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For a certain first order reaction 32 % of the reactant is left after 570 s . The rate constant of this reaction is × 10 - 3 s - 1 . (Round off to the Nearest Integer). [Given: log 10 2 = 0 . 301 , ln 10 = 2 . 303 ]

Correct answer

0

Step-by-step solution

For 1 st  order reaction, K = 2 . 303 t · log A 0 A t = 2 . 303 570 sec · log 100 32 = 1 . 999 × 10 - 3 sec - 1 ≈ 2 × 10 - 3 sec - 1

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