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JEE Main202127 Aug 2021Evening ShiftChemistryChemical KineticsActual

The first order rate constant for the decomposition of CaCO 3 at 700 K is 6 . 36 × 10 - 3 s - 1 and activation energy is 209 kJ mol - 1 . Its rate constant (in s - 1 at 600 K is x × 10 - 6 . The value of x is (Nearest integer) [Given R = 8 . 31 J K - 1 mol - 1 ; log 6 . 36 × 10 - 3 = - 2 . 19 , 10 - 4 . 79 = 1 . 62 × 10 - 5

Correct answer

0

Step-by-step solution

K 700 = 6 . 36 × 10 - 3   s - 1 K 600 = x × 10 - 6   s - 1 E a = 209   kJ / mol Applying : log K T 2   K T 1 = - E a 2 . 303 R 1   T 2 - 1   T 1 log K 700   K 600 = - E a 2 . 303 R 1 700 - 1 600 log 6 . 36 × 10 - 3   K 600 = + 209 × 1000 2 . 303 × 8 . 31 100 700 × 600 log 6 . 36 × 10 - 3 - logK 600 = 2 . 6 log 6 . 36 × 10 - 3 - log K 600 = 2 . 6 ⇒ logK 600 = - 2 . 19 - 2 . 6 = - 4 . 79 ⇒ K 600 = 10 - 4 . 79 = 1 . 62 × 1

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