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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, aA+bB→cC+dD,aA + bB \to cC + dD, the plot of logklogk v/s 1T\dfrac{1}{T} is given below:

For the reaction, aA + bB → cC + dD, the plot of logk v/s (1)/(T) is given below: The temperature at which the rate

The temperature at which the rate constant of the reaction is 10−4s−1{10}^{- 4}{s}^{- 1} is _____ K.K .
(Rounded-off to the nearest integer) [Given : The rat5e constant of the reaction is 10−5s−1{10}^{- 5}{s}^{- 1} at 500K500K.]

Correct answer

0

Step-by-step solution

logk=logA−Ea2.303RTlogk = logA - \dfrac{Ea}{2 . 303RT}

\left|Slope\right| = \dfrac{Ea}{2 . 303R} = 10, 000</p><p></p><p>\log \left(\dfrac{{k}_{2}}{{k}_{1}}\right) = \dfrac{Ea}{2 . 303R}\left(\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}}\right)</p><p></p><p>\log \left(\dfrac{{10}^{- 4}}{{10}^{- 5}}\right) = 10, 000\left[\dfrac{1}{500} - \dfrac{1}{{T}_{2}}\right]</p><p></p><p>\log \left(\dfrac{{10}^{- 4}}{{10}^{- 5}}\right) = 10, 000\left[\dfrac{1}{500} - \dfrac{1}{{T}_{2}}\right]</p><p></p><p>{T}_{2} = 526 . 31 ≃ 526K</p><p>Henceansweris(526)</p><p>Hence answer is \left(526\right)

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