Medium
The graph obtained between $\ln k$ ($k=$ Rate constant) on $y$-axis and $1/T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$
- A$166$
- B$332$
- C$765$
- D$382$
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