Rate laws, reaction order, activation energy and the Arrhenius equation, worked on one table of data and two temperatures

A rate law says how a reaction's speed depends on concentration: rate = k[A]ᵐ[B]ⁿ, where the orders m and n come from experiment, not from the balanced equation. Activation energy is the energy hill a collision must clear, and the Arrhenius equation says how the rate constant k climbs with temperature; from k at two temperatures you can compute the hill. Below, one table of initial rates gives the orders and k with every line shown, two temperatures give the activation energy, and the two wrong lines that cost the most marks are named.

What a rate law says, and what it does not

For a reaction A + B → products, the rate law has the shape rate = k[A]ᵐ[B]ⁿ. The square brackets are concentrations in mol/L. The exponents m and n are the reaction orders: m is the order in A, n is the order in B, and their sum is the overall order. The rate constant k is a number fixed for a given reaction at a given temperature.

Here is the sentence that costs the most marks in this unit: the orders come from experiment, not from the coefficients of the balanced equation. The equation tells you what reacts and what forms. It does not tell you how the rate responds when you change a concentration. Only measurements do that. The one exception, a single elementary step, is in the mechanisms section below.

The rate constant carries whatever units make the equation balance. A rate is in mol/(L·s), so k's units change with the overall order. Writing them out is a free check on the order you found.

In SCH4U this is three ideas of strand D: Reaction rates and factors affecting them, Rate laws and reaction order, and Activation energy, catalysts and reaction mechanisms. The SCH4U course map places them after the enthalpy ideas and before equilibrium.

Worked example: the orders by the method of initial rates

The method of initial rates runs the same reaction several times, changes one starting concentration at a time, and measures the rate at the very start. These data are ours, invented for teaching, not a measured reaction.

Experiment	[A] (mol/L)	[B] (mol/L)	Initial rate (mol/(L·s))
1	0.10	0.10	2.0 × 10⁻³
2	0.20	0.10	8.0 × 10⁻³
3	0.10	0.20	4.0 × 10⁻³

Order in A. Pick two runs where only [A] changed: experiments 1 and 2.

• [A] ratio: 0.20 ÷ 0.10 = 2 (dividing; [B] is 0.10 in both, so it drops out)
• Rate ratio: 8.0 × 10⁻³ ÷ 2.0 × 10⁻³ = 4 (dividing the rates)
• 2ᵐ = 4, so m = 2 (the order is the power that turns the concentration ratio into the rate ratio)

Order in B. Experiments 1 and 3, where only [B] changed.

• [B] ratio: 0.20 ÷ 0.10 = 2
• Rate ratio: 4.0 × 10⁻³ ÷ 2.0 × 10⁻³ = 2
• 2ⁿ = 2, so n = 1

The rate law. rate = k[A]²[B], second order in A, first order in B, third order overall.

The rate constant, from experiment 1.

• k = rate ÷ ([A]²[B]) (rearranging the rate law)
• [A]²[B] = (0.10)² × 0.10 = 1.0 × 10⁻³ mol³/L³ (squaring, then multiplying)
• k = 2.0 × 10⁻³ ÷ 1.0 × 10⁻³ = 2.0 L²/(mol²·s) (dividing; two significant figures, from the data)

The units check. mol/(L·s) ÷ (mol/L)³ = mol/(L·s) × L³/mol³ = L²/(mol²·s). That is the unit of a third-order rate constant, so the order and the unit agree. The rule to keep: first order overall, k is in s⁻¹; second order, L/(mol·s); third order, L²/(mol²·s).

The check against another run. Predict experiment 2: 2.0 × (0.20)² × 0.10 = 8.0 × 10⁻³ mol/(L·s), which is the measured rate. Experiment 3: 2.0 × (0.10)² × 0.20 = 4.0 × 10⁻³, also a match. A rate law that predicts every row is a rate law you can trust.

The common wrong line. "A + B has coefficients 1 and 1, so the rate law is k[A][B]." The data say otherwise: doubling [A] quadrupled the rate, which a first-order term cannot do. The second wrong line is comparing two runs in which both concentrations changed, say experiments 2 and 3. Their rate ratio mixes both effects and gives no single order.

Half-life of a first-order reaction

The half-life, t½, is the time for half of a reactant to be used up. For a first-order reaction it is

t½ = ln 2 ÷ k

and it does not depend on the starting concentration. That independence is the signature of first order: if every halving takes the same time, the reaction is first order.

Take a first-order reaction with k = 2.5 × 10⁻³ s⁻¹ (our value).

