Question

The rate of a first order reaction is 0.04 $molL^{ -1 }s^{ -1 }$ at 10 seconds and 0.03 $molL^{ -1 }s^{ -1 }$ at 20 seconds after initiation of the reaction. The half-life period of the reaction is:
A. 24.1 s
B. 34.1 s
C. 44.1 s
D. 54.1 s

Answer 1

Hint: In the question, it is given that the reaction follows first-order kinetics, so, you need to first find the rate constant k for the reaction and then use this value to find the half-life period of the reaction.

Complete step by step answer:

We know that for a first-order reaction -
$K\quad =\quad \dfrac { 2.303 }{ { t }_{ 2 }-{ t }_{ 1 } } .log\dfrac { { [A] }_{ 1 } }{ { [A] }_{ 2 } } $
Here,
${ t }_{ 2 }$ = final time; ${ [A] }_{ 2 }$ = final concentration
${ t }_{ 1 }$ = initial time; ${ [A] }_{ 1 }$ = initial concentration
So, we can now just insert values of given quantities to calculate rate constant -
$K\quad =\quad \dfrac { 2.303 }{ 20-10 } .log\dfrac { (0.04) }{ (0.03) } $
K = 0.2303 x 0.125
K = 0.02877 $s^{ -1 }$
Now we have the value of rate constant (K). We will put this value in the expression of half life period,
$t_{ 1/2 }\quad =\quad \dfrac { 0.693 }{ K }$
$t_{ 1/2 }\quad =\quad \dfrac { 0.693 }{ 0.02877 } $
$t_{ 1/2 }$ = 24.1 s

Therefore, we can conclude that the correct answer to this question is option A.

Note: We must know that the rate constant, K, is a proportionality constant that indicates the relationship between the molar concentration of reactants and the rate of a chemical reaction.