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Solve the following.
40% of a first order reaction is completed in 50 minutes. How much time will it take for the…
$40\%$ of a first order reaction is completed in $50\text{ minutes}$. How much time will it take for the completion of $80\%$ of this reaction?
Answer
Answer
Formulas used: First-order integrated rate law: k = 2.303/t*log10(A0/A) with A = 60, t = 50 min, A0 = 100, so k = 0.01022 min^-1 First-order integrated rate law: k = 2.303/t*log10(A0/A) with A = 20, k = 0.01022 min^-1, A0 = 100, so t = 157.5 min
Formula: $k = \dfrac{2\cdot303}{t}\log\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]}$
For $40\%$ completion: $[\mathrm{A}]_0 = 100$, $[\mathrm{A}] = 60$
$k = \dfrac{2\cdot303}{50\text{ min}}\log\dfrac{100}{60} = 1\cdot02\times10^{-2}\text{ min}^{-1}$
For $80\%$ completion: $[\mathrm{A}] = 20$
$t = \dfrac{2\cdot303}{1\cdot02\times10^{-2}\text{ min}^{-1}}\log\dfrac{100}{20} = 157\cdot5\text{ min}$
Final answer: 157.5 min