Reaction rates
Reaction rate measures how quickly composition changes. A balanced equation tells how species are related stoichiometrically; a rate law tells how the observed speed depends on concentration and temperature. Those are different kinds of information.
Defining a reaction rate
For aA → pP, a stoichiometrically normalized rate can be written:
rate = −(1/a)d[A]/dt = (1/p)d[P]/dt
The minus sign makes the reactant-based rate positive. Average rates use finite concentration changes; instantaneous rates use derivatives.
Units are usually concentration per time, such as mol L⁻¹ s⁻¹.
Rate laws come from experiments
A common empirical form is:
rate = k[A]m[B]n
m and n are reaction orders determined experimentally. The overall order is m+n. Doubling [A] doubles rate for first order, quadruples it for second order, and has no effect for zero order.
Only for an elementary reaction can the stoichiometric coefficients be used directly as rate-law exponents.
Integrated laws turn concentration into a clock
| Order | Linear plot | Half-life behavior |
|---|---|---|
| 0 | [A] vs t | depends on initial [A] |
| 1 | ln[A] vs t | t½ = ln2/k |
| 2 | 1/[A] vs t | depends inversely on initial [A] |
A straight line identifies the matching integrated rate law. First-order radioactive decay and many unimolecular processes have a constant half-life.
Temperature, collisions and mechanisms
Raising temperature usually increases k because a larger fraction of molecular encounters can cross the activation barrier. But collision frequency alone is not enough: orientation and reaction pathway matter.
A multistep mechanism can create a rate law that bears little resemblance to the overall stoichiometric equation. Intermediates are formed in one step and consumed in another; catalysts participate in steps but are regenerated overall.
Initial-rates method
To determine an exponent, compare experiments where one reactant concentration changes while others stay constant. If doubling [A] makes the initial rate four times larger, the reaction is second order in A because 2m = 4 gives m = 2.
Good kinetic data require controlled temperature, accurate timing and a measurable signal such as absorbance, pressure or conductivity.
Extracting a mechanism from kinetic evidence
Initial rates isolate concentration effects
If experiment 1 and experiment 2 differ only in [A], their rate ratio isolates the exponent on A. For example, doubling [A] while the rate rises by a factor of eight implies third order in A because 2³ = 8.
Repeating this for each reactant builds the empirical rate law. The rate constant k is then obtained from any experiment after units consistent with the overall order are chosen.
Mechanisms must reproduce the observed rate law
A proposed elementary-step mechanism is not accepted merely because its steps add to the overall equation. It must also explain the measured kinetics and any detected intermediates. Pre-equilibrium or steady-state approximations may be needed to eliminate intermediate concentrations from the predicted rate law.
This is why kinetics is mechanistic evidence rather than just arithmetic. A wrong mechanism can give the right stoichiometry but the wrong dependence on concentration.
Measuring rate in the laboratory
Absorbance is useful when a reactant or product has a distinctive spectrum; gas pressure works when gas moles change; conductivity follows ionic composition; mass loss can follow an escaping gas. The best signal is one that changes specifically and rapidly with reaction progress.
Temperature control is critical because k is temperature-sensitive. Comparing rates from experiments run only a few degrees apart can confuse concentration effects with Arrhenius effects.
Worked example: finding a rate law
Experiment 1 has [A] = 0.10 M and rate 2.0×10⁻³ M s⁻¹. Experiment 2 doubles [A] to 0.20 M while every other condition is fixed, and rate becomes 8.0×10⁻³ M s⁻¹.
The rate rose by four when [A] doubled, so 2m = 4 and m = 2. The reaction is second order in A. This conclusion comes from comparison of controlled experiments, not from the coefficient of A in the balanced equation.
Exercises
Order
Doubling [A] triples nothing else and makes the rate four times larger. What is the order in A?
Solution
Second order, because 2² = 4.
Half-life
A first-order process has k = 0.20 s⁻¹. Find t½.
Solution
t½ = ln2/0.20 = 3.47 s.
Balanced equation
Can reaction orders normally be read from the coefficients of the overall equation?
Solution
No. They must normally be determined experimentally unless the reaction is known to be elementary.