Question
Option 2 : 53.6 kJ/mol
Detailed Solution
सही उत्तर 53.6 kJ/mol है।
prepsnap.in • Part -AQuestion:
The rate constant of a reaction doubles when the temperature is raised from $300$ K to $310$ K. The activation energy of the reaction is approximately: (R = $8.314$ J/mol·K) / किसी अभिक्रिया का दर स्थिरांक ताप $300$ K से $310$ K बढ़ाने पर दोगुना हो जाता है। अभिक्रिया की सक्रियण ऊर्जा लगभग है: (R = $8.314$ J/mol·K)
Correct Answer: 53.6 kJ/mol
From $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$: $\log 2 = \frac{E_a}{2.303 \times 8.314}\left(\frac{10}{300 \times 310}\right)$, giving $E_a = \frac{0.3010 \times 2.303 \times 8.314 \times 300 \times 310}{10} = 53.6$ kJ/mol. / $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ से: $\log 2 = \frac{E_a}{2.303 \times 8.314}\left(\frac{10}{300 \times 310}\right)$, जिससे $E_a = \frac{0.3010 \times 2.303 \times 8.314 \times 300 \times 310}{10} = 53.6$ kJ/mol।