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Q.

The rate constant is doubled when temperature increases from 27°C to 37°C. Activation energy in kJ is

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a

34

b

54

c

100

d

50

answer is B.

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Detailed Solution

log⁡K2K1=Ea2.303R1T1−1T2If K2K1=2log⁡2=Ea2.303×8.3141300−1310Ea=3010×2.303×8.314300×31010     =53598.9 Jmol−1=54 kJ
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