The time for the half-life period of a certain reaction \[A\to \] products is $1$ hour. When the initial concentration of the reactant is $\text{A}$ is \[2.0\text{ }mol{{L}^{-1}}\], how much time does it take for its concentration to come from \[0.50\text{ to 0}\text{.25 }mol{{L}^{-1}}\] if it is a zero-order reaction?
A.$4$ h
B.$0.5$ h
C.$0.25$ h
D.$1$ h
Answer
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Hint:Use the formula for half-life of zero order reaction to find out the value of rate constant $\left( k \right)$, then using the rate constant find out the required time.
Formula used: \[{{t}_{{}^{1}/{}_{2}}}=\dfrac{\left[ {{A}_{0}} \right]}{2k}\], \[{{t}_{{}^{1}/{}_{2}}}\]= half –life time
\[\left[ {{A}_{0}} \right]\]= initial concentration
$k$= rate constant
\[\left[ ({{A}_{0}})-(A) \right]\]= amount of reactant decomposed
Complete step by step answer:
Given initial concentration of reactant is \[2.0\text{ }mol{{L}^{-1}}\] and half life time period $=1$ hour
So, using the above formula, \[{{t}_{{}^{1}/{}_{2}}}=\dfrac{\left[ {{A}_{0}} \right]}{2k}\]we get,
\[k=\dfrac{1}{2}\dfrac{\left[ {{A}_{0}} \right]}{{{t}_{{}^{1}/{}_{2}}}}\]
= \[\dfrac{1}{2}\times \dfrac{2}{1}\]
= 1 \[mol{{L}^{-1}}{{h}^{-1}}\]
Then using other equation, \[k=\dfrac{1}{t}\left[ ({{A}_{0}})-(A) \right]\]
\[t=\dfrac{1}{k}\left[ ({{A}_{0}})-(A) \right]\]
= \[\dfrac{1}{1mol{{L}^{-1}}{{h}^{-1}}}\left( 0.5-0.25 \right)mol{{L}^{-1}}\]
= \[\dfrac{1}{1mol{{L}^{-1}}{{h}^{-1}}}\left( 0.25 \right)mol{{L}^{-1}}\]
$=0.25\text{ h}$
So, the correct option is (C).
Note:
Zero Order Reaction: These reactions are typically found when a material is required for the reaction to precede such as surface or a catalyst. For zero order reaction, the graph of concentration data versus time is a straight line.
The integral form of zero order reaction is written as- \[\left[ A \right]=-kt+\left[ {{A}_{0}} \right]\]
So when we compare this equation to straight line equation, $y=mx+c$ so we get a graph of \[\left[ A \right]\]against t as straight line with slope equal to $\left( -k \right)$ and intercept equal to \[\left[ {{A}_{0}} \right]\]. Example of zero order reaction is the reaction of hydrogen with chlorine.
Rate Law for a zero order reaction is rate=k where k is the rate constant.
The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.
Half Life Time: The half- life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value. The half -life of a zero order reaction decreases as the initial concentration of the reactant in the reaction decreases.
The half-life of a First order reaction is independent of concentration and the half-life of a second order reaction decreases as the concentration increases.
Formula used: \[{{t}_{{}^{1}/{}_{2}}}=\dfrac{\left[ {{A}_{0}} \right]}{2k}\], \[{{t}_{{}^{1}/{}_{2}}}\]= half –life time
\[\left[ {{A}_{0}} \right]\]= initial concentration
$k$= rate constant
\[\left[ ({{A}_{0}})-(A) \right]\]= amount of reactant decomposed
Complete step by step answer:
Given initial concentration of reactant is \[2.0\text{ }mol{{L}^{-1}}\] and half life time period $=1$ hour
So, using the above formula, \[{{t}_{{}^{1}/{}_{2}}}=\dfrac{\left[ {{A}_{0}} \right]}{2k}\]we get,
\[k=\dfrac{1}{2}\dfrac{\left[ {{A}_{0}} \right]}{{{t}_{{}^{1}/{}_{2}}}}\]
= \[\dfrac{1}{2}\times \dfrac{2}{1}\]
= 1 \[mol{{L}^{-1}}{{h}^{-1}}\]
Then using other equation, \[k=\dfrac{1}{t}\left[ ({{A}_{0}})-(A) \right]\]
\[t=\dfrac{1}{k}\left[ ({{A}_{0}})-(A) \right]\]
= \[\dfrac{1}{1mol{{L}^{-1}}{{h}^{-1}}}\left( 0.5-0.25 \right)mol{{L}^{-1}}\]
= \[\dfrac{1}{1mol{{L}^{-1}}{{h}^{-1}}}\left( 0.25 \right)mol{{L}^{-1}}\]
$=0.25\text{ h}$
So, the correct option is (C).
Note:
Zero Order Reaction: These reactions are typically found when a material is required for the reaction to precede such as surface or a catalyst. For zero order reaction, the graph of concentration data versus time is a straight line.
The integral form of zero order reaction is written as- \[\left[ A \right]=-kt+\left[ {{A}_{0}} \right]\]
So when we compare this equation to straight line equation, $y=mx+c$ so we get a graph of \[\left[ A \right]\]against t as straight line with slope equal to $\left( -k \right)$ and intercept equal to \[\left[ {{A}_{0}} \right]\]. Example of zero order reaction is the reaction of hydrogen with chlorine.
Rate Law for a zero order reaction is rate=k where k is the rate constant.
The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.
Half Life Time: The half- life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value. The half -life of a zero order reaction decreases as the initial concentration of the reactant in the reaction decreases.
The half-life of a First order reaction is independent of concentration and the half-life of a second order reaction decreases as the concentration increases.
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