Q33 / 180 · KCET 2012 · Chemistry
A given sample of milk turns sour at room temperature $\left(27^{\circ} \mathrm{C}\right)$ in $5 \mathrm{~h}$. In a refrigerator at $-3^{\circ} \mathrm{C}$, it can be stored 10 times longer. The energy of activation for the souring of milk is
- A$2.303 \times 5 \mathrm{R} \mathrm{kJ} \cdot \mathrm{mol}^{-1}$
- B$2.303 \times 3 \mathrm{R} \mathrm{kJ} \cdot \mathrm{mol}^{-1}$
- Correct answerC$2.303 \times 2.7 \mathrm{R} \mathrm{kJ}^{-1} \mathrm{~mol}^{-1}$
- D$2.303 \times 10 \mathrm{R} \mathrm{kJ} \cdot \mathrm{mol}^{-1}$
Solution
Energy of activation is given by
$$
\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right]
$$
Here, $\quad \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{1}{10}$
$\mathrm{T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
$\mathrm{T}_{2}=-3^{\circ} \mathrm{C}=270 \mathrm{~K}$
$\therefore \quad \log \frac{1}{10}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{270-300}{270 \times 300}\right)$
$-1=-\frac{E_{a}}{2.303 R} \frac{30}{270 \times 300}$
$\mathrm{E}_{\mathrm{a}}=+2.303 \times 2700 \times \mathrm{R} \mathrm{J} \mathrm{mol}^{-1}$
$=+2.303 \times 2.7 \times \mathrm{R} \mathrm{kJ} \mathrm{mol}=1$
Hence, the correct answer is option: 'C'.