Ncert Exercises & Solutions

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

Calculate the rate constant.

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

$(C H_{3} C H N = N N C H {(C H_{3})}_{2 (g)} \to N_{2 (g)} + C_{6} H_{14 (g)}$
$A t t = 0 P_{0} 0 0$
$A t t = t P_{0} - p p p$
$P_{t} = (P_{0} - p) + p + p$

After time, t, total pressure,

$\Rightarrow P_{t} = P_{0} + P$
$\Rightarrow p = P_{t} - P_{0}$
$T h e r e f o r e, P_{0} - p = P_{0} (P_{t} - P_{0})$
$= 2 P_{0} - P_{t}$

For a first order reaction,

$K = \frac{2.303}{t} \log \frac{P_{0}}{P_{0} - p}$
$= \frac{2.303}{t} \log \frac{P_{0}}{2 P_{0} - P_{t}}$
$K = \frac{2.303}{360 s} \log \frac{35.0}{2 \times 35.0 - 54.0}$
$W h e n t = 360 s,$
$= 2.175 \times 10^{- 3} s^{- 1}$
$K = \frac{2.303}{720 s} \log \frac{35.5}{2 \times 35.0 - 63.0}$
$W h e n t = 720 s,$
$= 2.235 \times 720 s^{- 1}$

Hence, the average value of rate constant is

$k = \frac{(2.175 \times 10^{- 3}) + (2.235 \times 10^{- 3})}{2} s^{- 1}$
$= 2.21 \times 10^{- 3} s^{- 1}$

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