Question

the Half of a first order reaction is 60min.
What percent of reactant will be left behind after 120minutes?
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Answer 1

Given :

Half life of first order reaction = 75 minutes

To find :

concentration after 120 minutes of reaction

Formula used :

$\sf \: k \: = \: \dfrac{2.303}{t } \times log( \dfrac{a}{(a - x)} )$

Here,

k = rate constant

t = time

a = initial concentration

x = amount of reaction complete OR amount of reactant used

Solution:

First of all we will find rate constant of reaction

Part 1)

• Half life = 60 min
• a = a (let)
• x = (a/2) { as half of reactant is used}

$\sf \implies \: \: k \: = \: \dfrac{2.303}{60 \: min } \times log( \dfrac{a}{(a - \frac{a}{2} )} )$

$\sf \implies\: k \: = \: \dfrac{2.303}{60 \: min } \times log( \dfrac{a}{( \frac{a}{2} )} )$

$\sf \implies\: k \: = \: \dfrac{2.303}{60 \: min } \times log( \dfrac{2a}{a} )$

$\sf \implies\: k \: = \: \dfrac{2.303}{60 \: min } \times log( 2 ) \: \: \: equation \: 1$

Part 2)

• t = 120min
• a = a
• (a-x) = find

$\sf \implies \: \: k \: = \: \dfrac{2.303}{120 \: min } \times log( \dfrac{a}{(a - x)} )$

Put value of k from equation 1,

$\sf \implies \: \:\dfrac{2.303}{60 \: min } \times log( 2 ) \: = \: \dfrac{2.303}{120 \: min } \times log( \dfrac{a}{(a - x)} )$

$\sf \implies \: log( \dfrac{a}{(a - x)} ) \: = \dfrac{2.303}{60 \: min } \times log( 2 ) \: \times \: \dfrac{120min}{2.303 }$

$\sf \implies \: log( \dfrac{a}{(a - x)} ) \: = \dfrac{2.303}{ 2.303 } \times log( 2 ) \: \times \: \dfrac{120min}{60 \: min }$

$\sf \implies \: log( \dfrac{a}{(a - x)} ) \: = log( 2 ) \: \times \: 2$

$\sf \implies \: log( \dfrac{a}{(a - x)} ) \: = log( {2}^{2} ) \:$

$\sf \implies \: log( \dfrac{a}{(a - x)} ) \: = log( 4 ) \:$

$\sf \implies \: \dfrac{a}{(a - x)} = 4$

$\sf \implies \: ( a - x) = \dfrac{a}{4}$

Fraction left = (1/4)th of original

ANSWER : (1/4)th of original