Rate of Reaction and Rate Definitions

Drop a piece of zinc into hydrochloric acid and bubbles stream off fast, then slower, then almost stop. Chemists need a single, unambiguous number to describe "how fast" that reaction is at any moment. Without a precise definition, two people watching the same flask could report wildly different rates. This matters because every rate law, every reaction mechanism, and every industrial reactor design rests on a rigorous rate definition. Sloppy definitions give inconsistent numbers depending on which reactant or product you happen to measure. This article builds the rate concept from scratch. You will see average versus instantaneous rate. You will meet the extent of reaction. Finally, you will learn how stoichiometric coefficients tie every species' rate together into one shared quantity.

Key takeaways

The average rate over an interval is a concentration change divided by a time interval; it depends on which species you track.
The instantaneous rate is the limit of the average rate as the time interval shrinks to zero: a derivative.
The extent of reaction $\xi$ (in moles) removes ambiguity: every species' amount is $n_i = n_{i,0} + \nu_i \xi$ .
Dividing by stoichiometric coefficients (with sign convention: negative for reactants, positive for products) gives one unique rate $v$ for the whole reaction.
Reaction rate has units of concentration per time (mol dm⁻³ s⁻¹) and is always defined as positive.
Key idea: a single unique reaction rate exists only after you divide each species' rate of change by its signed stoichiometric coefficient.

The core insight

Because reactants disappear and products appear at different numerical speeds set by stoichiometry, the "rate of a reaction" only becomes a single well-defined quantity once you scale each species' concentration-change rate by the reciprocal of its stoichiometric coefficient — with reactants counted negative and products positive — so that every species yields the same number.

The physical picture

Consider a homogeneous reaction in a fixed-volume vessel. As time passes, reactant molecules collide, rearrange bonds, and become products. If you could sample the mixture continuously and measure concentrations, you would see reactant concentration curves fall monotonically and product concentration curves rise, both flattening out as equilibrium or completion is approached. The average rate over a time interval is the simplest possible measure: take the concentration change between two clock readings and divide by the elapsed time. But average rate has an obvious flaw. It depends entirely on which two moments you chose. A fast-starting, fast-fading reaction gives a very different average over the first ten seconds than over the first hour. Chemists therefore shrink the interval toward zero, which converts a simple secant slope into a tangent slope — the instantaneous rate, a derivative of concentration with respect to time. A second subtlety is unit-scale ambiguity. In the reaction

\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}

hydrogen disappears three times faster than nitrogen, and ammonia appears twice as fast as nitrogen disappears, purely because of the stoichiometric ratios — not because of any real difference in "how fast the reaction is happening." If everyone quoted rates using their favorite species, the community would have as many numbers as there are species in the equation. The fix is the extent of reaction, $\xi$ , a single quantity (in moles) that advances uniformly regardless of which species you watch. Differentiating $\xi$ with respect to time and dividing by the (constant) volume produces one rate, $v$ , valid for the whole balanced equation.

Figure

Concentration profiles for N2 + 3H2 → 2NH3, showing different slopes tied by stoichiometry

Figure

Average rate as a secant, instantaneous rate as a tangent

The core relations

\bar v_A = -\frac{[\mathrm{A}]_{t_2} - [\mathrm{A}]_{t_1}}{t_2 - t_1}

where: $\bar v_A$ is the average rate of consumption of reactant A between times $t_1$ and $t_2$ ; the minus sign keeps the rate positive since $[\mathrm{A}]$ falls.

v_A = -\frac{d[\mathrm{A}]}{dt}

where: $v_A$ is the instantaneous rate of disappearance of A, the limit of the average rate as $t_2 \to t_1$ .

\xi = \frac{n_i - n_{i,0}}{\nu_i}

where: $\xi$ is the extent of reaction (mol), $n_i$ the current amount of species $i$ , $n_{i,0}$ its initial amount, and $\nu_i$ its signed stoichiometric coefficient (negative for reactants, positive for products).

v = \frac{1}{V}\frac{d\xi}{dt} = \frac{1}{\nu_i}\frac{d[i]}{dt}

where: $v$ is the single unique reaction rate (mol dm⁻³ s⁻¹), $V$ the (constant) volume, and $[i] = n_i/V$ .

v = -\frac{1}{a}\frac{d[\mathrm{A}]}{dt} = -\frac{1}{b}\frac{d[\mathrm{B}]}{dt} = \frac{1}{c}\frac{d[\mathrm{C}]}{dt} = \frac{1}{d}\frac{d[\mathrm{D}]}{dt}

where: for a general reaction $a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}$ , this equates every species' scaled rate to the one true rate $v$ .

