Rate of reaction

Say a reaction is "fast" and a chemist will immediately ask: fast compared to what, and measured how? A rusting nail and a burning matchstick are both oxidations, but one takes years and the other seconds — "fast" alone tells you nothing you can put in a table or compare between two experiments. The rate of reaction turns that vague word into a precise, measurable number: how much a reactant's concentration falls, or a product's concentration rises, per unit time. This article builds the rate expression for a general balanced equation, shows why every species in that equation — even though their concentrations change at different numerical speeds — must be divided by its own stoichiometric coefficient to give one consistent rate, and then draws the crucial line between average rate (measured over a finite time interval, the slope of a chord) and instantaneous rate (measured at one instant, the slope of a tangent). Two worked examples, reusing real concentration-versus-time data, show exactly how the numbers are pulled out of a table and a graph.

The core insight

The rate of reaction is the coefficient-normalised rate of change of concentration of any reactant (with a negative sign) or product (positive) with time, measured over a finite interval as the average rate (slope of a chord) or at one instant as the instantaneous rate (slope of a tangent), and both are always reported as positive numbers in mol L−1s−1\text{mol L}^{-1}\text{s}^{-1}.

The core idea

The rate expression. For a general reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD, reactant concentrations fall while product concentrations rise, so a single, unambiguous rate is defined by normalising each species' rate of concentration change by its own stoichiometric coefficient:

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][A], [B][B], [C][C], [D][D] are molar concentrations of the four species, tt is time, and aa, bb, cc, dd are the balancing coefficients in front of each species. The negative sign on the reactant terms is not optional bookkeeping — since [A][A] and [B][B] are falling, d[A]/dtd[A]/dt and d[B]/dtd[B]/dt are themselves negative numbers, and the minus sign flips the whole term positive, so "rate" is always quoted as a positive quantity regardless of whether you happen to be tracking a reactant or a product.

Average rate versus instantaneous rate. Over a finite interval Δt\Delta t, the average rate is −Δ[reactant]/Δt-\Delta[\text{reactant}]/\Delta t or +Δ[product]/Δt+\Delta[\text{product}]/\Delta t, computed from two measured concentrations at two clock times. Graphically, on a plot of concentration against time, this is the slope of the chord joining those two points. Shrinking that interval to a single instant, Δt→0\Delta t \rightarrow 0, gives the instantaneous rate:

Instantaneous rate=−d[reactant]dt=d[product]dt\text{Instantaneous rate} = -\frac{d[\text{reactant}]}{dt} = \frac{d[\text{product}]}{dt}

which is the slope of the tangent drawn to the concentration–time curve at that one point. Because concentration–time curves bend rather than run straight — the reaction slows as reactant is consumed — the tangent at t=0t = 0 (the initial rate) is steeper than any chord drawn over a later, longer interval; average and instantaneous rate genuinely disagree except for the special, rare case of a perfectly straight-line (zero order) curve.

Units. Rate is always concentration divided by time: mol L−1s−1\text{mol L}^{-1}\text{s}^{-1} for fast reactions, or mol L−1min−1\text{mol L}^{-1}\text{min}^{-1} for slower ones — the unit is chosen to match the timescale of the reaction being studied, never fixed in advance.

Figure
A concentration versus time curve bending downward with a straight chord drawn between two points marked at ten and twenty minutes and a straight tangent line touching the curve at the ten minute point, showing the chord slope is the average rate and the tangent slope is the instantaneous rate
Average rate (chord) versus instantaneous rate (tangent) time t (min) concentration [H2O2] chord: average rate, t=10 to t=20 tangent at t=10: instantaneous rate t=10 min t=20 min t=30 min curve flattens: the reaction slows as reactant is used up
Derivation AWhy every species must be divided by its own coefficient

The goal is to show, using nothing beyond the definition of stoichiometry, why the raw rates of change d[A]/dtd[A]/dt, d[B]/dtd[B]/dt, d[C]/dtd[C]/dt and d[D]/dtd[D]/dt in a reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD are numerically different from one another, and why dividing each one by its own coefficient is exactly the operation that removes that difference and yields a single shared number called the rate of reaction.

Step 1. Fix what stoichiometry actually says. The coefficients a,b,c,da, b, c, d record the fixed ratio in which the reaction consumes and produces each species: for every aa moles of A consumed, exactly bb moles of B are consumed, cc moles of C are produced, and dd moles of D are produced, all in the same span of time, because the balanced equation is a statement about relative amounts, not absolute ones.

Step 2. Turn that fixed ratio into a statement about rates of change. If aa moles of A disappear in the same time that cc moles of C appear, then the rate at which A disappears and the rate at which C appears are locked in the same ratio a:ca:c. Formally, −d[A]dt:d[C]dt=a:c-\dfrac{d[A]}{dt} : \dfrac{d[C]}{dt} = a : c, which rearranges to −1ad[A]dt=1cd[C]dt-\dfrac{1}{a}\dfrac{d[A]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt}.

Step 3. Recognise why the raw derivatives alone cannot serve as "the" rate. If c>ac > a (more product formed per reaction event than reactant with coefficient aa consumed), then d[C]/dtd[C]/dt is numerically larger than −d[A]/dt-d[A]/dt, even though both describe the very same ongoing reaction happening at the very same underlying speed. Reporting rate as "d[C]/dtd[C]/dt" alone would make the reaction look faster just because you chose to watch C instead of A — an artefact of coefficients, not a real difference in how fast the reaction runs.

Step 4. Confirm the fix removes the artefact for all four species at once. Dividing every term by its own coefficient, −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}, gives one shared value no matter which species supplied the data — this shared value is what "rate of reaction" means, and it is exactly Equation (1) above.

