Q.

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C?R=8.314Jmol−1K−1

Moderate

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answer is 1.

(Detailed Solution Below)

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Detailed Solution

logk2k1=Ea2.303R1T1−1T2 k2=2k1, T1=20+273=293K or T2=35+273=308K R=8.314 J mol−1K−1 log2=Ea2.303×8.3141293−1308 0.3010=Ea19.147×15293×308 Ea=34673Jmol−1 or 34.7kJmol−1

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