Question

For a first order reaction, the time taken to reduce
the initial concentration to a factor of 1/4 is 10 minute
If the reduction in concentration is carried out to a
factor of 1/16

then time required will be

(1) 10 minutes
(2) 20 minutes
(3) 40 minutes
(4) 60 minutes​

Answer 1

Let initial Concentration be x.

Concentration after 10 minutes = (1/4 ) x

We know that , for first order reaction :-

$\boxed{ \rm \large \: k \times t = ln \dfrac{a}{b} }$

Where :- k = rate constant

t = time

a = initial Concentration

b = final concentration

$\rm \longrightarrow k \times t = ln \dfrac{a}{b} \\ \\ \rm \longrightarrow k \times 10 = ln \dfrac{x}{ (\frac{1}{4} )x}\\ \\ \rm \longrightarrow k \times 10 = ln \dfrac{4x}{ x}\\ \\ \rm \longrightarrow k \times 10 = ln \dfrac{4}{ 1}\\ \\ \rm \longrightarrow k \times 10 = ln 4\\ \\ \rm \longrightarrow k \times 10 = ln {2}^{2} \\ \\ \rm \longrightarrow k \times 10 =2 ln {2}\\ \\ \rm \longrightarrow k \times 5 = ln {2}\\ \\ \rm \longrightarrow k = \frac{ln {2}}{5}$

Again , let initial Concentration be y

Concentration after time t = (1/16)y

We have to find out time (t).

$\rm \longrightarrow \: \dfrac{ln2}{5} \times t = ln \dfrac{y}{ (\frac{1}{16})y } \\ \\ \\ \rm \longrightarrow \: \dfrac{ln2}{5} \times t = ln \dfrac{16}{1 }\\ \\ \\ \rm \longrightarrow \: \dfrac{ln2}{5} \times t = ln \: {2}^{4} \\ \\ \\ \rm \longrightarrow \: \dfrac{ln2}{5} \times t =4 \: ln \: {2}\\ \\ \\ \rm \longrightarrow \: t =4 \times 5\\ \\ \\ \rm \longrightarrow \: t =20 \: min$

Answer :

Option (2) 20 minutes is correct

Answer 2

Answer:

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Explanation:

Hope it helps you