(A) For a first-order reaction,the rate constant $K$ is given by the formula: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $[A]_0 = 1.6 \ M$,

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(A) For a first-order reaction,the rate constant $K$ is given by the formula: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $[A]_0 = 1.6 \ M$,$[A]_t = 0.4 \ M$,and $t = 12 \ hours$. Substituting the values: $K = \frac{2.303}{12} \log \frac{1.6}{0.4}$ $K = \frac{2.303}{12} \log 4$ Since $\log 4 = 2 \log 2 \approx 2 \times 0.3010 = 0.6020$: $K = \frac{2.303 \times 0.6020}{12} \approx 0.1155 \ hour^{-1}$ Rounding to three decimal places,$K \approx 0.116 \ hour^{-1}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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What is the rate constant of a first-order reaction if the time required to decrease the concentration of the reactant from $1.6 \ M$ to $0.4 \ M$ is $12 \ hours$?

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A
$0.116 \ hour^{-1}$
B
$0.167 \ hour^{-1}$
C
$0.4 \ hour^{-1}$
D
$0.78 \ hour^{-1}$

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What is the rate constant of a first-order reaction if the time required to decrease the concentration of the reactant from $1.6 \ M$ to $0.4 \ M$ is $12 \ hours$?

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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.....$s$
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