Question

the half life period of a first order reaction is 10 min.the time required for that concentration of the reactant to change from 0.08M to 0.02M is

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Answer 1

Given :

➝ Half life of a first order reaction = 10 minutes

➝ Initial concentration of reagent= 0.08 M

➝ Concentration of reagent after time (t) = 0.02 M

To find :

➝ Time required for that concentration of the reactant to change from 0.08M to 0.02M.

Solution :

For first order reaction :-

$\boxed{\rm k = \dfrac{ ln(2) }{t_ \frac{1}{2} } } \\ \\ \sf where : \rm k = rate \: constant \\ \rm t_ \frac{1}{2} = half \: life$

Now, to find the value of rate constant put the value of half life = 10 minutes

$\rm k = \dfrac{ ln(2) }{10 }$

Therefore, value of rate constant = (ln2)/10

We know that :-

$\boxed{ \rm k \times t = ln( \dfrac{a}{b} ) } \\ \\ \rm where : a = initial \: concentration \\ \\ \rm b= concentration \: after \: time \: t$

Here :-

➝ a = 0.08 M

➝ b = 0.02 M

$\rm \longrightarrow k \times t = ln( \dfrac{a}{b} ) \\ \\ \rm \longrightarrow \dfrac{ ln(2) }{10} \times t = ln( \dfrac{0.08}{0.02} )\\ \\ \rm \longrightarrow \dfrac{ ln(2) }{10} \times t = ln( \dfrac{8}{2} )\\ \\ \rm \longrightarrow \dfrac{ ln(2) }{10} \times t = ln( 4)\\ \\ \rm \longrightarrow \dfrac{ ln(2) }{10} \times t = ln{(2)}^{2} \\ \\ \rm \longrightarrow \dfrac{ ln(2) }{10} \times t =2 \times ln {(2)} \\ \\ \rm \longrightarrow \dfrac{ 1}{10} \times t =2 \\ \\ \rm \longrightarrow t =2 0$

Answer :

Time required for that concentration of the reactant to change from 0.08M to 0.02M = 20 minutes