(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

After how many seconds will the concentration of the reactants in a first order reaction be halved,if the decay constant is $1.155 \times 10^{-3} \, \sec^{-1}$?

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$A$ reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \ min^{-1}$. If we start with $[A] = 0.5 \ mol \ L^{-1}$,when would $[A]$ reach the value $0.05 \ mol \ L^{-1}$? (in $min$)

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If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........
(Take: $\log 2=0.30 ; \log 2.5=0.40)$

MediumJEE MAIN 2020

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Rate of first order reaction $A \rightarrow \text{product}$ is $0.01 \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate constant if the concentration of the reactant is $0.2 \ M$. (in $s^{-1}$)

MediumMHT CET 2022

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For a first order reaction,the slope of the graph of $\log_{10}[A]_t$ versus time is equal to:

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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

After how many seconds will the concentration of the reactants in a first order reaction be halved,if the decay constant is $1.155 \times 10^{-3} \, \sec^{-1}$?

$A$ reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \ min^{-1}$. If we start with $[A] = 0.5 \ mol \ L^{-1}$,when would $[A]$ reach the value $0.05 \ mol \ L^{-1}$? (in $min$)

If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........(Take: $\log 2=0.30 ; \log 2.5=0.40)$

Rate of first order reaction $A \rightarrow \text{product}$ is $0.01 \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate constant if the concentration of the reactant is $0.2 \ M$. (in $s^{-1}$)

For a first order reaction,the slope of the graph of $\log_{10}[A]_t$ versus time is equal to:

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If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........
(Take: $\log 2=0.30 ; \log 2.5=0.40)$