Rate constant 'k' of a reaction varies with temperature 'T' according to the equation: log k= logA-(E_(a))/(2.303R)((1)/(T)) where E_(a) is the activation energy. When a graph is plotted for log k vs (1)/(T), a straight line with a slope of -4250 K is obtained. Calculate 'E_(a)'for the reaction. [R=8.314JK^(-1)"mol"^(-1)]

Rate constant 'k' of a reaction varies with temperature 'T' according

1.	Rate constant 'k' of a reaction varies with temperature 'T' according to the equation: log k= logA-(E_(a))/(2.303R)((1)/(T)) where E_(a) is the activation energy. When a graph is plotted for log k vs (1)/(T), a straight line with a slope of -4250 K is obtained. Calculate 'E_(a)'for the reaction. [R=8.314JK^(-1)"mol"^(-1)]
Answer» SOLUTION :The equation `log k=log A -(E_(a))/(2.303R) ((1)/(T))` is of the type `y=mx +C` where m is the slope of the line. `therefore (-E_(a))/(2.303R)= -4250 K or E_(a)=2.303R xx 4250 K` or`E_(a)=2.303xx 8.314 JK^(-1)"MOL"^(-1) xx 4250K` `=81375"J mol"^(-1)=81.375 "kJ mol"^(-1)`