The half-life period of a first order reaction is $15 \ minutes$. The amount of substance left after one hour will be
- A$\frac{1}{4}$ of the original amount
- B$\frac{1}{8}$ of the original amount
- C$\frac{1}{16}$ of the original amount
- D$\frac{1}{32}$ of the original amount
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Similar Questions
What is the value of the rate constant of a first-order reaction if the slope of the graph between $\log_{10} \frac{[A]_0}{[A]_t}$ ($y$-axis) and time ($x$-axis) is $1 \times 10^{-3}$?
For a first order reaction $A \rightarrow P$,$t_{1/2}$ (half-life) is $10 \text{ days}$. The time required for $1/4^{th}$ conversion of $A$ (in days) is:
Medium
View SolutionThe 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 SolutionIn the following graphs,which graph$(s)$ belong to a first-order reaction?
Medium
View SolutionWhich one of the following formulas represents a first-order reaction?
Easy
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