Reaction Orders

Chemical reactions are classified as zero-order, first-order, second-order, higher-order, or mixed-order based on how their rate depends on reactant concentration. This subtopic walks through each order's rate law, the units of the rate constant k, and the graphical technique used to identify a reaction's order from concentration-vs-time data.

Key Takeaways

• Reactions are classified by kinetic order — zero-order, first-order, second-order, higher-order, or mixed-order — based on how rate depends on reactant concentration.

• Zero-order: rate = k[A]⁰[B]⁰, k in M/s; rate is independent of concentration but still changes with temperature (via k) or with a catalyst. [A] vs. t is linear, slope = −k.

• First-order: rate = k[A]¹, k in s⁻¹; rate is directly proportional to concentration. ln[A] vs. t is linear, slope = −k.

• Second-order: rate = k[A][B] or k[A]² or k[B]², k in M⁻¹s⁻¹; often implies a bimolecular collision mechanism. 1/[A] vs. t is linear, slope = +k.

• Higher-order reactions (exponent sum > 2) are rare, requiring simultaneous multi-molecule collisions; mixed-order reactions don't follow constant integer orders, often due to complex mechanisms or intermediates. Both are low-yield on the MCAT — definitions are enough.

Classifying Reactions by Kinetic Order

For a general reaction:

aA + bB → cC + dD

The exponents in the reaction's rate law — how the rate scales with each reactant's concentration — determine its order. A reaction can be zero-order, first-order, second-order, higher-order, or mixed-order, and each classification has its own characteristic rate-law form, rate-constant units, and graphical signature.

Zero-Order Reactions

A zero-order reaction is one in which the rate of product formation is independent of reactant concentration — changing [A] or [B] doesn't change the rate. The rate law is:

rate = k[A]⁰[B]⁰

Since anything raised to the zero power equals 1, the rate constant k (units: M/s) is the only factor determining the rate.

Temperature and catalysts: even though concentration doesn't affect a zero-order reaction's rate, the rate constant k itself is temperature-dependent — so raising the temperature still changes the rate. A catalyst works the same way: it lowers the activation energy, which increases k and speeds up the reaction, without concentration playing any role.

Graphical signature: plotting [A] versus time for a zero-order reaction gives a straight line. The slope of that line equals −k, reflecting a constant rate of reaction.

First-Order Reactions

A first-order reaction has a rate directly proportional to the concentration of a single reactant — double the reactant's concentration, and the rate doubles too. The rate law is:

rate = k[A]¹ (or rate = k[B]¹)

Here k has units of s⁻¹.

Graphical signature: plotting [A] versus time gives a curve, not a line. But plotting the natural logarithm of concentration — ln[A] — versus time yields a straight line, with slope −k. That the reaction only linearizes on the ln[A] plot (and not the raw [A] plot) is itself confirmation that the rate depends on concentration to the first power.

Second-Order Reactions

A second-order reaction has a rate proportional either to the product of two reactants' concentrations, or to the square of a single reactant's concentration. The rate law takes one of these forms:

rate = k[A]¹[B]¹, or rate = k[A]², or rate = k[B]²

Here k has units of M⁻¹s⁻¹.

Physical significance: a second-order rate law — particularly one that's first-order in each of two different reactants — often points to a mechanism where a physical collision between two reactant molecules is required for the reaction to proceed.

Graphical signature: plotting [A] versus time gives a curve. Plotting the inverse of concentration — 1/[A] — versus time yields a straight line, with slope equal to +k.

Higher-Order and Mixed-Order Reactions

Higher-order reactions are those where the sum of the exponents in the rate law exceeds two. These are uncommon, because they'd require three or more reactant molecules to collide simultaneously — a statistically unlikely event. The rate constant's units for a higher-order reaction vary with the overall order, but always combine powers of concentration (M) and time (s) to keep the rate law dimensionally consistent.

Mixed-order reactions don't follow simple integer orders with respect to their reactants throughout the entire reaction. This can happen because of complex reaction mechanisms, the formation of intermediates, or an order that shifts across different concentration ranges.

Higher-order and mixed-order reactions are not high-yield on the MCAT. A basic grasp of what each term means is sufficient — you don't need to go beyond the definitions above.

Common MCAT Mistakes

• Mixing up which plot linearizes for which order. [A] vs. time is linear only for zero-order; ln[A] vs. time is linear only for first-order; 1/[A] vs. time is linear only for second-order. Plotting the wrong transformation gives a curve, not a line.

• Assuming k always has the same units regardless of order. k's units change with overall reaction order: M/s for zero-order, s⁻¹ for first-order, M⁻¹s⁻¹ for second-order — the units themselves are a clue to the order.

• Assuming reaction order must match a stoichiometric coefficient. Order is an experimentally determined property of the rate law, unrelated to the coefficients in the balanced equation.

• Expecting higher-order or mixed-order reactions to show up often on the exam. They're uncommon in real chemistry and low-yield on the MCAT — a basic definition-level understanding is sufficient.

MCAT-Style Concept Check

Question: A plot of 1/[A] versus time for a reaction yields a straight line with a positive slope, while a plot of [A] versus time is curved. What is the order of this reaction with respect to A, and what are the units of k?

• A) Zero order; units of M/s

• B) First order; units of s⁻¹

• C) Second order; units of M⁻¹s⁻¹

• D) Second order; units of M/s

Answer: C

Explanation: A straight line only appears when 1/[A] is plotted against time for a second-order reaction, with slope equal to +k. Second-order rate constants carry units of M⁻¹s⁻¹, distinguishing them from the M/s units of a zero-order k or the s⁻¹ units of a first-order k.

FAQ

How do you tell if a reaction is first-order or second-order from a graph?

Plot the concentration data three ways: [A] vs. time (linear only if zero-order), ln[A] vs. time (linear only if first-order), and 1/[A] vs. time (linear only if second-order). Whichever transformation produces a straight line reveals the order.

What are the units of the rate constant k for a zero-order reaction?

M/s (molarity per second), since a zero-order rate law is rate = k, with no concentration terms contributing units.

Why are higher-order reactions rare?

A higher-order reaction (exponent sum greater than two) would require three or more reactant molecules to collide simultaneously with the correct orientation and energy — a statistically unlikely event compared to the two-molecule collisions behind first- and second-order reactions.

What does it mean for a reaction to be "mixed-order"?

A mixed-order reaction doesn't follow a single constant integer order with respect to its reactants throughout the whole reaction — its apparent order can shift due to a complex mechanism, intermediates, or the concentration range being observed.