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Heated to 499 degrees Celsius, cyclopropane isomerises to propene with k = 5.95 x 10^-4 per second. What fraction of the cyclopropane remains after 0.75 h?

AAbout 0.20 of it remains
BAbout 0.80 of it remains
CAbout 0.50 of it remains
DAbout 0.10 of it remains
Answer & Solution
Correct answer: A. About 0.20 of it remains
1. A rate constant in per second units on a rearrangement of one molecule signals first order kinetics. 2. Convert the time to seconds: 0.75 h x 3600 s/h = 2700 s. 3. Compute the exponent: kt = 5.95 x 10^-4 per second x 2700 s = 1.61. 4. The surviving fraction is [A]/[A]0 = e^(-1.61) = 0.20. 5. Cross-check with half-lives: t half = 0.693 / 5.95 x 10^-4 = 1165 s, and 2700 s is about 2.3 half-lives, giving (1/2)^2.3, near one fifth. 6. The value 0.80 reports the fraction that reacted rather than the fraction remaining. 7. The value 0.50 assumes exactly one half-life passed, and 0.10 overshoots by a decade. _Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.4 Integrated Rate Laws_
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