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.

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

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^(-) ]

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 rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction [R=8.314JK^(-1)mol^(-1)]

A graph plotted between log k versus 1//T for calculating activation energy is shown by

The rate of a reaction depends upon the temperature and is quantitatively expressed as k=Ae^((-Ea)/(RT) i) If a graph is plotted between log k and 1/T, write the expression for the slope of the reaction? ii) If at under different conditions E_(a1)andE_(a2) are the activation energy of two reactions. If E_(a1) = 40 J// "mol and" E_(a2) = 80 J//" mol" . Which of the two has a larger value of the rate constant?