Chapter 11, Problem 84AE

The reaction

H₂SeO₃(aq) + 6I^-(aq) + 4H⁺(aq) → Se(s) + 2I^-₃(aq) + 3H₂O(l)

was studied at 0°C, and the following data were obtained:

[H2SeO3]0 (mol/L)	[H+]0 (mol/L)	[I−]0(mol/L)	Initial Rate (mol/L  s)
1.0 × 10−4	2.0 × 10−2	2.0 × 10−2	1.66 × 10−7
2.0 × 10−4	2.0 × 10−2	2.0 × 10-2	3.33 × 10−7
3.0 × 10−4	2.0 × 10−2	2.0 × 10−2	4.99 × 10−7
1.0 × 10−4	4.0 × 10−2	2.0 × 10−2	6.66 × 10−7
1.0 × 10−4	1.0 × 10−2	2.0 × 10−2	0.42 × 10−7
1.0 × 10−4	2.0 × 10−2	4.0 × 10−2	13.2 × 10−7
1.0 × 10−4	1.0 × 10−2	4.0 × 10−2	3.36 × 10−7

These relationships hold only if there is a very small amount of I₃⁻ present. What is the rate law and the value of the rate constant? $\left(\text{Assume that rate }=-\frac{\Delta \left[{\text{H}}_{\text{2}}{\text{SeO}}_3\right]}{\Delta t}\right)$

Interpretation Introduction

Interpretation: The reaction of ${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\left(aq\right)$ with ${\text{I}}^{-},{\text{H}}^{+}$ and the data obtained after this reaction is given. The expression of rate law and the value of rate constant are to be calculated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

To determine: Theexpression of rate law and value of the rate constant for the given reaction.

Answer to Problem 84AE

Answer

The rate law of the given reaction is $\text{Rate}=k\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]{\left[{\text{H}}^{+}\right]}^{\text{2}}{\left[{\text{I}}^{-}\right]}^{\text{3}}$.

The value of rate constant is $\underset{\_}{5.2\times 10^5{L}^5/mo{l}^5\cdot s}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\left(aq\right)+{\text{6I}}^{-}\left(aq\right)+{\text{4H}}^{+}\left(aq\right)\stackrel{}{\to }\text{Se}\left(s\right){\text{+2I}}_{\text{3}}{}^{-}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)$

The rate law is represented as,

$\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z$ (1)

Where,

• $k$ is the rate constant.
• $\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]$ is the initial concentration of reactant ${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}$.
• $\left[{\text{H}}^{+}\right]$ is the initial concentration of reactant ${\text{H}}^{+}$.
• $\left[{\text{I}}^{-}\right]$ is the final concentration of reactant ${\text{I}}^{-}$.
• $\left[\text{B}\right]$ is the final concentration of reactant $\left(\text{V}\right)$.
• $X$ is the order of reactant ${\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}$.
• $Y$ is the order of reactant ${\text{H}}^{+}$.
• $Z$ is the order of reactant ${\text{I}}^{-}$.

Given table is,

Exp. No.	$\begin{array}{l}{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}_{\text{0}}\\ \left(\text{mol/L}\right)\end{array}$	$\begin{array}{c}{\left[{\text{H}}^{+}\right]}_{\text{0}}\\ \left(\text{mol/L}\right)\end{array}$	$\begin{array}{c}{\left[{\text{I}}^{-}\right]}_{\text{0}}\\ \left(\text{mol/L}\right)\end{array}$	$\begin{array}{c}\text{Initial}\text{rate}\\ \left(\text{mol/L}\cdot \text{s}\right)\end{array}$
$1$	$\text{1}\text{.0}\times {\text{10}}^{-\text{4}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{1}\text{.66}\times {\text{10}}^{-\text{7}}$
$2$	$\text{2}\text{.0}\times {\text{10}}^{-\text{4}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{3}\text{.33}\times {\text{10}}^{-\text{7}}$
$3$	$\text{3}\text{.0}\times {\text{10}}^{-\text{4}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{4}\text{.99}\times {\text{10}}^{-\text{7}}$
$4$	$\text{1}\text{.0}\times {\text{10}}^{-\text{4}}$	$\text{4}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{6}\text{.66}\times {\text{10}}^{-\text{7}}$
$5$	$\text{1}\text{.0}\times {\text{10}}^{-\text{4}}$	$\text{1}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{0}\text{.42}\times {\text{10}}^{-\text{7}}$
$6$	$\text{1}\text{.0}\times {\text{10}}^{-\text{4}}$	$\text{2}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{4}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{13}\text{.2}\times {\text{10}}^{-\text{7}}$
$7$	$\text{1}\text{.0}\times {\text{10}}^{-\text{4}}$	$\text{1}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{4}\text{.0}\times {\text{10}}^{-\text{2}}$	$\text{3}\text{.36}\times {\text{10}}^{-\text{7}}$

Substitute the values of experiment $1$ in the above equation.

