
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
513k+ 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.
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