Half-lives of first-order and zeroth order reactions are same. Ratio of rates at the start of nreaction is
- $0.693$
- $\frac{1}{0.693}$
- $2 \times 0.693$
- $\frac{2}{0693}$
Solution
$T_{50}($ first $)=\frac{0.693}{k_{1}}$
$\frac{a}{2 k_{0}}=\frac{0.693}{k_{1}}$
$\therefore \frac{k_{1}}{k_{0}}=\frac{2 \times 0.693}{a}$
$\frac{\left(\frac{d x}{d t}ight)_{1}}{\left(\frac{d x}{d t}ight)_{0}}=\frac{k_{1} a}{k_{0}}=2 \times 0.693$ /
Asked in: JEE-TOPICTESTS-CHEMISTRY