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  • For a reaction given below is a graph of <img alt="\mathrm{lnk\, Vs\, \frac{1}{T}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Blnk%5C%2C%20Vs%5C%2C%20%5Cfrac%7B1%7D%7BT%7D%7D"

For a reaction given below is a graph of \mathrm{lnk\, Vs\, \frac{1}{T}} The activation energy for the reaction will be equal to how many cal mol \mathrm{\left\{\right. given \left.R=2 \, \, \mathrm{cal\, \, k}^{-1} \mathrm{~mol}^{-1}\right\}.}

Option: 1

\mathrm{25\, \mathrm{cal} / \mathrm{mol}}


Option: 2

\mathrm{10\, \mathrm{cal} / \mathrm{mol}}


Option: 3

\mathrm{5\, \mathrm{cal} / \mathrm{mol}}


Option: 4

\mathrm{30\, \mathrm{cal} / \mathrm{mol}}


Answers (1)

best_answer

\mathrm{\begin{aligned} \text { slope k } & =A e^{-Ea/RT} \\ \ln k & =\frac{-E a}{R T}+\ln A \\ \text { slope } & =\frac{E a}{R}=\frac{25}{5}=5 \\ E a & =5 R \\ E a & =5 \times 2 \\ E a & =10 \, \, cal/mol \end{aligned}}

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sudhir kumar

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