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A first-order reaction has rate constant $k = 6.93\times 10^{-3}$ s$^{-1}$. Its half-life is:
A$100$ s, since $t_{1/2} = 0.693/k = 0.693/6.93\times 10^{-3}$
B$50$ s, taking half the correct calculation value here
C$1000$ s, applying the wrong exponential factor here
D$10$ s, the simple inverse without the natural log term
Answer & Solution
Correct answer: A. $100$ s, since $t_{1/2} = 0.693/k = 0.693/6.93\times 10^{-3}$
$t_{1/2} = 0.693/k = 0.693/0.00693 = 100$ s.
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