Common misconceptions

Common mistake
Wrong: A straight line on a plot of [A] vs. time always indicates a first-order reaction.
Right: A straight line on [A] vs. time indicates zero-order; first-order gives a straight line on ln[A] vs. time; second-order gives a straight line on 1/[A] vs. time.
A straight [A] vs. time graph means concentration drops at a constant rate — that's the definition of zero-order kinetics, where rate doesn't depend on concentration at all. First-order reactions produce exponential decay in [A], which only becomes linear when you take the natural log; so ln[A] vs. time is the first-order tell. Second-order produces a different curve that only linearizes as 1/[A] vs. time. Match the plot shape to the integrated rate law form, not to intuition about 'simplicity.'
Common mistake
Wrong: Half-life is always independent of initial concentration for any reaction order.
Right: Only first-order half-life is concentration-independent; zero-order and second-order half-lives depend on initial concentration.
The concentration-independence of half-life is a special property of first-order kinetics, not a universal rule. For first-order reactions, the t½ = 0.693/k formula has no concentration term — that's why it's constant. For zero-order reactions, t½ = [A]₀/2k, and for second-order, t½ = 1/(k[A]₀) — both explicitly depend on initial concentration. Generalizing the first-order rule to all orders is one of the most reliable wrong-answer traps on the MCAT.
Common mistake
Wrong: The second-order integrated rate law is ln[A] = ln[A]₀ − kt (same form as first-order but with different k).
Right: The second-order integrated rate law is 1/[A] = 1/[A]₀ + kt, which is distinct in form from the first-order law.
The second-order integrated rate law is 1/[A] = 1/[A]₀ + kt — it's a reciprocal relationship, not a logarithmic one. The first-order form uses ln[A] because integrating 1/[A] gives a natural log; the second-order form comes from integrating 1/[A]², which gives the reciprocal. These two equations have fundamentally different algebraic structures, and mixing them up will lead to wrong answers on both calculation questions and graph-identification questions.
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What the exam tests

  1. Know the three integrated rate law equations by form: zero-order uses [A] directly, first-order uses ln[A], and second-order uses 1/[A] — and be able to write or recognize each.
  2. Given a graph of concentration data over time, identify the reaction order by determining which plot (linear [A], ln[A], or 1/[A] vs. time) produces a straight line.
  3. Explain why first-order half-life is constant and concentration-independent, while zero-order and second-order half-lives change as the initial concentration changes.
  4. Perform calculations using integrated rate laws: given initial concentration, rate constant, and time, find remaining concentration; or given concentration data, solve for k or the time required to reach a target concentration.

Can you avoid these mistakes?

A reaction has the following data: [A] vs. time is curved, ln[A] vs. time is curved, but 1/[A] vs. time is a straight line with positive slope. What is the reaction order, and what does the slope of that line represent?
A first-order reaction has a half-life of 20 minutes regardless of starting concentration. A student claims this must also be true for a second-order reaction with the same rate constant. What's wrong with this reasoning, and how does the second-order half-life actually depend on concentration?
You have a zero-order reaction with k = 0.05 M/s and [A]₀ = 2.0 M. What is the concentration of A after 30 seconds? At what time does the reaction effectively stop (i.e., [A] = 0)?
For a first-order reaction, if ln[A]₀ = 3.0 and k = 0.1 min⁻¹, what is ln[A] after 10 minutes? What is the actual concentration [A] at that time?

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