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A first order reaction has rate constant $1 \times 10^{-2} \mathrm{~s}^{-1}$. What time will it take for $20 \mathrm{~g}$ of reactant to reduce to $5 \mathrm{~g}$ ?
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Verified Answer
The correct answer is:
$138.6 \mathrm{~S}$
(A)
$\mathrm{k}=1 \times 10^{-2} \mathrm{~s}^{-1}, \quad[\mathrm{~A}]_{0}=20 \mathrm{~g},[\mathrm{~A}]_{\mathrm{t}}=5 \mathrm{~g}$
For first order reaction,
$t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{t}}$
$\therefore t=\frac{2.303}{1 \times 10^{-2}} \log _{10} \frac{20}{5}$
$\therefore \mathrm{t}=2.303 \times 0.602 \times 10^{2}=138.6 \mathrm{~s}$
$\mathrm{k}=1 \times 10^{-2} \mathrm{~s}^{-1}, \quad[\mathrm{~A}]_{0}=20 \mathrm{~g},[\mathrm{~A}]_{\mathrm{t}}=5 \mathrm{~g}$
For first order reaction,
$t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{t}}$
$\therefore t=\frac{2.303}{1 \times 10^{-2}} \log _{10} \frac{20}{5}$
$\therefore \mathrm{t}=2.303 \times 0.602 \times 10^{2}=138.6 \mathrm{~s}$
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