
In first order reaction, the concentration of the reactant decreases from \[{\rm{0}}{\rm{.8}}\;{\rm{M}}\]to \[{\rm{7}}{\rm{.5}}\;{\rm{minutes}}\] \[{\rm{0}}{\rm{.4}}\;{\rm{M}}\] in \[{\rm{15}}\;{\rm{minutes}}\]. The time taken for the concentration to change from \[{\rm{0}}{\rm{.1}}\;{\rm{M}}\] to\[{\rm{0}}{\rm{.025}}\;{\rm{M}}\] is:
A) \[{\rm{30}}\;{\rm{minutes}}\]
B) \[{\rm{60}}\;{\rm{minutes}}\]
C) \[{\rm{7}}{\rm{.5}}\;{\rm{minutes}}\]
D) \[{\rm{15}}\;{\rm{minutes}}\]
Answer
556.8k+ views
Hint:We know that if in any reaction, our concentration becomes exactly half in the given time then that time is known as the half-life period of that reaction. The concentration decreases from \[{\rm{0}}{\rm{.8}}\;{\rm{M}}\]to \[{\rm{0}}{\rm{.4}}\;{\rm{M}}\], so actually in our question half-life order is given.
Complete answer:
Let a first order reaction in which A is converted to B. First at the start of the reaction the concentration is \[{\rm{0}}{\rm{.4}}\;{\rm{M}}\] but after \[{\rm{15}}\;{\rm{minutes}}\] its concentration becomes \[{\rm{0}}{\rm{.4}}\;{\rm{M}}\]
$
A \to B\\
t = 0\,\,\,\,\,\,\,\,0.{\rm{8}}\;{\rm{M}}\\
t = 15\,\,\,\,\,\,0.4\;{\rm{M}}
$
Since our concentration is exactly half in \[{\rm{15}}\;{\rm{minutes}}\]then in reaction,
${t_{\dfrac{1}{2}}} = 15\;{\rm{minutes}}$
Hence the correct answer is option is A.
Note:In these first order reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R. The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as ${t_{\dfrac{1}{2}}}$. ${t_{\dfrac{1}{2}}}$for a zero-order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa. There are various examples of first order reaction; the hydrolysis of acetylsalicylic acid and the reaction of t-butyl bromide with water to give t-butanol, hydrolysis of cisplatin.
Complete answer:
Let a first order reaction in which A is converted to B. First at the start of the reaction the concentration is \[{\rm{0}}{\rm{.4}}\;{\rm{M}}\] but after \[{\rm{15}}\;{\rm{minutes}}\] its concentration becomes \[{\rm{0}}{\rm{.4}}\;{\rm{M}}\]
$
A \to B\\
t = 0\,\,\,\,\,\,\,\,0.{\rm{8}}\;{\rm{M}}\\
t = 15\,\,\,\,\,\,0.4\;{\rm{M}}
$
Since our concentration is exactly half in \[{\rm{15}}\;{\rm{minutes}}\]then in reaction,
${t_{\dfrac{1}{2}}} = 15\;{\rm{minutes}}$
Hence the correct answer is option is A.
Note:In these first order reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R. The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as ${t_{\dfrac{1}{2}}}$. ${t_{\dfrac{1}{2}}}$for a zero-order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa. There are various examples of first order reaction; the hydrolysis of acetylsalicylic acid and the reaction of t-butyl bromide with water to give t-butanol, hydrolysis of cisplatin.
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