}}=3.0 \\ \frac{{\text{k}}_{Br}}{{\text{k}}_I}=\frac{1.0}{3.0} \\ {\left[\text{I}\right]}_0=2{\left[\text{Br}\right]}_0 \\ \frac{{\left[\text{Br}\right]}_0}{{\left[\text{I}\right]}_0}=\frac{1}{2} \end{gathered}$$

Substitute these values in equation 8 to obtain half lives

$$\begin{gathered} \frac{t_{1/2\left(I\right)}}{t_{1/2\left(Br\right)}}=\frac{k_{Br}\left[Br\right]}{k_I\left[I\right]}.....\left(8\right) \\ \frac{t_{1/2\left(I\right)}}{t_{1/2\left(Br\right)}}=\frac{1x1}{3.0x2 \\ } \\ \frac{t_{1/2\left(I\right)}}{t_{1/2\left(Br\right)}}=\frac{1}{6} \end{gathered}$$

Hence the ratio of half-lives are 1:6