(C) For a second-order reaction,the integrated rate equation is given by: $k = \frac{1}{t} \left( \frac{1}{[A]_t} - \frac{1}{[A]_0} \right)$.Given: $k

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$1.25 \times 10^4 \, \text{min}$

(C) For a second-order reaction,the integrated rate equation is given by: $k = \frac{1}{t} \left( \frac{1}{[A]_t} - \frac{1}{[A]_0} \right)$.Given: $k = 8 \times 10^{-5} \, M^{-1} \, \text{min}^{-1}$,$[A]_0 = 1 \, M$,and $[A]_t = 0.5 \, M$.Substituting the values: $8 \times 10^{-5} = \frac{1}{t} \left( \frac{1}{0.5} - \frac{1}{1} \right)$.$8 \times 10^{-5} = \frac{1}{t} (2 - 1)$.$8 \times 10^{-5} = \frac{1}{t} \times 1$.$t = \frac{1}{8 \times 10^{-5}} = \frac{10^5}{8} = 0.125 \times 10^5 = 1.25 \times 10^4 \, \text{min}$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

For a second-order reaction,the rate constant is $8 \times 10^{-5} \, M^{-1} \, \text{min}^{-1}$. In how much time will a $1 \, M$ solution decrease to $0.5 \, M$?

A
$8.665 \times 10^3 \, \text{min}$
B
$8 \times 10^3 \, \text{min}$
C
$1.25 \times 10^4 \, \text{min}$
D
$4 \times 10^5 \, \text{min}$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

The units for the rate constant and the rate of reaction are same for a reaction. What will be the order of the reaction?

EasyGUJCET 2018

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For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate equation can be expressed in two ways $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ and $+\frac{d[NO_2]}{dt} = k'[N_2O_5]$. $k$ and $k'$ are related as:

DifficultJEE MAIN 2014

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For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.
Rate $(mol \ L^{-1} \ s^{-1})$ $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$
$0.10$ $20$ $0.5$
$0.40$ $x$ $0.5$
$0.80$ $40$ $y$

What is the value of $x$ and $y$?

Rate $(mol \ L^{-1} \ s^{-1})$	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$
$0.10$	$20$	$0.5$
$0.40$	$x$	$0.5$
$0.80$	$40$	$y$

MediumJEE MAIN 2023

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The rate constant of esterification is given by $k = k^{\prime} [H_2O]$. If the rate constant $k = 2.0 \times 10^{-3} \ min^{-1}$,calculate $k^{\prime}$. (Assume $[H_2O] = 55.5 \ mol \ L^{-1}$)

Medium

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For the reaction ${H_{2(g)} + I_{2(g)} \to 2HI_{(g)}}$,the possible mechanism is given below:
$I_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2 I$ (fast step)
${2I + H_2 \xrightarrow{K_2} 2HI}$ (slow step)
Determine the rate law for the reaction.

Medium

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For a second-order reaction,the rate constant is $8 \times 10^{-5} \, M^{-1} \, \text{min}^{-1}$. In how much time will a $1 \, M$ solution decrease to $0.5 \, M$?

Similar Questions

The units for the rate constant and the rate of reaction are same for a reaction. What will be the order of the reaction?

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate equation can be expressed in two ways $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ and $+\frac{d[NO_2]}{dt} = k'[N_2O_5]$. $k$ and $k'$ are related as:

For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.Rate $(mol \ L^{-1} \ s^{-1})$$[A]$ $(mol \ L^{-1})$$[B]$ $(mol \ L^{-1})$$0.10$$20$$0.5$$0.40$$x$$0.5$$0.80$$40$$y$What is the value of $x$ and $y$?

The rate constant of esterification is given by $k = k^{\prime} [H_2O]$. If the rate constant $k = 2.0 \times 10^{-3} \ min^{-1}$,calculate $k^{\prime}$. (Assume $[H_2O] = 55.5 \ mol \ L^{-1}$)

For the reaction ${H_{2(g)} + I_{2(g)} \to 2HI_{(g)}}$,the possible mechanism is given below: $I_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2 I$ (fast step) ${2I + H_2 \xrightarrow{K_2} 2HI}$ (slow step) Determine the rate law for the reaction.

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For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.
Rate $(mol \ L^{-1} \ s^{-1})$ $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$
$0.10$ $20$ $0.5$
$0.40$ $x$ $0.5$
$0.80$ $40$ $y$

What is the value of $x$ and $y$?

For the reaction ${H_{2(g)} + I_{2(g)} \to 2HI_{(g)}}$,the possible mechanism is given below:
$I_2 \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} 2 I$ (fast step)
${2I + H_2 \xrightarrow{K_2} 2HI}$ (slow step)
Determine the rate law for the reaction.