For a first order reaction A → P , the temperature (T) dependent rate constant k was found to follow the equation log ⁡ k = - 2000 1 T + 6.0 . The pre-exponential factor A and the activation energy E a , respectively, are

log k = log A − E a 2.303 R T ...(1) Also given log k = 6.0 − ( 2000 ) 1 T …(2) On comparing equations, (1) and (2) log A = 6.0 ⇒ A = 10 6 s − 1 and E a 2.303 R = 2000 ; ⇒ E a = 2000 × 2.303 × 8.314 = 38.29 kJ mol − 1

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ChemistryChemical KineticsMiscellaneous/MixedEasy2 minPYQ_2020

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For a first order reaction $A \to P,$ the temperature (T) dependent rate constant $(k)$ was found to follow the equation $log k = - (2000) \frac{1}{T} + 6.0 .$ The pre-exponential factor A and the activation energy $E_{a},$ respectively, are

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Answer:

Solution:

$log k = log A - \frac{E_{a}}{2.303 R T}$ ...(1)

Also given $log k = 6.0 - (2000) \frac{1}{T}$ …(2)

On comparing equations, (1) and (2) $log A = 6.0 \Rightarrow A = 10^{6} s^{- 1}$

and $\frac{E_{a}}{2.303 R} = 2000;$

$\Rightarrow E_{a} = 2000 \times 2.303 \times 8.314$

$= 38.29 kJ mol^{- 1}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical KineticsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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