Question
Question: The rate constant of a first order reaction is \( 1.54 \times {10^{ - 3}}{\sec ^{ - 1}} \) . Calcula...
The rate constant of a first order reaction is 1.54×10−3sec−1 . Calculate its half-life period.
Solution
First we have to know what is the half-life period of first order reaction and the formula used for finding the half-life period of the reaction when the rate constant of the reaction is given.
Formula used
For first order reaction,
The relation of rate of reaction to the time and initial concentration and final concentration are as follows:
k=tln[A[A]O]
Where, k is rate of reaction, t is time, [A]o is initial concentration and A is final concentration left.
For half-life, (Half-life: The time at which reaction is half completed i.e, A=2[A]O )
t21=kln(2)=k0.693
Where, k is the rate of reaction and t21 is half-life time.
Complete step-by-step answer
We will start with the definition of the half-life:
Half-life of a reaction: The time required to decrease the initial concentration to the one-half of its initial concentration is called half-life of a reaction.
The given data in the question:
Reaction is a first order reaction.
Rate constant, k=1.54×10−3sec−1
So, now we will find the half-life of reaction when the rate constant is given with the formula:
For half-life,
t21=kln(2)=k0.693
Where, k is the rate of reaction and t21 is half-life time.
Here, rate constant is given. Hence, putting the value of the k=1.54×10−3sec−1 in the given formula;
t21=kln(2)=1.54×10−30.693
Solving,
t21=450sec
Hence, the half-life of the reaction is 450sec
The rate constant of a first order reaction is 1.54×10−3sec−1 its half-life period is 450sec .
Note
How we differentiate any reaction with the first order reaction or how to find the order of a reaction, for these we have to see the unit of the reaction, we can find out the reaction’s order by finding the value of n in the given formula (mol)1−nLn−1sec−1 , where n is the order of reaction.