Under the same reaction conditions, the initial concentration of 1.386 mol $d m^{- 3}$ of a substance becomes half in 40 s and 20 s 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 m o l^{- 1} d m^{3}

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1.0 m o l d m^{- 3}

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1.5 m o l d m^{- 3}

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2.0 m o l^{- 1} d m^{3}

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Solution

The correct option is A $0.5 m o l^{- 1} d m^{3}$
For the first order reaction, rate constant

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

For zero order reaction, 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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Similar questions

Under the same reaction conditions, initial concentration of 1.386 $m o l d m^{- 3}$ of a substance becomes half in 40 s amd 20 s through first order and zero order kinetics respectively. Ratio $(\frac{k_{1}}{k_{0}})$ of the rate constants for first order $(k_{1})$ and zero order $(k_{0})$ of the reaction is:

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