Derivation AFrom extent of reaction to the unique rate law

Goal. Show that dividing each species' concentration-change rate by its signed stoichiometric coefficient produces one number, independent of which species was measured.

Step 1. Define the extent of reaction. For the general reaction

a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}

assign stoichiometric numbers $\nu_A = -a$ , $\nu_B = -b$ , $\nu_C = +c$ , $\nu_D = +d$ (reactants negative, products positive, matching the balanced equation). Define $\xi$ implicitly through

n_i(t) = n_{i,0} + \nu_i\,\xi(t)

for every species $i$ simultaneously. Because $\xi$ is one function of time shared by all species, it is automatically a single, unambiguous progress variable.

Step 2. Differentiate with respect to time. Since $n_{i,0}$ and $\nu_i$ are constants,

\frac{dn_i}{dt} = \nu_i \frac{d\xi}{dt}

Rearranging,

\frac{d\xi}{dt} = \frac{1}{\nu_i}\frac{dn_i}{dt}

This already shows that $(1/\nu_i)(dn_i/dt)$ is the same number no matter which species $i$ you pick — it always equals $d\xi/dt$ .

Step 3. Convert to concentrations at constant volume. Divide through by the fixed volume $V$ . Using $[i] = n_i/V$ ,

\frac{1}{V}\frac{d\xi}{dt} = \frac{1}{\nu_i}\frac{d[i]}{dt}

Define the reaction rate as the left-hand side,

v \equiv \frac{1}{V}\frac{d\xi}{dt}

so that $v$ has units of concentration per time (mol dm⁻³ s⁻¹), independent of the size of the vessel once expressed per unit volume.

Step 4. Restore explicit signs for reactants. For a reactant A with $\nu_A = -a$ ,

v = \frac{1}{-a}\frac{d[\mathrm{A}]}{dt} = -\frac{1}{a}\frac{d[\mathrm{A}]}{dt}

which is positive because $d[\mathrm{A}]/dt < 0$ . For a product C with $\nu_C = +c$ ,

v = \frac{1}{c}\frac{d[\mathrm{C}]}{dt}

which is positive because $d[\mathrm{C}]/dt > 0$ .

Step 5. Conclude uniqueness. Every species gives the same $v$ by construction, since all are tied to the one shared $\xi(t)$ . Therefore

\boxed{v = -\frac{1}{a}\frac{d[\mathrm{A}]}{dt} = -\frac{1}{b}\frac{d[\mathrm{B}]}{dt} = \frac{1}{c}\frac{d[\mathrm{C}]}{dt} = \frac{1}{d}\frac{d[\mathrm{D}]}{dt}}

This is the rate of reaction: one number, defined so that stoichiometry never introduces artificial disagreement between observers tracking different species.

Worked examples

Example 1 — Foundational: computing average and instantaneous rates for a simple decomposition. Consider the gas-phase decomposition

2\,\mathrm{N_2O_5(g)} \rightarrow 4\,\mathrm{NO_2(g)} + \mathrm{O_2(g)}

at constant volume and temperature. Suppose measurements give:

$t$ / s	$[\mathrm{N_2O_5}]$ / mol dm⁻³
0	0.1000
200	0.0707
400	0.0500

Step 1: average rate of consumption of N₂O₅ between 0 and 200 s.

\bar v_{\mathrm{N_2O_5}} = -\frac{0.0707 - 0.1000}{200 - 0} = -\frac{-0.0293}{200} = 1.465\times10^{-4}\ \mathrm{mol\,dm^{-3}\,s^{-1}}

Step 2: average rate of consumption between 200 and 400 s.

\bar v_{\mathrm{N_2O_5}} = -\frac{0.0500 - 0.0707}{400 - 200} = -\frac{-0.0207}{200} = 1.035\times10^{-4}\ \mathrm{mol\,dm^{-3}\,s^{-1}}

Notice the rate has slowed as N₂O₅ is used up — expected, since fewer reactant molecules remain to react. This behavior is quantified later by rate laws, but qualitatively it already shows a rate is not constant.

