Question

For the decomposition of Azoisopropane to Hexane and Nitrogen at $543\mathrm{K}$. The following data is obtained.

t(sec)	P(mm of $\mathrm{Hg}$)
$0$	$35$
$360$	$54$
$720$	$63$

Calculate the rate constant.

Answer 1

Step 1: Given data

Temperature$=543\mathrm{K}$

t(sec)	P(mm of $\mathrm{Hg}$)
$0$	$35$
$360$	$54$
$720$	$63$

Step 2: Calculating rate constant

The equation for the decomposition of Azoisopropane to hexane and Nitrogen $543\mathrm{K}$ is $\underset{\mathrm{Azoisopropane}}{{\left({\mathrm{CH}}_3\right)}_2\mathrm{CHN}=\mathrm{NCH}{\left({\mathrm{CH}}_3\right)}_2}\to \underset{\mathrm{dinitrogen}}{{\mathrm{N}}_2}+\underset{\mathrm{Hexane}}{{\mathrm{C}}_6{\mathrm{H}}_{14}}$

From the decomposition reaction of Azoisopropane, Let the initial pressure of azoisopropane at $\mathrm{t}=0$ is ${\mathrm{P}}_{°}$, and after some time decomposition of azoisopropane occur and pressure of Azoisopropane is ${\mathrm{P}}_{°}-\mathrm{p}$, the pressure of nitrogen and hexane is $\mathrm{p}$

Time	Pressure of Azoisopropane	Pressure of nitrogen	Pressure of hexane
At $\mathrm{t}=0$	${\mathrm{P}}_{°}$	$0$	$0$
At $\mathrm{t}=\mathrm{t}$	${\mathrm{P}}_{°}-\mathrm{p}$	$\mathrm{p}$	$\mathrm{p}$

After a time $\mathrm{t}$, the total pressure, ${\mathrm{P}}_{\mathrm{t}}=({\mathrm{P}}_{°}-\mathrm{p})+\mathrm{p}+\mathrm{p}$

$$\begin{gathered} {\mathrm{P}}_{\mathrm{t}}={\mathrm{P}}_{°}+\mathrm{p} \\ \mathrm{p}={\mathrm{P}}_{\mathrm{t}}-{\mathrm{P}}_{°} \end{gathered}$$

Therefore, ${\mathrm{P}}_{°}-\mathrm{p}={\mathrm{P}}_{°}-({\mathrm{P}}_{\mathrm{t}}-{\mathrm{P}}_{°})$

$=2{\mathrm{P}}_{°}-{\mathrm{P}}_{\mathrm{t}}$

For the first-order reaction

$$\begin{gathered} =\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_{°}}{{\mathrm{P}}_{°}-\mathrm{p}} \\ =\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_{°}}{2{\mathrm{P}}_{°}-{\mathrm{P}}_{\mathrm{t}}} \end{gathered}$$

Where $\mathrm{t}=360\mathrm{s}$

$$\begin{gathered} \mathrm{K}=\frac{2.303}{360\mathrm{s}}\log \frac{35}{2\times 35-54} \\ \mathrm{K}=2.175\times {10}^{-3}{\mathrm{s}}^{-1} \end{gathered}$$

When $\mathrm{t}=720\mathrm{s}$

$$\begin{gathered} \mathrm{K}=\frac{2.303}{720\mathrm{s}}\log \frac{35}{2\times 35-63} \\ \mathrm{K}=2.235\times {10}^{-3}{\mathrm{s}}^{-1} \end{gathered}$$

Therefore, the average value of the rate constant is

$$\begin{gathered} \mathrm{K}=\frac{(2.175\mathrm{x}{10}^{-3})+(2.235\times {10}^{-3})}{2}{\mathrm{s}}^{-1} \\ \mathrm{K}=2.21\times {10}^{-3}{\mathrm{s}}^{-1} \end{gathered}$$

Therefore, the rate constant for the decomposition of Azoisopropane to Hexane and Nitrogen is $2.21\times {10}^{-3}{\mathrm{s}}^{-1}$.