Chapter 19.5, Problem 19.4WE

(a)

Interpretation Introduction

Interpretation:

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$ and time taken to drop the concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to $0.10M$ for the given reaction should be calculated.

Concept introduction:

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

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

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$, ${\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t$ is calculated

Answer to Problem 19.4WE

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$, ${\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t$ is $0.62M$

Explanation of Solution

Given,

$\begin{array}{c}\text{Initial}\text{concentration}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{,}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.75}\text{M}\\ \text{Rate}\text{constant,}\text{k}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{-5}{s}^{-1}\\ Time,t=3h=3\times 3600s=10800s\end{array}$

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$, ${\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t$ is calculated

$\begin{array}{c}\ln \frac{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t}{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}}=-kt\\ \ln \frac{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t}{0.75M}=-\left(\text{1}\text{.8}\text{×}{\text{10}}^{-5}{s}^{-1}\right)\times \left(10800s\right)\\ \frac{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t}{0.75M}=e^{-\left(\text{1}\text{.8}\text{×}{\text{10}}^{-5}{s}^{-1}\right)\times \left(10800s\right)}\\ =0.823\end{array}$

Therefore,

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$, ${\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t$ is,

$\begin{array}{c}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t=\left(0.823\right)\times \left(0.75\right)\\ =0.62M\end{array}$

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$, ${\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t$ is $0.62M$

(b)

Interpretation Introduction

Interpretation

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$ and time taken to drop the concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to $0.10M$ for the given reaction should be calculated.

Concept introduction:

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

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

Concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ after $3h$, ${\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t$ is calculated

Time taken to drop the concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to $0.10M$ is calculated

Answer to Problem 19.4WE

Time taken to drop the concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to $0.10M$ is $1.12\times {10}^{5}s$

Explanation of Solution

Given,

$\begin{array}{c}\text{Initial}\text{concentration}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{,}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}\text{=}\text{0}\text{.75}\text{M}\\ \text{Concentration}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{ after time t,}{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t\text{=}\text{0}\text{.10}\text{M}\\ \text{Rate}\text{constant,}\text{k}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{-5}{s}^{-1}\end{array}$

Time taken to drop the concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to $0.10M$ is calculated

$\begin{array}{c}\ln \frac{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_t}{{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}_{\text{0}}}=-kt\\ \ln \frac{0.10M}{0.75M}=-\left(\text{1}\text{.8}\text{×}{\text{10}}^{-5}{s}^{-1}\right)\times t\\ -2.015=-\left(\text{1}\text{.8}\text{×}{\text{10}}^{-5}{s}^{-1}\right)\times t\end{array}$

Therefore,

Time taken to drop the concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to $0.10M$ is,

$\begin{array}{c}t=\frac{2.015}{\text{1}\text{.8}\text{×}{\text{10}}^{-5}{s}^{-1}}\\ =1.12\times {10}^{5}s\end{array}$

Time taken to drop the concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to $0.10M$ is $1.12\times {10}^{5}s$