Chemistry / Unit 9 Chem 1660 worked example, 60 to 90 seconds
Solving the First-Order Integrated Rate Law for Time or Concentration
The idea
Solving the first-order integrated rate law for an unknown time or concentration means substituting the known rate constant and the two related concentration values into ln[A]t = ln[A]0 − kt, then rearranging to isolate whichever quantity is unknown, taking care that both concentrations are expressed in the same consistent unit before the natural logarithm is taken.
In practice
Cyclobutane decomposes at 500 °C with k = 9.2 × 10⁻³ s⁻¹. Treating the initial concentration as x, 80.0% decomposition leaves 0.200x remaining; substituting both expressions into the integrated law, taking the logarithm of the ratio, and solving for t gives about 175 seconds for that much decomposition to occur.
Where people slip
Substituting the percent decomposed, 0.800, for [A]t instead of the fraction remaining, 0.200, silently inverts the ratio inside the logarithm and gives a wrong time.
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