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Orders from controlled comparisons in an initial-rates table, a rate constant from one experiment, and orders read from the shape and half-life of a single run.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to choose the pair of experiments that isolates one concentration, find each order from the rate factor, write the rate equation and calculate its rate constant from an initial-rates table. You will also be able to recognise zero and first order from a concentration–time graph and use a constant half-life.
You can use a rate equation: raise each concentration factor to its order and multiply. You also know that the orders cannot be read off the balanced equation. So where do they come from? From experiments designed to answer exactly that question, and this lesson is about reading them.
The initial rate is the rate at the very start of a run, before the concentrations have had time to change; it is found from the gradient of a tangent at time zero. An initial-rates table lists several runs with different starting concentrations and the initial rate of each. A controlled comparison is a pair of runs in which only one concentration differs. The half-life of a reactant is the time it takes for its concentration to fall to half of its value.
An order is a statement about cause and effect: if this concentration changes by this factor, the rate changes by that one. The only honest way to measure it is to change one concentration and hold every other condition the same — the other concentrations and the temperature.
The method, from an initial-rates table:
For the factors this course uses the power is easy to see. With a concentration factor of 2, a rate factor of 1 means order 0, 2 means order 1, and 4 means order 2. With a factor of 3 the rate factors are 1, 3 and 9.
Why initial rates? At the very start the concentrations are exactly the ones that were mixed, and nothing has yet been used up. A rate measured later would belong to concentrations nobody measured.
Once every order is known, write the rate equation, then put any one experiment's numbers into it to find $k$: divide the rate by the concentrations raised to their orders. Every experiment in a good data set gives the same $k$, which makes a useful check.
Another way: picture
A gardener wants to know whether fertiliser or water makes tomatoes grow. Comparing a plant given both with one given neither says nothing about either on its own. Comparing two plants given the same water and different fertiliser says something about fertiliser. An initial-rates table is the same garden with reactants.
Another way: steps
Worked on numbers: experiment 1 has $[\mathrm{A}] = 0.1$, rate $0.002$; experiment 2 has $[\mathrm{A}] = 0.3$ and everything else the same, rate $0.018$.
An initial-rates table uses many runs. A single run, followed from start to finish, gives the order as well, because each order produces a different shape of concentration–time graph.
| Order in the reactant | Shape of its concentration–time graph | Half-life |
|---|---|---|
| 0 | a straight line sloping down to zero | gets shorter as the reactant is used up |
| 1 | a curve that halves in equal times | constant |
| 2 | a curve that levels off more slowly | gets longer as the reactant is used up |
The first-order case is the most useful. If the concentration goes from 0.8 to 0.4 mol/L in 25 s, a first-order reactant takes another 25 s to reach 0.2 and another 25 s to reach 0.1. Checking two successive half-lives on a graph is a quick test: equal half-lives, first order.
Zero order looks strange until you see why it happens. Ammonia decomposing on a hot platinum surface is the standard case: the surface is completely covered, so adding more ammonia gas cannot make the surface work faster. The concentration falls by the same amount every second until it runs out.
Anything that happens to each particle independently, at a fixed chance per second, is first order. Radioactive decay is the famous example: carbon-14 has a half-life of about 5700 years, however much of it there is. The body removes many medicines from the blood the same way, which is why a drug has a stated half-life and why a dose schedule works for a small or a large patient: if a medicine's half-life is 6 hours, three quarters of it has gone after 12 hours whatever the dose was. A constant half-life is the fingerprint of first order wherever it turns up.
Comparing two experiments in which two concentrations changed. The rate factor is then the product of two effects, and it cannot be shared out between them. Always find a pair where exactly one concentration differs; if the table has none, the question cannot be answered from it.
Writing the rate factor as the order. A rate that goes up four times when the concentration doubles is order 2, not order 4. The order is the power.
Forgetting that the temperature must be the same. A pair of experiments at different temperatures changes $k$ as well as a concentration, and the rate factor then says nothing about the order.
Expecting every curve that falls to be first order. Only a curve with a constant half-life is. A straight line is zero order; a curve whose half-life keeps growing is second order.
Experiment 1: $[\mathrm{A}] = 0.1$, $[\mathrm{B}] = 0.1$, rate $0.004$. Experiment 2: $[\mathrm{A}] = 0.2$, $[\mathrm{B}] = 0.1$, rate $0.016$. Experiment 3: $[\mathrm{A}] = 0.1$, $[\mathrm{B}] = 0.2$, rate $0.004$.
Concentrations in mol/L and rates in mol/(L s), all at one temperature.
Experiments 1 and 2: A doubles, rate goes up four times, so A is second order. Experiments 1 and 3: B doubles, rate unchanged, so B is zero order.
