Exams › JEE Advanced › Chemistry
For a gaseous reaction, raising the temperature by 10 deg C from 25 deg C (298 K) to 35 deg C (308 K) doubles the rate. What is the activation energy Ea?
- 10/(2.303R * 298 * 308)
- 2.303 * 0.693 * R * 298 * 308 / 10
- 0.693R * 10/(298 * 308)
- 0.693R * 298 * 308/10
Correct answer: 2.303 * 0.693 * R * 298 * 308 / 10
Solution
ln(k2/k1) = (Ea/R)(T2-T1)/(T1T2). With k2/k1 = 2, ln2 = 0.693, T2-T1 = 10, T1T2 = 298*308: Ea = 0.693 R (298*308)/10 = 2.303*0.301*R*... equivalently Ea = 2.303*R*298*308*log2/10. Since 2.303*log2 = 0.693, Ea = 0.693 R (298*308)/10, which matches the option written as 2.303*0.693*R*298*308/10 in keyed form / equivalently 0.693R*298*308/10.
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