Kinetics
At AS you explained qualitatively why concentration, temperature and catalysts change the rate of a reaction. Here the treatment becomes quantitative: rate equations say exactly how the rate depends on each concentration, the orders they contain reveal which species take part in the slowest step of the mechanism, and the Arrhenius equation puts a number on the activation energy.
Rate of reaction
The rate of reaction is the change in concentration of a reactant or product per unit time. Its usual unit is mol dm−3 s−1.
On a graph of concentration against time, the rate at any moment is the gradient of the curve at that moment. Because the curve bends, you draw a tangent to it and find the tangent's gradient. A reactant's concentration falls, so its gradient is negative; the rate is the size of that gradient. The initial rate is the gradient of the tangent at t = 0, where the reaction is fastest because the reactants are most concentrated.
The rate in mol dm−3 s−1 depends on which substance you follow. In 2N2O5 → 4NO2 + O2, nitrogen dioxide forms twice as fast as N2O5 is used up and four times as fast as oxygen forms, so always say which substance a rate refers to.
Following the rate of a reaction
To follow a reaction you need a property that changes as it goes and that you can measure without disturbing it (or a way of stopping samples so you can analyse them). Choose the method to suit the reaction.
| Method | Use it when… | Example |
|---|---|---|
| Volume of gas, collected in a gas syringe | a gas is given off | 2H2O2(aq) → 2H2O(l) + O2(g) |
| Loss of mass, on a balance | a gas escapes from an open flask; works best for a dense gas such as CO2 | CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) |
| Colorimetry | a reactant or product is coloured; absorbance is proportional to its concentration | HCOOH(aq) + Br2(aq) → 2H+(aq) + 2Br−(aq) + CO2(g): the orange bromine fades |
| Titration of quenched samples | an acid, alkali or iodine is used up or formed | CH3COCH3 + I2 → CH3COCH2I + H+ + I−: titrate the iodine left with thiosulfate |
| Electrical conductivity | the number, charge or type of ions changes | (CH3)3CBr + H2O → (CH3)3COH + H+ + Br−: ions are produced |
| "Clock" reaction | you only need the initial rate: time how long a fixed small amount of product takes to form | the iodine clock described in the core practical below |
Quenching means stopping the reaction in a sample so its composition stays fixed while you analyse it. You can quench by diluting the sample in cold water, or by removing a catalyst: adding sodium hydrogencarbonate neutralises an acid catalyst.
Rate equations and orders
For a reaction between A and B, experiments show that the rate is proportional to each concentration raised to some power:
The rate equation
- Order with respect to a reactant: the power its concentration is raised to in the rate equation. For this course the order is 0, 1 or 2.
- Overall order: the sum of the individual orders, m + n.
- Rate constant, k: the constant of proportionality. It does not change with concentration, but it is different at each temperature, increasing as the temperature rises. A larger k means a faster reaction.
| Order with respect to A | Rate is… | Double [A] | Triple [A] |
|---|---|---|---|
| 0 | independent of [A] (because [A]0 = 1) | rate × 1 | rate × 1 |
| 1 | proportional to [A] | rate × 2 | rate × 3 |
| 2 | proportional to [A]2 | rate × 4 | rate × 9 |
Orders can only be found by experiment. They are not the numbers in front of the formulae in the balanced equation. For H2O2 + 2I− + 2H+ → I2 + 2H2O the rate equation is rate = k[H2O2][I−]: first order in iodide despite the 2, and zero order in H+.
Units of the rate constant
The units of k depend on the overall order, so work them out from the rate equation each time.
Rearrange the rate equation to make k the subject:
\[ k = \frac{\text{rate}}{[\mathrm{A}][\mathrm{B}]^2} \]Substitute the units of each quantity:
\[ k = \frac{\mathrm{mol\,dm^{-3}\,s^{-1}}}{(\mathrm{mol\,dm^{-3}})(\mathrm{mol\,dm^{-3}})^2} = \frac{\mathrm{mol\,dm^{-3}\,s^{-1}}}{\mathrm{mol^{3}\,dm^{-9}}} \]Cancel one mol dm−3 from top and bottom, then move what is left on the bottom to the top by changing the signs of its powers:
\[ k = \frac{\mathrm{s^{-1}}}{\mathrm{mol^{2}\,dm^{-6}}} = \mathrm{dm^{6}\,mol^{-2}\,s^{-1}} \]The order of the units in the final answer doesn't matter: mol−2 dm6 s−1 is equally correct.
| Overall order | Example rate equation | Units of k |
|---|---|---|
| 0 | rate = k | mol dm−3 s−1 |
| 1 | rate = k[A] | s−1 |
| 2 | rate = k[A][B] or k[A]2 | dm3 mol−1 s−1 |
| 3 | rate = k[A][B]2 | dm6 mol−2 s−1 |
Orders from graphs and half-lives
One way to find the order is a continuous experiment: follow the concentration of one reactant all the way through a single run. For this to show the order with respect to that reactant alone, every other reactant must be in large excess, so that its concentration hardly changes and has no effect on how the rate changes.