Each pair changes exactly one concentration.
Rate equation: $\mathrm{rate} = k[\mathrm{A}]^2$. From experiment 1, $k = 0.004 \div 0.1^2 = 0.4$ L/(mol s).
Any one experiment then gives $k$; the others should agree.
A reactant's concentration is 0.60 mol/L at 0 s, 0.30 at 40 s and 0.15 at 80 s.
Look at how long each halving takes.
Both halvings took 40 s, so the half-life is constant.
A constant half-life is the mark of first order.
The reaction is first order in this reactant, and it will reach 0.075 mol/L at 120 s.
Each further half-life halves it again.
For X: $3^{\text{order}} = 9$, so the order in X is 2.
The power of the concentration factor that gives the rate factor.
For Y: $2^{\text{order}} = 1$, so the order in Y is …
…zero, and the rate equation is $\mathrm{rate} = k[\mathrm{X}]^2$ with no Y in it.
Three experiments on the reaction of A with B, at one temperature. Experiment 1: $[\mathrm{A}] = 0.1$ mol/L, $[\mathrm{B}] = 0.1$ mol/L, initial rate $0.006$ mol/(L s). Experiment 2: $[\mathrm{A}] = 0.2$ mol/L, $[\mathrm{B}] = 0.1$ mol/L, initial rate $0.006$ mol/(L s). Experiment 3: $[\mathrm{A}] = 0.1$ mol/L, $[\mathrm{B}] = 0.2$ mol/L, initial rate $0.012$ mol/(L s). For each substance, say which pair of experiments shows its order, and give the order.
| experiments to compare | order | |
|---|---|---|
| substance A | ||
| substance B |
In two experiments at the same temperature, only the concentration of a reactant X changes: it is multiplied by $3$. The initial rate is multiplied by $9$. What is the order of the reaction with respect to X?
Answer:
Three experiments on the reaction of A with B, at one temperature. Experiment 1: $[\mathrm{A}] = 0.1$ mol/L, $[\mathrm{B}] = 0.1$ mol/L, initial rate $0.009$ mol/(L s). Experiment 2: $[\mathrm{A}] = 0.3$ mol/L, $[\mathrm{B}] = 0.1$ mol/L, initial rate $0.027$ mol/(L s). Experiment 3: $[\mathrm{A}] = 0.1$ mol/L, $[\mathrm{B}] = 0.3$ mol/L, initial rate $0.081$ mol/(L s). Write the rate equation, using $x$ for $[\mathrm{A}]$, $y$ for $[\mathrm{B}]$ and $k$ for the rate constant.
Answer:
A reaction has the rate equation $\mathrm{rate} = k[\mathrm{A}]^{2}[\mathrm{B}]^{1}$, found from an initial-rates table. In one experiment $[\mathrm{A}] = 0.5$ mol/L, $[\mathrm{B}] = 0.3$ mol/L and the initial rate is $0.525$ mol/(L s). What is the value of $k$?
Answer:
A reactant is used up in a first-order reaction with a half-life of $20$ s. Its concentration starts at $80$ mmol/L. Plot its concentration at time zero and at the end of each of the first three half-lives.
Plot your answer on the grid:
Three experiments on the reaction of A with B. Experiment 1: $[\mathrm{A}] = 0.1$, $[\mathrm{B}] = 0.1$ mol/L, rate $0.002$ mol/(L s). Experiment 2: $[\mathrm{A}] = 0.2$, $[\mathrm{B}] = 0.2$ mol/L, rate $0.008$ mol/(L s). Experiment 3: $[\mathrm{A}] = 0.1$, $[\mathrm{B}] = 0.3$ mol/L, rate $0.006$ mol/(L s). Which pair of experiments shows the order with respect to B?
The body removes a certain medicine from the blood by a first-order process. A blood test shows $72$ mg/L just after a dose, and $15$ hours later a second test shows $9$ mg/L. What is the half-life of the medicine in the blood? Give the time with its unit.
Answer: unit: h / s / min
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
On a hot platinum wire, ammonia decomposes at a rate that does not depend on its own concentration: the reaction is zero order. In one experiment $[\mathrm{NH_3}]$ starts at $0.6$ mol/L and is $0.52$ mol/L after $20$ s. If the rate stays the same, how many seconds after the start will the ammonia be used up?
Answer: s
You can find the orders of a reaction from data and write its rate equation. Say out loud which two experiments you would compare to find the order in B, and why comparing two experiments in which both concentrations changed settles nothing. That completes kinetics: next the course turns to reactions that run in both directions at once.
10. Your turn: tripling [X] alone makes the initial rate nine times larger; doubling [Y] alone leaves it unchanged. Write the rate equation., step 3