JEE Main 2026 — Chemical Kinetics Question with Solution
JEE Main 2026 (22 January Shift 1)
Question
The temperature at which the rate constants of the given below two gaseous reactions become equal is K. (Nearest integer)
Given:
Given:
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Show full solutionCorrect answer: 1303
Correct answer
1303
Step-by-step explanation
To find the temperature at which ,
set the rate constant equations equal: .
Dividing both sides by : .
Taking natural logarithm: .
Since : .
Solving for T: K.
Rounding to the nearest integer gives T = 1303 K.
set the rate constant equations equal: .
Dividing both sides by : .
Taking natural logarithm: .
Since : .
Solving for T: K.
Rounding to the nearest integer gives T = 1303 K.
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This is a previous-year question from JEE Main 2026, covering the Chemical Kinetics chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.