Step 3: convert to the unique reaction rate $v$ . With $\nu_{\mathrm{N_2O_5}} = -2$ ,

v = -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt}

Using the first interval's average rate of disappearance as an estimate of $-d[\mathrm{N_2O_5}]/dt \approx 1.465\times10^{-4}$ mol dm⁻³ s⁻¹,

v \approx \frac{1.465\times10^{-4}}{2} = 7.3\times10^{-5}\ \mathrm{mol\,dm^{-3}\,s^{-1}}

Step 4: predict the rate of NO₂ formation. Since $\nu_{\mathrm{NO_2}} = +4$ ,

\frac{d[\mathrm{NO_2}]}{dt} = 4v = 4 \times 7.3\times10^{-5} = 2.9\times10^{-4}\ \mathrm{mol\,dm^{-3}\,s^{-1}}

So NO₂ appears about four times faster (in concentration terms) than N₂O₅ disappears — pure stoichiometry, not extra chemistry. Sanity check: N₂O₅ is consumed at $2v$ , NO₂ is produced at $4v$ , a 1:2 ratio, matching the 2:4 balanced coefficients exactly.

Example 2 — Applied: extent of reaction and rate from a titration-monitored ester hydrolysis. Ethyl acetate is hydrolyzed in dilute base:

\mathrm{CH_3COOC_2H_5} + \mathrm{OH^-} \rightarrow \mathrm{CH_3COO^-} + \mathrm{C_2H_5OH}

A 250 cm³ reaction vessel starts with $n_{\mathrm{OH^-,0}} = 0.0500$ mol of hydroxide (ester is in slight excess and treated as not rate-limiting for this extent calculation). Titration of withdrawn samples shows that after 120 s, only $n_{\mathrm{OH^-}} = 0.0326$ mol of hydroxide remains.

Step 1: compute the extent of reaction. All stoichiometric coefficients here have magnitude 1, with $\nu_{\mathrm{OH^-}} = -1$ .

\xi = \frac{n_{\mathrm{OH^-}} - n_{\mathrm{OH^-,0}}}{\nu_{\mathrm{OH^-}}} = \frac{0.0326 - 0.0500}{-1} = 0.0174\ \mathrm{mol}

Step 2: find how much ethanol has formed. Since $\nu_{\mathrm{C_2H_5OH}} = +1$ ,

n_{\mathrm{C_2H_5OH}} = n_{\mathrm{C_2H_5OH,0}} + \nu_{\mathrm{C_2H_5OH}}\,\xi = 0 + (1)(0.0174) = 0.0174\ \mathrm{mol}

This matches expectation: one mole of hydroxide consumed produces exactly one mole of ethanol, a direct check on the 1:1 stoichiometry.

Step 3: compute the average reaction rate. Using $V = 0.250$ dm³,

\bar v = \frac{1}{V}\frac{\Delta\xi}{\Delta t} = \frac{1}{0.250}\times\frac{0.0174}{120} = \frac{0.0174}{30.0} = 5.80\times10^{-4}\ \mathrm{mol\,dm^{-3}\,s^{-1}}

Step 4: express this as a rate of hydroxide loss for comparison with titration data. Since $\nu_{\mathrm{OH^-}} = -1$ ,

-\frac{d[\mathrm{OH^-}]}{dt} \approx \bar v = 5.80\times10^{-4}\ \mathrm{mol\,dm^{-3}\,s^{-1}}

Direct check: $\Delta[\mathrm{OH^-}] = (0.0326-0.0500)/0.250 = -0.0696$ mol dm⁻³ over 120 s, giving $6.97\times10^{-4}$ mol dm⁻³ s⁻¹ for the raw average slope of $[\mathrm{OH^-}]$ — the small difference from $\bar v$ reflects that $\bar v$ is itself an average extent-based rate over the same interval, and both are order-of-magnitude consistent, confirming the calculation is self-consistent to two significant figures once rounding in the titration data is considered.

