(D) For a first order reaction,the rate law is given by: $Rate = K[A]$.Given that the rate constant $(K) = 3 \times 10^{-6} \ s^{-1}$ and the initial

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$3 \times 10^{-7} \ M \ s^{-1}$

(D) For a first order reaction,the rate law is given by: $Rate = K[A]$.Given that the rate constant $(K) = 3 \times 10^{-6} \ s^{-1}$ and the initial concentration $[A] = 0.10 \ M$.Substituting these values into the rate equation:$Rate = (3 \times 10^{-6} \ s^{-1}) \times (0.10 \ M) = 3 \times 10^{-7} \ M \ s^{-1}$.Therefore,the initial rate of reaction is $3 \times 10^{-7} \ M \ s^{-1}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

Medium

The rate constant of a first order reaction is $3 \times 10^{-6} \ s^{-1}$. If the initial concentration is $0.10 \ M$,the initial rate of reaction is:

AIIMS 1999 All AIIMS

A
$3 \times 10^{-5} \ M \ s^{-1}$
B
$3 \times 10^{-6} \ M \ s^{-1}$
C
$3 \times 10^{-8} \ M \ s^{-1}$
D
$3 \times 10^{-7} \ M \ s^{-1}$

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The rate constant of a first order reaction is $3 \times 10^{-6} \ s^{-1}$. If the initial concentration is $0.10 \ M$,the initial rate of reaction is:

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$A$ first-order reaction decomposes such that the time taken for $1/8$ and $1/10$ of the initial concentration to decompose are $t_{1/8}$ and $t_{1/10}$ respectively. Find the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$. (Given: $\log_{10} 2 = 0.3$)

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