(A) For a first order reaction,the rate constant $k$ is given by the formula:$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$Given values are:$t = 5 \ min

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(A) For a first order reaction,the rate constant $k$ is given by the formula:$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$Given values are:$t = 5 \ min$$[A]_0 = 0.6 \ mol \ L^{-1}$$[A] = 0.2 \ mol \ L^{-1}$Substituting the values in the formula:$k = \frac{2.303}{5} \log \frac{0.6}{0.2}$$k = \frac{2.303}{5} \log 3$Using $\log 3 = 0.4771$:$k = \frac{2.303 \times 0.4771}{5}$$k = \frac{1.0988}{5} = 0.21976 \ min^{-1} \approx 0.219 \ min^{-1}$

Class 12 Chemistry Chemical Kinetics First Order reaction

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In a first order reaction,the concentration of the reactant is reduced from $0.6 \ mol \ L^{-1}$ to $0.2 \ mol \ L^{-1}$ in $5 \ min$. What is the rate constant of the reaction (in $min^{-1}$)? $(\log 3 = 0.4771)$

AP EAMCET 2012 All AP EAMCET

A
$0.219$
B
$0.325$
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$0.421$
D
$0.522$

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Chemical Kinetics

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In a first order reaction,the concentration of the reactant is reduced from $0.6 \ mol \ L^{-1}$ to $0.2 \ mol \ L^{-1}$ in $5 \ min$. What is the rate constant of the reaction (in $min^{-1}$)? $(\log 3 = 0.4771)$

Similar Questions

The rate equation for a first-order reaction is given by $[R] = [R]_0 e^{-kt}$. $A$ straight line with a positive slope is obtained by plotting: ($[R]_0 =$ initial concentration of reactant,$[R] =$ concentration of reactant at time $t$)

$99 \%$ of a first order reaction was completed in $64 \text{ min}$. In how many minutes will $99.9 \%$ of the reaction be complete?

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:

At $27^{\circ} C$ temperature,the time required for $75 \%$ completion of a first order reaction is $20 \ s$. What will be its rate constant?

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