Question

The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to disappear?

Answer 1

It will take 32.8 minutes for 20% of the reactant to disappear.

Explanation :

The half life of the reaction, T1/2 = 1.7 hr = 102 minutes

we know that ,

k = 0.693/(T1/2)

= 0.693/102

= 0.0068

let the Initial volume, A₀ = 100

=> final volume, At = 100 - 20 = 80

we know that,

k = (2.303/t)  x log(A₀/At)

=> 0.0068 = (2.303/t) x log(100/80)

=> 0.0068 = (2.303/t) x log(1.25)

=> 0.0068 = (2.303/t) x 0.0969

=> t = 2.303 x 0.0969/0.0068

=> t = 32.8 minutes

Hence it will take 32.8 minutes for 20% of the reactant to disappear.

Answer 2

Answer:32.9 min

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