(A) For a first-order reaction,the integrated rate equation is given by: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.Given,$k = 70 \, s^{-1}$ and $

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(A) For a first-order reaction,the integrated rate equation is given by: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.Given,$k = 70 \, s^{-1}$ and $[A]_t = \frac{1}{18} [A]_0$,which implies $\frac{[A]_0}{[A]_t} = 18$.Substituting the values: $t = \frac{2.303}{70} \log(18)$.Since $\log(18) \approx 1.255$,we get $t = \frac{2.303 \times 1.255}{70}$.$t = \frac{2.890}{70} \approx 0.0413 \, s$.

Class 12 Chemistry Chemical Kinetics First Order reaction

Medium

In a first-order reaction,the rate constant $k = 70 \, s^{-1}$. How much time is required for the concentration to become $\frac{1}{18}$ of the initial concentration (in $, s$)?

A
$0.0413$
B
$0.0521$
C
$0.0385$
D
$0.0612$

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Chemical Kinetics

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First Order reaction

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In a first-order reaction,the rate constant $k = 70 \, s^{-1}$. How much time is required for the concentration to become $\frac{1}{18}$ of the initial concentration (in $, s$)?

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$A_{(g)} \rightarrow 2B_{(g)}$. Initially,$2 \ mol$ of $A$ are taken in a $5 \ L$ vessel. After $20 \ min$,$[A]_t = \frac{[B]_t}{2}$. Find the half-life time of $A$ in the first-order reaction in $\min$.

For the first order reaction $A \rightarrow 2B$,$1 \ mole$ of reactant $A$ gives $0.2 \ moles$ of $B$ after $100 \ minutes$. The half-life of the reaction is $..... \ min$. [Use: $\ln 2 = 0.69, \ln 10 = 2.3$]

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What is the value of the rate constant for a first-order reaction,if it takes $15 \ min$ for the consumption of $20 \%$ of the reactants?

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