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Solve the following.
20% of a first order reaction is completed in five minutes. How much time will the 60% reaction…
$20\%$ of a first order reaction is completed in five minutes. How much time will the $60\%$ reaction take to complete? Calculate the half-life period ($t_{1/2}$) for the above reaction.
Answer
Answer
Formulas used: First-order integrated rate law: k = 2.303/t*log10(A0/A) with A = 80, t = 5 min, A0 = 100, so k = 0.0446 min^-1 First-order integrated rate law: k = 2.303/t*log10(A0/A) with A = 40, k = 0.0446 min^-1, A0 = 100, so t = 20.5 min Half-life, first order: t_half = 0.693/k with k = 0.0446 min^-1, so t_half = 15.5 min
For a first order reaction $k=\dfrac{2.303}{t}\log\dfrac{a}{a-x}$
When $20\%$ is completed in 5 min: $k=\dfrac{2.303}{5\text{ min}}\log\dfrac{100}{80}=0.0446\text{ min}^{-1}$
Time for $60\%$ completion: $t=\dfrac{2.303}{0.0446\text{ min}^{-1}}\log\dfrac{100}{40}=20.5\text{ min}$
Half-life: $t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{0.0446\text{ min}^{-1}}=15.5\text{ min}$
Final answer: 20.5 min