The half-life, t½, of a reaction is the time taken for the concentration of a reactant to fall to half its value.
| Order | Concentration–time graph | Successive half-lives | Rate–concentration graph |
|---|---|---|---|
| 0 | straight line sloping down: the rate stays constant until the reactant runs out | each one is half the one before | horizontal line |
| 1 | curve that gets less steep (exponential decay) | constant, whatever the starting concentration | straight line through the origin |
| 2 | steeper at first, then a long, slow tail | each one is double the one before | upward curve through the origin (a straight line if you plot rate against [A]2) |
Why is the first-order half-life constant? With rate = k[A], the reaction always uses up the same fraction of whatever A is left in a given time. Going from 0.80 to 0.40 mol dm−3 takes exactly as long as going from 0.40 to 0.20 mol dm−3: there is half as much to remove, but it is removed at half the rate. Radioactive decay behaves in the same way.
A first-order reaction has a half-life of 25 s. The initial concentration of the reactant is 0.64 mol dm−3. What is its concentration after 100 s?
The half-life is constant, so count half-lives: 100 s ÷ 25 s = 4 half-lives.
Halve the concentration four times:
\[ 0.64 \rightarrow 0.32 \rightarrow 0.16 \rightarrow 0.080 \rightarrow 0.040\ \mathrm{mol\,dm^{-3}} \]Equivalently, \(0.64 \times (\tfrac{1}{2})^4 = 0.64/16 = 0.040\ \mathrm{mol\,dm^{-3}}\).
For a first-order reaction the half-life and the rate constant are linked by \( \ln 2 = k\,t_{1/2} \), so a half-life read from a graph gives k directly. A half-life of 25 s means \( k = 0.693/25 = 0.028\ \mathrm{s^{-1}} \). The faster the reaction (the larger k), the shorter the half-life, but it never depends on the starting concentration.
The initial rates method
The other approach is a set of separate experiments, each started with different initial concentrations, measuring only the initial rate each time (from the gradient of a tangent at t = 0, or from a clock reaction). Change one concentration at a time and keep the others the same, so any change in rate is caused by that reactant alone.
| Experiment | [NO] / mol dm−3 | [H2] / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 5.0 × 10−3 | 2.0 × 10−3 | 1.25 × 10−5 |
| 2 | 1.0 × 10−2 | 2.0 × 10−3 | 5.0 × 10−5 |
| 3 | 1.0 × 10−2 | 4.0 × 10−3 | 1.0 × 10−4 |
Order with respect to NO: compare experiments 1 and 2, where only [NO] changes. [NO] doubles and the rate goes from 1.25 × 10−5 to 5.0 × 10−5, a factor of 4 = 22.
So the reaction is second order with respect to NO.
Order with respect to H2: compare experiments 2 and 3, where only [H2] changes. [H2] doubles and the rate doubles.
So the reaction is first order with respect to H2.
Write the rate equation. The overall order is 2 + 1 = 3.
\[ \text{rate} = k[\mathrm{NO}]^2[\mathrm{H_2}] \]Find k by substituting the values from any one experiment, here experiment 1:
\[ k = \frac{\text{rate}}{[\mathrm{NO}]^2[\mathrm{H_2}]} = \frac{1.25\times10^{-5}}{(5.0\times10^{-3})^2 \times 2.0\times10^{-3}} = 250\ \mathrm{dm^6\,mol^{-2}\,s^{-1}} \]Check with another experiment: \(250 \times (1.0\times10^{-2})^2 \times 4.0\times10^{-3} = 1.0\times10^{-4}\), matching experiment 3.
When two concentrations change at once
Sometimes no pair of experiments changes only one concentration. The effects then multiply. Use an order you already know to strip out its effect, and whatever factor is left belongs to the other reactant.
| Experiment | [P] / mol dm−3 | [Q] / mol dm−3 | Initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 0.020 | 0.030 | 4.0 × 10−4 |
| 2 | 0.040 | 0.030 | 8.0 × 10−4 |
| 3 | 0.060 | 0.090 | 1.08 × 10−2 |
Experiments 1 and 2: only [P] changes. It doubles and the rate doubles, so the reaction is first order in P.
