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 For a first order reaction A \rightarrow B having initial concentration of reactant A_0  and rate constant 3.5 \mathrm{~min}^{-1}. If time taken to convert remaining reactant amount from \mathrm{A_0 \! \; to\; \frac{A_0}{2}} is 10 minutes, than what will be the time taken to convert remaining reactant amount from \frac{A_0}{4}\: to\: \frac{A_0}{8} ?

Option: 1

2.5 minutes


Option: 2

10.5 minutes


Option: 3

40.0 minutes


Option: 4

10.0 minutes


Answers (1)

best_answer

When reactant amount changes from \mathrm{A_0\; to \; \frac{A_0}{2}}, amount gets half, which means time taken is half life.

\Rightarrow t_{\frac{1}{2}}=10 minutes, similarly when reactant amount changes from \frac{A_0}{4}\: \; \text{to} \; \; \frac{A_0}{8}, again the reactant amount gets half, thus time taken will be equal to half life. This is because t_{\frac{1}{2}} of the first order reaction is independent of reactant amount/concentration.

Thus time taken to convert reactant amount from \mathrm{\frac{A_0}{4}\; to \; \frac{A_0}{8}} is also 10 minutes.

NOTE: For a first order reaction, if reactant is halved the time taken will be equal to half life of the reaction, as half life of first order reaction is independent of reactant amount.

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