 # Ethyl chloride vapor decomposes by the first-order reaction: C_{2}H_{5}{Cl}\r eiraszero11cu 2021-12-17 Answered
Ethyl chloride vapor decomposes by the first-order reaction: ${C}_{2}{H}_{5}\left\{Cl\right\}⇒{C}_{2}{H}_{4}+HCl$. The activation energy is $249k\frac{j}{m}ol$, and the frequency factor is $1.6×{10}^{14}{s}^{-1}$. Find the value of the rate constant at 710 K. What fraction of the ethyl chloride decomposes in 15 minutes at this temperature? Find the temperature at which the rate of the reaction would be twice as fast.
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Rate constant at 710 K can be calculated using the one-point Arrhenius equation.
$k=Ae\left(-\frac{{E}_{a}}{RT}\right)$
Given are $A=1.6×{10}^{14}{s}^{-1}$
${E}_{a}=249k\frac{J}{mol}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}249000\frac{J}{mol}$
$R=8.314\frac{J}{mol\cdot K}$
$T=710K$
Plugging in
$k=\left(1.6×{10}^{14}{\left\{s\right\}}^{-1}\right){e}^{\left(-d\frac{\left\{249000\frac{J}{mol}\right\}}{8.314\frac{\left\{J\right\}}{\left\{mol\right\}\cdot \left\{K\right\}}\cdot \left\{710K\right\}}\right)}=7.67×{10}^{-5}{\text{s}}^{-1}$

We have step-by-step solutions for your answer! Maria Lopez

Use the first-order integrated rate law to calculate the fraction of ethyl chloride that decomposed after 15 mins.
$\mathrm{ln}\left(\frac{\left[A{\right]}_{t}}{\left[A{\right]}_{o}}\right)=-kt$
Isolating the ratio of the remaining ethyl chloride $\frac{{\left[A\right]}_{t}}{{\left[A\right]}_{o}}$
$\frac{{\left[A\right]}_{t}}{{\left[A\right]}_{o}}={e}^{-kt}$
plugging in $k=7.67×{10}^{-5}{\left\{s\right\}}^{-1}$ and $t=15mins\cdot \frac{60seconds}{1min}=900seconds$.

The fraction of remaining ethyl chloride is 0.9333. The fraction of ethyl chloride that decomposed is $1-0.9333=0.067$.

We have step-by-step solutions for your answer! Don Sumner

The temperature where the rate of of reaction is twice than the rate of reaction at 710 K can be calculated using the two-point Arrhenius equation.
$\mathrm{ln}\left(\frac{{k}_{2}}{{k}_{1}}\right)=\frac{{E}_{a}}{R}\cdot \left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)$
If we assume ${k}_{1}$ is the rate constant at ${T}_{1}=710K$ then ${k}_{2}$  is the rate constant at ${T}_{2}=?$ where ${k}_{2}=2\cdot {k}_{1}.$
$R=8.314\frac{J}{mol\cdot K}$
Isolate ${T}_{2}$
$\mathrm{ln}\left(\frac{{k}_{2}}{{k}_{1}}\right)=\frac{{E}_{a}}{R}\cdot \left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)$
$\mathrm{ln}\left(\frac{{k}_{2}}{{k}_{1}}\right)\cdot \frac{R}{{E}_{a}}=\left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)$
$\frac{1}{{T}_{2}}=\frac{1}{{T}_{1}}-\mathrm{ln}\left(\frac{{k}_{2}}{{k}_{1}}\right)\cdot \frac{R}{{E}_{a}}$
${T}_{2}=\frac{1}{\frac{1}{{T}_{1}}-\mathrm{ln}\left(\frac{{k}_{2}}{{k}_{1}}\right)\cdot \frac{R}{{E}_{a}}}$
plugging in
${T}_{2}=\frac{1}{\frac{1}{710K}-\mathrm{ln}\left(\frac{2\cdot {k}_{1}}{{k}_{1}}\right)\cdot \frac{8.314\frac{J}{mol\cdot K}}{249000J/mol}}$
${T}_{2}=721.86K$
Therefore, ${T}_{2}=720K$

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