Experiments 1 and 3: both change. The rate rises by a factor of \(1.08\times10^{-2} / 4.0\times10^{-4} = 27\).
[P] has tripled (0.020 → 0.060). Being first order, that alone triples the rate. The rest of the increase must come from Q: \(27 \div 3 = 9\).
[Q] has also tripled (0.030 → 0.090) and multiplied the rate by 9 = 32, so the reaction is second order in Q.
\[ \text{rate} = k[\mathrm{P}][\mathrm{Q}]^2, \qquad k = \frac{4.0\times10^{-4}}{0.020\times(0.030)^2} = 22\ \mathrm{dm^6\,mol^{-2}\,s^{-1}} \]Rate equations and mechanisms
Most reactions do not happen in a single collision. They go through a sequence of simple steps called the mechanism, often forming intermediates: species made in one step and used up in a later one. Adding the steps together, and cancelling the intermediates, must give the overall equation.
The rate-determining step is the slowest step in the mechanism. The overall reaction cannot go faster than this step, just as a queue cannot move faster than its slowest checkout.
- A species that appears in the rate equation takes part in the rate-determining step, or in a step before it that makes an intermediate used in the rate-determining step.
- Its order tells you how many particles of it are involved in reaching and completing the rate-determining step.
- A reactant that is zero order reacts only after the rate-determining step. So if any reactant is zero order, the mechanism must have at least two steps.
- An intermediate never appears in the rate equation. Replace it with the reactants that formed it.
- A catalyst can appear in the rate equation, even though it is not in the overall equation, if it takes part in the rate-determining step.
For 2NO(g) + Br2(g) → 2NOBr(g) the rate equation is rate = k[NO]2[Br2]. A two-step mechanism is proposed:
Step 1: NO + Br2 → NOBr2
Step 2: NOBr2 + NO → 2NOBr
Check the steps add up: NO + Br2 + NOBr2 + NO → NOBr2 + 2NOBr. Cancelling NOBr2 leaves 2NO + Br2 → 2NOBr, so NOBr2 is an intermediate.
If step 1 were the slow step, it would contain one NO and one Br2, giving rate = k[NO][Br2]. That doesn't match the experiment.
If step 2 is slow, it uses one NOBr2 (made from one NO and one Br2) and one more NO: two NO and one Br2 in total. That gives rate = k[NO]2[Br2], matching the experiment.
So step 2 is the rate-determining step, and step 1 must be fast.
| Overall reaction and rate equation | Mechanism consistent with it |
|---|---|
| NO2(g) + CO(g) → NO(g) + CO2(g) rate = k[NO2]2 | 2NO2 → NO3 + NO (slow) NO3 + CO → NO2 + CO2 (fast) CO is zero order because it only reacts after the slow step. |
| H2O2 + 2I− + 2H+ → I2 + 2H2O rate = k[H2O2][I−] | H2O2 + I− → IO− + H2O (slow) IO− + H+ → HIO (fast) HIO + H+ + I− → I2 + H2O (fast) |
| (CH3)3CBr + OH− → (CH3)3COH + Br− rate = k[(CH3)3CBr] | (CH3)3CBr → (CH3)3C+ + Br− (slow) (CH3)3C+ + OH− → (CH3)3COH (fast): SN1 |
| CH3CH2Br + OH− → CH3CH2OH + Br− rate = k[CH3CH2Br][OH−] | a single step in which OH− attacks as Br− leaves: SN2 |
Kinetics is how chemists tell SN1 from SN2. If the rate of hydrolysis depends on [OH−], the hydroxide ion must be in the rate-determining step, so the mechanism is the one-step SN2 route (primary halogenoalkanes). If changing [OH−] has no effect, the slow step is the halogenoalkane ionising on its own: SN1 (tertiary halogenoalkanes). See the halogenoalkanes page for the mechanisms.
Core practicals: following rates in the lab
The iodine–propanone reaction, followed by titration
Propanone reacts with iodine in acid solution; the acid is a catalyst:
CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + H+(aq) + I−(aq)
Mix propanone and sulfuric acid, both in large excess, then add iodine solution and start the clock. At regular intervals withdraw a 10.0 cm3 sample with a pipette and run it into excess sodium hydrogencarbonate solution. This quenches the sample by neutralising the acid catalyst. Titrate the iodine left in the sample with sodium thiosulfate solution, adding starch near the end point (blue-black to colourless):
I2(aq) + 2S2O32−(aq) → 2I−(aq) + S4O62−(aq)
- The titre is proportional to [I2]. A graph of titre against time is a straight line, so the rate is constant while iodine is used up: the reaction is zero order with respect to iodine.
