Sum

**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?

Advertisement Remove all ads

#### Solution

**Given: **

Half-life t_{1/2 }= 1.7 hours, [A]_{0 }= 100%, [A]_{t }= 100 - 20 = 80%

**To find: **

Time for 20% of reactant to react = t

**Formulae: **

**i. **t_{1/2 }= `0.693/"k"`

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

**Calculation:**

t_{1/2 }= `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.

Concept: Integrated Rate Law

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads