Answer
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Hint: To answer this question,recall the concept of integrated rate law. It is used to provide a relationship between the rate of the reaction and the concentrations of the reactants participating in it. It can be only determined experimentally and not from the balanced chemical equation. By substituting the appropriate values in the half-time formula (given), we can calculate the half-life of the given first-order reaction.
Formula used:
${t_{1/2}} = \dfrac{{\ln 2}}{{\text{k}}}$ where ${{\text{t}}_{1/2}}$ is the half-life and k is the rate constant.
Complete step by step answer:
From the formula of half life we know that ${t_{1/2}} = \dfrac{{\ln 2}}{{\text{k}}}$.
Substituting the given values in this equation we have,
\[{t_{1/2}} = \dfrac{{{\text{ln}}2}}{{\text{k}}} = \dfrac{{0.693}}{{1.7 \times {{10}^{ - 5}}}}\]
Simplifying this, we get:
\[{{\text{t}}_{1/2}} = 40764\;{\text{sec}}\]
Converting this into hours by diving with 3600, we get:
\[{{\text{t}}_{1/2}} = 11.3\;{\text{hrs}}\]
Hence, the correct option is option C.
Note:
The overall order of any reaction is the sum of the partial components of a chemical reaction. Rate$ = {\text{ k}}{\left[ {\text{A}} \right]^x}{\left[ {\text{B}} \right]^y}\;$then, the overall order can be written as \[x + y\]. As we said that the order provides relationship between rate and concentration of components:
1.If the reaction is a zero-order reaction, change in the reactant concentration will have no effect on the reaction rate.
2.In case of first order, doubling the reactant concentration will double the reaction rate.
3.In case of second-order, doubling the concentration of the reactants will quadruple the overall reaction rate.
4.In case of third order, the overall rate increases by eight times when the reactant concentration is doubled.
5.Radioactivity of elements are also considered as first-order reactions as they are only dependent on the initial concentration of the reactant.
Formula used:
${t_{1/2}} = \dfrac{{\ln 2}}{{\text{k}}}$ where ${{\text{t}}_{1/2}}$ is the half-life and k is the rate constant.
Complete step by step answer:
From the formula of half life we know that ${t_{1/2}} = \dfrac{{\ln 2}}{{\text{k}}}$.
Substituting the given values in this equation we have,
\[{t_{1/2}} = \dfrac{{{\text{ln}}2}}{{\text{k}}} = \dfrac{{0.693}}{{1.7 \times {{10}^{ - 5}}}}\]
Simplifying this, we get:
\[{{\text{t}}_{1/2}} = 40764\;{\text{sec}}\]
Converting this into hours by diving with 3600, we get:
\[{{\text{t}}_{1/2}} = 11.3\;{\text{hrs}}\]
Hence, the correct option is option C.
Note:
The overall order of any reaction is the sum of the partial components of a chemical reaction. Rate$ = {\text{ k}}{\left[ {\text{A}} \right]^x}{\left[ {\text{B}} \right]^y}\;$then, the overall order can be written as \[x + y\]. As we said that the order provides relationship between rate and concentration of components:
1.If the reaction is a zero-order reaction, change in the reactant concentration will have no effect on the reaction rate.
2.In case of first order, doubling the reactant concentration will double the reaction rate.
3.In case of second-order, doubling the concentration of the reactants will quadruple the overall reaction rate.
4.In case of third order, the overall rate increases by eight times when the reactant concentration is doubled.
5.Radioactivity of elements are also considered as first-order reactions as they are only dependent on the initial concentration of the reactant.
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