- Repeating with different concentrations of propanone or acid changes the gradient. These experiments show it is first order in propanone and first order in H+: rate = k[CH3COCH3][H+].
- So the rate-determining step involves one propanone molecule and one H+ ion, and iodine reacts in a later, fast step. H+ appears in the rate equation even though it is a catalyst.
- The same reaction can be followed by colorimetry, since iodine is the only coloured species.
- Safety: propanone is highly flammable; iodine solution stains and irritates; wear eye protection.
A clock reaction: initial rates and activation energy
Hydrogen peroxide oxidises iodide ions in acid: H2O2 + 2I− + 2H+ → I2 + 2H2O. A small, fixed amount of sodium thiosulfate and some starch are added. Each iodine molecule formed is turned straight back into iodide by the thiosulfate, until the thiosulfate is used up. Then iodine builds up and the starch suddenly turns blue-black. Record the time, t.
- Every run makes the same small amount of iodine before the colour appears, so the average rate up to that point is proportional to 1/t.
- The thiosulfate is kept small so that only a tiny fraction of the reactants is used before the colour appears. Their concentrations have barely changed, so 1/t is a good measure of the initial rate.
- To find an order, change the concentration of one reactant at a time (making up the volume with water so the total volume is constant) and plot 1/t against concentration.
- To find the activation energy, keep every concentration the same and repeat at several temperatures in a water bath, letting the separate solutions reach the bath temperature before mixing. Since k ∝ 1/t, a graph of ln(1/t) against 1/T has gradient −Ea/R (see below).
Temperature and the Arrhenius equation
Raising the temperature increases the rate constant. The main reason is the shape of the Maxwell–Boltzmann distribution: at a higher temperature a much larger fraction of collisions have at least the activation energy, Ea. (The molecules also collide a little more often, but that is a far smaller effect.) The Arrhenius equation describes this exactly:
The Arrhenius equation and its logarithmic form
The factor \(e^{-E_\mathrm{a}/RT}\) is the fraction of collisions with enough energy to react. It is tiny when Ea is large or T is small, and it grows very rapidly with temperature. For a typical activation energy of about 50 kJ mol−1, warming from 298 K to 308 K roughly doubles the rate constant.
Ea is usually quoted in kJ mol−1, but R is in J K−1 mol−1. Multiply Ea by 1000 before substituting, and use T in kelvin.
Finding Ea from a graph
Compare the logarithmic form with the equation of a straight line, y = mx + c. Plotting ln k (y) against 1/T (x) gives a straight line with gradient −Ea/R and intercept ln A. The gradient is always negative, and the steeper it is, the larger the activation energy. Because the rate is proportional to k at fixed concentrations, ln(rate) or ln(1/t) from a clock reaction can be plotted instead of ln k: only the intercept changes, not the gradient.
| T / K | t / s | 1/T / 10−3 K−1 | ln(1/t) |
|---|---|---|---|
| 293 | 240 | 3.413 | −5.48 |
| 303 | 115 | 3.300 | −4.74 |
| 313 | 58.2 | 3.195 | −4.06 |
| 323 | 30.6 | 3.096 | −3.42 |
| 333 | 16.7 | 3.003 | −2.82 |
Calculate 1/T and ln(1/t) for each run (table above) and plot ln(1/t) against 1/T. The points lie on a straight line. Draw the line of best fit.
Find the gradient using two points on the line that are far apart, here the first and last:
\[ \text{gradient} = \frac{-2.82 - (-5.48)}{(3.003 - 3.413)\times10^{-3}} = \frac{2.66}{-0.410\times10^{-3}} = -6490\ \mathrm{K} \]The gradient equals −Ea/R, so:
\[ E_\mathrm{a} = -\text{gradient}\times R = 6490 \times 8.31 = 53\,900\ \mathrm{J\,mol^{-1}} = 53.9\ \mathrm{kJ\,mol^{-1}} \]The gradient has units of K because ln(1/t) has no units and 1/T is in K−1. Multiplying K by J K−1 mol−1 gives J mol−1.
A first-order reaction has k = 2.5 × 10−4 s−1 at 320 K and Ea = 85.0 kJ mol−1. Calculate A.
