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QJEEM21Ch6ea3: JEE Main 2021 Chemistry - Class 12 JEEM21Ch6ea3 Solution: For the reaction, aA + bB cC + dD, the plot of log k vs is…

For the reaction, aA + bB →\to cC + dD, the plot of log k vs 1T{1 \over T} is given below :

JEE Main 2021 (Online) 25th February Morning Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 104 English
The temperature at which the rate constant of the reaction is 10-4 s-1 is _________ K. (Rounded off to the nearest integer)

[Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]

Step-by-Step Verified Solution

log⁡10K=log⁡10A−Ea2.303RT{\log _{10}}K = {\log _{10}}A - {{{E_a}} \over {2.303RT}}

Slope =Ea2.303R=−10000= {{{E_a}} \over {2.303R}} = - 10000

log⁡10K2K1=Ea2.303R×[1T1−1T2]{\log _{10}}{{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}} \times \left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]

log⁡1010−410−5=10000×[1500−1T]{\log _{10}}{{{{10}^{ - 4}}} \over {{{10}^{ - 5}}}} = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]

⇒\Rightarrow 1=10000×[1500−1T]1 = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]

⇒\Rightarrow 110000=1500−1T{1 \over {10000}} = {1 \over {500}} - {1 \over T}

⇒\Rightarrow 1T=1500−110000{1 \over T} = {1 \over {500}} - {1 \over {10000}}

⇒\Rightarrow 1T=20−110000=1910000{1 \over T} = {{20 - 1} \over {10000}} = {{19} \over {10000}}

⇒\Rightarrow T=1000019=526T = {{10000} \over {19}} = 526 K

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