The time taken for $90\% $ of a first order reaction to complete is approximately:
A: $1.1$ times that of half-life
B: $2.2$ times that of half-life
C: $3.3$ times that of half-life
D: $4.4$ times that of half-life
Answer
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Hint: Rate law of the reaction relates the reaction rate with the concentration or partial pressure of the reactants. First order reaction is that reaction in which rate of the reaction depends linearly on the concentration of only one reactant.
Formula used: For first order reaction,
$k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}} = \dfrac{{0.693}}{{{t_{1/2}}}}$
Where ${t_{1/2}}$ is the time taken for the completion of half of the reaction, $a$ is initial concentration, $x$ is the amount of substance that has reacted and $t$ is the time taken for the consumption of $x$ amount.
Complete step by step answer:
In this question we have to find the time taken for the completion of $90\% $ of the reaction. Reaction is $90\% $ completed this means if the initial mass of reactant is $100g$ then after time $t$, $90g$ of the reactant is used (because the reaction is $90\% $ completed).
This means in this case $a$ is equal to $100$ and $x$ is equal to $90$. Rate law of first order reaction that is:
$\dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}} = \dfrac{{0.693}}{{{t_{1/2}}}}$
Substituting the values,
$\dfrac{{2.303}}{t}\log \dfrac{{100}}{{100 - 90}} = \dfrac{{0.693}}{{{t_{1/2}}}}$
$t = \dfrac{{2.303 \times {t_{1/2}}}}{{0.693}}\log \dfrac{{100}}{{100 - 90}}$
Solving the value of $\log \dfrac{{100}}{{100 - 90}}$ ,
$t = \dfrac{{2.303}}{{0.693}}{t_{1/2}}$
Solving this we get:
$t = 3.3{t_{1/2}}$
Therefore time needed for the completion of $90\% $ of a first order reaction is $3.3$ times that of half-life.
Hence, the required answer is C.
Note:
Time needed for the completion of $90\% $ of the reaction is different for different order reactions. For first order reactions it is $3.3$ times that of half-life but it is not the same for other higher order reactions. Second order reaction is the reaction in which rate is either proportional to the square of concentration of reactant or is proportional to the product of concentration of two reactants.
Formula used: For first order reaction,
$k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}} = \dfrac{{0.693}}{{{t_{1/2}}}}$
Where ${t_{1/2}}$ is the time taken for the completion of half of the reaction, $a$ is initial concentration, $x$ is the amount of substance that has reacted and $t$ is the time taken for the consumption of $x$ amount.
Complete step by step answer:
In this question we have to find the time taken for the completion of $90\% $ of the reaction. Reaction is $90\% $ completed this means if the initial mass of reactant is $100g$ then after time $t$, $90g$ of the reactant is used (because the reaction is $90\% $ completed).
This means in this case $a$ is equal to $100$ and $x$ is equal to $90$. Rate law of first order reaction that is:
$\dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}} = \dfrac{{0.693}}{{{t_{1/2}}}}$
Substituting the values,
$\dfrac{{2.303}}{t}\log \dfrac{{100}}{{100 - 90}} = \dfrac{{0.693}}{{{t_{1/2}}}}$
$t = \dfrac{{2.303 \times {t_{1/2}}}}{{0.693}}\log \dfrac{{100}}{{100 - 90}}$
Solving the value of $\log \dfrac{{100}}{{100 - 90}}$ ,
$t = \dfrac{{2.303}}{{0.693}}{t_{1/2}}$
Solving this we get:
$t = 3.3{t_{1/2}}$
Therefore time needed for the completion of $90\% $ of a first order reaction is $3.3$ times that of half-life.
Hence, the required answer is C.
Note:
Time needed for the completion of $90\% $ of the reaction is different for different order reactions. For first order reactions it is $3.3$ times that of half-life but it is not the same for other higher order reactions. Second order reaction is the reaction in which rate is either proportional to the square of concentration of reactant or is proportional to the product of concentration of two reactants.
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