Under the same reaction condition, initial concentration of $1.386 {mol dm}^{- 3}$ of a substance becomes half in $40$ seconds and $20$ seconds through first order and zero order kinetics respectively. Ratio $(k_{1} / k_{o})$ of the rate constant for first order $(k_{1})$ and zero order $(k_{0})$ of the reaction is:

0.5 {mol}^{- 1} {dm}^{3}

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1.0 {mol} {dm}^{- 3}

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1.5 {mol} {dm}^{- 3}

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2.0 {mol}^{- 1} {dm}^{3}

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Solution

The correct option is A $0.5 {mol}^{- 1} {dm}^{3}$
The values of rate constants $k_{o}, k_{1}$ for zero order and first order reaction, respectively, are given by the following equation:
$k_{o} = \frac{A_{o}}{2 \times t_{1 / 2}} k_{1} = \frac{0.693}{t_{1 / 2}}$
Substituting the various values we get,
$k_{o} = \frac{1.386 {mol litre}^{- 1}}{2 \times 20 s} . . (i) k_{1} = \frac{0.693}{40 s} . . . (i i)$
Dividing (ii) by (i), we get
$\frac{k_{1}}{k_{o}} = \frac{0.693}{40} \times \frac{2 \times 20}{1.386} {mol}^{- 1} litre = \frac{0.693}{1.386} {mol}^{- 1} litre = 0.5 {mol}^{- 1} {dm}^{3}$

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Under the same reaction condition, initial concentration of 1.386 mol dm−3 of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics respectively. Ratio (k1/ko) of the rate constant for first order (k1) and zero order (k0) of the reaction is: