\[C_{2}H_{5}(Cl)\Rightarrow C_{2}H_{4}+HCl\]

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.**

**Expert Answer**

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 $1-0.9333=0.067$.

T**emperature $T_{2}$ 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:

\[T_{2}=824K\]