Question

The rate of a reaction decreased by $3.555$ times when the temperature was changed from ${40}^{°}\mathrm{C}\mathrm{to}30°\mathrm{C}$. The activation energy $(\mathrm{in}\mathrm{KJ}{\mathrm{mol}}^{-1})$ of the reaction is……… $[\mathrm{Take};\mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1},\mathrm{In}3.555=1.268]$

Answer 1

Given:

Rate of Reaction, ${\mathrm{K}}_{{40}^{°}\mathrm{C}}=\mathrm{K};\mathrm{and}{\mathrm{K}}_{{30}^{°}\mathrm{C}}=\mathrm{K}/3.555$

$\mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1},\mathrm{In}3.555=1.268$

The expression used in the calculation of Activation Energy:

$\mathrm{In}\{3.555\}={\mathrm{E}}_{\mathrm{a}}/\mathrm{R}\{1/{\mathrm{T}}_2-{\mathrm{T}}_1\}$

Putting values in the above expression:

$\begin{array}{rcl}\mathrm{In}\{3.555\}& =& {\mathrm{E}}_{\mathrm{a}}/\mathrm{R}\{1/303-1/313\}\\ {\mathrm{E}}_{\mathrm{a}}& =& \{1.268\times 8.314\times 313\times 303\}/10\\ & =& 99980.7\mathrm{J}{\mathrm{mol}}^{-1}\\ & =& 99.98\mathrm{kJ}{\mathrm{mol}}^{-1}\\ & =& 100\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

Final Answer: Therefore, the value of activation energy is $100\mathrm{kJ}{\mathrm{mol}}^{-1}$