Chapter 17, Problem 17B.4P

Interpretation Introduction

Interpretation:

From the data given the order of the reaction, rate constant and the molar concentration of ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}\text{CBr}$ remaining after $\text{43}\text{.8h}$ for the reaction ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}\text{CBr(aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\stackrel{}{\to }{{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}\text{COH(aq)}\text{+}\text{HBr(aq)}$

Explanation of Solution

For the first order reaction:

    $\text{ln}\left(\frac{\text{[A]}}{{\text{[A]}}_{\text{0}}}\right)\text{=}\text{-kt}$

Plot a graph between $\text{ln}\left(\frac{\text{[A]}}{{\text{[A]}}_{\text{0}}}\right)\text{against}\text{t}$ to see if a straight line is obtained.

  $\begin{array}{l}\text{t/h}\text{0}\text{3}\text{.15}\text{6}\text{.20}\text{10}\text{.00}\text{18}\text{.30}\text{30}\text{.80}\\ {\text{[A]/(10}}^{\text{-2}}{\text{moldm}}^{\text{-3}}\text{)}\text{10}\text{.39}\text{8}\text{.96}\text{7}\text{.76}\text{6}\text{.39}\text{3}\text{.53}\text{2}\text{.07}\\ \frac{\text{[A]}}{{\text{[A]}}_{\text{0}}}\text{1}\text{0}\text{.862}\text{0}\text{.747}\text{0}\text{.615}\text{0}\text{.340}\text{0}\text{.199}\\ \text{ln}\frac{\text{[A]}}{{\text{[A]}}_{\text{0}}}\text{0}\text{-0}\text{.148}\text{-0}\text{.292}\text{-0}\text{.486}\text{-1}\text{.080}\text{-1}\text{.613}\\ \left(\frac{\text{1}}{{\text{[A]}}_{\text{0}}}\right){\text{/(dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\text{)}\text{9}\text{.62}\text{11}\text{.16}\text{12}\text{.89}\text{15}\text{.65}\text{28}\text{.3}\text{48}\text{.3}\end{array}$

Below is the graph:

Figure.1

From the graph it is clear that it is a first order reaction.

The rate constant is negative of the slope of the first order plot:

$\begin{array}{l}\text{k}\text{=}\text{0}{\text{.0542h}}^{\text{-1}}\text{=}\text{1}{\text{.51×10}}^{\text{-5}}{\text{s}}^{\text{-1}}\\ \\ \text{At}\text{43}\text{.8h}\\ \\ \text{ln}\left(\frac{\text{[A]}}{{\text{[A]}}_{\text{0}}}\right)\text{=}\text{-(0}{\text{.0542h}}^{\text{-1}}\text{)×(43}\text{.8h)}\text{=}\text{-2}\text{.359}\\ \\ \text{[A]}\text{=}\text{(10}{\text{.39×10}}^{\text{-2}}\text{mol}{\text{dm}}^{\text{-3}}\text{)}\text{×}{\text{e}}^{\text{-2}\text{.359}}\text{=}\text{9}{\text{.82×10}}^{\text{-3}}\text{mol}{\text{dm}}^{\text{-3}}\end{array}$