A first order reaction has a rate constant of . The time required for 40 g of this reactant to reduce to 10 g will be [Given that 2 = 0.3010] [2019]

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Solution

Q.
A first order reaction has a rate constant of $2.303 \times 10^{- 3}$ $s^{- 1}$ . The time required for 40 g of this reactant to reduce to 10 g will be

[Given that $\log_{10}$ 2 = 0.3010] [2019]

1 230.3 s

2 301 s

3 2000 s

4 602 s

Ans.
(4)

For a first order reaction, $k = \frac{2.303}{t} \log \frac{{[A]}_{0}}{{[A]}_{t}}$

$2.303 \times 10^{- 3} = \frac{2.303}{t} \log \frac{40}{10}$

$t = \frac{1}{10^{- 3}} \log 2^{2} = \frac{2}{10^{- 3}} \log 2 = \frac{2}{10^{- 3}} \times 0.3010 = 602$ $s$

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