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Q.

The ratio of rate constants at 27oC and 37oC is Q10. What should be the energy of activation of a reaction for which Q10=2.5?

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a

71 kJ

b

212 kJ

c

35 kJ

d

12.1 kJ

answer is A.

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Detailed Solution

log⁡Q10   or  log⁡2.5=Ea2.303R1T1−1T2 log⁡2.5=Ea2.303RT2−T1T1T2log 2.5=Ea2.303×8.314×10-310300×310 Ea=71 kj
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