The rate constants for decomposition of acetaldehyde have been measured over the temperature range $700 - 1000 K$ . The data has been analyzed by plotting In k vs $\frac{10^{3}}{T}$ graph. The value of activation energy for the reaction is _____ ${kJmol}^{- 1}$ . (Nearest integer) (Given : $R = 8.31 J K^{- 1} {mol}^{- 1}$ )

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answer is 154.

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Detailed Solution

$lnk = lnA - \frac{Ea}{10^{3} RT} \times 10^{3} = lnA + \frac{10^{3}}{T} [- \frac{Ea}{10^{3} RT}] From the graph \begin{aligned} \frac{- Ea}{10^{3} \times R} = - 18 .5 \\ Ea = 153 .735 kJ / mol \\ \approx 154 \end{aligned}$

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