(D) For a first order reaction,the integrated rate equation is: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$Given: $k = 2.303 \times 10^{-3} \; s^{-

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(D) For a first order reaction,the integrated rate equation is: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$Given: $k = 2.303 \times 10^{-3} \; s^{-1}$,$[A]_0 = 40 \; g$,$[A]_t = 10 \; g$Substituting the values: $t = \frac{2.303}{2.303 \times 10^{-3}} \log \frac{40}{10}$$t = \frac{1}{10^{-3}} \log 4$$t = 1000 \times \log(2^2) = 1000 \times 2 \times \log 2$$t = 2000 \times 0.3010 = 602 \; s$

Class 12 Chemistry Chemical Kinetics First Order reaction

Medium

$A$ first order reaction has a rate constant of $2.303 \times 10^{-3} \; s^{-1}$. The time required for $40 \; g$ of this reactant to reduce to $10 \; g$ will be.....$s$
[Given that $\log_{10} 2 = 0.3010$]

NEET 2019 All NEET

A
$230.3$
B
$301$
C
$2000$
D
$602$

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Chemical Kinetics

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First Order reaction

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What is the half-life of a first order reaction if the time required to decrease the concentration of the reactant from $1.0 \ M$ to $0.25 \ M$ is $10 \ hours$ (in $hours$)?

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$A$ first order reaction has a rate constant of $2.303 \times 10^{-3} \; s^{-1}$. The time required for $40 \; g$ of this reactant to reduce to $10 \; g$ will be.....$s$[Given that $\log_{10} 2 = 0.3010$]

Similar Questions

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 $(t_{0.5})$?

Calculate the rate constant for the first order reaction,$A \rightarrow B$ if the rate of reaction is $5.4 \times 10^{-6} \ mol \ dm^{-3} \ s^{-1}$ and $[A]=0.3 \ M$.

For a first-order reaction,the time taken for the concentration of the reactant to reach $3/4$ of its initial value is $t_{1/4}$. If the rate constant for the reaction is $K$,then $t_{1/4}$ can be expressed as: (in $/K$)

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What is the half-life of a first order reaction if the time required to decrease the concentration of the reactant from $1.0 \ M$ to $0.25 \ M$ is $10 \ hours$ (in $hours$)?

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$A$ first order reaction has a rate constant of $2.303 \times 10^{-3} \; s^{-1}$. The time required for $40 \; g$ of this reactant to reduce to $10 \; g$ will be.....$s$
[Given that $\log_{10} 2 = 0.3010$]