Question

The decomposition of iodoethane in the gas phase proceeds according to the following equation:

${\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{5}}\mathit{I}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{4}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{H}\mathit{I}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

At $\mathbf{600}\mathbf{.}\mathit{k}\mathbf{,}\mathbf{}\mathit{k}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{;}$at $\mathbf{720}\mathbf{.}\mathit{K}\mathbf{,}\mathbf{}\mathit{k}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$. What is the rate constant for this first- order decomposition at $\mathbf{325}\mathbf{°}\mathit{C}$? If the initial pressure of iodoethane is $\mathbf{894}$ torr at $\mathbf{245}\mathbf{°}\mathit{C}$what is the pressure of iodoethane after three half-lives?

Short answer

The rate constant for first-order decomposition at $325°C$ is $1.3\times {10}^{-5}{s}^{-1}$ . The pressure of iodoethane after three half-lives is $112torr$ .

Step-by-step solution

Determine the value of  Ea

The formula we will use to determine the rate constant is:

$$\begin{gathered} \ln \left(\frac{k_2}{k_1}\right)=\frac{E_A}{RT}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ \ln \left(\frac{k_2}{1.7\times {10}^{-2}}\right)=\frac{210000}{8.314}\left(\frac{1}{720}-\frac{1}{598k}\right) \\ \ln \left(\frac{k_2}{1.7\times {10}^{-2}}\right)=-7.157 \\ {k}_2=1.3\times {10}^{-5} \end{gathered}$$

Given,

$$\begin{gathered} R=8.314JKmo{l}^{-1} \\ {T}_1=720K \\ {T}_2=600K \\ {K}_1=1.7\times {10}^{-7} \\ {K}_2=7.2\times {10}^{-4} \end{gathered}$$

Determining the activation energy $E_a$.

$$\begin{gathered} \ln \left(\frac{7.2\times {10}^{-4}}{1.7}\right)=\frac{E_A}{8.314}\left(\frac{1}{720}-\frac{1}{660}\right) \\ 3.16=E_a\times 1.51\times {10}^{-5} \\ {E}_a=210000\frac{J}{mol} \end{gathered}$$

Determine rate constant for first order decomposition at  325°C .

Substituting the values in rate constant formula:

$$\begin{gathered} \ln \left(\frac{k_2}{k_1}\right)=\frac{E_A}{RT}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ \ln \left(\frac{k_2}{1.7\times {10}^{-2}}\right)=\frac{210000}{8.314}\left(\frac{1}{720}-\frac{1}{598k}\right) \\ \ln \left(\frac{k_2}{1.7\times {10}^{-2}}\right)=-7.157 \\ {k}_2=1.3\times {10}^{-5} \end{gathered}$$

Determine the pressure of iodoethane after three half-lives.

Determining the pressure of iodoethane:

Given that the three half-lives passed, and the half-life is independent of initial concentration.

So, ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$of the initial iodoethane is remaining after three half-lives.

$P_{C_2{H}_{5}I}=\frac{1}{8}\left(894\right)=112torr$