Chapter 13, Problem 13.32QE

Interpretation Introduction

Interpretation:

The instantaneous rate for the change of other species has to be calculated and also the instantaneous rate of reaction has to be calculated.

Answer to Problem 13.32QE

Instantaneous rate of appearance of ${\text{H}}_{\text{2}}\text{O}$, ${\text{Mn}}^{\text{2+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$ is $0.018M\text{/s}$, $0.012M\text{/s}$, and $0.030M\text{/s}$ respectively.  Instantaneous rate of disappearance of ${\text{HNO}}_{\text{2}}$, and ${\text{H}}^{\text{+}}$, is $0.030M\text{/s}$, and $0.006M\text{/s}$ respectively.  Rate of the reaction is $0.0060M\text{/s}$.

Explanation of Solution

The reaction that is given in the problem statement is shown below;

    ${\text{2MnO}}_{\text{4}}{}^{-}+{\text{5HNO}}_{\text{2}}+{\text{H}}^{+}\to 2{\text{Mn}}^{2+}+{\text{5NO}}_{\text{3}}{}^{-}+3{\text{H}}_{\text{2}}\text{O}$

From the above equation, it is found that the stoichiometric relationship between ${\text{MnO}}_{\text{4}}{}^{-}$ and other species is given as follows;

    $\begin{array}{l}2{\text{mol MnO}}_{\text{4}}{}^{-}=3\text{mol}{\text{H}}_{\text{2}}\text{O}\\ 2{\text{mol MnO}}_{\text{4}}{}^{-}=2\text{mol}{\text{Mn}}^{\text{2+}}\\ 2{\text{mol MnO}}_{\text{4}}{}^{-}=5\text{mol}{\text{NO}}_{\text{3}}{}^{-}\\ 2{\text{mol MnO}}_{\text{4}}{}^{-}=1\text{mol}{\text{OH}}^{+}\\ 2{\text{mol MnO}}_{\text{4}}{}^{-}=5\text{mol}{\text{HNO}}_{\text{2}}\end{array}$

The instantaneous rate of formation of ${\text{H}}_{\text{2}}\text{O}$ can be calculated using the mole relationship as follows;

    $\begin{array}{c}\frac{\Delta [{\text{H}}_{\text{2}}\text{O}]}{\Delta t}=\frac{\Delta [{\text{MnO}}_{\text{4}}{}^{-}]}{\Delta t}\times \left(\frac{3\text{mol}{\text{H}}_{\text{2}}\text{O}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =-\left(-0.012M\text{/s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{3\text{mol}{\text{H}}_{\text{2}}\text{O}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.012\text{mol/L}\cdot \text{s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{3\text{mol}{\text{N}}_{\text{2}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.018\text{mol/L}\cdot \text{s}\right){\text{H}}_{\text{2}}\text{O}\\ =0.018M\text{/s}\end{array}$

Therefore, the appearance rate of water will be $0.018M\text{/s}$.

The instantaneous rate of formation of ${\text{Mn}}^{\text{2+}}$ can be calculated using the mole relationship as follows;

    $\begin{array}{c}\frac{\Delta [{\text{Mn}}^{\text{2+}}]}{\Delta t}=\frac{\Delta [{\text{MnO}}_{\text{4}}{}^{-}]}{\Delta t}\times \left(\frac{2\text{mol}{\text{Mn}}^{\text{2+}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =-\left(-0.012M\text{/s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{2\text{mol}{\text{Mn}}^{\text{2+}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.012\text{mol/L}\cdot \text{s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{2\text{mol}{\text{Mn}}^{\text{2+}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.012\text{mol/L}\cdot \text{s}\right){\text{Mn}}^{\text{2+}}\\ =0.012M\text{/s}\end{array}$

Therefore, the appearance rate of ${\text{Mn}}^{\text{2+}}$ will be $0.012M\text{/s}$.

The instantaneous rate of formation of ${\text{NO}}_{\text{3}}{}^{-}$ can be calculated using the mole relationship as follows;

    $\begin{array}{c}\frac{\Delta [{\text{NO}}_{\text{3}}{}^{-}]}{\Delta t}=\frac{\Delta [{\text{MnO}}_{\text{4}}{}^{-}]}{\Delta t}\times \left(\frac{5\text{mol}{\text{NO}}_{\text{3}}{}^{-}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =-\left(-0.012M\text{/s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{5\text{mol}{\text{NO}}_{\text{3}}{}^{-}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.012\text{mol/L}\cdot \text{s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{5\text{mol}{\text{NO}}_{\text{3}}{}^{-}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.03\text{mol/L}\cdot \text{s}\right){\text{NO}}_{\text{3}}{}^{-}\\ =0.03M\text{/s}\end{array}$

Therefore, the appearance rate of ${\text{NO}}_{\text{3}}{}^{-}$ will be $0.03M\text{/s}$.

The instantaneous rate of disappearance of ${\text{HNO}}_{\text{2}}$ can be calculated using the mole relationship as follows;

    $\begin{array}{c}\frac{\Delta [{\text{HNO}}_{\text{2}}]}{\Delta t}=-\frac{\Delta [{\text{MnO}}_{\text{4}}{}^{-}]}{\Delta t}\times \left(\frac{5\text{mol}{\text{HNO}}_{\text{2}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =-\left(-0.012M\text{/s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{5\text{mol}{\text{HNO}}_{\text{2}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =-\left(-0.012\text{mol/L}\cdot \text{s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{5\text{mol}{\text{HNO}}_{\text{2}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.03\text{mol/L}\cdot \text{s}\right){\text{HNO}}_{\text{2}}\\ =0.03M\text{/s}\end{array}$

Therefore, the instantaneous disappearance rate of ${\text{HNO}}_{\text{2}}$ will be $0.03M\text{/s}$.

The instantaneous rate of disappearance of ${\text{H}}^{\text{+}}$ can be calculated using the mole relationship as follows;

    $\begin{array}{c}\frac{\Delta [{\text{H}}^{\text{+}}]}{\Delta t}=-\frac{\Delta [{\text{MnO}}_{\text{4}}{}^{-}]}{\Delta t}\times \left(\frac{1\text{mol}{\text{H}}^{\text{+}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =-\left(-0.012M\text{/s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{1\text{mol}{\text{H}}^{\text{+}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =-\left(-0.012\text{mol/L}\cdot \text{s}\right){\text{MnO}}_{\text{4}}{}^{-}\times \left(\frac{1\text{mol}{\text{H}}^{\text{+}}}{2\text{mol}{\text{MnO}}_{\text{4}}{}^{-}}\right)\\ =\left(0.006\text{mol/L}\cdot \text{s}\right){\text{H}}^{\text{+}}\\ =0.006M\text{/s}\end{array}$

Therefore, the instantaneous disappearance rate of ${\text{H}}^{\text{+}}$ will be $0.006M\text{/s}$.

Rate of the reaction can be found out by dividing the rate of appearance or disappearance of any species involved in the reaction by its coefficient.  Therefore, the rate of the reaction can be calculated using the disappearance of ${\text{H}}^{\text{+}}$ as shown below;

    $\begin{array}{c}\text{Rate}\text{of}\text{reaction}=\frac{0.0060M\text{/s}}{1}\\ =0.0060M\text{/s}\end{array}$

Therefore, the rate of the reaction is $0.0060M\text{/s}$.