The rate of a reaction depends upon the temperature and is quantitatively expressed as K=Ae^(-E_(a)//RT) (i) If a graph between log K and 1/t write the expression for the slope of the reaction ? (ii) If at under different condition E_(a1)and E_(a2) are the activation energy of two reactions . If E_(a1)=40j/mol and E_(a_(2))=80j/mol. Which of the two has larger value of the rate constasnt ?

The rate of a reaction depends upon the temperature and is quantitativ

1.	The rate of a reaction depends upon the temperature and is quantitatively expressed as K=Ae^(-E_(a)//RT) (i) If a graph between log K and 1/t write the expression for the slope of the reaction ? (ii) If at under different condition E_(a1)and E_(a2) are the activation energy of two reactions . If E_(a1)=40j/mol and E_(a_(2))=80j/mol. Which of the two has larger value of the rate constasnt ?
Answer» Solution : In `k=-(E_(a))/(RT) + In A` This equation is of the tyep y = MX+ C SLOPE` = -(E_(a))/( R)` (ii) Smaller the ACTIVATION a=energy activation energy , GREATER is the value of rate cosntant . `therefor E _(1)> K_(2)`