The rate of a reaction doubles when its temperature changes from 300 K to 310 K…
Chemistry · JEE Main · NTA Exams — Chemical Kinetics
The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be (R = 8.314 JK–1 mol–1 and log 2 = 0.301)
- 53.6 kJ mol–1
- 48.6 kJ mol–1
- 58.5 kJ mol–1
- 60.5 kJ mol–1
Answer
(A) 53.6 kJ mol –1
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- A reactant (A) forms two products A k _ 1 B, Activation energy E_ a_ 1 A k _ 2 C, Activation energy E _ a _ 2…
- For the reaction, 3A + 2B → C + D The differential rate law can be written as :
- For the reaction, 2 ~N _ 2 O _ 5 4 NO _ 2 + O _ 2 , the rate equation can be expressed in two ways – d [ ~N _…
- Given below are two statements. One is labelled as Assertion and the other is labelled as Reason. Assertion…
- The rate of a chemical reaction generally increases rapidly even for small temperature rise because of rapid…
- The rate coefficient (k) for a particular reactions is 1.3 × 10 –4 M –1 s –1 at 100 0 C, and 1.3 ×10 –3 M –1…
- Rate of a reaction can be expressed by Arrhenius equation as k = A e –E/RT In this equation, E represents
- The rate of a reaction 2X + Y → Products is given by d [ Y ] dt = k [X] 2 [Y] if X is present in large…
More Chemical Kinetics questions · Browse all practice questions