Chemistry / Unit 9 Chem 1666 worked example, 60 to 90 seconds
Solving the Second-Order Integrated Rate Law
The idea
Solving the second-order integrated rate law for a later concentration means substituting the known initial concentration, the rate constant, and the elapsed time into 1/[A]t = kt + 1/[A]0, then taking the reciprocal of the resulting sum to recover the concentration itself, since this equation is linear in 1/[A] rather than in [A] directly, the same distinction Chem 1663's plot check relies on.
In practice
Butadiene at 0.200 M dimerizes with k = 5.76 × 10⁻² L/mol/min. After 10.0 min, 1/[A]t = (5.76 × 10⁻²)(10.0) + 1/0.200 = 5.576 L/mol, so [A]t = 1/5.576, about 0.179 mol/L of butadiene remaining after that time, compared with the 0.200 M initially present.
Where people slip
Adding the rate constant term to 1/[A]0 rather than to [A]0 itself is the step most often reversed — the second-order law is linear in the reciprocal of concentration, never in concentration directly.
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