Question

The first order reaction is completed 90% in 40 minutes. Calculate its half-life period. (log 10 =1)

Answer 1

Hint:In first order reaction the reaction rate is dependent upon only one reactant. The formula used for calculating the rate constant is \[K = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}\], where K is rate constant, t is time, a is initial concentration.

Complete step by step answer:
Given,
90% reaction is completed in 40 minutes.
log 10 = 1
The reaction is said as a first order reaction when the reaction rate depends on the concentration of only one reactant. The unit referring the first order reaction is ${s^{ - 1}}$.
The rate constant for the first order reaction is given by the formula as shown below.
\[K = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}\]
Where,
K is the rate constant.
t is the time
$a$ is the initial concentration.
$a - x$ is the left out concentration.
Let, the initial concentration be 100.
Substitute the values in above equation
\[ \Rightarrow K = \dfrac{{2.303}}{{40}}\log \dfrac{{100}}{{100 - 90}}\]
\[ \Rightarrow K = \dfrac{{2.303}}{{40}}\log \dfrac{{100}}{{10}}\]
And hence on solving we have,
\[ \Rightarrow K = \dfrac{{2.303}}{{40}}\log 10\]
\[ \Rightarrow K = \dfrac{{2.303}}{{40}} \times 1\]
Now on doing the simplification, we have
\[ \Rightarrow K = 5.757 \times {10^{ - 2}}{\min ^{ - 1}}\]
Thus, the rate constant value of first order reaction is \[5.757 \times {10^{ - 2}}{\min ^{ - 1}}\].
The half-life of the reaction is defined as the time required to decrease the amount of reactant consumed by one half.
The half-life of first order reaction is given as shown below.
\[{t_{1/2}} = \dfrac{{0.693}}{k}\]
Where,
K is the rate constant.
To calculate the half-life of first order reaction, substitute the value of K in the equation.
\[ \Rightarrow {t_{1/2}} = \dfrac{{0.693}}{{5.757 \times {{10}^{ - 2}}}}\]
\[ \Rightarrow {t_{1/2}} = 10.3\min \]
Thus, the half-life of first order reaction is 10.3 min.

Note:
As the half-life of first order reaction is inversely proportional to the rate constant. So, for a shorter half-life, the reaction will be fast and the rate constant value will be larger. For, longer value of half-life the reaction will be slow and the rate constant value will be low.