What is the time needed to reduce the initial concentration of reactant to 10% in a first order reaction if its half life time is 10 minutes?

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TestHub · Accepted Answer

For a first order reaction, the rate constant $k$ is given by:

$k = \frac{2.303 lo g _{10} 2}{t _{1/2}}$

Given $t_{1/2} = 10$ minutes, we have:

$k = \frac{2.303 \times 0.3010}{10} = 0.0693 min^{- 1}$

The time $t$ required for the concentration to reduce to $10%$ of its initial value is given by:

$t = \frac{2.303}{k} lo g_{10} (\frac{[ A ] _{0}}{[ A ] _{t}})$

Here, $[A]_{t} = \frac{10}{100} [A]_{0} = 0.1 [A]_{0}$ .

$t = \frac{2.303}{0.0693} lo g_{10} (\frac{[ A ] _{0}}{0.1 [ A ] _{0}})$

$t = \frac{2.303}{0.0693} lo g_{10} (10)$

$t = \frac{2.303}{0.0693} \times 1 \approx 33.23 minutes$

The closest value is $33 minute$ .

Answer: 33 minute

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