Rate constant k for a reaction varies with temperature according to the equation log k ="constant"-(E_(a))/(2.303R)*(1)/(T) where E_(a) is the energy of activation for the reaction. When a graph is plotted for log k vs 1//T, a straight line with a slope of -6670 K is obtained. Calculate energy of activation for this reaction. State the units. ""[R=8.314JK^(-1)"mol"^(-1)]

Rate constant k for a reaction varies with temperature according to th

1.	Rate constant k for a reaction varies with temperature according to the equation log k ="constant"-(E_(a))/(2.303R)*(1)/(T) where E_(a) is the energy of activation for the reaction. When a graph is plotted for log k vs 1//T, a straight line with a slope of -6670 K is obtained. Calculate energy of activation for this reaction. State the units. ""[R=8.314JK^(-1)"mol"^(-1)]
Answer» Solution :The following data is provided SLOPE `= -6670K, R=8.314 JK^(-1)"mol"^(-1)` Slope `= -(E_(a))/(2.303R) or E_(a)= -"Slope" xx 2.303 xx R` SUBSTITUTING the VALUES, we have `E_(a)= -(-6670)K xx 2.303 xx 8.314JK^(-1)"mol"^(-1)=+127.714"KJ mol"^(-1)`