For the reaction shown in the image,the half-life does not depend on the concentration of the reactant. After $10 \, \text{min}$,the volume of $N_2$ gas evolved is $20 \, \text{L}$ and after the completion of the reaction,it is $100 \, \text{L}$. Hence,the rate constant is:
- A$\frac{2.303}{10} \log \frac{5}{4} \, \text{min}^{-1}$
- B$\frac{2.303}{10} \log 5 \, \text{min}^{-1}$
- C$\frac{2.303}{10} \log 15 \, \text{min}^{-1}$
- D$\frac{2.303}{10} \log 20 \, \text{min}^{-1}$
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Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3 \text{ hours}$. The percentage of sucrose remaining after $6 \text{ hours}$ is . . . . . . . (Nearest integer) (Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$)
The following results were obtained during kinetic studies of the reaction $2A + B \to$ products:
Experiment $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ min^{-1})$ $I$ $0.10$ $0.20$ $6.93 \times 10^{-3}$ $II$ $0.10$ $0.25$ $6.93 \times 10^{-3}$ $III$ $0.20$ $0.30$ $1.386 \times 10^{-2}$
The time (in minutes) required to consume half of $A$ is:
| Experiment | $[A]$ $(mol \ L^{-1})$ | $[B]$ $(mol \ L^{-1})$ | Initial rate $(mol \ L^{-1} \ min^{-1})$ |
|---|---|---|---|
| $I$ | $0.10$ | $0.20$ | $6.93 \times 10^{-3}$ |
| $II$ | $0.10$ | $0.25$ | $6.93 \times 10^{-3}$ |
| $III$ | $0.20$ | $0.30$ | $1.386 \times 10^{-2}$ |
The time (in minutes) required to consume half of $A$ is:
DifficultJEE MAIN 2019
View Solution$N_{2}O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$
In the above first order reaction,the initial concentration of $N_{2}O_{5}$ is $2.40 \times 10^{-2} \ mol \ L^{-1}$ at $318 \ K$. The concentration of $N_{2}O_{5}$ after $1 \ hour$ was $1.60 \times 10^{-2} \ mol \ L^{-1}$. The rate constant of the reaction at $318 \ K$ is $..... \times 10^{-3} \ min^{-1}$. (Nearest integer)
[Given: $\log 3 = 0.477, \log 5 = 0.699$]
In the above first order reaction,the initial concentration of $N_{2}O_{5}$ is $2.40 \times 10^{-2} \ mol \ L^{-1}$ at $318 \ K$. The concentration of $N_{2}O_{5}$ after $1 \ hour$ was $1.60 \times 10^{-2} \ mol \ L^{-1}$. The rate constant of the reaction at $318 \ K$ is $..... \times 10^{-3} \ min^{-1}$. (Nearest integer)
[Given: $\log 3 = 0.477, \log 5 = 0.699$]
$A$ first order reaction is given as $A \rightarrow \text{products}$. Its integrated rate equation is:
The rate constant of a first order reaction whose half-life is $480 \ s$,is
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