Rate of a Chemical Reaction

Some reactions are instantaneous, like a firecracker exploding, and some barely move, like rusting iron. Chemical kinetics puts a number on that difference. Thermodynamics tells you whether a reaction can happen at all, and equilibrium tells you how far it goes — but neither says how fast. The rate of a reaction fills that gap: it is simply how quickly a reactant's concentration falls, or a product's concentration rises, per unit time. Once you can measure and express that rate correctly — for an interval, at an instant, and for a whole balanced equation at once — every later idea in kinetics, from rate laws to the Arrhenius equation, builds directly on this one definition.

The core insight

The rate of a reaction is the (always positive) change in concentration of a reactant or product per unit time, and dividing each species' rate of change by its own stoichiometric coefficient gives one unambiguous rate for the whole balanced equation.

The core idea

This idea opens the Chemical Kinetics chapter. Thermodynamics told you whether a reaction is feasible, and equilibrium told you how far it eventually goes, but neither one is concerned with the clock — kinetics is. A reaction is tracked in the laboratory by taking samples (or, better, a continuous physical measurement such as colour, pressure or conductivity) and plotting the concentration of a chosen species against time. Over a finite stretch of that time, this gives the average rate:

Average rate=Δ[reactant or product]Δt\text{Average rate} = \frac{\Delta[\text{reactant or product}]}{\Delta t}

where: Δ[ ]\Delta[\ ] is the change in molar concentration over the interval and Δt\Delta t is the length of that interval. Because a reactant's concentration falls while a product's rises, a negative sign is attached to the reactant term so that rate is always quoted as a positive number:

Rate=−Δ[R]Δt=Δ[P]Δt\text{Rate} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t}

where: RR denotes a reactant and PP a product. Shrinking the interval Δt\Delta t toward zero turns the average rate into the instantaneous rate, the rate at one specific moment:

Instantaneous rate=−d[R]dt=d[P]dt\text{Instantaneous rate} = -\frac{d[R]}{dt} = \frac{d[P]}{dt}

where: d[ ]/dtd[\ ]/dt is the derivative of concentration with respect to time — the slope of the concentration-time curve at a single point. Graphically, the average rate over an interval is the slope of the secant line joining the curve's two end points; as those two points are brought closer together, the secant approaches the tangent at one point, and the tangent's slope is the instantaneous rate. This is why instantaneous rate is written as a derivative rather than a simple ratio of two readings, and why a reactant's concentration-time curve is generally not a straight line — its slope, and hence the rate, keeps changing.

Figure
Concentration time curve for a reactant showing a secant line for average rate over an interval and a tangent line for instantaneous rate at one point
Average rate (secant) versus instantaneous rate (tangent) time t concentration of R secant = average rate tangent = instantaneous rate t1 t2

For a general reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD, each species changes concentration at a rate scaled by its own coefficient, so the raw rates of A, B, C and D are all different numbers. Dividing every term by its coefficient removes this mismatch and gives one unified rate for the reaction as a whole:

Rate=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

where: a,b,c,da, b, c, d are the stoichiometric coefficients of A,B,C,DA, B, C, D in the balanced equation. Rate carries units of concentration/time, most commonly mol L−1s−1\text{mol L}^{-1}\text{s}^{-1} (also mol L−1min−1\text{mol L}^{-1}\text{min}^{-1}, or atm s−1\text{atm s}^{-1} for gas-phase reactions measured through pressure).

Derivation AWhy the rates of different species must be divided by their coefficients

The goal is to show that the unified rate expression is not an arbitrary extra rule but a direct, necessary consequence of stoichiometry — that dividing each species' rate of concentration change by its own coefficient is the only way to get one single, self-consistent number for "the rate of the reaction," for a general reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD.

Step 1. Stoichiometry fixes a strict mole ratio for how fast species are consumed or formed relative to each other: for every aa moles of AA that react, exactly cc moles of CC are produced, because that is what the balanced equation states. This ratio holds at every instant of the reaction, not just at the end.

Step 2. Convert the mole ratio into a rate ratio. If aa moles of AA disappear in the same time that cc moles of CC appear, then CC must be forming c/ac/a times as fast, in moles per unit time, as AA is disappearing — so d[C]/dt−d[A]/dt=ca\dfrac{d[C]/dt}{-d[A]/dt} = \dfrac{c}{a} at every instant.

Step 3. Rearrange this ratio so the coefficients sit on the same side as their own species: −1ad[A]dt=1cd[C]dt-\dfrac{1}{a}\dfrac{d[A]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt}. The same argument applied to any pair among AA, BB, CC, DD gives the full unified expression, −1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt-\dfrac{1}{a}\dfrac{d[A]}{dt} = -\dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt} = \dfrac{1}{d}\dfrac{d[D]}{dt}.

Step 4. Read the result. Each term, once divided by its coefficient, equals the same single quantity — the rate of the reaction as written. This is why comparing raw, undivided rates of two species (as in the common trap below) is meaningless unless their coefficients happen to be equal.

