Question

The reaction

$\left(C{H}_3\right)CBr+O{H}^{-}\to {\left(C{H}_3\right)}_{3}COH+B{r}^{-}$

in a certain solvent is first order with respect to${\left(C{H}_3\right)}_{3}CBr$and zero order with respect to $O{H}^{-}$. In several experiments the rate constant k was determined at different temperatures. A plot of $n\left(k\right)$versus $\frac{1}{T}$was constructed that resulted in a straight line with a slope of $-1.10\times {10}^{4}K$and a $y$intercept of $33.5$. Assume that $k$has units of$s^{-1}$ .

a. Determine the activation energy for this reaction.

b. Determine the value of the frequency factor$A$ .

c. Calculate the value of k at$25°C$ .

Short answer

$$\begin{gathered} a.E_A=91.45kJ/mo{l}^{-1} \\ b.A=3.54\times {10}^{14}{s}^{-1} \\ c.K_2=0.034s^{-1} \end{gathered}$$

Step-by-step solution

Determine the activation energy.

We have

$$\begin{gathered} k=A+e{}^{\frac{E_A}{RT}} \\ \ln \left(k\right)=\frac{E_A}{R}\times \frac{1}{T} \\ {E}_A=1.10\times {10}^{4}K \\ {E}_A=91.45kJmo{l}^{-1} \end{gathered}$$

So, the activation energy for the given equation is$E_A=91.45kJ/mo{l}^{-1}$

Determine the value of the frequency factor .

We have

$$\begin{gathered} k=A+e{}^{\frac{E_A}{RT}} \\ \ln \left(k\right)=\frac{E_A}{R}\times \frac{1}{T}+\ln \left(A\right) \\ \ln \left(A\right)=33.5 \\ A=3.54\times {10}^{14}{s}^{-1} \end{gathered}$$

So, the frequency factor A in the given equation is$E_A=91.45kJ/mo{l}^{-1}$ .

Determine the value of k.

We have,

$k=A+e{}^{\frac{E_A}{RT}}$

$$\begin{gathered} k=3.54\times {10}^{14}{s}^{-1}\times e^{\frac{91.45\times {10}^{3}Jmol}{8.314J{K}^{-1}\times 298.15K}} \\ k=0.034s^{-1} \end{gathered}$$

So, the value of k at $25°C$is $k=0.034s^{-1}$.