You know how to measure a rate. Experiments show that the rate depends on concentrations, and in a simple way: as a power of each concentration. That relationship is the rate law.
🔑 Key idea
A rate law has the form $\text{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n$. The exponents $m$ and $n$ are the orders with respect to A and B, their sum is the overall order, and $k$ is the rate constant. Orders are found by experiment, not read from the balanced equation.
💡 Why it works
Why can't you read the orders from the coefficients? Most reactions do not happen in one collision. They run through a series of simpler steps, and the rate law reflects the slowest step and what comes before it, not the overall recipe. For example, $\mathrm{2N_2O_5 \to 4NO_2 + O_2}$ has the measured rate law $\text{rate} = k[\mathrm{N_2O_5}]$: first order, not second. And for $\mathrm{NO_2 + CO \to NO + CO_2}$ below about $500\ \text{K}$, $\text{rate} = k[\mathrm{NO_2}]^2$: carbon monoxide does not appear at all.
What each order means: with order 0, changing the concentration does nothing to the rate; with order 1, doubling it doubles the rate; with order 2, doubling it multiplies the rate by $2^2 = 4$. Orders are usually 0, 1 or 2, though fractions such as $\tfrac{1}{2}$ occur.
Rate against concentration with $k = 1$. Trace each curve: from $[\mathrm{A}] = 1$ to $2$, the zero-order rate stays put, the first-order rate doubles, and the second-order rate quadruples.
The units of $k$ make the rate come out in $\text{M/s}$, so they depend on the overall order:
In general the units of k are M^(1−n)·s⁻¹ for overall order n.
Overall order
Rate law example
Units of k
0
$k$
$\text{M/s}$
1
$k[\mathrm{A}]$
$\text{s}^{-1}$
2
$k[\mathrm{A}]^2$ or $k[\mathrm{A}][\mathrm{B}]$
$\text{M}^{-1}\,\text{s}^{-1}$
3
$k[\mathrm{A}]^2[\mathrm{B}]$
$\text{M}^{-2}\,\text{s}^{-1}$
Worked examples
✏️ Worked example: Reading a rate law
For $\mathrm{2NO + 2H_2 \to N_2 + 2H_2O}$ the measured rate law is $\text{rate} = k[\mathrm{NO}]^2[\mathrm{H_2}]$. Give the orders and the units of $k$. What happens to the rate if $[\mathrm{NO}]$ triples?
Second order in NO, first order in $\mathrm{H_2}$, third order overall. Notice that $\mathrm{H_2}$ has coefficient 2 but order 1.
Tripling $[\mathrm{NO}]$ multiplies the rate by $3^2 = 9$.
Answer: Orders 2 and 1 (overall 3), $k$ in $\text{M}^{-2}\,\text{s}^{-1}$, and the rate becomes 9 times larger.
✏️ Worked example: Using a rate law
$\text{rate} = k[\mathrm{A}][\mathrm{B}]^2$ with $k = 0.50\ \text{M}^{-2}\,\text{s}^{-1}$. Find the rate when $[\mathrm{A}] = 0.20\ \text{M}$ and $[\mathrm{B}] = 0.10\ \text{M}$.
$[\mathrm{B}]^2 = (0.10)^2 = 0.010\ \text{M}^2$.
$$\text{rate} = (0.50)(0.20)(0.010) = 1.0 \times 10^{-3}\ \text{M/s}$$ The units check: $\text{M}^{-2}\,\text{s}^{-1} \times \text{M}^3 = \text{M/s}$.
Answer: $1.0 \times 10^{-3}\ \text{M/s}$.
⚠️ Common mistake
Writing the rate law from the balanced equation, as in $\text{rate} = k[\mathrm{N_2O_5}]^2$ for $\mathrm{2N_2O_5 \to 4NO_2 + O_2}$. The coefficients describe the overall recipe, not the collisions that set the speed. Check: if a problem gives you only the balanced equation and no data or mechanism, you cannot know the orders.
The rate constant depends on temperature (and on a catalyst), but not on concentration. That is why one value of $k$ works for every experiment run at the same temperature.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.