Q80 / 180 · NEET 2013 (All India) · Chemistry
What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^{\circ} \mathrm{C}$ to $35^{\circ} \mathrm{C} ?\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
- A$342 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$269 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerC$34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$15.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Given, initial temperature,
$T_1=20+273=293 \mathrm{~K}$
Final temperature
$\begin{aligned}
T_2 & =35+273 \\
& =308 \mathrm{~K} \\
R & =8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
\end{aligned}$
Since, rate becomes double on raising temperature,
$\therefore r_2=2 r_1 \text { or } \frac{r_2}{r_1}=2$
As rate constant, $k \propto r$
$\therefore \frac{k_2}{k_1}=2$
From Arrnhenius equation, we know that
$\begin{aligned}
& \log \frac{k_2}{k_1}=-\frac{E_2}{2.303 R}\left[\frac{T_1-T_2}{T_1 T_2}\right] \\
& \log 2=-\frac{E_2}{2.303 \times 8.314}\left[\frac{293-308}{293 \times 308}\right] \\
& 0.3010=-\frac{E_2}{2.303 \times 8.314}\left[\frac{-15}{293 \times 308}\right]
\end{aligned}$
$\begin{aligned}
\therefore E_2 & =\frac{0.3010 \times 2.303 \times 8.314 \times 293 \times 308}{15} \\
& =34673.48 \mathrm{~J} \mathrm{~mol}^{-1}=34.7 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
Hence, the correct answer is option: 'C'.