Topic 11 · page 2 of 21
The rate constant and its units
The rate constant k turns concentrations into a rate. Its value depends on temperature and catalyst but not on concentration, and its units depend on the overall order, so you derive them each time.
Read first: Rate, rate equation and order
What the rate constant is 11.1(v)
In rate = k[A][B], the concentrations change as the reaction goes on, and so does the rate. The ratio rate ÷ ([A][B]) does not change. That ratio is k.
A large k means a fast reaction at that temperature. Two things change k:
- temperature: k increases when the temperature rises (page 15 shows how)
- a catalyst: it gives a route with a lower activation energy, so k increases
Units of k 11.1(v)
The units of k depend on the overall order, because k has to make the right-hand side of the equation come out in mol dm⁻³ s⁻¹. Never memorise them. Derive them in three moves: rearrange for k, substitute units, cancel.
MethodDeriving the units of k
- Rearrange the rate equation to make k the subject.
- Replace rate with mol dm⁻³ s⁻¹ and each concentration with mol dm⁻³.
- Cancel, and write the answer with positive and negative powers of mol, dm and s.
Worked exampleUnits of k for rate = k[A][B]²
- k = rate ÷ ([A][B]²)
- Units = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = mol dm⁻³ s⁻¹ ÷ mol³ dm⁻⁹
- mol¹⁻³ dm⁻³⁺⁹ s⁻¹ = dm⁶ mol⁻² s⁻¹
Doing this for each overall order gives the pattern in the table. You can use it as a check on your derivation.
| Overall order | Rate equation (example) | Units of k |
|---|---|---|
| 0 | rate = k | mol dm⁻³ s⁻¹ |
| 1 | rate = k[A] | s⁻¹ |
| 2 | rate = k[A][B] | dm³ mol⁻¹ s⁻¹ |
| 3 | rate = k[A][B]² | dm⁶ mol⁻² s⁻¹ |
The units can be used backwards. If a question gives k in dm³ mol⁻¹ s⁻¹, the reaction is second order overall. If it gives k in s⁻¹, it is first order overall.
Calculating with k 11.1(v)
Worked exampleFinding k from one experiment
A reaction has rate = k[A][B]. In one experiment [A] = 0.10 mol dm⁻³, [B] = 0.20 mol dm⁻³ and the rate is 3.0 × 10⁻³ mol dm⁻³ s⁻¹. Calculate k, with units.
- k = rate ÷ ([A][B]) = 3.0 × 10⁻³ ÷ (0.10 × 0.20)
- = 3.0 × 10⁻³ ÷ 0.020 = 0.15
- The reaction is second order overall, so k = 0.15 dm³ mol⁻¹ s⁻¹
Worked exampleFinding a rate from k
A reaction has rate = k[A]²[B] with k = 0.50 dm⁶ mol⁻² s⁻¹ at a certain temperature. Calculate the rate when [A] = 0.20 mol dm⁻³ and [B] = 0.10 mol dm⁻³.
- rate = 0.50 × (0.20)² × 0.10
- = 0.50 × 0.040 × 0.10 = 2.0 × 10⁻³ mol dm⁻³ s⁻¹
Notice that k is calculated at one temperature. If the temperature changes, the value of k changes too, and a k from one temperature cannot be used at another.
Check yourself
- Derive the units of k for the rate equation rate = k[A]²[B].
- A reaction has rate = k[A]. Show that the units of k are s⁻¹.
- For the reaction with rate = k[X][Y], [X] = 0.040 mol dm⁻³ and [Y] = 0.025 mol dm⁻³ give a rate of 2.0 × 10⁻⁵ mol dm⁻³ s⁻¹. Calculate k, with units.
- A reaction has rate = k[A]²[B], with k = 25 dm⁶ mol⁻² s⁻¹. Calculate the rate when [A] = 0.020 mol dm⁻³ and [B] = 0.040 mol dm⁻³.
- State whether k increases, decreases or stays the same when (a) the concentration of A is doubled, (b) the temperature is raised, (c) a catalyst is added.
- The rate constant of a reaction is 3.0 × 10⁻³ mol dm⁻³ s⁻¹. State the overall order of the reaction and explain how you know.
Answers cover these until you have tried the questions
- k = rate ÷ ([A]²[B]). Units = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹.
- k = rate ÷ [A]. Units = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³). The mol dm⁻³ cancels, leaving s⁻¹.
- k = 2.0 × 10⁻⁵ ÷ (0.040 × 0.025) = 2.0 × 10⁻⁵ ÷ 1.0 × 10⁻³ = 0.020. The reaction is second order overall, so k = 0.020 dm³ mol⁻¹ s⁻¹.
- rate = 25 × (0.020)² × 0.040 = 25 × 4.0 × 10⁻⁴ × 0.040 = 4.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
- (a) Stays the same: concentration changes the rate, not k. (b) Increases. (c) Increases, because the catalyst provides a route with a lower activation energy.
- Zero order overall. The units of k are the same as the units of rate, so no concentration term is left on the right-hand side: rate = k.