At $T \ K$,if the rate constant for a zero order reaction is $2.5 \times 10^{-3} \ M \ s^{-1}$,the time required for the initial concentration of reactant $R$ to fall from $0.10 \ M$ to $0.075 \ M$ at the same temperature in seconds is:
- A$25$
- B$5$
- C$10$
- D$20$
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Similar Questions
The rates of a certain reaction $(dc/dt)$ at different times are as follows:
Time $(sec)$ Rate $(mole \ litre^{-1} \ sec^{-1})$ $0$ $2.8 \times 10^{-2}$ $10$ $2.78 \times 10^{-2}$ $20$ $2.81 \times 10^{-2}$ $30$ $2.79 \times 10^{-2}$
The reaction is:
| Time $(sec)$ | Rate $(mole \ litre^{-1} \ sec^{-1})$ |
|---|---|
| $0$ | $2.8 \times 10^{-2}$ |
| $10$ | $2.78 \times 10^{-2}$ |
| $20$ | $2.81 \times 10^{-2}$ |
| $30$ | $2.79 \times 10^{-2}$ |
The reaction is:
Medium
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The half-life period for a certain zero-order reaction is $10 \text{ min}$. How much time is required for this reaction to complete $100\%$ (in $\text{ min}$)?
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View SolutionIn a zero-order reaction,if the initial concentration of the reactant is doubled,the time required for half the reactant to be consumed:
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