1.

Graph between log k and 1//T[where K is rate constant in s^(-1) and T is the temperature (in K) is a straight line with Hence, E_(a) will be

Answer»

`2.303 xx 2` CAL
`2/(2.303) cal`
2 cal
None of these

Solution :log k = log A`-(E_(a))/(2.303 R) = TAN phi= 1/(2.303)` (Given)
`E_(a) = 2.303 R xx` Slope
`=2.303 R xx 1/(2.303) = R=2 Cal`


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