Medium
The rate constant for the decomposition of a hydrocarbon is given as $K = (4.5 \times 10^{11} \ s^{-1}) e^{-(28000 \ K) / T}$. The activation energy of the reaction (in $J/mol$) is:
- A$28000$
- B$28000 \times (8.314)^2$
- C$28000 \times 8.314$
- D$28000 / 8.314$
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