1.

If we plot a graph between log K and (1)/(T) by Arrhenius equation , the slope is

Answer»

`-(E_(a))/(R)`
`+(E_(a))/(R)`
`-(E_(a))/(2.303R)`
`+(E_(a))/(2.303 R)`

Solution :ln k = ln `- (E_(a))/(RT)` is Arrhenius EQUATION . Thus plots of ln k VS 1/T will give SLOPE = `-E_(a)//RT` or `-E_(a)// 2.303R`.


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