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Second-Order Integrated Rate Law

1[A]=1[A]0+kt\frac{1}{[\mathrm{A}]} = \frac{1}{[\mathrm{A}]_0} + kt

Worked example: 0.100 M second-order, k = 0.500 L/(mol·s), 10.0 s → 0.0666667 M — press Try an example to run it live, then adjust anything.

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Second-Order Integrated Rate Law explained

[A]0t[A]k

When two molecules of the same reactant must find each other, the rate goes as [A]², and integrating gives a straight line only if you plot the reciprocal concentration against time. The slope of that line is k, in L/(mol·s). Starting at 0.100 M with k = 0.500 L/(mol·s), after 10 s the reciprocal has climbed from 10 to 10 + 5 = 15, so [A] = 0.0667 M.

The three integrated laws form the standard diagnostic kit: plot [A] versus t, ln[A] versus t, and 1/[A] versus t, and whichever comes out straight tells you the order. Second-order kinetics has a distinctive personality — it starts fast and then drags, because losing reactant hurts the rate twice over. Gas-phase NO₂ decomposition and many radical recombinations follow it. Be careful with the "pseudo" cases: a reaction that is genuinely second order overall but run with a huge excess of one partner behaves like first order in the other, which is exactly how kineticists tame a two-variable problem into a one-variable measurement.

Second-Order Integrated Rate Law formula

1[A]=1[A]0+kt\frac{1}{[\mathrm{A}]} = \frac{1}{[\mathrm{A}]_0} + kt
Where
  • [A][\mathrm{A}]= Concentration at time t (M)
  • [A]0[\mathrm{A}]_0= Initial concentration (M)
  • kk= Rate constant in L/(mol·s) (L/(mol·s))
  • tt= Elapsed time (s)

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Part of this formula set

The integrated rate laws

5 formulasZero-, first- and second-order concentration-versus-time laws and their half-lives — and how to read the order off a straight line.