Topic 11 · page 1 of 21

Rate, rate equation and order

A rate equation links the rate of a reaction to the concentrations of the reactants. The power of each concentration is its order, and the orders can only be found by experiment.

12 min read11.1(i)11.1(ii)11.1(iii)11.1(iv)

Topic 9 treated the rate of reaction as something you could compare: faster or slower. This topic turns it into a number with an equation. Everything later in the topic, from graphs to mechanisms, rests on the four terms on this page.

Rate of reaction 11.1(i)

Concentration is written in square brackets, so [A] means the concentration of A in mol dm⁻³. For a reactant that is used up:

rate=−Δ[A]Δt \text{rate} = -\frac{\Delta[\text{A}]}{\Delta t}

The unit is mol dm⁻³ s⁻¹. If time is measured in minutes, the unit is mol dm⁻³ min⁻¹, and you convert to seconds by dividing by 60.

The rate at one moment is the gradient of the tangent to a concentration–time curve at that moment (Topic 9). A reactant graph falls, so its gradient is negative and the rate is minus the gradient. The rate also falls as the reaction goes on, because the reactant is used up.

The rate equation 11.1(ii)

Experiments show that the rate depends on the concentration of some reactants and not on others. The rate equation records this.

Only concentrations of species that affect the rate appear. A substance with order 0 would be raised to the power 0, which equals 1, so it simply drops out of the equation.

The reaction between nitrogen monoxide and hydrogen is a standard example.

2 NO(g)+2 HX2(g)→NX2(g)+2 HX2O(g) \ce{2NO(g) + 2H2(g) -> N2(g) + 2H2O(g)}

Experiment gives

rate=k[NO]2[H2] \text{rate} = k[\text{NO}]^2[\text{H}_2]

Order and overall order 11.1(iii)11.1(iv)

For rate = k[NO]²[H₂]:

The order tells you what happens to the rate when you change that concentration, with everything else kept the same.

Order with respect to A Double [A] Triple [A] Halve [A]
0 rate unchanged rate unchanged rate unchanged
1 rate × 2 rate × 3 rate × ½
2 rate × 4 rate × 9 rate × ¼

The effects of several reactants multiply.

Worked exampleChanging two concentrations at once

For rate = k[NO]²[H₂], [NO] is doubled and [H₂] is tripled. By what factor does the rate change?

  1. NO is second order: doubling gives 2² = 4.
  2. H₂ is first order: tripling gives 3.
  3. The factors multiply: 4 × 3 = 12, so the rate is 12 times faster.

Check yourself

  1. The rate equation for a reaction is rate = k[NO]²[O₂]. State the order with respect to each reactant and the overall order.
  2. Write the rate equation for a reaction that is first order with respect to A, second order with respect to B and zero order with respect to C.
  3. For rate = k[A][B]², the concentration of A is doubled and the concentration of B is halved. By what factor does the rate change?
  4. The concentration of bromine falls from 0.0200 to 0.0150 mol dm⁻³ in 50 s. Calculate the mean rate of loss of bromine over this time, in mol dm⁻³ s⁻¹.
  5. Explain why the order of a reaction cannot be deduced from its balanced equation.
  6. When the concentration of X is tripled, the rate of a reaction becomes nine times faster. State the order with respect to X.

Answers cover these until you have tried the questions

  1. Second order with respect to NO, first order with respect to O₂, and third order overall (2 + 1).
  2. rate = k[A][B]². Substance C is zero order, so it does not appear.
  3. A is first order: ×2. B is second order, halved: (½)² = ¼. Overall 2 × ¼ = ×½, so the rate is halved.
  4. Δ[Br₂] = 0.0150 − 0.0200 = −0.0050 mol dm⁻³. Rate = 0.0050 ÷ 50 = 1.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
  5. A balanced equation shows only the overall change in amounts. The rate depends on the steps by which the reaction happens, and the slowest step may involve some reactants and not others, or more than one particle of a reactant. Only experiment shows which concentrations affect the rate, and by how much.
  6. Tripling gives ×9 = 3², so the reaction is second order with respect to X.