Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
67 views
in Chemistry by (90.6k points)
closed by
Activation energy of a chemical reaction can be determined by
A. evaluting rate constant at standard temperature
B. evaluting velocities of reaction at two different temperatures
C. evaluating rate constants at two different temperatures
D. changing concentration of reactants

1 Answer

0 votes
by (88.6k points)
selected by
 
Best answer
Correct Answer - C
According to Arrhenius equation
`k=Ae^(-E_(a)//RT)`
Taking natural logarithm of both sides, we get
`Ink=InA-(E_(a))/(RT)`
At temperature `T_(1))` , equation becomes
`Ink_(1)=InA-(E_(a))/(RT_(1)`
At temperature `T_(2))` , equation becomes
`Ink_(1)=InA-(E_(a))/(RT_(1)`
(Since `A` is constant for a given reaction). `K_(1)` and `k_(2)` are the value of rate constant at temperature `T_(1)` and `T_(2)` respectively.
Subtracting equation `(1)` from equation `(2)` , we obtain
`Ink_(2)-Ink_(1)=(-(E_(a))/(RT_(2)))-(-(E_(a))/(RT_(1)))`
`In(k_(2))/(k_(1))=-(E_(a))/(R)((1)/(T_(2))-(1)/(T_(1)))`
or log`(k_(2))/(k_(1))=-(E_(a))/(2.303R)((1)/(T_(2))-(1)/(T_(1)))`
or log`(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` Thus, we can evaluate the activation energy of a chemical reaction by knowing rate constants `(k_(1) and k_(2))` at two differences temperatures `(T_(1) and T_(2))` .

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...