 # Ethyl chloride vapor decomposes by the first-order reaction shown below. The activation energy is 249kj/mol, 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. $C_{2}H_{5}(Cl)\Rightarrow C_{2}H_{4}+HCl$

This question aims to find the temperature where the reaction rate is twice that at 710K. The Arrhenius equation is $k = Ae^(\dfrac{-E_{a}}{RT})$, where A is the frequency or pre-exponential factor and $e^(\dfrac{-E_{a}}{RT})$ shows the fraction of collisions that have enough energy to control the activation barrier (i.e., have energy greater than or equal to activation energy Ea at temperature T. This equation can be used to understand how the rate of a chemical reaction depends on temperature.

One point Arrhenius equation is used to calculate the rate constant at $710\:K$.

$k=Ae(-\dfrac{E_{a}}{RT})$

The constant $A$ is given as $1.6\times 10^{14}s^{-1}$.

$E_{a}=249k\dfrac{J}{mol}=249000\dfrac{J}{mol}$

$R=8.314 \dfrac{J}{mol.K}$

$T=710K$

Plug the values into the equation.

$k=(1.6\times 10^{14} s^{-1})e^(-d\dfrac{249k\dfrac{J}{mol}}{8.314 \dfrac{J}{mol.K}\times 710K})$

$k=7.67\times 10^{-5}s^{-1}$

To find the fraction of ethyl chloride that decomposes after $15$ minutes, use the first-order integrated rate law.

$\ln(\dfrac{[A]_{t}}{[A]_{o}})=-kt$

$\dfrac{[A]_{t}}{[A]_{o}}=e^{-kt}$

Plug the values of $k=7.67\times 10^{-5}s^{-1}$ and $t=15\:min=900\:s$.

$\dfrac{[A]_{t}}{[A]_{o}}=e^{-(7.67\times 10^{-5}s^{-1})(900\:s)}$

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

The temperature at which the reaction rate is twice the reaction rate at $710\: K$ can be calculated using the two-point Arrhenius equation.

$\ln(\dfrac{k_{2}}{k_{1}})=\dfrac{E_{a}}{R}.(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$

Assume $k_{1}$ is the rate constant at $T_{1}=710K$ and $k_{2}$ is the rate constant at $T_{2}$ which is unknown where $k_{2}=2.k_{1}$.

$R=8.314 \dfrac{J}{mol.K}$

$\ln(\dfrac{k_{2}}{k_{1}})=\dfrac{E_{a}}{R}.(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$

$\ln(\dfrac{k_{2}}{k_{1}}).\dfrac{E_{a}}{R}=(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$

$\dfrac{1}{T_{2}}=\dfrac{1}{T_{2}}-\ln(\dfrac{k_{2}}{k_{1}}).\dfrac{R}{E_{a}}$

$T_{2}=\dfrac{1}{\dfrac{1}{T_{1}}-\ln{k_{2}}{k_{1}}.\dfrac{R}{E_{a}}}$

Plug values into the equation to find $T_{2}$.

$T_{2}=721.86K$

Therefore, the temperature is $T_{2}=720K$.

## Numerical Result

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

Temperature $T_{2}$ at which the rate of the reaction would be twice as fast is:

$T_{2}=720K$

## Example

Ethyl chloride vapors are decomposed by a first order reaction:

$C_{2}H_{5}(Cl)\Rightarrow C_{2}H_{4}+HCl$.

Activation energy is $260k \dfrac{J}{mol}$ and the frequency factor is $1.8\times 10^{14}s^{-1}. Determine the value of the rate constant at$810\:K$. What fraction of ethyl chloride will decompose in$15$minutes at this temperature? Find the temperature at which the reaction rate would be twice as fast. Solution One point Arrhenius equation is used to calculate the rate constant at$810\:K$. $k=Ae(-\dfrac{E_{a}}{RT})$ The constant$A$is given as$1.8\times 10^{14}s^{-1}$. $E_{a}=260k\dfrac{J}{mol}=260000\dfrac{J}{mol}$ $R=8.314 \dfrac{J}{mol.K}$ $T=810K$ Plug the values into the equation. $k=(1.8\times 10^{14} s^{-1})e^(-d\dfrac{260k\dfrac{J}{mol}}{8.314 \dfrac{J}{mol.K}\times 810K})$ $k=2.734\times 10^{-3}s^{-1}$ To find the fraction of ethyl chloride that decomposes after$15$minutes, use the first order integrated rate law. $\ln(\dfrac{[A]_{t}}{[A]_{o}})=-kt$ $\dfrac{[A]_{t}}{[A]_{o}}=e^{-kt}$ Plug the values of$k=2.734\times 10^{-3}s^{-1}$and$t=15\:min=900\:s$. $\dfrac{[A]_{t}}{[A]_{o}}=e^{-(2.734\times 10^{-3}s^{-1})(900\:s)}$ The fraction of the remaining ethyl chloride is$0.0853$. The fraction of remaining ethyl chloride is$1-0.0853=0.914$. The temperature at which the reaction rate is twice the reaction rate at$810\: K$can be calculated using the two-point Arrhenius equation. $\ln(\dfrac{k_{2}}{k_{1}})=\dfrac{E_{a}}{R}.(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$ Assume$k_{1}$is the rate constant at$T_{1}=810K$and$k_{2}$is the rate constant at$T_{2}$which is unknown where$k_{2}=2.k_{1}$. $R=8.314 \dfrac{J}{mol.K}$ $\ln(\dfrac{k_{2}}{k_{1}})=\dfrac{E_{a}}{R}.(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$ $\ln(\dfrac{k_{2}}{k_{1}}).\dfrac{E_{a}}{R}=(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$ $\dfrac{1}{T_{2}}=\dfrac{1}{T_{2}}-\ln(\dfrac{k_{2}}{k_{1}}).\dfrac{R}{E_{a}}$ $T_{2}=\dfrac{1}{\dfrac{1}{T_{1}}-\ln{k_{2}}{k_{1}}.\dfrac{R}{E_{a}}}$ Plug values into the equation to find$T_{2}$. $T_{2}=824.8K$ Therefore, the temperature is$T_{2}=824K$. The fraction of the remaining ethyl chloride is$0.0853$. The fraction of remaining ethyl chloride is$1-0.0853=0.914\$.

Temperature is calculated as:

$T_{2}=824K$