(D) For a first-order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ $k = \frac{2.303}{40} \log \frac{0.1}

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$7.50 \times 10^{-4} \, M \, min^{-1}$

(D) For a first-order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ $k = \frac{2.303}{40} \log \frac{0.1}{0.005} = \frac{2.303}{40} \log 20$ $k = \frac{2.303 \times 1.3010}{40} \approx 0.0749 \, min^{-1}$ Now,the rate of reaction is given by: $Rate = k[X] = 0.0749 \times 0.01$ $Rate = 7.49 \times 10^{-4} \, M \, min^{-1} \approx 7.50 \times 10^{-4} \, M \, min^{-1}$

Class 12 Chemistry Chemical Kinetics First Order reaction

Difficult

In a first-order reaction,the concentration of reactant $X$ decreases from $0.1 \, M$ to $0.005 \, M$ in $40 \, min$. What will be the rate of reaction when the concentration of $X$ is $0.01 \, M$?

A
$1.73 \times 10^{-4} \, M \, min^{-1}$
B
$3.74 \times 10^{-4} \, M \, min^{-1}$
C
$3.47 \times 10^{-5} \, M \, min^{-1}$
D
$7.50 \times 10^{-4} \, M \, min^{-1}$

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Chemical Kinetics

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First Order reaction

For a first order reaction $A \rightarrow B$, the data is provided in the table below. Find $x$ in minutes. (Nearest integer)
$t/\text{min}$ $[A]/M$
$0$ $0.6500$
$x$ $0.0650$
$20$ $0.00065$

$t/\text{min}$	$[A]/M$
$0$	$0.6500$
$x$	$0.0650$
$20$	$0.00065$

DifficultJEE MAIN 2026

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For which of the following reactions is the half-life period independent of the initial concentration of the reactant?

Medium

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Which is the correct relation between rate constant and half-life of a first-order reaction?

EasyMHT CET 2022

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The half-life period of a first-order chemical reaction is $6.93 \, \text{min}$. The time required for the completion of $99 \%$ of the chemical reaction will be ........ $\text{min}$. $(\log 2 = 0.301)$

DifficultAIEEE 2009

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For a certain first order reaction,$32 \%$ of the reactant is left after $570 \ s$. The rate constant of this reaction is ........... $\times 10^{-3} \ s^{-1}$. (Round off to the Nearest Integer).
$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$

MediumJEE MAIN 2021

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In a first-order reaction,the concentration of reactant $X$ decreases from $0.1 \, M$ to $0.005 \, M$ in $40 \, min$. What will be the rate of reaction when the concentration of $X$ is $0.01 \, M$?

Similar Questions

For a first order reaction $A \rightarrow B$, the data is provided in the table below. Find $x$ in minutes. (Nearest integer)$t/\text{min}$$[A]/M$$0$$0.6500$$x$$0.0650$$20$$0.00065$

For which of the following reactions is the half-life period independent of the initial concentration of the reactant?

Which is the correct relation between rate constant and half-life of a first-order reaction?

The half-life period of a first-order chemical reaction is $6.93 \, \text{min}$. The time required for the completion of $99 \%$ of the chemical reaction will be ........ $\text{min}$. $(\log 2 = 0.301)$

For a certain first order reaction,$32 \%$ of the reactant is left after $570 \ s$. The rate constant of this reaction is ........... $\times 10^{-3} \ s^{-1}$. (Round off to the Nearest Integer).$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$

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For a first order reaction $A \rightarrow B$, the data is provided in the table below. Find $x$ in minutes. (Nearest integer)
$t/\text{min}$ $[A]/M$
$0$ $0.6500$
$x$ $0.0650$
$20$ $0.00065$

For a certain first order reaction,$32 \%$ of the reactant is left after $570 \ s$. The rate constant of this reaction is ........... $\times 10^{-3} \ s^{-1}$. (Round off to the Nearest Integer).
$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$