The gaseous reaction ${A_2} \to 2A$ is first order in ${A_2}$ . After $12.3$ minutes $65\% $ of ${A_2}$ remains undecomposed. How long will it take to decompose $90\% $ of ${A_2}$ ? what is the half life of the reaction?
Answer
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Hint:First order reaction is a reaction that depends on the concentration of only one reactant, it is a unimolecular reaction. In first order reaction other reactants can be present, but each will be zero order.
Complete step by step answer:
In the question it is given that the reaction is a first order and we have to determine time taken to decompose $90\% $ of ${A_2}$ . We can use a first order reaction formula to calculate the time.
The formula is, $k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {A\,} \right]}_o}}}{{\left[ {{A_2}} \right]}}$
Before calculating time we have to know first rate constant $\left( k \right)$ . Now solve the question
Rate constant $\left( k \right)$ of ${A_2}$ remain undecomposed, by using above formula,
For $65\% $ of ${A_2}$ remains,
By putting the value of known quantity, we get rate constant $\left( k \right)$
$k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {A\,} \right]}_o}}}{{\left[ {{A_2}} \right]}}$
On solving above equation, we get
$k = \dfrac{{2.303}}{{12.3\min }}{\log _{10}}\dfrac{{100}}{{65}}$
$k = 0.03503{\min ^{ - 1}}$
Now, we know the value of the rate constant $\left( k \right)$ . We can calculate time to decompose $90\% $ of ${A_2}$ , using the same formula as we used to calculate the rate constant.
For $90\% $ of ${A_2}$ decomposes.
$t = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{{{\left[ {A\,} \right]}_o}}}{{\left[ {{A_2}} \right]}}$
By putting the value of known quantity, we get time required to decompose $90\% $ of ${A_2}$ .
$t = \dfrac{{2.303}}{{0.03505{{\min }^{ - 1}}}}{\log _{10}}\dfrac{{100}}{{10}}$
On solving above equation, we get
$t = 65.8\min $
It will take $65.7\min $ to decompose $90\% $ of ${A_2}$ .
We have to also calculate half life for the reaction, we can calculate half life of a reaction using formula.
The formula for half life is, \[{t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}\]
By putting the value of rate constant, we get half life of the reaction in the above formula of half life.
\[{t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}\]
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{0.03503}}$
${t_{\dfrac{1}{2}}} = 19.8{\min ^{ - 1}}$ .
Note:
It is to be noted that a first order reaction depends on the concentration of a reactant whereas the half life of a reaction does not depend on the concentration of reactant. Half life is of species is the time it takes for the concentration on reactant to reduce to half of its initial concentration.
Complete step by step answer:
In the question it is given that the reaction is a first order and we have to determine time taken to decompose $90\% $ of ${A_2}$ . We can use a first order reaction formula to calculate the time.
The formula is, $k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {A\,} \right]}_o}}}{{\left[ {{A_2}} \right]}}$
Before calculating time we have to know first rate constant $\left( k \right)$ . Now solve the question
Rate constant $\left( k \right)$ of ${A_2}$ remain undecomposed, by using above formula,
For $65\% $ of ${A_2}$ remains,
By putting the value of known quantity, we get rate constant $\left( k \right)$
$k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {A\,} \right]}_o}}}{{\left[ {{A_2}} \right]}}$
On solving above equation, we get
$k = \dfrac{{2.303}}{{12.3\min }}{\log _{10}}\dfrac{{100}}{{65}}$
$k = 0.03503{\min ^{ - 1}}$
Now, we know the value of the rate constant $\left( k \right)$ . We can calculate time to decompose $90\% $ of ${A_2}$ , using the same formula as we used to calculate the rate constant.
For $90\% $ of ${A_2}$ decomposes.
$t = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{{{\left[ {A\,} \right]}_o}}}{{\left[ {{A_2}} \right]}}$
By putting the value of known quantity, we get time required to decompose $90\% $ of ${A_2}$ .
$t = \dfrac{{2.303}}{{0.03505{{\min }^{ - 1}}}}{\log _{10}}\dfrac{{100}}{{10}}$
On solving above equation, we get
$t = 65.8\min $
It will take $65.7\min $ to decompose $90\% $ of ${A_2}$ .
We have to also calculate half life for the reaction, we can calculate half life of a reaction using formula.
The formula for half life is, \[{t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}\]
By putting the value of rate constant, we get half life of the reaction in the above formula of half life.
\[{t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}\]
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{0.03503}}$
${t_{\dfrac{1}{2}}} = 19.8{\min ^{ - 1}}$ .
Note:
It is to be noted that a first order reaction depends on the concentration of a reactant whereas the half life of a reaction does not depend on the concentration of reactant. Half life is of species is the time it takes for the concentration on reactant to reduce to half of its initial concentration.
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