Question

Dehydration of tertiary butyl alcohol follows a first order reaction, ${\left(C{H}_3\right)}_{3}C-OH\left(g\right)\to {\left(C{H}_3\right)}_{2}C=C{H}_2\left(g\right)+H_{2}O\left(g\right)$, the rate constant at ${300}^{o}C$ is $2.27\times {10}^{-8}{s}^{-1}$. Calculate the rate constant at ${400}^{0}C$ if the energy of activation for the reaction is $58$ $kcal$.

Answer 1

By Arrhenius equation we have

$K={Ae}^{{-E}_A/RT}$

where $K$= Rate constant

$E_A$= Activation Energy

$R$= Rate Constant

$T$= Temperature

Taking log on both sides we have

$\log K=\log A-\frac{E_A}{RT}$

Writing the above equation at temperature $300°C$ i.e. $573k$,

$\log K_1=\log A-\frac{E_A}{R\left(573\right)}$ $\left(1\right)$

Similarly, writing equation at temperature $400°C$ i.e. $673K$,

$\log K_2=\log A-\frac{E_A}{R\left(673\right)}$ $\left(2\right)$

Using $\left(2\right)$ and $\left(1\right)$ we get

$\log \left(\frac{K_2}{K_1}\right)=\frac{{-E}_A}{R}[\frac{1}{673}-\frac{1}{573}]$

putting the value of $E_A$ and $K_1$ from both the question we get

$\Rightarrow \log \left(\frac{K_2}{2.27\times {10}^{-8}}\right)=\frac{-58\times {10}^3\times 4.18}{8.314}[\frac{1}{673}-\frac{1}{573}]$

$\Rightarrow K_2=4.36\times {10}^{-5}$ $\{using{K}_1=2.27\times {10}^{-8}{E}_A=58kcal=58\times {10}^3\times 4.185J\}$