Chapter 7, Problem 1OCE

(a)

Interpretation Introduction

Interpretation:

The order of the reaction with respect to each reactant has to be determined.

Concept Introduction:

According to the rate law, the rate of the reaction is directly proportional to the initial concentration of the reactant of the reaction.  The overall order of the reaction is the sum of the order of all the reaction in the chemical reaction.  The unit for the concentration is $\text{mol}\cdot {\text{L}}^{-1}$ and that for the rate of the reaction is $\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}$.  The unit of rate constant for a reaction depends upon the overall order of the reaction.

Answer to Problem 1OCE

The order of reaction with respect to $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}$ and ${\text{H}}_3{\text{O}}^{+}$ is $0$ and $1$ respectively.

Explanation of Solution

The given chemical equation is shown below.

  $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\left(\text{aq}\right)+{\text{3H}}_3{\text{O}}^{+}\left(\text{aq}\right)\to {\text{Fe}}^{2+}\left(\text{aq}\right)+3{\text{Hphen}}^{+}\left(\text{aq}\right)+3{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Suppose the order of the reaction with respect to $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}$ and ${\text{H}}_3{\text{O}}^{+}$ is $a$ and $b$ respectively.

Therefore, the generic rate law expression for the given chemical reaction is shown below.

  $\text{R}=k_r{\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]}^a{\left[{\text{H}}_3{\text{O}}^{+}\right]}^b$        (1)

Where,

• $\text{R}$ is the initial rate of the reaction.
• $k_r$ is the rate constant.
• $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ is the concentration of $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}$.
• $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of ${\text{H}}_3{\text{O}}^{+}$.

The value of $\text{R}$ for first experiment is $9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}$.

The value of $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ for first experiment is $7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ for first experiment is $0.5\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $\text{R}$, $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ and $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (1).

  $9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}=k_{\text{r}}{\left(7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\right)}^a{\left(0.5\text{mol}\cdot {\text{L}}^{-1}\right)}^b$        (2)

The value of $\text{R}$ for second experiment is $9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}$.

The value of $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ for second experiment is $7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ for second experiment is $0.05\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $\text{R}$, $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ and $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (1).

  $9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}=k_{\text{r}}{\left(7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\right)}^a{\left(0.05\text{mol}\cdot {\text{L}}^{-1}\right)}^b$        (3)

The value of $\text{R}$ for third experiment is $4.5\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}$.

The value of $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ for third experiment is $3.75\times {10}^{-2}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ for third experiment is $0.05\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $\text{R}$, $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ and $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (1).

  $4.5\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}=k_{\text{r}}{\left(3.75\times {10}^{-2}\text{mol}\cdot {\text{L}}^{-1}\right)}^a{\left(0.05\text{mol}\cdot {\text{L}}^{-1}\right)}^b$        (4)

Divide equation (2) and equation (3) to calculate the value of $b$.

    $\begin{array}{c}\frac{9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}}{9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}}=\frac{k_{\text{r}}{\left(7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\right)}^a{\left(0.5\text{mol}\cdot {\text{L}}^{-1}\right)}^b}{k_{\text{r}}{\left(7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\right)}^a{\left(0.05\text{mol}\cdot {\text{L}}^{-1}\right)}^b}\\ 1={\left(10\right)}^b\\ {\left(10\right)}^0={\left(10\right)}^b\\ b=0\end{array}$

Therefore, the order with respect to $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}$ is $0$.

Divide equation (3) and equation (4) to calculate the value of $a$.

    $\begin{array}{c}\frac{9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}}{4.5\times {10}^{-5}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}}=\frac{k_{\text{r}}{\left(7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\right)}^a{\left(0.05\text{mol}\cdot {\text{L}}^{-1}\right)}^b}{k_{\text{r}}{\left(3.75\times {10}^{-2}\text{mol}\cdot {\text{L}}^{-1}\right)}^a{\left(0.05\text{mol}\cdot {\text{L}}^{-1}\right)}^b}\\ 0.2={\left(0.2\right)}^a\\ a=1\end{array}$

Therefore, the order with respect to ${\text{H}}_3{\text{O}}^{+}$ is $1$.

Thus, the order of reaction with respect to $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}$ and ${\text{H}}_3{\text{O}}^{+}$ is $0$ and $1$ respectively.

(b)

Interpretation Introduction

Interpretation:

The rate law and the rate constant for the given reaction have to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 1OCE

The rate law for given reaction is $\text{R}=k_r{\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]}^0\left[{\text{H}}_3{\text{O}}^{+}\right]$ and the rate constant for the given reaction is $\underset{\_}{1.8\times 10^{-5}{s}^{-1}}$.

Explanation of Solution

The order of reaction with respect to $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}$ and ${\text{H}}_3{\text{O}}^{+}$ is $0$ and $1$ respectively.

Substitute the value of $a$ and $b$ in equation (1).

  $\text{Rate}=k_r{\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]}^0\left[{\text{H}}_3{\text{O}}^{+}\right]$

Thus, the expression for rate law for the given chemical reaction is shown below.

  $\text{Rate}=k_r{\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]}^0\left[{\text{H}}_3{\text{O}}^{+}\right]$        (5)

The value of $\text{R}$ for first experiment is $9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}$.

The value of $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ for first experiment is $7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}$.

The value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ for first experiment is $0.5\text{mol}\cdot {\text{L}}^{-1}$.

Substitute the value of $\text{R}$, $\left[\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}\right]$ and $\left[{\text{H}}_3{\text{O}}^{+}\right]$ for $k_{\text{r}}$ in equation (5).

