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  4. A first order reaction has a rate constant of 2.303× 10-3 s-1. The time required for 40g of this reactant to reduce to 10 g will be- [Given that log10 2 = 0.3010

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Q. A first order reaction has a rate constant of $2.303\times 10^{-3}$ s$^{-1}$. The time required for 40g of this reactant to reduce to 10 g will be-
[Given that $log_{10}$ 2 = 0.3010

NEETNEET 2019Chemical Kinetics

Solution:

For a first order reaction; $t_{1/2} = \frac{0.693}{K}$
$t_{1/2} = \frac{0.693}{2.303 \times 10^{-3}} = 301 \,s$
The time required for $40\, g$ of reactant to reduce to $10g$
$t_{75\%} = 2 \times t_{1/2}$
$t_{75\%} = 2 \times 301 = 602 \,s$