The slope of the line graph of `log k` versus `1//T` for the reaction `N_(2)O_(5) rarr2NO_(2) + 1//2O_(2)` is `-5000`.Calculate the energy of activation of the reaction (in `kJ K^(-1) mol^(-1)`).

The slope of the line in the graph of logk (k = rate constant) versus 1/T for a reaction is - 5841 K. Calculate the energy of activation for this reaction. [R = 8.314 JK^(-1) mol^(-) ]

Rate constant k of a reaction varies with temperature according to equation: log k= constant - (E_a)/( 2.303 R).(1)/(T ) What is the activation energy for the reaction. When a graph is plotted for log k versus 1/T a straight line with a slope-6670 K is obtained. Calculate energy of activation for this reaction (R=8.314 JK^(-1) mol^(-) 1)

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

The slope of a line in the graph of log k versus 1/T for a reaction is -5841 K. Calculate energy of activation for the reaction.

The slope of the line in the graph of log k (k = rate constant) versus 1/T " is " - 5841 . Calcu late the activation energy of the reaction.