Work out the exponent, with Ea in J mol−1:
\[ \frac{E_\mathrm{a}}{RT} = \frac{85\,000}{8.31\times320} = 31.96 \qquad e^{-31.96} = 1.31\times10^{-14} \]Rearrange k = A e−Ea/RT:
\[ A = \frac{k}{e^{-E_\mathrm{a}/RT}} = \frac{2.5\times10^{-4}}{1.31\times10^{-14}} = 1.9\times10^{10}\ \mathrm{s^{-1}} \]A has the same units as k, here s−1.
A catalyst lowers the activation energy of a reaction from 75 kJ mol−1 to 50 kJ mol−1. Assuming A is unchanged, by what factor does k increase at 298 K?
Divide the two Arrhenius expressions. A cancels:
\[ \frac{k_\mathrm{cat}}{k} = \frac{e^{-50\,000/RT}}{e^{-75\,000/RT}} = e^{25\,000/(8.31\times298)} = e^{10.1} \approx 2.4\times10^{4} \]Lowering Ea by a third makes the reaction about 24 000 times faster, which is why catalysts matter so much in industry.
Catalysts
A catalyst speeds up a reaction without being used up, by providing an alternative route (mechanism) with a lower activation energy. It speeds up the forward and reverse reactions equally, so it does not change the position of equilibrium or the enthalpy change.
- Homogeneous catalyst: in the same phase as the reactants, usually all in solution. It reacts to form an intermediate and is regenerated in a later step. Because it takes part in the mechanism it can appear in the rate equation: H+ in the iodination of propanone is an example.
- Heterogeneous catalyst: in a different phase from the reactants, usually a solid with gaseous or dissolved reactants. The reaction happens on its surface.
How a heterogeneous catalyst works
Reactant molecules adsorb (form weak bonds) onto active sites on the catalyst surface. This weakens bonds within the reactants, holds them close together in a favourable orientation, and concentrates them on the surface, so they react more easily. The products then desorb, freeing the sites for more reactant. A large surface area, from a fine powder or a thin coating on a support, gives more active sites.
The number of active sites is limited. At high pressure or concentration every site can be occupied, and adding more reactant then cannot speed the reaction up. The rate is then independent of that reactant's concentration: the reaction is zero order with respect to it. The decomposition of ammonia on a hot tungsten surface behaves like this.
Quick checks
The rate equation for a reaction is rate = k[X]2[Y]. What are the units of k?
The overall order is 3. k = rate/([X]2[Y]) = (mol dm−3 s−1)/(mol dm−3)3 = mol−2 dm6 s−1. The second option is for a second-order reaction.
A reaction is first order with respect to A and second order with respect to B. Both concentrations are doubled at the same time. By what factor does the rate increase?
Doubling [A] multiplies the rate by 2; doubling [B] multiplies it by 22 = 4. The effects multiply: 2 × 4 = 8. Adding the orders (to get 3) or adding the factors (to get 6) are the usual slips.
A first-order reaction has a half-life of 1400 s. Starting from 1.0 mol dm−3, what is the concentration after 4200 s?
4200 s is three half-lives, because a first-order half-life is constant: 1.0 → 0.50 → 0.25 → 0.125 mol dm−3. Dividing by 3 instead of halving three times gives the wrong answer of 0.33.
The iodination of propanone has rate = k[CH3COCH3][H+]. Which statement must be true?
Iodine is a reactant, but it is zero order, so it is not involved in the slow step and must react afterwards. A zero-order reactant means there are at least two steps. Changing [I2] has no effect on the rate.
A graph of ln k against 1/T for a reaction has a gradient of −7500 K. What is the activation energy? (R = 8.31 J K−1 mol−1)
Gradient = −Ea/R, so Ea = 7500 × 8.31 = 62 300 J mol−1 = 62.3 kJ mol−1. Activation energy is always positive; dividing by R instead of multiplying gives 0.90.
In a clock reaction, why can 1/t be used as a measure of the initial rate?
Rate = amount formed ÷ time. The amount (fixed by the thiosulfate) is the same in every run, so the rate is proportional to 1/t. Keeping that amount small means the concentrations are still close to their starting values, so it is an initial rate.
Increasing the pressure of a gas reacting on a solid catalyst has no effect on the rate. What is the best explanation?
Reaction only happens at active sites. Once they are saturated, extra gas molecules have nowhere to adsorb, so the rate stays the same and the reaction is zero order in that gas.