LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
841 views
in Chemistry by (75.8k points)

Rate constant for a reaction are 2.5 x 10-4 atm and 1 atm at temperature 327°C and 527°C respectively. Calculate activation energy in KJ.

(Given R = 8.314 J mol-1K-1)

(1) 166 KJ

(2) -257 KJ

(3) 120 KJ

(4) 70 KJ

Please log in or register to answer this question.

1 Answer

+1 vote
by (65.3k points)

Correct option: (1) 166 KJ

Explanation: 

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

...