• t½ = 0.693 ÷ 2.5 × 10⁻³ s⁻¹ (ln 2 = 0.693)
• t½ = 277 s, which is 2.8 × 10² s to two significant figures, or about 4.6 minutes (dividing; s⁻¹ in the denominator leaves seconds)
• After one half-life, half the reactant is left; after two, at about 555 s, a quarter (each half-life halves what remains)

A second-order half-life does depend on the starting concentration, so a run whose half-life stays constant as it goes is first order, and one whose half-life grows is not.

The common wrong line. t½ = 1/k, or 0.5/k. The ln 2 comes from the integrated rate law, and leaving it out gives 400 s or 200 s here instead of 277 s.

Activation energy: the hill, and what a catalyst does

Picture an energy diagram. On the left floor sit the reactants. On the right floor sit the products. Between them is a hill, and at its top is the activated complex, the short-lived arrangement halfway between reactant and product. Activation energy, Ea, is the climb from the reactants' floor to the top of the hill.

Two separate heights live on that diagram. Ea is the hill. ΔH is the difference between the two floors, the quantity the enthalpy and Hess's law page computes. A reaction can be strongly exothermic and still have a high hill, which is why some fuels sit in air for years without burning.

Collision theory explains why the hill matters. A reaction needs a collision between particles, with enough energy to clear Ea and with the right orientation. Raising the concentration makes collisions more frequent. Raising the temperature makes collisions more frequent and, far more importantly, makes a larger fraction of them energetic enough.

A catalyst offers a different path with a lower hill. It changes neither ΔH, because the floors are where they were, nor the position of equilibrium, because it lowers the hill for the reverse reaction by the same amount.

The common wrong line. "A catalyst gives the molecules energy." It does not add energy; it provides a route that needs less.

Worked example: activation energy from two temperatures with the Arrhenius equation

The Arrhenius equation is k = A e^(−Ea/RT). A is a constant for the reaction, R is the gas constant, and T is the temperature in kelvin. Taking the natural log of each side gives a straight line, ln k = −(Ea/R)(1/T) + ln A, and writing that line at two temperatures and subtracting gives the two-point form:

ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)

The problem (our data). A first-order reaction has k₁ = 2.5 × 10⁻³ s⁻¹ at 300 K and k₂ = 2.0 × 10⁻² s⁻¹ at 320 K. R = 8.314 J/(mol·K). Find Ea.

• k₂/k₁ = 2.0 × 10⁻² ÷ 2.5 × 10⁻³ = 8.0 (dividing; the units cancel)
• ln 8.0 = 2.079 (taking the natural log)
• 1/T₁ = 1/300 K = 3.3333 × 10⁻³ K⁻¹ (taking the reciprocal)
• 1/T₂ = 1/320 K = 3.1250 × 10⁻³ K⁻¹
• 1/T₁ − 1/T₂ = 2.083 × 10⁻⁴ K⁻¹ (subtracting; keep extra digits here, because this small difference is where rounding hurts)
• Ea = R × ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) (rearranging the two-point form)
• Ea = 8.314 J/(mol·K) × 2.079 ÷ 2.083 × 10⁻⁴ K⁻¹ = 17.29 ÷ 2.083 × 10⁻⁴ J/mol (multiplying first; K cancels)
• Ea = 8.298 × 10⁴ J/mol = 83 kJ/mol (dividing, then converting; two significant figures, from the rate constants)

The sense check. The rate constant rose eightfold for a 20 K rise, a steep dependence on temperature, and 83 kJ/mol is a typical activation energy for a reaction that is slow at room temperature. Plug it back as a last check: k₁ × e^((Ea/R)(1/300 − 1/320)) returns 2.0 × 10⁻² s⁻¹.

The common wrong line. Temperatures in degrees Celsius, 27 and 47. Then 1/27 − 1/47 = 0.0158, and Ea comes out near 1.1 kJ/mol, about 76 times too small. A number that small would mean the reaction barely cares about temperature, which the eightfold jump says it does. The second wrong line is ln(k₁/k₂) with the temperatures left in their original order, which gives a negative Ea, and a negative hill is impossible.

Reaction mechanisms and the rate-determining step

A reaction mechanism is the sequence of elementary steps, single collisions or rearrangements, that add up to the overall equation. Three rules carry the idea:

• The slowest step sets the overall rate. It is called the rate-determining step.
• For an elementary step, and only for one, the orders equal the coefficients. A step in which two molecules of X collide has rate = k[X]².
• An intermediate is made in one step and used up in a later one, so it never appears in the overall equation.

