Law of mass action

The statement. The Law of Mass Action says that at a given temperature, the rate of a chemical reaction is directly proportional to the product of the active masses of the reacting substances, each active mass raised to a power equal to the number of molecules of that substance taking part in the reaction. "Active mass" is not everyday mass in grams — it is molar concentration, written in square brackets and measured in $\text{mol L}^{-1}$.

Why concentration, not mass in grams. Doubling the grams of a substance without changing the volume of the vessel doubles its concentration too, so the two often move together — but it is concentration (moles per litre) that drives collision frequency, not the raw quantity of material. For a gas, active mass may equally be expressed as partial pressure, since pressure and concentration are proportional at fixed temperature.

Solids and pure liquids drop out. A pure solid or pure liquid reacts only at its exposed surface, and the "concentration" of the substance within that solid or liquid phase does not change as reaction proceeds — it stays essentially constant no matter how much of the solid is present. Because it never varies, its active mass is folded into the rate constant rather than written as a separate variable term.

From statement to equation. For a general reaction

the law reads

where: $[A]$ and $[B]$ are the active masses (molar concentrations) of reactants $A$ and $B$, $a$ and $b$ are the numbers of molecules of each participating in the reaction, and $k$ is the rate constant (or velocity constant), a fixed value for a given reaction at a given temperature that converts the proportionality into an equality. This single equation is the ancestor of the rate law $\text{Rate} = k[\text{reactant}]^n$ used throughout kinetics — with the caveat, sharpened later under order of reaction, that for a multi-step reaction the exponents must be measured experimentally and need not equal the stoichiometric coefficients of the overall balanced equation.

The collision picture. Why should packing in more active mass speed things up? More moles per litre means more reactant particles occupying the same volume, which raises the chance that particles meet and collide within any given time interval. More collisions per second means more opportunities for an effective, rate-producing encounter — so rate rises with concentration, and rises faster when a species appears with a higher power in the expression, since that species' particles must feature in more than one "slot" of a productive collision.

Question. For the elementary gas-phase reaction $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$, write the Law of Mass Action rate expression, then find the factor by which the rate changes if $[\text{NO}]$ is tripled while $[\text{O}_2]$ is halved.

Step 1. Identify the active masses and their powers. The reactants are NO and $\text{O}_2$, with 2 molecules of NO and 1 molecule of $\text{O}_2$ taking part, so the Law of Mass Action gives $\text{Rate} \propto [\text{NO}]^2[\text{O}_2]^1$, i.e. $\text{Rate} = k[\text{NO}]^2[\text{O}_2]$.

Step 2. Write the original rate for reference: $R_1 = k[\text{NO}]^2[\text{O}_2]$.

Step 3. Substitute the new concentrations. $[\text{NO}]$ becomes $3[\text{NO}]$ and $[\text{O}_2]$ becomes $\tfrac{1}{2}[\text{O}_2]$, so $R_2 = k(3[\text{NO}])^2\left(\tfrac{1}{2}[\text{O}_2]\right) = k \times 9[\text{NO}]^2 \times \tfrac{1}{2}[\text{O}_2] = \tfrac{9}{2}\,k[\text{NO}]^2[\text{O}_2]$.

Step 4. Compare to the original: $R_2 = \tfrac{9}{2}R_1 = 4.5\,R_1$. The rate becomes 4.5 times the original rate — tripling NO (squared power) contributes a factor of $9$, and halving $\text{O}_2$ (power 1) contributes a factor of $\tfrac{1}{2}$, and the two combine multiplicatively because the expression is a product, not a sum.

Competitive stretch

If instead $[\text{NO}]$ were doubled and $[\text{O}_2]$ were also doubled, the new rate would be $k(2[\text{NO}])^2(2[\text{O}_2]) = 8\,k[\text{NO}]^2[\text{O}_2] = 8R_1$ — the rate rises by the product of each factor raised to its own power ($2^2 \times 2^1 = 8$), a quick check worth doing before assuming "doubling everything just doubles the rate."

• Treating active mass as mass in grams or as number of moles alone, instead of molar concentration ($\text{mol L}^{-1}$) — active mass always means $\dfrac{\text{moles}}{\text{volume}}$.
• Writing a solid or pure liquid reactant as an explicit $[\text{solid}]^n$ term in the rate expression — its constant active mass is absorbed into $k$, not shown separately.
• Assuming the powers in the expression must equal the stoichiometric coefficients of the overall balanced equation for any given reaction — that is guaranteed only for an elementary step; for multi-step reactions the true powers are experimental orders.
• Forgetting that the expression is a product of active-mass terms, not a sum — changing two concentrations at once multiplies their individual rate factors together rather than adding them.

Ask the tutor: if the jump from "active mass raised to a power" to the full rate law, or the solids-drop-out rule, still feels shaky, ask your AI tutor to walk through two or three more elementary reactions and build the expression with you.