JEE Main 202125 FebruaryMorning Shift
- Exam
- JEE Main
- Year
- 2021
- Date
- 25 February
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Chemical Kinetics
JEE Main 2021 · 25 February · Morning Shift · Chemistry · Chemical Kinetics
For the reaction, aA + bB → cC + dD, the plot of logk v/s (1)/(T)...
For the reaction, the plot of v/s is given below:
The temperature at which the rate constant of the reaction is is _____
(Rounded-off to the nearest integer) [Given : The rat5e constant of the reaction is at .]
Correct answer
0
Step-by-step solution
\left|Slope\right| = \dfrac{Ea}{2 . 303R} = 10, 000\log \left(\dfrac{{k}_{2}}{{k}_{1}}\right) = \dfrac{Ea}{2 . 303R}\left(\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}}\right)\log \left(\dfrac{{10}^{- 4}}{{10}^{- 5}}\right) = 10, 000\left[\dfrac{1}{500} - \dfrac{1}{{T}_{2}}\right]\log \left(\dfrac{{10}^{- 4}}{{10}^{- 5}}\right) = 10, 000\left[\dfrac{1}{500} - \dfrac{1}{{T}_{2}}\right]{T}_{2} = 526 . 31 ≃ 526K