$\begin{array}{l}\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z\\ \text{1}\text{.66}\times {\text{10}}^{-\text{7}}=k{\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right)}^X{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z\end{array}$ (2)

Substitute the values of experiment $2$ in equation (1).

$\begin{array}{l}\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z\\ \text{3}\text{.33}\times {\text{10}}^{-\text{7}}=k{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{4}}\right)}^X{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z\end{array}$ (3)

Take the ratio of equation (2) and (3).

$\begin{array}{c}\frac{\text{1}\text{.66}\times {\text{10}}^{-\text{7}}}{\text{3}\text{.33}\times {\text{10}}^{-\text{7}}}=\frac{k{\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right)}^X{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z}{k{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{4}}\right)}^X{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z}\\ \text{0}\text{.50}=\text{0}{\text{.50}}^X\\ X=\underset{\_}{1}\end{array}$

The value of $X$ is $\text{1}$.

Substitute the values of experiment $4$ and the value of $X$ in equation (1).

$\begin{array}{l}\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z\\ \text{6}\text{.66}\times {\text{10}}^{-\text{7}}=k\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right){\left(\text{4}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z\end{array}$ (4)

Substitute the values of experiment $5$ and the value of $X$ in equation (1).

$\begin{array}{l}\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z\\ \text{0}\text{.42}\times {\text{10}}^{-\text{7}}=k\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right){\left(\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z\end{array}$ (5)

Take the ratio of equation (4) and (5).

$\begin{array}{c}\frac{\text{6}\text{.66}\times {\text{10}}^{-\text{7}}}{\text{0}\text{.42}\times {\text{10}}^{-\text{7}}}=\frac{k\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right){\left(\text{4}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z}{k{\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right)}^X{\left(\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Y{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z}\\ \text{16}={\text{4}}^Y\\ Y=\underset{\_}{2}\end{array}$

The value of $X$ is $\text{1}$.

The value of $Y$ is $\text{2}$.

Substitute the values of experiment $6$ and the value of $X,Y$ in equation (1).

$\begin{array}{l}\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z\\ \text{3}\text{.36}\times {\text{10}}^{-\text{7}}=k\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right){\left(\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^{\text{2}}{\left(\text{4}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z\end{array}$ (6)

Substitute the values of experiment $7$ and the value of $X,Y$ in equation (1).

$\begin{array}{l}\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z\\ \text{0}\text{.42}\times {\text{10}}^{-\text{7}}=k\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right){\left(\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^{\text{2}}{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z\end{array}$ (7)

Take the ratio of equation (6) and (7).

$\begin{array}{c}\frac{\text{3}\text{.36}\times {\text{10}}^{-\text{7}}}{\text{0}\text{.42}\times {\text{10}}^{-\text{7}}}=\frac{k\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right){\left(\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^{\text{2}}{\left(\text{4}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z}{k{\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right)}^X{\left(\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^{\text{2}}{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^Z}\\ \text{8}={\text{2}}^Z\\ Z=\underset{\_}{3}\end{array}$

The rate law of the given reaction is $\text{Rate}=k\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]{\left[{\text{H}}^{+}\right]}^{\text{2}}{\left[{\text{I}}^{-}\right]}^{\text{3}}$.

The value of $X$ is $\text{1}$.

The value of $Y$ is $\text{2}$.

The value of $Z$ is $\text{3}$.

Substitute the values of $X,Y$ and $Z$ in equation (1) to get the rate law.

$\begin{array}{c}\text{Rate}=k{\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]}^X{\left[{\text{H}}^{+}\right]}^Y{\left[{\text{I}}^{-}\right]}^Z\\ =k\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]{\left[{\text{H}}^{+}\right]}^{\text{2}}{\left[{\text{I}}^{-}\right]}^{\text{3}}\end{array}$

The value of rate constant is $\underset{\_}{5.2\times 10^5{L}^5/mo{l}^5\cdot s}$.

Substitute the values of experiment 1 in equation (1) to get the rate constant.

$\begin{array}{c}\text{Rate}=k\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]{\left[{\text{H}}^{+}\right]}^{\text{2}}{\left[{\text{I}}^{-}\right]}^{\text{3}}\\ \text{1}\text{.66}\times {\text{10}}^{-\text{7}}\text{mol/L}\cdot \text{s}=\left[k\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right){\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^{\text{2}}{\left(\text{2}\text{.0}\times {\text{10}}^{-\text{2}}\right)}^{\text{3}}\right]\text{mol/L}\\ k=\underset{\_}{5.2\times 10^5{L}^5/mo{l}^5\cdot s}\end{array}$

Conclusion

Conclusion

The rate law of the given reaction is $\text{Rate}=k\left[{\text{H}}_{\text{2}}{\text{SeO}}_{\text{3}}\right]{\left[{\text{H}}^{+}\right]}^{\text{2}}{\left[{\text{I}}^{-}\right]}^{\text{3}}$. The value of rate constant is $\underset{\_}{5.2\times 10^5{L}^5/mo{l}^5\cdot s}$