(B) For a first order reaction,the rate constant $K$ is given by the formula:$K = \frac{2.303}{(t_2 - t_1)} \log \frac{[A_1]}{[A_2]}$Substituting the

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(B) For a first order reaction,the rate constant $K$ is given by the formula:$K = \frac{2.303}{(t_2 - t_1)} \log \frac{[A_1]}{[A_2]}$Substituting the given values:$K = \frac{2.303}{(1600 - 800)} \log \frac{1.45}{0.88}$$K = \frac{2.303}{800} \log(1.6477)$$K = \frac{2.303}{800} \times 0.2169$$K = 6.24 \times 10^{-4} \ s^{-1}$

Class 12 Chemistry Chemical Kinetics First Order reaction

Medium

The decomposition of dinitrogen pentoxide $(N_2O_5)$ follows first order rate law. What will be the rate constant from the given data?
At $t = 800 \ s$,$[N_2O_5] = 1.45 \ mol \ L^{-1}$
At $t = 1600 \ s$,$[N_2O_5] = 0.88 \ mol \ L^{-1}$

A
$3.12 \times 10^{-4} \ s^{-1}$
B
$6.24 \times 10^{-4} \ s^{-1}$
C
$2.84 \times 10^{-4} \ s^{-1}$
D
$8.14 \times 10^{-4} \ s^{-1}$

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Chemical Kinetics

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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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$N_2O_5 \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$ is a first order reaction. $k = 5 \times 10^{-4} \ s^{-1}$. The initial concentration of $N_2O_5 = 0.2 \ mol \ L^{-1}$. After how much time will its concentration be $25 \%$ of the initial concentration (in $s$)?

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If benzene diazonium chloride undergoes first order decomposition at $T \ K$ with a rate constant of $6.93 \times 10^{-2} \ min^{-1}$,the time for completion of $90 \%$ of the reaction (in $min$) is (nearest integer) $(\log 2 = 0.30, \log 3 = 0.477)$.

MediumTS EAMCET 2023

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Consider a general first order reaction $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$. If the initial pressure is $200 \ mm$ and after $20 \ minutes$ it is $250 \ mm$,then the half-life period of the reaction (in minutes) is. $(\log 2 = 0.30, \log 3 = 0.48, \log 4 = 0.60)$

MediumTS EAMCET 2025

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The decomposition of dinitrogen pentoxide $(N_2O_5)$ follows first order rate law. What will be the rate constant from the given data?At $t = 800 \ s$,$[N_2O_5] = 1.45 \ mol \ L^{-1}$At $t = 1600 \ s$,$[N_2O_5] = 0.88 \ mol \ L^{-1}$

Similar Questions

$20 \%$ 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:

$N_2O_5 \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$ is a first order reaction. $k = 5 \times 10^{-4} \ s^{-1}$. The initial concentration of $N_2O_5 = 0.2 \ mol \ L^{-1}$. After how much time will its concentration be $25 \%$ of the initial concentration (in $s$)?

If benzene diazonium chloride undergoes first order decomposition at $T \ K$ with a rate constant of $6.93 \times 10^{-2} \ min^{-1}$,the time for completion of $90 \%$ of the reaction (in $min$) is (nearest integer) $(\log 2 = 0.30, \log 3 = 0.477)$.

Consider a general first order reaction $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$. If the initial pressure is $200 \ mm$ and after $20 \ minutes$ it is $250 \ mm$,then the half-life period of the reaction (in minutes) is. $(\log 2 = 0.30, \log 3 = 0.48, \log 4 = 0.60)$

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The decomposition of dinitrogen pentoxide $(N_2O_5)$ follows first order rate law. What will be the rate constant from the given data?
At $t = 800 \ s$,$[N_2O_5] = 1.45 \ mol \ L^{-1}$
At $t = 1600 \ s$,$[N_2O_5] = 0.88 \ mol \ L^{-1}$