Medium
The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$. Calculate the activation energy $E_a$.
(N/A) The given equation is $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$ $(i)$.
The Arrhenius equation is given by $k = A e^{-E_a / RT}$ $(ii)$.
Comparing equations $(i)$ and $(ii)$,we get $\frac{E_a}{RT} = \frac{28000 \ K}{T}$.
Therefore,$E_a = R \times 28000 \ K$.
Using $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,we have $E_a = 8.314 \ J \ K^{-1} \ mol^{-1} \times 28000 \ K$.
$E_a = 232792 \ J \ mol^{-1} = 232.792 \ kJ \ mol^{-1}$.
The Arrhenius equation is given by $k = A e^{-E_a / RT}$ $(ii)$.
Comparing equations $(i)$ and $(ii)$,we get $\frac{E_a}{RT} = \frac{28000 \ K}{T}$.
Therefore,$E_a = R \times 28000 \ K$.
Using $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,we have $E_a = 8.314 \ J \ K^{-1} \ mol^{-1} \times 28000 \ K$.
$E_a = 232792 \ J \ mol^{-1} = 232.792 \ kJ \ mol^{-1}$.
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