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

Sum

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

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Solution

Given:

Half-life t1/2 = 1.7 hours, [A]= 100%, [A]= 100 - 20 = 80%

To find:

Time for 20% of reactant to react = t

Formulae:

i. t1/2 = `0.693/"k"`

ii. `"t" = 2.303/"K" "log"_10 ["A"]_0/["A"]_"t"`

Calculation:

t1/2 = `0.693/"k"`

`"k" = 0.693/"t"_"1/2" = 0.693/(1.7 "h") = 0.4076` h-1

t = `2.303/"k" "log"_10 ["A"]_0/["A"]_"t" = 2.303/(0.4076 "h"^-1) "log" 100/80`

`"t" = 2.303/(0.4076  "h"^-1) xx 0.0969 = 0.5475  "h" xx (60  "min")/(1 "h") = 32.9` min

The time required for 20% of reaction to react is 32.9 min.

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APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercise | Q 4.2 | Page 137
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