1.

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. ?

Answer»


SOLUTION :`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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