(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 \ min$,so $k = \frac{0.693}{10} = 0.0

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$0.0693 \times 3 \ M \ min^{-1}$

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 10 \ min$,so $k = \frac{0.693}{10} = 0.0693 \ min^{-1}$.After $20 \ min$ (which is $2 \times t_{1/2}$),the concentration $[A]$ will be reduced to $\frac{1}{4}$ of the initial concentration.$[A] = \frac{12}{4} = 3 \ M$.The rate of reaction is given by $Rate = k[A]$.$Rate = 0.0693 \times 3 \ M \ min^{-1}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

Difficult

Half-life period of a first-order reaction is $10 \ min$. Starting with an initial concentration of $12 \ M$,the rate after $20 \ min$ is:

KCET 2015 All KCET

A
$0.0693 \ M \ min^{-1}$
B
$0.693 \times 3 \ M \ min^{-1}$
C
$0.0693 \times 3 \ M \ min^{-1}$
D
$0.0693 \times 4 \ M \ min^{-1}$

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Half-life period of a first-order reaction is $10 \ min$. Starting with an initial concentration of $12 \ M$,the rate after $20 \ min$ is:

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For the reaction ${N_2}{O_{5(g)}} \to 2N{O_{2(g)}} + \frac{1}{2}{O_{2(g)}}$,the rate constant is $2.3 \times 10^{-2} \ s^{-1}$. Which of the following equations represents the variation of $[{N_2}{O_5}]$ with time?

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