JEE MainChemistryChemical Kinetics
Match List I with List II. List I List II A. Slope of k vs 1 T plot I. - E_a 2.303R B. Slope of k vs 1 T plot II. Fraction of molecules having kinetic energy E_a C. Intercept of k vs 1 T plot III. - E_a R D. Factor e^ -E_a/RT IV. A Choose the correct answer from the options given below:
Options
- AA-I, B-III, C-IV, D-II
- BA-III, B-I, C-II, D-IV
- CA-I, B-III, C-II, D-IV
- DA-III, B-I, C-IV, D-II
Correct answer
D. A-III, B-I, C-IV, D-II
Step-by-step solution
The Arrhenius equation is given by k = A e^ -E_a/RT . Taking the natural logarithm, we get k = A - E_a RT . Comparing this with the equation of a straight line y = mx + c , a plot of k vs 1 T has a slope of - E_a R and an intercept of A . Thus, A matches with III and C matches with IV. Converting to base 10 logarithm, we get k = A - E_a 2.303RT . A plot of k vs 1 T has a slope of - E_a 2.303R . Thus, B matches with I. The exponential factor e^ -E_a/RT represents the fraction of molecules that possess kinetic energy