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Recipe

Solve a reaction rate law -d[A]/dt=k[A]^n for [A](t)

Use when: the rate law is -\dfrac{d[A]}{dt}=k[A]^n for some order n, and you know [A](0)=[A]_0

Goal: an explicit formula for [A](t), plus (if needed) whether the half-life depends on the starting concentration

Taught in Chemical reaction rates

Quick recipe

  1. Separate the variables
  2. Integrate, using a different antiderivative for n=1 than for every other n
  3. Apply the initial condition and solve for [A](t) explicitly
  4. Find the half-life and check whether it depends on [A]_0
  1. 1

    Rewrite the rate law as \dfrac{d[A]}{[A]^n}=-k\,dt.

    Example

    For n=2: \dfrac{d[A]}{[A]^2}=-k\,dt.

    Why: the rate law already factors into a pure function of [A] times a constant — exactly the separable shape from earlier in the course.

  2. 2

    Integrate both sides of the separated equation from Step 1; which antiderivative rule applies to the left side depends on the exponent n.

    Example

    n=2: \displaystyle\int[A]^{-2}\,d[A]=-[A]^{-1}, so -\dfrac{1}{[A]}=-kt+C.

    • n=1 — \displaystyle\int\frac{d[A]}{[A]}=\ln[A], a logarithm.
    • n\ne1 — \displaystyle\int[A]^{-n}\,d[A]=\dfrac{[A]^{1-n}}{1-n}, an ordinary power rule.

    Why: n=1 is the one exponent the power rule for integration can't handle ([A]^0/0 is undefined), so it needs its own case, exactly like \int x^{-1}\,dx in Calculus I.

  3. 3

    Use [A](0)=[A]_0 to pin down the constant, then solve for [A].

    Example

    n=2, [A]_0=8, k=0.2: at t=0, -1/8=C, so \dfrac{1}{[A]}=0.2t+\dfrac18, giving [A](t)=\dfrac{8}{1+1.6t}

    • n=1 — [A](t)=[A]_0e^{-kt}, ordinary exponential decay.
    • n\ne1 — [A]^{1-n}=[A]_0^{1-n}+(n-1)kt, solved for [A] by raising both sides to the power \frac{1}{1-n}.

    Why: the general n\ne1 formula collapses to whatever exponent-specific answer direct substitution gives — it isn't a separate technique, just the same algebra one level more general.

  4. 4

    Set [A](t_{1/2})=[A]_0/2 and solve for t_{1/2}.

    Example

    n=2, k=0.2, [A]_0=8: the formula gives t_{1/2}=\dfrac{8^{-1}(2^1-1)}{1\cdot0.2}=\dfrac{1/8}{0.2}=0.625 — matching a direct check of [A](t)=\dfrac{8}{1+1.6t}=4 at t=0.625.

    • n=1 — t_{1/2}=\ln2/k, the same for every starting concentration.
    • n\ne1 — t_{1/2}=\dfrac{[A]_0^{1-n}(2^{n-1}-1)}{(n-1)k}, which depends explicitly on [A]_0.

    Why: this is exactly the "shorter half-life at higher concentration" pattern the section's second worked example derives, reached here from the general Step 3 formula instead of a separate derivation from scratch.

Common mistakes

  • Applying the n\ne1 integration formula when n=1, which divides by zero (1-n=0).
  • Stopping at the implicit power form [A]^{1-n}=\dots without finishing the algebra to solve for [A](t) itself.
  • Assuming half-life is always constant — true only for n=1 — and carrying that assumption into a second- or third-order reaction.
  • Losing track of the sign or root choice when raising [A]^{1-n}=[A]_0^{1-n}+(n-1)kt to the power 1/(1-n) — concentration is never negative, so only the physically meaningful root counts.