Question

The rate of a first-order reaction is $0.04mollitr{e^{ - 1}}{\sec ^{ - 1}}$at 10 minute and $0.03mollitr{e^1}$ at 20 minutes after initiation. Find the half-life of the reaction.
A.$2.406\min $
B.$24.06\min $
C.$240.6\min $
D.$0.204\min $

Answer 1

Hint:The half-existence of a chemical reaction can be characterized as the time taken for the grouping of an offered reactant to arrive at half of its underlying focus (for example the time taken for the reactant fixation to arrive at half of its underlying worth.

Complete step by step answer:
Let us first write the equation for the first order reaction of the rate law expression. And it is given as:
First order reaction = $K \times [A]$
As we were given that the $K \times {[A]_{10}}$ i.e., at 10 minutes the value is 0.04
$ \Rightarrow 0.04 = K \times {[A]_{10}}.....(1)$
As well at the 20miuntes we write it as:
$ \Rightarrow 0.03 = K \times {[A]_{20}}.....(2)$
Now let us divide both the equations:
\[ \Rightarrow \dfrac{{{{[A]}_{10}}}}{{{{[A]}_{20}}}} = \dfrac{{0.04}}{{0.03}} = \dfrac{4}{3}.........(3)\].
As we got the above let us calculate the value of t at 10 minutes and it is given as;
The expression of rate of expression at 10 minutes is t = \[\dfrac{{2.303}}{{10}}\log \dfrac{{{{[A]}_{10}}}}{{{{[A]}_{20}}}}\]
And as we said t = 10 minutes
$ \Rightarrow 10 = \dfrac{{2.303}}{K}\log \dfrac{4}{3}.....from(3)$
This leaves us to the value of $K = \dfrac{{2.303}}{{10}}\log \dfrac{4}{3} = 0.0288{\min ^{ - 1}}$
Now, let us calculate what we need that is the half life and it is done as:
${t_{1/2}} = \dfrac{{0.693}}{K} = \dfrac{{0.693}}{{0.0288}} = 24.06{\min ^{ - 1}}$

Hence, the correct option is B.

Note:
The rate steady is signified by k and is otherwise called reaction rate consistent or reaction rate coefficient. It is subject to the temperature. There are two potential approaches to figure rate consistent and they are Utilizing the Arrhenius condition. Utilizing the molar centralizations of the reactants and the request for the reaction.