Worked example

For the reaction 2N2O5(g)→4NO2(g)+O2(g)2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g) + \text{O}_2(g), the concentration of N2O5\text{N}_2\text{O}_5 falls from 1.60×10−2 mol L−11.60\times10^{-2}\ \text{mol L}^{-1} to 1.20×10−2 mol L−11.20\times10^{-2}\ \text{mol L}^{-1} over 300 s. Find the average rate of disappearance of N2O5\text{N}_2\text{O}_5, the average rate of reaction, and the average rate of appearance of NO2\text{NO}_2.

Step 1. Find the change in concentration of N2O5\text{N}_2\text{O}_5: Δ[N2O5]=1.20×10−2−1.60×10−2=−0.40×10−2 mol L−1\Delta[\text{N}_2\text{O}_5] = 1.20\times10^{-2} - 1.60\times10^{-2} = -0.40\times10^{-2}\ \text{mol L}^{-1}, negative because it is being consumed.

Step 2. Compute the rate of disappearance of N2O5\text{N}_2\text{O}_5 directly: −Δ[N2O5]Δt=0.40×10−2300=1.33×10−5 mol L−1s−1-\dfrac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = \dfrac{0.40\times10^{-2}}{300} = 1.33\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}.

Step 3. Apply Equation (1) to get the rate of reaction, dividing by the coefficient of N2O5\text{N}_2\text{O}_5, which is 2: Rate=12×1.33×10−5=6.67×10−6 mol L−1s−1\text{Rate} = \dfrac{1}{2}\times1.33\times10^{-5} = 6.67\times10^{-6}\ \text{mol L}^{-1}\text{s}^{-1}.

Step 4. Recover the rate of appearance of NO2\text{NO}_2 by multiplying the rate of reaction by the NO2\text{NO}_2 coefficient, which is 4: 4×6.67×10−6=2.67×10−5 mol L−1s−14\times6.67\times10^{-6} = 2.67\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}. Rate of disappearance of N2O5=1.33×10−5 mol L−1s−1\text{N}_2\text{O}_5 = 1.33\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}, rate of reaction =6.67×10−6 mol L−1s−1= 6.67\times10^{-6}\ \text{mol L}^{-1}\text{s}^{-1}, rate of appearance of NO2=2.67×10−5 mol L−1s−1\text{NO}_2 = 2.67\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}.

Competitive stretch

The decomposition of H2O2\text{H}_2\text{O}_2 gives [H2O2]=0.50 mol L−1[\text{H}_2\text{O}_2] = 0.50\ \text{mol L}^{-1} at t=0t=0, 0.40 mol L−10.40\ \text{mol L}^{-1} at t=10 mint=10\ \text{min}, and 0.32 mol L−10.32\ \text{mol L}^{-1} at t=20 mint=20\ \text{min}. The average rate over the first interval is (0.50−0.40)/10=0.010 mol L−1min−1(0.50-0.40)/10 = 0.010\ \text{mol L}^{-1}\text{min}^{-1}; over the second interval it is (0.40−0.32)/10=0.008 mol L−1min−1(0.40-0.32)/10 = 0.008\ \text{mol L}^{-1}\text{min}^{-1}. The rate has fallen even though the time interval is identical each time, because the concentration–time curve is not a straight line — it flattens as H2O2\text{H}_2\text{O}_2 is used up, so the chord over the second interval is less steep than the chord over the first, exactly as sketched in the figure above.

Common traps

  • Forgetting to divide by the stoichiometric coefficient — the rate of disappearance of N2O5\text{N}_2\text{O}_5 in 2N2O5→4NO2+O22\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2 is not numerically equal to the rate of reaction unless every coefficient happens to be 1.
  • Dropping or misplacing the negative sign on reactant terms — rate must always come out positive, even though the reactant's own concentration is decreasing.
  • Treating average rate and instantaneous rate as interchangeable — a chord and a tangent give different numbers except for a perfectly linear (zero order) concentration–time graph.
  • Assuming rate stays constant through a reaction — for almost every reaction, rate falls as reactant is consumed, which is precisely why the average rate over an early interval exceeds the average rate over a later one of equal length.

Ask the tutor: if a concentration–time graph or a tricky stoichiometric-coefficient rate problem is confusing you, ask your AI tutor to work through a fresh example with you step by step.

Why does dividing by the stoichiometric coefficient matter if I only ever care about one species anyway

Because "rate of reaction" is meant to be a single number that describes how fast the whole reaction is proceeding, comparable between different reactions and different textbooks. If you quote a raw d[C]/dtd[C]/dt without normalising, the number you report depends on which species you happened to pick and what its coefficient is, so two people measuring the exact same reaction could report different "rates" just by tracking different species. Dividing by the coefficient removes that arbitrariness and gives one consistent value, which is also the value used later when rate is linked to a rate law and rate constant.

Is average rate ever exactly equal to instantaneous rate

Yes, but only in the special case where the concentration-time graph is a perfectly straight line over the interval you are considering — you will meet this case later as zero order kinetics, where rate does not depend on concentration at all. For almost every real reaction the curve bends, so the chord (average rate) and the tangent at any single point on that chord (instantaneous rate) have different slopes, and the gap between them narrows only as you shrink the interval toward zero.

How do I actually read an instantaneous rate off a graph in an exam

Draw the tangent line to the curve by eye at the time point you are asked about, touching the curve at exactly that one point and not crossing through it like a chord would. Then pick two convenient points on that tangent line itself (not on the curve), read off the rise in concentration and the run in time between them, and divide rise by run — that ratio is the slope of the tangent and therefore the instantaneous rate, reported as a positive number with concentration-over-time units.

Before this

Where this leads