Question

For the decomposition of azo-isopropane to hexane and nitrogen at 543K, the following data are obtained.

Time in seconds	Pressure in ${\text{mm}}$ of ${\text{Hg}}$
$0$	$35.0$
$360$	$54.0$
$720$	$63.0$

Calculate the rate constant.

Answer 1

Hint: Rate constant is a proportionality constant that appears in a rate law. It is independent of the concentrations but depends on other factors, most notably temperature. Order of a reaction is the sum of powers of concentration terms in the rate equation. In first order reaction, reaction rate is determined by one concentration term.

Complete step by step answer:
The decomposition reaction of azo-isopropane to hexane is given below:
${\left( {{\text{C}}{{\text{H}}_3}} \right)_2}{\text{CHN = NCH}}{\left( {{\text{C}}{{\text{H}}_3}} \right)_2} \to {{\text{C}}_6}{{\text{H}}_{14}} + {{\text{N}}_2}$
Let the initial pressure of azo-isopropane, ${{\text{P}}_{{\text{azo}}}}$ be ${{\text{P}}_{\text{i}}}$, pressure of ${{\text{N}}_2}$ be ${{\text{P}}_{{{\text{N}}_2}}}$, pressure of hexane be${{\text{P}}_{{{\text{C}}_6}{{\text{H}}_{14}}}}$.
Now at time, ${\text{t}} = 0$,
${{\text{P}}_{{\text{azo}}}} = {{\text{P}}_{\text{i}}},{{\text{P}}_{{{\text{N}}_2}}} = 0,{{\text{P}}_{{{\text{C}}_6}{{\text{H}}_{14}}}} = 0$ since there is no decomposition occurs at time, ${\text{t}} = 0$.
But at time ${\text{t}}$,
${{\text{P}}_{{\text{azo}}}} = {{\text{P}}_{\text{i}}} - {\text{p}},{{\text{P}}_{{{\text{N}}_2}}} = {\text{p}},{{\text{P}}_{{{\text{C}}_6}{{\text{H}}_{14}}}} = {\text{p}}$
We know that the total pressure will be the sum of the pressures at time ${\text{t}}$.
i.e. Total pressure, ${{\text{P}}_{\text{t}}} = \left( {{{\text{P}}_{\text{i}}} - {\text{p}}} \right) + {\text{p}} + {\text{p}}$
${{\text{P}}_{\text{t}}}{\text{ = }}{{\text{P}}_{\text{i}}} + {\text{p}} \Leftrightarrow {\text{p = }}{{\text{P}}_{\text{t}}} - {{\text{P}}_{\text{i}}}$
Thus we can write ${{\text{P}}_{{\text{azo}}}}$ as ${{\text{P}}_{{\text{azo}}}} = {{\text{P}}_{\text{i}}} - \left( {{{\text{P}}_{\text{t}}} - {{\text{P}}_{\text{i}}}} \right) = 2{{\text{P}}_{\text{i}}} - {{\text{P}}_{\text{t}}}$
Here, the rate is expressed with respect to pressure.
We know that for first order reaction, rate constant, ${\text{k}} = \dfrac{{2.303}}{{\text{t}}}\log \dfrac{{{{\left[ {\text{A}} \right]}_0}}}{{{{\left[ {\text{A}} \right]}_{\text{t}}}}}$, where ${\left[ {\text{A}} \right]_0}$ is the initial value(here, it is initial pressure) and ${\left[ {\text{A}} \right]_{\text{t}}}$ is the final value (here, it is final pressure).
Substituting the values, we get
${\text{k}} = \dfrac{{2.303}}{{\text{t}}}\log \dfrac{{{{\text{P}}_{\text{i}}}}}{{2{{\text{P}}_{\text{i}}} - {{\text{P}}_{\text{t}}}}}$
When time, ${\text{t}} = 360{\text{s}}$, ${{\text{P}}_{\text{i}}} = 35.0{\text{mm}}$, ${{\text{P}}_{\text{t}}} = 54.0$
${\text{k}} = \dfrac{{2.303}}{{360{\text{s}}}}\log \dfrac{{35.0}}{{2 \times 35.0 - 54.0}}$
$ {\text{k}} = 6.39 \times {10^{ - 3}}\log 2.18 \\
{\text{k}} = 6.39 \times {10^{ - 3}} \times 0.34 \\
{\text{k}} = 2.17 \times {10^{ - 3}}{{\text{s}}^{ - 1}} \\ $
When time, ${\text{t}} = 720{\text{s}}$, ${{\text{P}}_{\text{i}}} = 35.0{\text{mm}}$, ${{\text{P}}_{\text{t}}} = 63.0$
${\text{k}} = \dfrac{{2.303}}{{720{\text{s}}}}\log \dfrac{{35.0}}{{2 \times 35.0 - 63.0}}$
On simplification, we get
$ {\text{k}} = 3.19 \times {10^{ - 3}}\log \dfrac{{35.0}}{7} \\
{\text{k}} = 3.19 \times {10^{ - 3}}\log 5 \\
{\text{k}} = 3.19 \times {10^{ - 3}} \times 0.69 \\$
By solving, we get
${\text{k}} = 2.23 \times {10^{ - 3}}{{\text{s}}^{ - 1}}$
Thus we get both the rate constants almost equal.
Now we can calculate the average rate constant.
i.e. ${{\text{k}}_{{\text{avg}}}} = \dfrac{{2.17 \times {{10}^{ - 3}}{{\text{s}}^{ - 1}} + 2.23 \times {{10}^{ - 3}}{{\text{s}}^{ - 1}}}}{2} = 2.20 \times {10^{ - 3}}{{\text{s}}^{ - 1}}$

Thus, the rate constant is $2.20 \times {10^{ - 3}}{{\text{s}}^{ - 1}}$.

Note: In first order reaction, the rate of reaction is proportional to the concentration of reacting substance. The unit of first order reaction is inverse of time. Half-life can also be determined from the value of rate constant. It is the time required for half of the compound to decompose. It is inversely proportional to the rate constant.