The activation energy of one of the reactions in a biochemical process is $532611 J mol^{- 1}$ . When the temperature falls from 310 K to 300 K , the change in rate constant observed is $k_{300} = X \times 1 0^{- 3} \times k_{310}$ . The value of $X$ is $____$ .
[Given: ln10 = 2.3 ; R=8.3 J K^-1 mol^-1 ]

Question

The activation energy of one of the reactions in a biochemical process is $532611 J mol^{- 1}$ . When the temperature falls from 310 K to 300 K , the change in rate constant observed is $k_{300} = X \times 1 0^{- 3} \times k_{310}$ . The value of $X$ is $____$ .

[Given: ln10 = 2.3 ; R=8.3 J K^-1 mol^-1 ]

TestHub · Accepted Answer

ℓ n ( k 1 ​ k 2 ​ ​ ) = R E a ​ ​ ( T 1 ​ 1 ​ − T 2 ​ 1 ​ ) ℓ n ( k 1 ​ k 2 ​ ​ ) = 8.3 532611 ​ × ( 310 × 300 10 ​ ) where k 2 ​ is at 310 K & k 1 ​ is at 310 K ℓ n ( k 1 ​ k 2 ​ ​ ) = 6.9 = 3 × ln 10 ℓ n k 1 ​ k 2 ​ ​ = ℓ n 1 0 3 k 2 ​ = k 1 ​ × 1 0 3 k 1 ​ = k 2 ​ × 1 0 − 3 So X = 1

Chemistry - Chemical Kinetics Question with Solution | TestHub

Doubts & Discussion

JEE Main (2026)