JEE Main202622 January 2026Morning ShiftChemistryChemical KineticsActual
The temperature at which the rate constants of the given below two gaseous reactions become equal is _ _ _ _ K. (Nearest integer) X Y k ₁=10⁶ e^ -30000 ~T P Q k ₂=10⁴ e^ -24000 ~T Given: 10=2.303
Correct answer
0
Step-by-step solution
To find the temperature at which k₁ = k₂ , set the rate constant equations equal: 10^6 e^ -30000/T = 10^4 e^ -24000/T . Dividing both sides by 10^4 : 100 = e^ -24000/T + 30000/T = e^ 6000/T . Taking natural logarithm: (100) = 6000 T . Since (100) = 2 (10) = 2 2.303 = 4.606 : 4.606 = 6000 T . Solving for T: T = 6000 4.606 = 1302.5 K. Rounding to the nearest integer gives T = 1303 K.