(B) For a first-order reaction, the relationship between the rate constant $(k)$ and the half-life $(t_{1/2})$ is given by:$k = \frac{0.693}{t_{1/2}}$

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$0.5 \times 10^{-3} \text{ s}^{-1}$

(B) For a first-order reaction, the relationship between the rate constant $(k)$ and the half-life $(t_{1/2})$ is given by:$k = \frac{0.693}{t_{1/2}}$Given $t_{1/2} = 1386 \text{ s}$.Substituting the value:$k = \frac{0.693}{1386} \text{ s}^{-1}$$k = \frac{693 \times 10^{-3}}{1386} \text{ s}^{-1}$$k = 0.5 \times 10^{-3} \text{ s}^{-1}$Thus, the rate constant is $5 \times 10^{-4} \text{ s}^{-1}$ or $0.5 \times 10^{-3} \text{ s}^{-1}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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The half-life period of a first-order reaction is $1386 \text{ s}$. The rate constant of the reaction is:

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A
$5.5 \times 10^{-2} \text{ s}^{-1}$
B
$0.5 \times 10^{-3} \text{ s}^{-1}$
C
$5 \times 10^{-2} \text{ s}^{-1}$
D
$5 \times 10^{-3} \text{ s}^{-1}$

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The half-life period of a first-order reaction is $1386 \text{ s}$. The rate constant of the reaction is:

Similar Questions

The time required for $90\%$ completion of a certain first order reaction is $1 \text{ hour}$. Calculate the time required for $99.9\%$ completion of the same reaction.

$A$ first-order reaction takes $69.3 \ min$ to complete $50\%$ of the reaction. How much time (in $min$) will it take to complete $80\%$ of the reaction?

$A$ reactant undergoes $90\%$ decomposition in $366 \text{ minutes}$. What is the half-life of this first-order reaction?

The decomposition of $N_2O_5$ occurs as $2N_2O_5 \to 4NO_2 + O_2$ and follows first-order kinetics. Hence,

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: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

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