Worked example

For the reaction 2N2O5(g)→4NO2(g)+O2(g)2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g), the rate of disappearance of N2O5N_2O_5 is 2.5×10−3 mol L−1s−12.5\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}. Find the rate of the reaction and the rate of formation of NO2NO_2 and O2O_2.

Step 1. Write the unified rate expression using the coefficients 2, 4 and 1:

Rate=−12d[N2O5]dt=14d[NO2]dt=d[O2]dt\text{Rate} = -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}

Step 2. Compute the rate of the reaction from the given rate of disappearance of N2O5N_2O_5:

Rate=12(2.5×10−3)=1.25×10−3 mol L−1s−1\text{Rate} = \frac{1}{2}(2.5\times10^{-3}) = 1.25\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}

Step 3. Find the rate of formation of NO2NO_2 by equating it to the reaction rate through its own coefficient:

14d[NO2]dt=1.25×10−3  ⟹  d[NO2]dt=5.0×10−3 mol L−1s−1\frac{1}{4}\frac{d[NO_2]}{dt} = 1.25\times10^{-3} \implies \frac{d[NO_2]}{dt} = 5.0\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}

Step 4. Find the rate of formation of O2O_2, whose coefficient is 1, so its rate of formation equals the reaction rate directly:

d[O2]dt=1.25×10−3 mol L−1s−1\frac{d[O_2]}{dt} = 1.25\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}

The reaction proceeds at 1.25×10−3 mol L−1s−11.25\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}, NO2NO_2 forms at 5.0×10−3 mol L−1s−15.0\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}, and O2O_2 forms at 1.25×10−3 mol L−1s−11.25\times10^{-3}\ \text{mol L}^{-1}\text{s}^{-1} — NO2NO_2 forms exactly four times as fast as O2O_2, matching the 4:1 ratio of their coefficients.

Competitive stretch

In BrO3−(aq)+5Br−(aq)+6H+(aq)→3Br2(l)+3H2O(l)BrO_3^-(aq) + 5Br^-(aq) + 6H^+(aq) \rightarrow 3Br_2(l) + 3H_2O(l), [Br−][Br^-] falls from 0.1800.180 to 0.146 mol L−10.146\ \text{mol L}^{-1} in 40 s. Step 1. Δ[Br−]=0.146−0.180=−0.034 mol L−1\Delta[Br^-] = 0.146 - 0.180 = -0.034\ \text{mol L}^{-1} over Δt=40 s\Delta t = 40\ \text{s}, so the average rate of disappearance of Br−Br^- is 0.034/40=8.5×10−4 mol L−1s−10.034/40 = 8.5\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}. Step 2. Dividing by the coefficient 5 gives the reaction rate, 15(8.5×10−4)=1.7×10−4 mol L−1s−1\frac{1}{5}(8.5\times10^{-4}) = 1.7\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}, which also equals the rate of consumption of BrO3−BrO_3^- since its coefficient is 1. This confirms the same coefficient-division logic used above, now applied in reverse — from a raw concentration change to the single reaction rate.

Common traps

  • Reporting a species' own rate of change as "the rate of the reaction" without dividing by its stoichiometric coefficient — the two are equal only when the coefficient is 1.
  • Dropping or misplacing the negative sign for reactants — rate must always come out positive, whether computed from a reactant or a product.
  • Confusing average rate (a secant slope over an interval) with instantaneous rate (a tangent slope at a point) — they are numerically different unless the rate happens to be constant.
  • Using the wrong units, especially in gas-phase problems given in pressure (atm s−1\text{atm s}^{-1}) instead of molarity per time.
  • Treating rate as constant across a whole experiment — because concentration keeps falling, rate itself generally keeps changing, which is exactly why the instantaneous rate at a chosen instant can differ noticeably from the average rate over the surrounding interval.

Ask the tutor: if you're unsure whether a problem wants an average or instantaneous rate, or how to handle an awkward stoichiometric ratio, ask the AI tutor to walk through your specific reaction step by step.

Why do we bother dividing by the stoichiometric coefficient instead of just quoting each species' own rate

Because different species in the same reaction change concentration at different speeds whenever their coefficients differ, so quoting a raw rate is ambiguous — a student could report five different numbers for the same reaction and none would be wrong on their own. Dividing by the coefficient scales every species down to one shared value, the rate of the reaction itself, which is what rate laws and rate constants are actually built on later in the chapter.

Is average rate ever equal to instantaneous rate

Yes, but only in special cases. If the concentration-time graph happens to be a straight line over the interval you are averaging across, the secant and the tangent coincide everywhere on it, so the average rate equals the instantaneous rate at every point in that interval. This actually happens for a zero-order reaction, which you will meet later in this chapter; for most reactions the curve bends and the two rates differ.

Why is rate always written as a positive number even for reactants that are disappearing

By convention, "rate" describes how fast the reaction is progressing, not the sign of any one concentration's change. Since a reactant's concentration change Δ[R]\Delta[R] is itself negative, the minus sign in −Δ[R]/Δt-\Delta[R]/\Delta t cancels it, leaving a positive rate — consistent with the positive rate you get directly from a product's increasing concentration. Both routes must give the same positive number for the same reaction.

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Where this leads