If a reation follows the Arrhensis equation the plot lnk vs`1/(RT)` gives straight line with a gradient (-y) unti .The energy required to active the reactant is :
If a reaction follows the Arrhenius equation, the plot lnk vs 1/RT gives a straight-line having intercept 'In a' unit on positive y-axis. The maximum value of rate constant k is :
In accordance to Arrhenius equation, the plot of log k against (1)/(T) is a straight line. The slope of the line is equal to
For first order gaseous reaction , log k when plotted atgains (1)/(t) gives a straight line with a slop of -8000. Calcualte the activation energy of the reaction
Rate constant k of a reaction varies with temperature according to the equation logk = 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 versus 1/T , a straight line with a slope - 6670 K is obtained. Calculate the energy of activation for this reaction
Find the equation of a straight line: with slope -1/3 and y -intercept -4.
Which straight line gives the activation energy for a reaction?
