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Under the same reaction conditions, initial concentration of $1.386 \mathrm{~mol} \mathrm{dm}^{-3}$ of a substance becomes half in $40 \mathrm{~s}$ and 20 s through first-order and zero-order kinetics respectively. Ratio $\left(\frac{k_{1}}{k_{0}}\right)$ of the rate constants for first-order $\left(k_{1}\right)$ and zero-order $\left(\mathrm{k}_{0}\right)$ of the reactions is
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$0.5 \mathrm{~mol}^{-1} \mathrm{dm}^{3}$
For First order reaction
$\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}_{1}} \Rightarrow \mathrm{k}_{1}=\frac{0.693}{40} \mathrm{~s}^{-1}\quad_{-} \text {(i) }$
For Zero order reaction
$\mathrm{t}_{1 / 2}=\frac{[\mathrm{R}]_{0}}{2 \mathrm{k}_{0}} \Rightarrow \mathrm{k}_{0}=\frac{1.386}{2 \times 20} \mathrm{~mol} / \mathrm{dm}^{3} / \mathrm{s} \quad_{-} \text {(ii) }$
$\underline{\text { Using (i) and (ii) }}$
$\frac{\mathrm{k}_{1}}{\mathrm{k}_{0}}=\frac{0.693 / 40}{1.386 / 2 \times 20}=\frac{0.693}{1.386}=0.5 \mathrm{~mol}^{-1}-\mathrm{dm}^{3}$
$\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}_{1}} \Rightarrow \mathrm{k}_{1}=\frac{0.693}{40} \mathrm{~s}^{-1}\quad_{-} \text {(i) }$
For Zero order reaction
$\mathrm{t}_{1 / 2}=\frac{[\mathrm{R}]_{0}}{2 \mathrm{k}_{0}} \Rightarrow \mathrm{k}_{0}=\frac{1.386}{2 \times 20} \mathrm{~mol} / \mathrm{dm}^{3} / \mathrm{s} \quad_{-} \text {(ii) }$
$\underline{\text { Using (i) and (ii) }}$
$\frac{\mathrm{k}_{1}}{\mathrm{k}_{0}}=\frac{0.693 / 40}{1.386 / 2 \times 20}=\frac{0.693}{1.386}=0.5 \mathrm{~mol}^{-1}-\mathrm{dm}^{3}$
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