For the decomposition of azoisopropane to hexane and nitrogen at $543$ $K ,$ the following data are obtained.
$t$ $(sec)$
$P(m m \text { of } H g)$
$0$
$35.0$
$360$
$54.0$
$720$
$63.0$
Calculate the rate constant.
For the decomposition of azoisopropane to hexane and nitrogen at $543$ $K ,$ the following data are obtained.
| $t$ $(sec)$ | $P(m m \text { of } H g)$ |
| $0$ | $35.0$ |
| $360$ | $54.0$ |
| $720$ | $63.0$ |
Calculate the rate constant.
The decomposition of azoisopropane to hexane and nitrogen at $543 K$ is represented by the following equation.
After time, $t,$ total pressure$P _{t}=\left( P _{0}-p\right)+p+p$
$\Rightarrow P _{t}= P _{0}+p$
$\Rightarrow p= P _{t}- P _{0}$
Therefore, $P_{0}-p=P_{0}-\left(P_{t}-P_{0}\right)$
$=2 P_{0}-P_{t}$
For a first order reaction,
$k=\frac{2.303}{t} \log \frac{ P _{0}}{ P _{0}-p}$
$=\frac{2.303}{t} \log \frac{P_{0}}{2 P_{0}-P_{t}}$
When $t=360 \,s$,
$=2.175 \times 10^{-3} \,s ^{-1}$
$k=\frac{2.303}{360 s } \log \frac{35.0}{2 \times 35.0-54.0}$
When $t=720\, s$
$=2.235 \times 10^{-3} \,s ^{-1}$
$k=\frac{2.303}{720 s } \log \frac{35.0}{2 \times 35.0-63.0}$
Hence, the average value of rate constant is
$k=\frac{\left(2.175 \times 10^{-3}\right)+\left(2.235 \times 10^{-3}\right)}{2} s ^{-1}$
$=2.21 \times 10^{-3} \,s ^{-1}$
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