alleryStart learning →Cyclobutane decomposes by a first order path with k = 9.2 x 10^-3 per second. How long does it take for 80.0 percent of a sample to decompose?
AAbout 75 seconds
BAbout 175 seconds
CAbout 350 seconds
DAbout 500 seconds
Answer & Solution
Correct answer: B. About 175 seconds
1. Use the first order law in the form t = ln([A]0/[A]) divided by k.
2. If 80.0 percent has gone, then 20.0 percent remains, so [A] is 0.200 times [A]0.
3. The starting concentration cancels, leaving the ratio [A]0/[A] equal to 1/0.200, which is 5.00.
4. The natural logarithm of 5.00 is 1.609.
5. Divide 1.609 by 9.2 x 10^-3 per second to get 175 seconds.
6. About 75 seconds would correspond to roughly half decomposing, not four fifths.
7. About 350 seconds doubles the correct answer, as if 96 percent had to decompose.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.4 Integrated Rate Laws_
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