What is the slope of the straight line for the graph drawn between Ink and y, where k is the rate constant of a reaction at temperature T?

The slope of the graph drawn between ln k and 1/T as per Arrhenius equaiton gives the value (R= gas constant, E_(a)= Activation energy) .

(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. (b) Rate constant 'K' of a reaction varies with temperature 'T' according to the equation : logK=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) )

In photoelectric the slope of the straight line graph between stopping potential and frequency of incident radiation gives the ratio of Plancks constant to

The graph between log k versus 1//T is a straight line.

The graph between log k versus 1//T is a straight line.

The slope of straight line graph between ln k us (1)/(T) is equal to 2.4 xx 10^(4)K . Calculate the activation energy of the reaction at 400 K. (K represents rate constant)

The graph drawn between rate and time is a straight line parallel to x-axis.The order of the reaction is