Question

When the temperature is raised from $27\mathrm{to}42°\mathrm{C}$, there are five times as many molecules with energy above the reaction threshold. Its activation energy is _______ $\mathrm{J}/\mathrm{mol}$. $\left(\mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1},\mathrm{Take}\ln 5=1.6094\right)$

Answer 1

Given:

$\begin{array}{rcl}{\mathrm{T}}_1& =& 300\mathrm{K}\\ {\mathrm{T}}_2& =& 315\mathrm{K}\\ \mathrm{R}& =& 8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\\ \ln 5& =& 1.6094\end{array}$

According to the question:

${\mathrm{K}}_{{\mathrm{T}}_2}=5{\mathrm{K}}_{{\mathrm{T}}_2}$

$\mathrm{K}$ will increase five times as molecules are stimulated five times more.

$$\begin{gathered} \mathrm{In}\left(\frac{{\mathrm{K}}_{{\mathrm{T}}_2}}{{\mathrm{K}}_{{\mathrm{T}}_1}}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \\ \mathrm{In}5=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{15}{300\times 315}\right) \\ {\mathrm{E}}_{\mathrm{a}}=\frac{1.6094\times 8.314\times 300\times 315}{15} \end{gathered}$$

Therefore, the value of the activation energy is, ${\mathrm{E}}_{\mathrm{a}}=84297.47\mathrm{Joules}/\mathrm{mole}$.