The decomposition of a hydrocarbon follows the equation k=(4.5×1011s-1)e-28000 K /T. What will be the value of activation energy?

Question

The decomposition of a hydrocarbon follows the equation k=(4.5×1011s-1)e-28000 K /T. What will be the value of activation energy? (A) 669 KJ mol-1 (B) 232.79 KJ mol-1 (C) 4.5×1011 KJ mol-1 (D) 28000 KJ mol-1

Tardigrade Admin · Accepted Answer

Arrhenius equation, k=Ae-Ea/ RT Given equation is k =(4.5×1011s-1) e-28000 K /T Comparing both the equations, we get -(Ea/RT) =-(28000 K /T) Ea =28000 K× R=28000 K×8.314 J K-1 mol-1 =232.79 KJ mol-1

Q. The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 1 0^{11} s^{- 1}) e^{- 28000 K / T} .$ What will be the value of activation energy?

$669 K J$ $m o l^{- 1}$

$232.79 K J$ $m o l^{- 1}$

$4.5 \times 1 0^{11} K J$ $m o l^{- 1}$

$28000 K J$ $m o l^{- 1}$

Q. The decomposition of a hydrocarbon follows the equation k=(4.5×1011s−1)e−28000K/T. What will be the value of activation energy?

669KJ mol−1

232.79KJ mol−1

4.5×1011KJ mol−1

28000KJ mol−1

Arrhenius equation, k=Ae−Ea/RT Given equation is k=(4.5×1011s−1) e−28000K/T Comparing both the equations, we get −RTEa​​ =−T28000K​ Ea​ =28000K×R=28000K×8.314JK−1 mol−1 =232.79KJ mol−1

Q. The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 1 0^{11} s^{- 1}) e^{- 28000 K / T} .$ What will be the value of activation energy?

$669 K J$ $m o l^{- 1}$

$232.79 K J$ $m o l^{- 1}$

$4.5 \times 1 0^{11} K J$ $m o l^{- 1}$

$28000 K J$ $m o l^{- 1}$