Reaction Orders
High-Yield Summary
- Reactions are classified zero-, first-, second-, higher-, or mixed-order based on how rate depends on concentration.
- Zero-order: rate = k (independent of [A]/[B]); [A] vs. t is linear, slope = −k.
- First-order: rate = k[A]; ln[A] vs. t is linear, slope = −k.
- Second-order: rate = k[A][B] or k[A]²/k[B]²; 1/[A] vs. t is linear, slope = +k.
- Higher-order (exponent sum > 2) and mixed-order reactions are rare/low-yield — definitions are enough.
Reaction Order Reference Table
| Order (Rate Law) | k Units / Linear Plot (slope) |
|---|---|
| Zero-order: rate = k[A]⁰[B]⁰ | k in M/s; [A] vs. t linear, slope = −k |
| First-order: rate = k[A] | k in s⁻¹; ln[A] vs. t linear, slope = −k |
| Second-order: rate = k[A][B] or k[A]² or k[B]² | k in M⁻¹s⁻¹; 1/[A] vs. t linear, slope = +k |
| Higher-order (exponent sum > 2) | Units vary; rare — requires 3+ molecules colliding simultaneously |
Must-Know Points
- For a zero-order reaction, k itself is still temperature-dependent — raising temperature or adding a catalyst still speeds up the reaction, even though concentration doesn't matter.
- A second-order rate law that's first-order in two different reactants (rate = k[A][B]) often signals a mechanism requiring a bimolecular collision.
- Mixed-order reactions can arise from complex mechanisms, intermediates, or an order that shifts across concentration ranges.
Common MCAT Trap
- Plotting [A] vs. time is only linear for a ZERO-order reaction — a curved [A]-vs-t plot doesn't mean 'no order,' it means you need to try ln[A] (first-order) or 1/[A] (second-order) instead.
- Zero-order rate constant units (M/s) are easy to confuse with first-order (s⁻¹) or second-order (M⁻¹s⁻¹) — mismatched units are a fast way to catch an order error.
- Higher-order and mixed-order reactions are low-yield — don't over-invest beyond knowing the definitions.
Quick Recall
If plotting ln[A] vs. time gives a straight line but [A] vs. time doesn't, what order is the reaction with respect to A?
What are the units of k for a second-order reaction, and what does that order often imply mechanistically?
For a zero-order reaction, does raising the temperature change the rate?