Under the same reaction conditions, the initial concentration of 1.386 mol $d m^{- 3}$ of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively.

The ratio $(k_{1} / k_{0}$ ) of the rate constants for first order $(k_{1})$ and zero order $(k_{0})$ of the reaction is

0.5

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1.0

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1.5

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2.0

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Solution

The correct option is A $0.5$
For the first order reaction,the rate constant

$k_{1} = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{40}$

For zero order reaction, the rate constant

$k_{0} = \frac{A_{0}}{2 \times t_{1 / 2}} = \frac{1.386}{2 \times 20}$

Hence,
$\frac{k_{1}}{k_{0}} = \frac{0.693}{1.386} = 0.5$

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