Question

t1/2 of a first order reaction is 75 min. Calculate time in which 75% of the reaction will take place​

Answer 1

Explanation:

Answer is 109.4 minutes

Answer 2

Given :

Half life of first order reaction = 75 minutes

To find :

Time in which 75% of reaction is completed

Formula used :

$\sf \: k \times t \: = \: ln( \dfrac{a}{a - x} )$

Here,

k = rate constant

t = time

a = initial concentration

x = amount of reaction complete

Solution:

First of all we will find rate constant of reaction

Part 1)

Half life = 75min

a = a

x = (a/2)

$\sf \implies \: k \times t_{ \frac{1}{2} } = ln( \dfrac{a}{a - \frac{a }{2} } ) \\ \\ \sf \implies \: k \times \: t_{ \frac{1}{2} } \: = \: ln( \dfrac{a}{ \frac{2a - a}{2} } ) \\ \\ \sf \implies \: k \times \: t_{ \frac{1}{2} } \: = \: ln( \dfrac{a}{ \frac{a}{2} } ) \\ \\ \sf \implies \: k \times \: t_{ \frac{1}{2} } \: = \: ln( \dfrac{2a}{a} ) \\ \\ \sf \implies \: k \times \: t_{ \frac{1}{2} } \: = \: ln(2) \\ \\ \sf \implies \: k \: = \: \dfrac{1}{t_{ \frac{1}{2} }} \times ln(2) \\ \\ \sf \implies \: k \: = \: \dfrac{ \: \: ln(2) \: }{t_{ \frac{1}{2} }} \: \: \: \: \: \: \: equation \: 1$

Part 2)

t = find

a = a .... let

x = 75% of a = (3/4)a

$\sf \implies \: k \times \: t_{ \frac{3}{4} } \: = \: ln( \dfrac{a}{a - \frac{3a}{4} } ) \\ \\ \sf \implies \: k \times \: t_{ \frac{3}{4} } \: = \: ln( \dfrac{a}{ \frac{(4a) - (3a)}{4} } ) \\ \\ \sf \implies \: k \times \: t_{ \frac{3}{4} } \: = \: ln( \dfrac{a}{ \frac{a}{4} } ) \\ \\ \sf \implies \: k \times \: t_{ \frac{3}{4} } \: = \: ln( \dfrac{4a}{a} ) \\ \\ \sf \implies \: k \times \: t_{ \frac{3}{4} } \: = \: ln(4) \\ \\ \sf \implies \: k \times \: t_{ \frac{3}{4} } \: = \:2( \: ln(2) \: ) \\ \\ \sf \: put \: value \: of \: k \: from \: equation \: 1 \\ \sf \implies \: \: \dfrac{ \: \: ln(2) \: }{t_{ \frac{1}{2} }} \: \times \: t_{ \frac{3}{4} } \: = \:2( \: ln(2) \: ) \\ \\ \sf \implies \:t_{ \frac{3}{4} } \: = \: 2 \times ln(2) \times \dfrac{ t_{ \frac{1}{2} } }{ ln(2) } \\ \\ \sf \implies \: \bold{t_{ \frac{3}{4} } \: = \: 2 \times t_{ \frac{1}{2}}} \\ \\ \sf \implies \: \: t_{\frac{3}{4}} \: = \: 2 \times 75 \: min \\ \\ \sf \implies \: \: t_{\frac{3}{4}} \: = \: \: \bold{150 \: min}$

ANSWER : 150 minutes