The rate of a reaction decreased by

Question:

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. The activation energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of the reaction is______.

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

$\mathrm{E}_{\mathrm{a}}=99.98 \frac{\mathrm{kJ}}{\mathrm{mole}}$

Solution:

$\ell\left(\frac{\mathrm{K}_{\mathrm{T}_{2}}}{\mathrm{~K}_{\mathrm{T}_{1}}}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

$\mathrm{T}_{1}=303 \mathrm{~K} ; \mathrm{T}_{2}=313 \mathrm{~K}$

$\frac{\mathrm{K}_{\mathrm{T}_{2}}}{\mathrm{~K}_{\mathrm{T}_{1}}}=3.555$

$\ell \mathrm{n}(3.555)=\frac{\mathrm{E}_{\mathrm{a}}}{8.314}\left[\frac{1}{303}-\frac{1}{313}\right]$

$\mathrm{E}_{\mathrm{a}}=99980.715$

 

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