Question
The temperature at which the rate constants of the given below two gaseous reactions become equal is \_\_\_\_ K. (Nearest integer) X Y k _ 1 =10^ 6 e^ -30000 ~T P Q k _ 2 =10^ 4 e^ -24000 ~T Given: 10=2.303
JEE Main
Chemistry
Chemical Kinetics
2026
JEE Main 2026 (22 January Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
To find the temperature at which k_1 = k_2, 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.
Related: Chemistry — Chemical Kinetics · All PYQ Banks