Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of $C_2{H}_{5}I$ in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, $A\underset{}{\to }B+C$ the rate constant is,

$\begin{array}{c}\ln \frac{{\left[A\right]}_t}{{\left[A\right]}_0}=-kt\\ where,{\left[A\right]}_0=\text{concentration}\text{of}\text{reactant}atstartingtime\left(t=0\right)\\ {\left[A\right]}_t=\text{concentration}\text{of}\text{reactant}afterttime\\ k=r\text{ate}\text{constant}\\ \text{t}\text{=}\text{time}\end{array}$

Answer to Problem 5PPB

$\begin{array}{cc}Time\left(\min \right)& \left[C_2{H}_{5}I\right]\left(M\right)\\ 0& 0.45\\ 10& 0.39\\ 20& 0.34\\ 30& 0.30\\ 40& 0.26\end{array}$

Explanation of Solution

Concentration of $C_2{H}_{5}I$ after $10\min$,${\left[C_2{H}_{5}I\right]}_t$ is calculated below

Given,

$\begin{array}{c}\text{Initial}\text{concentration}\text{of}{C}_2{H}_{5}I\text{,}{\left[C_2{H}_{5}I\right]}_{\text{0}}\text{=}\text{0}\text{.45}\text{M}\\ \text{Rate}\text{constant,}\text{k}\text{=}\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\\ Time,t=10\min \end{array}$

Concentration of $C_2{H}_{5}I$ after $10\min$, ${\left[C_2{H}_{5}I\right]}_t$ is calculated

$\begin{array}{c}\ln \frac{{\left[C_2{H}_{5}I\right]}_t}{{\left[C_2{H}_{5}I\right]}_{\text{0}}}=-kt\\ \ln \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(10\min \right)\\ \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=e^{-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(10\min \right)}\\ =0.869\end{array}$

Therefore,

Concentration of $C_2{H}_{5}I$ after $10\min$, ${\left[C_2{H}_{5}I\right]}_t$ is,

$\begin{array}{c}{\left[C_2{H}_{5}I\right]}_t=\left(0.869\right)\times \left(0.45\right)\\ =0.39M\end{array}$

Concentration of $C_2{H}_{5}I$ after $10\min$, ${\left[C_2{H}_{5}I\right]}_t$ is $0.39M$

Concentration of $C_2{H}_{5}I$ after $20\min$,${\left[C_2{H}_{5}I\right]}_t$ is calculated below

Given,

$\begin{array}{c}\text{Initial}\text{concentration}\text{of}{C}_2{H}_{5}I\text{,}{\left[C_2{H}_{5}I\right]}_{\text{0}}\text{=}\text{0}\text{.45}\text{M}\\ \text{Rate}\text{constant,}\text{k}\text{=}\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\\ Time,t=20\min \end{array}$

Concentration of $C_2{H}_{5}I$ after $20\min$, ${\left[C_2{H}_{5}I\right]}_t$ is calculated

$\begin{array}{c}\ln \frac{{\left[C_2{H}_{5}I\right]}_t}{{\left[C_2{H}_{5}I\right]}_{\text{0}}}=-kt\\ \ln \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(20\min \right)\\ \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=e^{-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(20\min \right)}\\ =0.756\end{array}$

Therefore,

Concentration of $C_2{H}_{5}I$ after $20\min$, ${\left[C_2{H}_{5}I\right]}_t$ is,

$\begin{array}{c}{\left[C_2{H}_{5}I\right]}_t=\left(0.756\right)\times \left(0.45\right)\\ =0.34M\end{array}$

Concentration of $C_2{H}_{5}I$ after $20\min$, ${\left[C_2{H}_{5}I\right]}_t$ is $0.34M$

Concentration of $C_2{H}_{5}I$ after $30\min$,${\left[C_2{H}_{5}I\right]}_t$ is calculated below

Given,

$\begin{array}{c}\text{Initial}\text{concentration}\text{of}{C}_2{H}_{5}I\text{,}{\left[C_2{H}_{5}I\right]}_{\text{0}}\text{=}\text{0}\text{.45}\text{M}\\ \text{Rate}\text{constant,}\text{k}\text{=}\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\\ Time,t=30\min \end{array}$

Concentration of $C_2{H}_{5}I$ after $30\min$, ${\left[C_2{H}_{5}I\right]}_t$ is calculated

$\begin{array}{c}\ln \frac{{\left[C_2{H}_{5}I\right]}_t}{{\left[C_2{H}_{5}I\right]}_{\text{0}}}=-kt\\ \ln \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(30\min \right)\\ \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=e^{-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(30\min \right)}\\ =0.657\end{array}$

Therefore,

Concentration of $C_2{H}_{5}I$ after $30\min$, ${\left[C_2{H}_{5}I\right]}_t$ is,

$\begin{array}{c}{\left[C_2{H}_{5}I\right]}_t=\left(0.657\right)\times \left(0.45\right)\\ =0.30M\end{array}$

Concentration of $C_2{H}_{5}I$ after $30\min$, ${\left[C_2{H}_{5}I\right]}_t$ is $0.30M$

Concentration of $C_2{H}_{5}I$ after $40\min$,${\left[C_2{H}_{5}I\right]}_t$ is calculated below

Given,

$\begin{array}{c}\text{Initial}\text{concentration}\text{of}{C}_2{H}_{5}I\text{,}{\left[C_2{H}_{5}I\right]}_{\text{0}}\text{=}\text{0}\text{.45}\text{M}\\ \text{Rate}\text{constant,}\text{k}\text{=}\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\\ Time,t=40\min \end{array}$

Concentration of $C_2{H}_{5}I$ after $40\min$, ${\left[C_2{H}_{5}I\right]}_t$ is calculated

$\begin{array}{c}\ln \frac{{\left[C_2{H}_{5}I\right]}_t}{{\left[C_2{H}_{5}I\right]}_{\text{0}}}=-kt\\ \ln \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(40\min \right)\\ \frac{{\left[C_2{H}_{5}I\right]}_t}{0.45M}=e^{-\left(\text{1}\text{.4}\text{×}{\text{10}}^{-2}{\min }^{-1}\right)\times \left(40\min \right)}\\ =0.5712\end{array}$

Therefore,

Concentration of $C_2{H}_{5}I$ after $40\min$, ${\left[C_2{H}_{5}I\right]}_t$ is,

$\begin{array}{c}{\left[C_2{H}_{5}I\right]}_t=\left(0.5712\right)\times \left(0.45\right)\\ =0.26M\end{array}$

Concentration of $C_2{H}_{5}I$ after $40\min$, ${\left[C_2{H}_{5}I\right]}_t$ is $0.26M$

Conclusion

The missing concentration of $C_2{H}_{5}I$ in the table for the given reaction were calculated,

$\begin{array}{cc}Time\left(\min \right)& \left[C_2{H}_{5}I\right]\left(M\right)\\ 0& 0.45\\ 10& 0.39\\ 20& 0.34\\ 30& 0.30\\ 40& 0.26\end{array}$