For the decomposition of azoisopropane to hexane and nitrogen at 543 K,

Question:

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

Calculate the rate constant.

Solution:

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, $t$, total pressure, $\mathrm{P}_{t}=\left(\mathrm{P}_{0}-p\right)+p+p$

$\Rightarrow \mathrm{P}_{t}=\mathrm{P}_{0}+p$

$\Rightarrow p=\mathrm{P}_{\mathrm{t}}-\mathrm{P}_{0}$

Therefore, $\mathrm{P}_{\mathrm{o}}-p=\mathrm{P}_{\mathrm{o}}-\left(\mathrm{P}_{\mathrm{t}}-\mathrm{P}_{\mathrm{o}}\right)$

= 2P0 − Pt

For a first order reaction,

$k=\frac{2.303}{t} \log \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-p}$

$=\frac{2.303}{t} \log \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{t}}$

When $t=360 \mathrm{~s}, k=\frac{2.303}{360 \mathrm{~s}} \log \frac{35.0}{2 \times 35.0-54.0}$

= 2.175 × 10−3 s−1

When $t=720 \mathrm{~s}, k=\frac{2.303}{720 \mathrm{~s}} \log \frac{35.0}{2 \times 35.0-63.0}$

= 2.235 × 10−3 s−1

Hence, the average value of rate constant is

$k=\frac{\left(2.175 \times 10^{-3}\right)+\left(2.235 \times 10^{-3}\right)}{2} \mathrm{~s}^{-1}$

= 2.21 × 10−3 s−1

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

 

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Comments

Pradeep
Oct. 7, 2022, 2:55 p.m.
It seems that excess data is given in question.The answer can be obtained by pressure value at 360 seconds OR pressure value at 720 seconds.
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