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Question: The decomposition of a hydrocarbon follows the equation \(k = (4.5 \times 10^{11}s^{- 1})e^{- 28000K...

The decomposition of a hydrocarbon follows the equation k=(4.5×1011s−1)e−28000K/Tk = (4.5 \times 10^{11}s^{- 1})e^{- 28000K/T}. What will be the value of activation energy?

A

669kJmol−1669kJmol^{- 1}

B

232.79kJmol−1232.79kJmol^{- 1}

C

4.5×1011KJmol−14.5 \times 10^{11}KJmol^{- 1}

D

28000kJmol−128000kJmol^{- 1}

Answer

232.79kJmol−1232.79kJmol^{- 1}

Explanation

Solution

Arrhenius equation, k=Ae−Ea/RTk = Ae^{- Ea/RT}

Given equation is

k=(4.5×1011s−1)e−28000K/Tk = (4.5 \times 10^{11}s^{- 1})e^{- 28000K/T}

Comparing both the equation we get

−EaRT=−28000KT- \frac{E_{a}}{RT} = - \frac{28000K}{T}

Ea=28000×K×R=28000K×8.314JK−1mol−1=232.79kJmol−1E_{a} = 28000 \times K \times R = 28000K \times 8.314JK^{- 1}mol^{- 1} = 232.79kJmol^{- 1}