The standard textbook example is NO₂(g) + CO(g) → NO(g) + CO₂(g) at low temperature. It is measured to be second order in NO₂ and zero order in CO, which the coefficients would never predict. The proposed mechanism is a slow first step, 2 NO₂ → NO₃ + NO, with rate = k[NO₂]², followed by a fast step, NO₃ + CO → NO₂ + CO₂. Add the steps and NO₃ and one NO₂ cancel, leaving the overall equation. NO₃ is the intermediate. The slow step's rate law matches the measured one, which is what makes the mechanism acceptable; it does not prove it.

That is why the coefficient rule fails for an overall reaction and works for a single step: the overall equation is a sum of steps, and only the slow one shows up in the rate.

Rate laws and activation energy on one screen

• Rate law: rate = k[A]ᵐ[B]ⁿ; the orders come from experiment; overall order = m + n.
• Method of initial rates: compare two runs where one concentration changed; concentration ratio raised to the order equals the rate ratio.
• Units of k: first order s⁻¹; second L/(mol·s); third L²/(mol²·s).
• First-order half-life: t½ = ln 2 ÷ k, the same at every concentration.
• Arrhenius, two-point form: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), T in kelvin, R = 8.314 J/(mol·K).
• Catalyst: lowers Ea; changes neither ΔH nor the position of equilibrium.

Try the two problems below before reading their answers. Attempting first, then checking, is chapter 3 of the free playbook.

Try one cold

Cover everything above.

(a) For X + Y → products, three runs (our data): (1) [X] = 0.20, [Y] = 0.10 mol/L, rate 1.5 × 10⁻⁴ mol/(L·s); (2) [X] = 0.40, [Y] = 0.10, rate 3.0 × 10⁻⁴; (3) [X] = 0.20, [Y] = 0.30, rate 1.35 × 10⁻³. Find the orders, the rate law and k with its units.

(b) A rate constant doubles between 298 K and 308 K. With R = 8.314 J/(mol·K), find Ea.

Check

• (a) Runs 2 and 1: [X] doubled, the rate went 3.0 ÷ 1.5 = 2 times, so first order in X. Runs 3 and 1: [Y] tripled, the rate went 1.35 × 10⁻³ ÷ 1.5 × 10⁻⁴ = 9 times, and 3² = 9, so second order in Y. Rate = k[X][Y]², third order overall. k = 1.5 × 10⁻⁴ ÷ (0.20 × 0.10²) = 1.5 × 10⁻⁴ ÷ 2.0 × 10⁻³ = 0.075 L²/(mol²·s).
• (b) ln 2 = 0.6931; 1/298 − 1/308 = 1.0896 × 10⁻⁴ K⁻¹; Ea = 8.314 × 0.6931 ÷ 1.0896 × 10⁻⁴ = 5.763 ÷ 1.0896 × 10⁻⁴ = 5.29 × 10⁴ J/mol, about 53 kJ/mol. This is the source of the old rule of thumb that a 10 K rise doubles a rate: it holds for an Ea near 50 kJ/mol at room temperature, not for every reaction.

If (a) gave you second order in X, you compared runs where both concentrations moved. If (b) gave you a tiny or negative number, check the kelvin and the order of the subtraction.

Where this sits in the course

These are ideas 13, 14 and 15 of 24 in SCH4U: Reaction rates and factors affecting them, Rate laws and reaction order, and Activation energy, catalysts and reaction mechanisms. They come after Hess's law and enthalpies of formation, which gives the ΔH on the energy diagram, and before Dynamic equilibrium and the equilibrium constant, where the forward and reverse rates become equal; the equilibrium and Le Chatelier page picks up there, including why a catalyst does not move the equilibrium.

Rates of gas reactions lean on the prerequisite review, Gas laws and stoichiometry, for a gas's concentration from its pressure; the ideal gas law page works that step. Every idea of the three science courses has a page like this one.

How to practise it

Mix the four problem types in one set: an initial-rates table, a units question, a half-life and an Arrhenius problem. Rohrer and Taylor (2007) found that students who practised problem types shuffled together did worse during practice and much better on a test a week later than students who practised each type in its own block, because a mixed set makes you decide which method a problem needs. Sweller and Cooper (1985) found that studying a worked example and then solving a matching problem teaches a procedure faster than solving problems alone, which is why each method here is one example and one sibling.

Write the unit check on every k, and the kelvin conversion on every temperature, even when you are sure. Those two lines catch most of the marks this unit takes. Mixed practice has the schedule, and how to study chemistry treats this strand as what it is: a few procedures, each a short list of lines.