The half-life period of a first order reaction is 50 minutes. How much time is needed for concentration of the reactants to be equal to 1/8 of the initial concentration?
A. 100 min
B. 150 min
C. 200 min
D. 50 min
Answer
534.3k+ views
Hint: There is a formula for the first order chemical reaction and it is as follows.
\[K=\dfrac{0.693}{{{t}_{1/2}}}\]
Here, K = rate constant of the chemical reaction
${{t}_{1/2}}$ = half-life of the chemical reaction
Complete answer:
- In the question it is asked to calculate the time needed for concentration of the reactants to equal 1/8th of the initial concentration.
- There is a relationship between the concentration and time of the first order chemical reaction and it is as follows.
\[\log \left( \dfrac{1}{C} \right)=Kt\to (i)\]
Here C = concentration of the reactant
K = rate constant
t = time required to complete the reaction
- First, we have to calculate the rate constant of the first order chemical reaction and it is as follows.
\[K=\dfrac{0.693}{{{t}_{1/2}}}\]
Here, K = rate constant of the chemical reaction
${{t}_{1/2}}$ = half-life of the chemical reaction = 50 min
\[\begin{align}
& K=\dfrac{0.693}{{{t}_{1/2}}} \\
& K=\dfrac{0.693}{50} \\
\end{align}\]
- We have to substitute the ‘K’ value in the equation (i) to get the time required to complete the chemical reaction.
\[\begin{align}
& \log \left( \dfrac{1}{C} \right)=Kt \\
& \log \left( \dfrac{1}{1/8} \right)=\dfrac{0.693}{50}\times t \\
& t=150\min \\
\end{align}\]
- Therefore, the time required for the concentration of the reactants to be equal to 1/8 of the initial concentration is 150 min.
So, the correct option is B.
Note:
First we have to find the rate constant of the first order chemical reaction and later we have to use the rate constant value to calculate the time required to complete the first order chemical reaction till certain concentration.
\[K=\dfrac{0.693}{{{t}_{1/2}}}\]
Here, K = rate constant of the chemical reaction
${{t}_{1/2}}$ = half-life of the chemical reaction
Complete answer:
- In the question it is asked to calculate the time needed for concentration of the reactants to equal 1/8th of the initial concentration.
- There is a relationship between the concentration and time of the first order chemical reaction and it is as follows.
\[\log \left( \dfrac{1}{C} \right)=Kt\to (i)\]
Here C = concentration of the reactant
K = rate constant
t = time required to complete the reaction
- First, we have to calculate the rate constant of the first order chemical reaction and it is as follows.
\[K=\dfrac{0.693}{{{t}_{1/2}}}\]
Here, K = rate constant of the chemical reaction
${{t}_{1/2}}$ = half-life of the chemical reaction = 50 min
\[\begin{align}
& K=\dfrac{0.693}{{{t}_{1/2}}} \\
& K=\dfrac{0.693}{50} \\
\end{align}\]
- We have to substitute the ‘K’ value in the equation (i) to get the time required to complete the chemical reaction.
\[\begin{align}
& \log \left( \dfrac{1}{C} \right)=Kt \\
& \log \left( \dfrac{1}{1/8} \right)=\dfrac{0.693}{50}\times t \\
& t=150\min \\
\end{align}\]
- Therefore, the time required for the concentration of the reactants to be equal to 1/8 of the initial concentration is 150 min.
So, the correct option is B.
Note:
First we have to find the rate constant of the first order chemical reaction and later we have to use the rate constant value to calculate the time required to complete the first order chemical reaction till certain concentration.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

