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Graph between log k and 1//T [k rate con...

Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be

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The correct Answer is:
2
`2.303logK=-E_a/(RT)+2.303logÅ`
Thus, `-E_a/(2.303R)=tan theta=-1/2.303`
`:. E_a=R=2` cal.
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