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(C) For a first order reaction,the half-life $(t_{1/2})$ is independent of the initial concentration.Given that the concentration drops from $0.8 \ mo

Vedclass · Accepted Answer

$30$ minutes

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(C) For a first order reaction,the half-life $(t_{1/2})$ is independent of the initial concentration.Given that the concentration drops from $0.8 \ mol \ L^{-1}$ to $0.4 \ mol \ L^{-1}$ (which is half) in $15$ minutes,the half-life is $t_{1/2} = 15$ minutes.To drop the concentration from $0.1 \ mol \ L^{-1}$ to $0.025 \ mol \ L^{-1}$:Step $1$: $0.1 \ mol \ L^{-1} \rightarrow 0.05 \ mol \ L^{-1}$ (one half-life,$15$ minutes).Step $2$: $0.05 \ mol \ L^{-1} \rightarrow 0.025 \ mol \ L^{-1}$ (another half-life,$15$ minutes).Total time required $= 15 + 15 = 30$ minutes.

In a first order reaction,if the concentration of reactant drops from $0.8 \ mol \ L^{-1}$ to $0.4 \ mol \ L^{-1}$ in $15$ minutes,what is the time required to drop the concentration from $0.1 \ mol \ L^{-1}$ to $0.025 \ mol \ L^{-1}$?

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