For the decomposition of azoisopropane to hex | vedclass

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(N/A) The decomposition reaction is: $(CH_3)_2CHN=NCH(CH_3)_2(g) \rightarrow C_6H_{14}(g) + N_2(g)$.Let $P_0$ be the initial pressure of azoisopropane

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(N/A) The decomposition reaction is: $(CH_3)_2CHN=NCH(CH_3)_2(g) \rightarrow C_6H_{14}(g) + N_2(g)$.
Let $P_0$ be the initial pressure of azoisopropane at $t=0$. At time $t$,let $p$ be the decrease in pressure of azoisopropane.
Total pressure $P_t = (P_0 - p) + p + p = P_0 + p$.
Thus,$p = P_t - P_0$.
The pressure of azoisopropane at time $t$ is $P_0 - p = P_0 - (P_t - P_0) = 2P_0 - P_t$.
For a first-order reaction: $k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}$.
At $t = 360 \ s$: $k = \frac{2.303}{360} \log \frac{35.0}{2(35.0) - 54.0} = \frac{2.303}{360} \log \frac{35.0}{16.0} \approx 2.175 \times 10^{-3} \ s^{-1}$.
At $t = 720 \ s$: $k = \frac{2.303}{720} \log \frac{35.0}{2(35.0) - 63.0} = \frac{2.303}{720} \log \frac{35.0}{7.0} = \frac{2.303}{720} \log 5 \approx 2.235 \times 10^{-3} \ s^{-1}$.
Average $k = \frac{2.175 \times 10^{-3} + 2.235 \times 10^{-3}}{2} = 2.21 \times 10^{-3} \ s^{-1}$.

$t \ (sec)$	$P \ (mm \ of \ Hg)$
$0$	$35.0$
$360$	$54.0$
$720$	$63.0$

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

$t \ (sec)$ $P \ (mm \ of \ Hg)$

$0$ $35.0$

$360$ $54.0$

$720$ $63.0$

Calculate the rate constant.

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For the decomposition of azoisopropane to hexane and nitrogen at $543 \ K,$ the following data are obtained. $t \ (sec)$ $P \ (mm \ of \ Hg)$ $0$ $35.0$ $360$ $54.0$ $720$ $63.0$ Calculate the rate constant.