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

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

            

          Calculate the rate constant.

Answers (1)

best_answer

Decomposition is represented by the equation-

\left(\mathrm{CH}_3\right)_2 \mathrm{CHN}=\mathrm{NCH}\left(\mathrm{CH}_3\right)_{2(\mathrm{~g})} \longrightarrow \mathrm{N}_{2(\mathrm{~g})}+\mathrm{C}_6 \mathrm{H}_{14(\mathrm{~g})}

\begin{tabular}{lccc} At $t=0$ & $\mathrm{P}_0$ & 0 & 0 \\ At $t=t$ & $\mathrm{P}_0-p$ & $p$ & $p$ \end{tabular}


After t time, the total pressure p_T=p_0-p+(p+p)=p_0+p

                                               So, p=p_t-p_0

thus, p_0-p=2 p_0-p_t

for first order reaction,

\begin{aligned} k & =\frac{2.303}{t} \log \frac{p_0}{p_0-p} \\ & =\frac{2.303}{360} \log \frac{p_0}{2 p_0-p_t}\end{aligned}


now putting the values of pressure,

when t =360sec

\begin{aligned} & =\frac{2.303}{360} \log \frac{35}{2 * 35-54} \\ & =2.175 \times 10^{-3} \mathrm{~s}^{-1}\end{aligned}

when t = 270sec

\begin{aligned} & =\frac{2.303}{270} \log \frac{35}{2 * 35-54} \\ & =2.235 \times 10^{-3} s^{-1}\end{aligned}

So, k_{a v g}=k_1+k_2 / 2
              =2.21 \times 10^{-3} \mathrm{~s}^{-1}

Posted by

manish

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