
In a first order reaction, the concentration of the reactant decreases from 0.6 M to 0.3 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025M in minutes is:
A.1.2
B.12
C.30
D.3
Answer
587.4k+ views
Hint: To solve this question, we will have to understand the relation between the initial and final concentrations of the reactant in the first observation. Then using this relationship, we can determine the value of time required for the second observation. Also, the time required for the reactants in a given reaction to reduce to half of its initial value is known as the half – life of the reaction. This value is constant at any given moment in the reaction.
Complete step by step answer:
In the given question, the data provided to us is as follows:
The given reaction is a first order reaction
The concentration of the reactant decreases from 0.6 M to 0.3 M in 15 minutes.
The concentration of the reactant decreases from 0.1 M to 0.025 M in ‘x’ minutes.
Now in the first observation made, we can see that the concentration of the reactant becomes half after 30 minutes, i.e. from 0.6 M to 0.3 M.
By keeping this concept in mind, let us move to the next part of the observation. The concentration of the reactant again decreases. This time, it decreases from 0.1 M to 0.025 M. This means that the value of the value of the concentration is reduced by 4 times.
While calculating half – life, the value of the concentration is reduced by 2 times. Hence, when the value of the concentration is reduced by 4 times, the total time for this to happen would be equivalent to 2 times the value of half - life.
Hence, the time taken for the concentration to change from 0.1 M to 0.025M in minutes is:
\[\begin{array}{*{20}{l}}
{ = {\text{ }}2{\text{ }}\left( {half - life} \right)} \\
{ = {\text{ }}2{\text{ }}\left( {15} \right)} \\
{ = {\text{ }}30{\text{ }}minutes}
\end{array}\]
Hence, Option C is the correct option.
Note:
This question can be solved by another approach:
We can use the formula to calculate the time required by using the rate law for the first order reaction. The rate law for first order reactions is: \[\ln \dfrac{{{{[A]}_0}}}{{[A]}} = kt\]. From the initial observations, we can substitute the corresponding values and obtain the value of ‘k’. then after that, using this value of ‘k’ you can find the time required for the second observation.
Complete step by step answer:
In the given question, the data provided to us is as follows:
The given reaction is a first order reaction
The concentration of the reactant decreases from 0.6 M to 0.3 M in 15 minutes.
The concentration of the reactant decreases from 0.1 M to 0.025 M in ‘x’ minutes.
Now in the first observation made, we can see that the concentration of the reactant becomes half after 30 minutes, i.e. from 0.6 M to 0.3 M.
By keeping this concept in mind, let us move to the next part of the observation. The concentration of the reactant again decreases. This time, it decreases from 0.1 M to 0.025 M. This means that the value of the value of the concentration is reduced by 4 times.
While calculating half – life, the value of the concentration is reduced by 2 times. Hence, when the value of the concentration is reduced by 4 times, the total time for this to happen would be equivalent to 2 times the value of half - life.
Hence, the time taken for the concentration to change from 0.1 M to 0.025M in minutes is:
\[\begin{array}{*{20}{l}}
{ = {\text{ }}2{\text{ }}\left( {half - life} \right)} \\
{ = {\text{ }}2{\text{ }}\left( {15} \right)} \\
{ = {\text{ }}30{\text{ }}minutes}
\end{array}\]
Hence, Option C is the correct option.
Note:
This question can be solved by another approach:
We can use the formula to calculate the time required by using the rate law for the first order reaction. The rate law for first order reactions is: \[\ln \dfrac{{{{[A]}_0}}}{{[A]}} = kt\]. From the initial observations, we can substitute the corresponding values and obtain the value of ‘k’. then after that, using this value of ‘k’ you can find the time required for the second observation.
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