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Question

At a certain temperature, the first order rate constant k1 is found to be smaller than the second order rate constant k2. If the energy of activation E1 of the first order reaction is greater than energy of activation E2 of the second order reaction, then with increase in temperature:

A
k1 will increase faster than k2, but always will remain less than k2
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B
k2 will increase faster than k1
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C
k1 will increase faster than k2 and becomes equal to k2
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D
k1 will increase faster than k2 and becomes greater than k2
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Solution

The correct option is A k1 will increase faster than k2, but always will remain less than k2
The relationship between the energy of activation of the reaction and the rate constant of the reaction is d(lnk)dT=EaRT2
With increase in the energy of activation, the increase in the rate constant of the reaction will be faster with increase in temperature.
Hence, k1 will increase faster than k2, but always will remain less than k2.

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