  $\begin{array}{c}9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}=k_{\text{r}}{\left(7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\right)}^0\left(0.5\text{mol}\cdot {\text{L}}^{-1}\right)\\ k_{\text{r}}=\frac{9.0\times {10}^{-6}\text{mol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}}{{\left(7.5\times {10}^{-3}\text{mol}\cdot {\text{L}}^{-1}\right)}^0\left(0.5\text{mol}\cdot {\text{L}}^{-1}\right)}\\ =1.8\times {10}^{-5}{\text{s}}^{-1}\end{array}$

Thus, the rate constant for the given reaction is $\underset{\_}{1.8\times 10^{-5}{s}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The activation energy for the given reaction has to be determined.

Concept Introduction:

The Arrhenius equation is the relationship between the temperature and rate constants and also involves the activation energy.  The mathematical form of Arrhenius equation involving the two rate constants $k_{\text{r1}}$ and $k_{\text{r2}}$ at temperature $T_1$ and $T_2$ respectively is shown below.

    $\ln \frac{k_{\text{r2}}}{k_{\text{r1}}}=\frac{E_{\text{a}}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Answer to Problem 1OCE

The activation energy of the given reaction is $\underset{\_}{33.11kJ\cdot mo{l}^{-1}}$.

Explanation of Solution

The Arrhenius equation for the calculation of activation energy is shown below.

  $\ln \frac{k_{\text{r2}}}{k_{\text{r1}}}=\frac{E_{\text{a}}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$        (6)

Where,

• $E_{\text{a}}$ is the activation energy.
• $k_{\text{r1}}$ is the rate constant of a reaction pathway without catalyst.
• $k_{\text{r2}}$ is the rate constant of a reaction pathway with catalyst.
• $T_{\text{1}}$ is the temperature at which rate constant is $k_{\text{r1}}$.
• $T_{\text{2}}$ is the temperature at which rate constant is $k_{\text{r2}}$.
• $R$ is the gas constant.

The value of $k_{\text{r1}}$ is $5.4\times {10}^{-3}{\text{s}}^{-1}$.

The value of $k_{\text{r2}}$ is $2.2\times {10}^{-2}{\text{s}}^{-1}$.

The value of $T_{\text{1}}$ is $298\text{K}$.

The value of $T_{\text{2}}$ is $333\text{K}$.

The value of $R$ is $\text{8}\text{.3145}\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$.

Substitute the value of $k_{\text{r1}}$, $k_{\text{r2}}$, $T_{\text{1}}$, $T_{\text{2}}$ and $R$ for $E_{\text{a}}$ in equation (6).

  $\begin{array}{c}\ln \frac{2.2\times {10}^{-2}{\text{s}}^{-1}}{5.4\times {10}^{-3}{\text{s}}^{-1}}=\frac{E_{\text{a}}}{\text{8}\text{.3145}\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}}\times \left(\frac{1}{298\text{K}}-\frac{1}{333\text{K}}\right)\\ \ln 4.074=\frac{E_{\text{a}}}{\text{8}\text{.3145}\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}}\times \left(\frac{333\text{K}-\text{298}\text{K}}{298\text{K}\times 333\text{K}}\right)\\ 1.4046=\frac{E_{\text{a}}}{\text{8}\text{.3145}\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}}\times \left(\frac{35}{99234\text{K}}\right)\\ E_{\text{a}}=1.4046\times \text{8}\text{.3145}\text{J}\cdot {\text{mol}}^{-1}\times \left(\frac{99234}{35}\right)\end{array}$

On further calculation the activation energy for the given reaction is shown below.

  $\begin{array}{c}{E}_{\text{a}}=33111.6829\text{J}\cdot {\text{mol}}^{-1}\times \frac{1\text{kJ}}{1000\text{J}}\\ =33.11\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the activation energy of the given reaction is $\underset{\_}{33.11kJ\cdot mo{l}^{-1}}$.

(d)

Interpretation Introduction

Interpretation:

The time in which the concentration of reduces to half of the initial value at has to be determined.

Concept Introduction:

The half-life of the particular chemical reaction is the time in which exactly half of the reactant gets consumed.  The mathematical expression for the half-life for a reaction follows first order kinetics is shown below.

  $t_{1/2}=\frac{\ln 2}{k_{\text{r}}}$

Answer to Problem 1OCE

The half-life of the given reaction at $25°\text{C}$ is $\underset{\_}{3.85\times 10^{4}s}$.

Explanation of Solution

The order of the reaction with respect to $\text{Fe}{\left(\text{phen}\right)}_3{}^{2+}$ is $1$.

The relation between the half-life and the rate constant for the first order at $25°\text{C}$ is shown below.

    $t_{1/2}=\frac{\ln 2}{k_{\text{r}}}$        (7)

Where,

• $t_{1/2}$ is the half-life.
• $k_{\text{r}}$  is the order rate constant.

The value of $k_{\text{r}}$ is $1.8\times {10}^{-5}{\text{s}}^{-1}$.

Substitute the value of $k_{\text{r}}$ in equation (7).

    $\begin{array}{c}{t}_{1/2}=\frac{\ln 2}{1.8\times {10}^{-5}{\text{s}}^{-1}}\\ =\frac{0.693}{1.8\times {10}^{-5}{\text{s}}^{-1}}\\ =3.85\times {10}^4\text{s}\end{array}$

Thus, the half-life of the given reaction at $25°\text{C}$ is $\underset{\_}{3.85\times 10^{4}s}$.