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.

For first order reaction half-life period Is Independent of initial concentration and Inversely proportional to the rate constant. The slope of the line in a graph of logK Vs 1/T for a reaction is -5841. Calculate energy of activation for this reaction. [R=8.314JK^-1mol^-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^(-) ]

i) Write the mathematics expressions relating the variation of the rate constant of a reaction with temperatures. ii) How can you graphically find the activation energy of the reaction from the above expression? iii) 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.

Give the plot of ln k versus 1/T.

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.