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A first order reaction is $50 \%$ completed in $1.26 \times 10^{14} \mathrm{~s}$. How much time would it takes for 100\% completion?
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The correct answer is:
Infinite
$k$-for first order. $\frac{0.693}{t_{1 / 2}}=\frac{0.693}{1.26 \times 10^{14}}$ for $100 \%$ completion.
$\because t=\frac{2.303}{k} \log \frac{a}{a-x}$
$t=\frac{2.303 \times 1.26 \times 10^{14}}{0.693} \log \frac{100}{0}$
$\therefore t=\infty$
$\because t=\frac{2.303}{k} \log \frac{a}{a-x}$
$t=\frac{2.303 \times 1.26 \times 10^{14}}{0.693} \log \frac{100}{0}$
$\therefore t=\infty$
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