(A) For a first-order reaction,the integrated rate equation is given by:$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:$k = 6 \ min^{-1}$$[A]_0 =

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(A) For a first-order reaction,the integrated rate equation is given by:$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:$k = 6 \ min^{-1}$$[A]_0 = 0.5 \ mol \ L^{-1}$$[A]_t = 0.05 \ mol \ L^{-1}$Substituting the values:$6 = \frac{2.303}{t} \log \frac{0.5}{0.05}$$6 = \frac{2.303}{t} \log(10)$Since $\log(10) = 1$,we have:$6 = \frac{2.303}{t}$$t = \frac{2.303}{6} \approx 0.384 \ min$

Class 12 Chemistry Chemical Kinetics First Order reaction

Difficult

$A$ reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \ min^{-1}$. If we start with $[A] = 0.5 \ mol \ L^{-1}$,when would $[A]$ reach the value $0.05 \ mol \ L^{-1}$? (in $min$)

A
$0.38$
B
$0.15$
C
$3$
D
$3.84$

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Chemical Kinetics

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First Order reaction

$A$ reaction is $50\%$ complete in $2 \ hours$ and $75\%$ complete in $4 \ hours$. The order of reaction is

EasyNEET 2013

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DifficultMHT CET 2026

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For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

DifficultJEE MAIN 2018

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For a first-order reaction,the time required for $93.75\%$ completion is $4$ times the half-life of the reaction. What is the relationship between the time required for $93.75\%$ completion and the half-life $(t_{0.5})$?

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The reaction $A_{(g)} \to 2B_{(g)} + C_{(g)}$ obeys the rate law,$r = K[A]$ where $K = 0.023 \ s^{-1}$. If $2.5 \ moles$ of $A$ were taken in a $5 \ L$ flask,how many moles of $A$ would remain left after $50 \ seconds$?

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$A$ reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \ min^{-1}$. If we start with $[A] = 0.5 \ mol \ L^{-1}$,when would $[A]$ reach the value $0.05 \ mol \ L^{-1}$? (in $min$)

Similar Questions

$A$ reaction is $50\%$ complete in $2 \ hours$ and $75\%$ complete in $4 \ hours$. The order of reaction is

The rate constant of the decomposition of hydrogen peroxide is $0.0204 \text{ min}^{-1}$. Calculate the half-life of the reaction. (in $\text{ min}$)

For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

For a first-order reaction,the time required for $93.75\%$ completion is $4$ times the half-life of the reaction. What is the relationship between the time required for $93.75\%$ completion and the half-life $(t_{0.5})$?

The reaction $A_{(g)} \to 2B_{(g)} + C_{(g)}$ obeys the rate law,$r = K[A]$ where $K = 0.023 \ s^{-1}$. If $2.5 \ moles$ of $A$ were taken in a $5 \ L$ flask,how many moles of $A$ would remain left after $50 \ seconds$?

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