Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction (k=0.02303) minute ( ^-1 ).

Question

Accepted Answer

Correct answer: A. (a) : For first order reaction, the equation is k= 2.303t log ⁡ [A ]_0 [A ] 0.02303= 2.30360 log ⁡ [A ]_0 [A ] ⇒ 0.02303 602.303= log ⁡ [A ]_0 [A ] ⇒0.6= log ⁡ [A ]_0 [A ] ⇒ antilog0.6 = [A ]_0 [A ] ⇒ [A ]_0 [A ] ≈4 Percentage of the reactant left after 60 min = [A ] [A ]_0 100= 1 1004=25 \%

$Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction (k = 0.02303) minute (^{- 1}) .$

Solution

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