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Question

The rate of a first order reaction gets doubled when the temperature is doubled from T1 to 2 T1 (in K). Ea is nearly constant activation energy at 1386 kcal. What is the value of T1?
(given R=2 calK−1mol−1,ln2=0.693)

A
250 K
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B
500 K
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C
1000 K
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D
750 K
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Solution

The correct option is B 500 K
According to the Arrhenius equation,
k=AeEaRT
Taking logarithm we will get,
lnk1=lnAEaRT1.....(1)
lnk2=lnAEaR(2 T)1.......(2)
Subtracting equation (1) from (2) we get
lnk2k1=EaR(1T112 T1 )ln 2=13862(1T112 T1)T1=13860.693×4=500 K

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