Chapter 17, Problem 17.28P

Interpretation Introduction

Interpretation:

Rate law and rate constant of the given reaction has to be determined.

    ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(g)\to {\text{CO}}_{\text{2}}(g)+{\text{HCHO}}_{\text{2}}(g)$

Concept Introduction:

Rate of a reaction depends on the species concentration that takes place in the reaction.  Chemical kinetic experiments are used in order to deduce the rate of reaction.  Rate law of a reaction can only be identified by experimental methods that are adopted.

Answer to Problem 17.28P

Rate law of the reaction is $\ln P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}=\ln P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-kt$ and rate constant of the reaction is $2.9\times {10}^{-5}{\text{s}}^{-1}$.

Explanation of Solution

Given reaction is:

    ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(g)\to {\text{CO}}_{\text{2}}(g)+{\text{HCHO}}_{\text{2}}(g)$

Rate of the reaction is measured using the change in pressure.  The total pressure that is reached after $2.00\times {10}^4\text{seconds}$ is given as follows:

$\begin{array}{c}{P}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}\text{/Torr}\\ \text{(at}t=0)\end{array}$	$\begin{array}{c}{P}_{\text{total}}^{\text{o}}\text{/Torr}\\ \text{(at}t=2.00\times {10}^{4}s)\end{array}$
$5.0$	$7.2$
$7.0$	$10$
$8.4$	$12$

By Dalton’s law of partial pressure, the total pressure at any time of the reaction can be given by the equation shown below:

  $P_{\text{Total}}=P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}+P_{{\text{CO}}_{\text{2}}}+P_{{\text{HCHO}}_{\text{2}}}$        (1)

Quantity of oxalic acid that undergoes the reaction is directly proportional to $P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}$.

For one mole of ${\text{HCHO}}_{\text{2}}$ that is formed, the equation can be written as follows:

  $P_{{\text{HCHO}}_{\text{2}}}=P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}$

For one mole of ${\text{CO}}_{\text{2}}$ that is formed, the equation can be written as follows:

  $P_{{\text{CO}}_{\text{2}}}=P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}$

Substituting these values in equation (1) as shown below:

  $\begin{array}{c}{P}_{\text{Total}}=P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}+P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}+P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}\\ =2P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}\end{array}$

Rearranging the above equation in terms of $P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}$.

  $P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}=2P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-P_{\text{Total}}$

Solving for $P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}$, the table obtained is:

$\begin{array}{c}{P}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}\text{/Torr}\\ \text{(at}t=0)\end{array}$	$\begin{array}{c}{P}_{\text{total}}^{\text{o}}\text{/Torr}\\ \text{(at}t=2.00\times {10}^{4}s)\end{array}$	$\begin{array}{c}{P}_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}\\ \text{(at}t=2.00\times {10}^{4}s)\end{array}$
$5.0$	$7.2$	$2.8$
$7.0$	$10$	$4.0$
$8.4$	$12$	$4.8$

Considering the reaction to be first order, the rate law of the reaction can be given as follows:

    $\ln P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}=\ln P_{{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}^{\text{o}}-kt$        (2)

Substituting the values in equation (2), the rate constant for the reaction is calculated as shown below:

    $\begin{array}{c}\ln (2.8)=\ln (5.0)-k(2.00\times {10}^4\text{s)}\\ k=-\frac{1}{2.00\times {10}^4\text{s}}\ln \frac{2.8}{5.0}\\ =-\frac{1}{2.00\times {10}^4\text{s}}\left(-0.5798\right)\\ =0.2899\times {10}^{-4}{\text{s}}^{-1}\\ =2.9\times {10}^{-5}{\text{s}}^{-1}\end{array}$

Thus the rate constant of the reaction is $2.9\times {10}^{-5}{\text{s}}^{-1}$.