Question

Consider the following reaction:

CH3X + Y → CH3Y + X

At 25^·C the following two experiments were run, yielding the following data:

Experiment 1: [Y]0 = 3.0 M

Experiment 2: [Y]0 = 4.5 M

Experiments were also run at 85^·C. The value of the rate constant at 85^·C was found to be 7.88 × 10⁸ (with the

time in units of hours), where [CH₃X]₀ = 1.0 × 10^-2 Mand [Y]₀ = 3.0 M.

a. Determine the rate law and the value of kfor this reaction at 25^·C.

b. Determine the half-life at 85^·C

c. Determine E_afor the reaction.

d. Given that the C-X bond energy is known to be about 325 kJ/mol, suggest a mechanism that explains the results in parts a and c.

Short answer

1. Rate = k[CH₃X] where k = 0.90 h^-1 at 25^·C.
2. Half-life at 85^·C = 8.80 × 10^-10 hour.
3. E_a= 3.0 × 10⁵ j/mol
4. Possible mechanism are

CH₃X→CH₃ + X (slow)

CH₃ + Y→CH₃Y (fast)

Step-by-step solution

Determining rate law of the equation

$Rate=k{\left[C{H}_{3}X\right]}^x{\left[Y\right]}^y$

For experiment 1, [Y] is in large excess, so its concentration will be constant.

role="math" localid="1663828536840" $$\begin{gathered} Rate=k\text{'}{\left[C{H}_{3}X\right]}^x \\ Wherek\text{'}=k{\left(3.0M\right)}^y \end{gathered}$$

A plot of ln(CH₃X) versus t is linear (x=1). The integrated rate law is:

$$\begin{gathered} LN\left[C{H}_{3}X\right]=-\left(0.093\right)t-3.99 \\ k\text{'}=0.93h^{-1} \end{gathered}$$

For experiment 2, [Y] is again constant, with Rate = k”[CH3X]x, where k”= k(4.5 M)^y.

The natural log plot is linear again with an integrated rate law:

role="math" localid="1663828617040" $$\begin{gathered} \left[C{H}_{3}X\right]=-\left(0.093\right)t-5.40 \\ k\text{'}\text{'}=0.93h^{-1} \end{gathered}$$

Dividing the rate-constant values:

$$\begin{gathered} \frac{k\text{'}}{k\text{'}\text{'}}=\frac{0.93}{0.9}=\frac{k{\left(3.0\right)}^y}{k{\left(4.5\right)}^y} \\ 1={\left(0.67\right)}^y \\ y=0 \end{gathered}$$

Reaction is first order in CH₃X and zero order in Y. The overall rate law is:

$Rate=k\left[C{H}_{3}X\right],wherek=0.93h^{-1}at{25}^{0}C$

Determining half-life

$t_{1/2}=\left(\ln 2\right)/k=0.693\left(7.88\times {10}^8{h}^{-1}\right)=8.80\times {10}^{-10}hour$

Determining Ea for the reaction

$$\begin{gathered} \ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \\ \ln \left(\frac{7.88\times {10}^8}{0.93}\right)=\frac{E_a}{8.314}\left[\frac{1}{298}-\frac{1}{358}\right] \\ {E}_a=3\times {10}^{5}J/mol \end{gathered}$$

Finding possible mechanismFrom part a, the reaction is first order in CH3X and zero order in Y.

From part c, the activation energy is close to C-X bond energy. So, a possible mechanism for this

CH₃X→CH₃ + X (slow)

CH₃ + Y→CH₃Y (fast)