The rate constant for a first order reaction is $20 \, min^{-1}$. The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is $........ \times 10^{-2} \, min$. (Nearest integer) (Given: $\ln 10 = 2.303, \log 2 = 0.3010$)
- A$16$
- B$15$
- C$17$
- D$14$
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The initial rate for a first order reaction is $0.6932 \times 10^{-2} \ mol \ L^{-1} \ min^{-1}$ and the initial concentration of the reactant is $0.1 \ M$. Then $t_{1/2}$ is equal to ...... $min$.
Medium
View Solution$A$ first order reaction takes $40 \ min$ for $20 \%$ decomposition. Calculate its rate constant.
The rate constant for a first order reaction is $60 \ s^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?
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The following results are obtained in one pseudo first order reaction:
Time $(s)$ $0$ $30$ $60$ $90$ Concentration $(mol \ L^{-1})$ $0.551$ $0.312$ $0.173$ $0.085$
$(a)$ Calculate the average rate of reaction between $30$ and $60$ seconds.
$(b)$ Calculate the rate constant $(k)$ of this first order reaction.
| Time $(s)$ | $0$ | $30$ | $60$ | $90$ |
| Concentration $(mol \ L^{-1})$ | $0.551$ | $0.312$ | $0.173$ | $0.085$ |
$(a)$ Calculate the average rate of reaction between $30$ and $60$ seconds.
$(b)$ Calculate the rate constant $(k)$ of this first order reaction.
Medium
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