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Question

Chemistry Question on kinetics equations

The decomposition of a hydrocarbon follows the equation k=(4.5×1011s1)e28000K/T.k=\left(4.5\times10^{11}s^{-1}\right)e^{-28000 K /T}. What will be the value of activation energy?

A

669KJ669 \, KJ mol1mol^{-1}

B

232.79KJ232.79 \, KJ mol1mol^{-1}

C

4.5×1011KJ4.5\times10^{11} \,KJ mol1mol^{-1}

D

28000KJ28000 \, KJ mol1mol^{-1}

Answer

232.79KJ232.79 \, KJ mol1mol^{-1}

Explanation

Solution

Arrhenius equation, k=AeEa/RTk=Ae^{-Ea/ RT} Given equation is k=(4.5×1011s1)k =\left(4.5\times10^{11}s^{-1}\right) e28000K/Te^{-28000 K /T} Comparing both the equations, we get EaRT-\frac{E_{a}}{RT} =28000KT=-\frac{28000\, K }{T} EaE_{a} =28000K×R=28000K×8.314JK1=28000 K\times R=28000 K\times8.314 J K^{-1} mol1 mol^{-1} =232.79KJ=232.79 \,KJ mol1mol^{-1}