(A) For a first-order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given: $t = 100 \ min$,$[A]_t = \frac{1}

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$k = 2.08 \times 10^{-2} \ min^{-1}, t_{1/2} = 33.3 \ min$

(A) For a first-order reaction,the integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given: $t = 100 \ min$,$[A]_t = \frac{1}{8} [A]_0$,so $\frac{[A]_0}{[A]_t} = 8$.Substituting the values: $k = \frac{2.303}{100} \log(8) = \frac{2.303}{100} \times 0.903 = 2.08 \times 10^{-2} \ min^{-1}$.The half-life time is calculated as $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{2.08 \times 10^{-2}} \approx 33.3 \ min$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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$A$ radioactive substance decomposes such that after $100 \ min$,its concentration becomes $1/8$ of the original concentration. Calculate the rate constant $(k)$ and half-life time $(t_{1/2})$.

A
$k = 2.08 \times 10^{-2} \ min^{-1}, t_{1/2} = 33.3 \ min$
B
$k = 1.04 \times 10^{-2} \ min^{-1}, t_{1/2} = 66.6 \ min$
C
$k = 2.08 \times 10^{-2} \ min^{-1}, t_{1/2} = 16.6 \ min$
D
$k = 4.16 \times 10^{-2} \ min^{-1}, t_{1/2} = 33.3 \ min$

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First Order reaction

$20 \%$ of a first order reaction was found to be completed at $10:00 \ am$. At $11:30 \ am$ on the same day,$20 \%$ of the reaction was found to be remaining. The half-life period in minutes of the reaction is

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The reaction $L \rightarrow M$ starts with $10 \, gL^{-1}$. After $30$ and $90$ minutes,$5 \, gL^{-1}$ and $1.25 \, gL^{-1}$ remain,respectively. The order of the reaction is:

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$A$ reaction has a half-life of $1 \, \text{min}$. The time required for $99.9 \, \%$ completion of the reaction is ......... $\text{min}$. (Round off to the nearest integer)
[Use: $\ln 2 = 0.69, \ln 10 = 2.3$]

DifficultJEE MAIN 2021

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For a reaction,$X_{(g)} \to Y_{(g)} + Z_{(g)}$,the half-life period is $10 \ min$. In what period of time would the concentration of $X$ be reduced to $10 \%$ of its original concentration? $........... \ min$

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In a first order reaction,the concentration of the reactant is reduced to $1/8$ of the initial concentration in $75 \ minutes$. The $t_{1/2}$ of the reaction (in minutes) is $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

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$A$ radioactive substance decomposes such that after $100 \ min$,its concentration becomes $1/8$ of the original concentration. Calculate the rate constant $(k)$ and half-life time $(t_{1/2})$.

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$20 \%$ of a first order reaction was found to be completed at $10:00 \ am$. At $11:30 \ am$ on the same day,$20 \%$ of the reaction was found to be remaining. The half-life period in minutes of the reaction is

The reaction $L \rightarrow M$ starts with $10 \, gL^{-1}$. After $30$ and $90$ minutes,$5 \, gL^{-1}$ and $1.25 \, gL^{-1}$ remain,respectively. The order of the reaction is:

$A$ reaction has a half-life of $1 \, \text{min}$. The time required for $99.9 \, \%$ completion of the reaction is ......... $\text{min}$. (Round off to the nearest integer)[Use: $\ln 2 = 0.69, \ln 10 = 2.3$]

For a reaction,$X_{(g)} \to Y_{(g)} + Z_{(g)}$,the half-life period is $10 \ min$. In what period of time would the concentration of $X$ be reduced to $10 \%$ of its original concentration? $........... \ min$

In a first order reaction,the concentration of the reactant is reduced to $1/8$ of the initial concentration in $75 \ minutes$. The $t_{1/2}$ of the reaction (in minutes) is $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

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$A$ reaction has a half-life of $1 \, \text{min}$. The time required for $99.9 \, \%$ completion of the reaction is ......... $\text{min}$. (Round off to the nearest integer)
[Use: $\ln 2 = 0.69, \ln 10 = 2.3$]