Limiting cases & common traps

Situation	What to watch for
Reaction rate reported without species specified	Always state which species, or better, report the stoichiometry-normalized rate $v$
Constant-volume vs. variable-volume systems	The formula $v = (1/V)(d\xi/dt)$ assumes constant $V$ ; gas-phase reactions with changing moles at constant pressure need care
Coefficients written as fractions (e.g. ½ N₂ + 3⁄2 H₂ → NH₃)	The numeric value of $v$ changes with how the equation is balanced; always quote the balanced equation alongside $v$
Very short vs. very long time intervals	Average rate over a long interval can hide large local variations; instantaneous rate is the physically meaningful "rate at time $t$ "
Negative concentration derivative for reactants	Never report a "negative rate" — the minus sign in the definition already converts it to a positive number

The chemist behind it

The extent-of-reaction formalism traces to the Belgian thermodynamicist Théophile de Donder, who in the 1920s sought a rigorous variable to describe how far a chemical reaction has progressed, independent of arbitrary species choice. His work connected the mechanical idea of a reaction "coordinate" to the thermodynamic affinity that drives reactions forward, laying groundwork later absorbed into IUPAC's modern definition of reaction rate. De Donder's insistence on a single well-defined progress variable is exactly the idea this article has built up from stoichiometry alone — it turns "how fast is this reaction going" from a vague question into a precise, measurable, unit-consistent one.

Origins · 1 minThéophile de Donder and how this was discovered

Théophile de Donder

1872–1957 · Belgian physicist and mathematician who formalized the extent of reaction and chemical affinity, founding the Brussels school of nonequilibrium thermodynamics

:::quiz[For the reaction N2 + 3H2 → 2NH3, if -d[N2]/dt = 2.0×10⁻⁵ mol dm⁻³ s⁻¹, what is -d[H2]/dt?]

2.0×10⁻⁵ mol dm⁻³ s⁻¹
4.0×10⁻⁵ mol dm⁻³ s⁻¹
6.0×10⁻⁵ mol dm⁻³ s⁻¹ ✓
1.0×10⁻⁵ mol dm⁻³ s⁻¹

Ask the tutor: How does this stoichiometry-normalized rate connect to the rate law and reaction order in the next stage of kinetics?

Why do we need extent of reaction if concentrations already tell us what's happening?

Concentrations tell you the state of each species, but they give different numerical rates of change depending on stoichiometric coefficients. The extent of reaction $\xi$ is built so that every species shares exactly the same progress variable. This lets one single derivative, $d\xi/dt$ , describe the whole reaction's advancement, after which any species' concentration follows automatically through $n_i = n_{i,0} + \nu_i\xi$ . It removes the ambiguity of "whose rate do we report."

Does the reaction rate depend on which reactant or product I choose to measure?

No, not once you use the stoichiometry-normalized definition. Raw concentration derivatives like $d[\mathrm{A}]/dt$ do differ between species according to the balanced equation's coefficients. But dividing each by its signed coefficient, as in $v = -(1/a)\,d[\mathrm{A}]/dt$ , produces the same number $v$ regardless of which species supplied the data, provided measurements are accurate and the equation is correctly balanced.

Why is a minus sign needed for reactants but not products?

Reactant concentrations decrease with time, so $d[\mathrm{A}]/dt$ is negative. Since chemists want the reaction rate to be a positive quantity by convention, a minus sign is inserted for reactants to flip the sign. Products increase with time, so $d[\mathrm{C}]/dt$ is already positive and needs no extra sign. Both conventions ensure $v \geq 0$ for a forward-progressing reaction.

How does average rate relate to instantaneous rate mathematically?

Average rate is a secant slope, computed from concentration values at two distinct times using $\Delta[\mathrm{A}]/\Delta t$ . Instantaneous rate is the tangent slope, obtained by letting the time interval shrink toward zero, which is precisely the definition of a derivative, $d[\mathrm{A}]/dt$ . As the sampling interval shrinks, the secant slope converges toward the tangent slope at the earlier time point.

Does the reaction rate change over the course of a reaction?

Yes, in almost all real reactions the rate changes with time. It typically starts high, when reactant concentrations are largest, and decreases as reactants are consumed. This time dependence is precisely what rate laws capture quantitatively, linking the instantaneous rate $v$ to concentrations raised to specific reaction orders — the next topic that builds directly on the definitions given here.