Reaction Rates and Rate Laws: How Concentration Controls Speed

Measuring how fast a reaction goes

A reaction rate is the change in concentration of a reactant or product per unit time, usually in mol L⁻¹ s⁻¹. Reactant concentrations fall as a reaction runs, so their change is negative. A minus sign is placed in front of reactant terms so that every rate is reported as a positive number.

An average rate is the change in concentration over a time interval divided by the length of that interval. An instantaneous rate is the slope of the tangent to the concentration–time curve at one moment. The initial rate is the instantaneous rate at time zero, before products build up. For most reactions the rate falls as time passes, because the reactants are being used up.

Concentration is one of several factors that set the rate. Temperature, catalysts, the surface area of a solid reactant, and the chemical nature of the reactants also matter. This guide holds temperature constant and focuses on concentration; the effect of temperature is the subject of the activation-energy guide.

One reaction, one rate: dividing by the coefficients

Different species in the same reaction change at different speeds, in proportion to their coefficients. To give a single rate for the whole reaction, divide each species' rate of change by its coefficient. For aA + bB → cC + dD, the rate of reaction is −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt = (1/d)Δ[D]/Δt.

Worked example 1. In the decomposition 2 N₂O₅(g) → 4 NO₂(g) + O₂(g), nitrogen dioxide is forming at 6.0 × 10⁻⁵ mol L⁻¹ s⁻¹ at one instant. The rate of reaction is (1/4)(6.0 × 10⁻⁵) = 1.5 × 10⁻⁵ mol L⁻¹ s⁻¹. N₂O₅ is used up at twice the rate of reaction, 3.0 × 10⁻⁵ mol L⁻¹ s⁻¹, and O₂ forms at the rate of reaction, 1.5 × 10⁻⁵ mol L⁻¹ s⁻¹. All four statements describe the same event.

The same bookkeeping works for average rates. Suppose, as an illustration, that [N₂O₅] falls from 0.0200 mol/L to 0.0169 mol/L during the first 300 s. The average rate of N₂O₅ consumption is (0.0200 − 0.0169)/300 s = 1.0 × 10⁻⁵ mol L⁻¹ s⁻¹. The subtraction leaves only two significant figures, so the result has two. Over the same interval NO₂ forms at twice the unrounded average rate, 2 × 1.03 × 10⁻⁵ = 2.1 × 10⁻⁵ mol L⁻¹ s⁻¹.

Writing a rate law: rate constant and reaction orders

A rate law expresses the rate as a product of concentrations raised to powers: rate = k[A]ᵐ[B]ⁿ. The rate constant k is specific to one reaction at one temperature and does not depend on concentration. The exponents m and n are the reaction orders with respect to A and B, and their sum is the overall order. Orders are usually 0, 1 or 2, but fractional and negative orders are also observed.

Each order tells you how the rate responds to a change in one concentration. If a reaction is zero order in A, changing [A] has no effect on the rate. If it is first order, doubling [A] doubles the rate. If it is second order, doubling [A] multiplies the rate by 2² = 4. For rate = k[A]²[B], doubling [A] while halving [B] changes the rate by a factor of 2² × ½ = 2.

Orders must be measured. OpenStax section 12.3 gives the reaction NO₂(g) + CO(g) → NO(g) + CO₂(g) as an example: its rate law is rate = k[NO₂]², second order in NO₂ and zero order in CO, even though each reactant has a coefficient of 1.

Method of initial rates, step by step

The method of initial rates runs the same reaction several times at one temperature, changing one starting concentration at a time, and measures the rate at the very start of each run.

Worked example 2. The data here are illustrative, built for practice. For A + 2 B → products, three runs give: trial 1, [A] = 0.0100 mol/L and [B] = 0.0100 mol/L, initial rate 2.40 × 10⁻⁶ mol L⁻¹ s⁻¹; trial 2, [A] = 0.0150 mol/L and [B] = 0.0100 mol/L, rate 5.30 × 10⁻⁶ mol L⁻¹ s⁻¹; trial 3, [A] = 0.0100 mol/L and [B] = 0.0300 mol/L, rate 7.30 × 10⁻⁶ mol L⁻¹ s⁻¹.

Only [A] changes between trials 1 and 2, by a factor of 1.50, and the rate rises by a factor of 2.208. Solve 2.208 = 1.50ᵐ with logarithms: m = ln 2.208 / ln 1.50 = 1.95. Measured rates carry experimental error, so a value this close to a whole number is rounded: m = 2. Only [B] changes between trials 1 and 3, by a factor of 3.00, and the rate rises by 3.04, so n = ln 3.04 / ln 3.00 = 1.01, which rounds to 1. The rate law is rate = k[A]²[B], third order overall. Notice that B has a coefficient of 2 in the equation but is first order.

Now find k from each trial: k = rate / ([A]²[B]). The three values are 2.40, 2.36 and 2.43 L² mol⁻² s⁻¹, with a mean of 2.40, so report k = 2.4 L² mol⁻² s⁻¹. Use it to predict the rate when [A] = 0.0200 mol/L and [B] = 0.0250 mol/L: rate = 2.4 × (0.0200)² × 0.0250 = 2.4 × 10⁻⁵ mol L⁻¹ s⁻¹.

• Pick two trials in which only one concentration changes.
• Divide the larger rate by the smaller and the matching concentrations the same way. Then solve: rate ratio = (concentration ratio) raised to the power of the order.
• Round each order to a sensible value (an integer, or occasionally ½) and repeat for the other reactants.
• Calculate k from every trial and average the results, rather than trusting a single run.

Units of k reveal the overall order

The rate always has units of concentration per time, so the units of k must make the rate law balance. With concentrations in mol/L and time in seconds, k has units of (mol/L)¹⁻ⁿ s⁻¹, where n is the overall order. Reading the units of a given rate constant is a quick way to tell the overall order of a reaction.

• Zero order: mol L⁻¹ s⁻¹.
• First order: s⁻¹, with no concentration unit at all.
• Second order: L mol⁻¹ s⁻¹.
• Third order: L² mol⁻² s⁻¹, as in worked example 2.

Why the balanced equation cannot give the orders

Most reactions happen through a sequence of simpler events called elementary steps, which together make up the mechanism. For a single elementary step, and only for one, the rate law can be written from the step's coefficients: a step in which two NO₂ molecules collide is second order in NO₂. The number of particles that collide in a step is its molecularity.

The slowest step in a mechanism limits the overall rate and is called the rate-determining step. OpenStax section 12.6 explains the NO₂ + CO rate law this way. A slow first step, NO₂ + NO₂ → NO₃ + NO, is followed by a fast step, NO₃ + CO → NO₂ + CO₂. The two steps add to the overall equation. CO enters only after the slow step, so its concentration does not appear in the rate law. A balanced overall equation describes what is consumed and produced, not how the reaction happens, which is why orders must be measured.

Rate-law slips and how to catch them

Most errors in kinetics problems come from reading the balanced equation as if it were a rate law, or from treating every species' rate as the same number.

• Taking reaction orders from the coefficients of an overall equation. Orders come from data or from a known mechanism.
• Forgetting to divide by coefficients when converting the rate of one species into the rate of another.
• Leaving a minus sign off a reactant rate and reporting a negative reaction rate.
• Comparing two trials in which more than one concentration changed at once.
• Calculating k without units. Its units confirm whether the overall order is right.
• Rounding a measured order of 1.95 to 1.9 instead of 2. Data from experiments rarely give exact integers.