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The graph between log k and 1/T[K is rat...

The graph between log k and 1/T[K is rate constant (`sec^(-1)`) and T the temperature (K)] is a straight line with `OX=5 and theta=tan^(-1)(-1/2.303)`.Calculate the value of `E_a` is cal. ?

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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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