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A rate equation as a constant times concentrations raised to measured orders, what each order does to the rate, and why the balanced equation cannot supply the orders.
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By the end of this lesson you will be able to calculate a rate from a rate equation, predict the factor by which a rate changes when a concentration changes for orders zero, one and two, give the units of a rate constant from the overall order, and explain why the orders in a rate equation are measured rather than read from the balanced equation.
You can measure a rate, read one from the gradient of a curve, and explain in terms of collisions why a more concentrated mixture reacts faster. What the collision model cannot tell you is how much faster. Doubling a concentration might double the rate, quadruple it, or — surprisingly often — do nothing at all. The rate equation is where that number lives.
A rate equation (or rate law) states how the rate of one reaction depends on the concentrations of the substances in it, at a fixed temperature. The order with respect to a substance is the power its concentration is raised to in the rate equation; the overall order is the sum of the separate orders. The rate constant, $k$, is the number that turns the concentrations into a rate. It is constant only at one temperature, and its units depend on the overall order. Square brackets, as in $[\mathrm{A}]$, mean the concentration of A in mol/L.
For a reaction whose rate depends on substances A and B, the rate equation has the form
$$\mathrm{rate} = k[\mathrm{A}]^m[\mathrm{B}]^n$$
where $m$ is the order in A and $n$ is the order in B. In the reactions this course meets, each order is 0, 1 or 2.
What an order does. If $[\mathrm{A}]$ is multiplied by some factor while everything else stays the same, the rate is multiplied by that factor raised to the power $m$.
Where the orders come from. They are found by experiment, and only by experiment. The balanced equation says how many particles of each substance take part in the overall change; it does not say how the change happens. Most reactions go in several steps, and the slowest step decides the rate. A substance that only turns up after the slow step has order zero, however large its coefficient.
The rate constant carries everything the concentrations do not: the temperature, the activation energy, the orientation requirement. Raise the temperature and $k$ gets bigger; the concentrations and orders do not change at all.
Another way: picture
Think of a toll road with one narrow booth. How many cars get through per minute depends on the queue at the booth, not on the cars already past it heading for the car park. A substance that joins the reaction after the slow step is a car already past the booth: it arrives, it is counted, and adding more of it does not make the queue move faster.
Another way: steps
To predict a new rate from a rate equation:
Here are four measured rate equations beside their balanced equations.
| Balanced equation | Measured rate equation | Orders match coefficients? |
|---|---|---|
| $\mathrm{2N_2O_5 \rightarrow 4NO_2 + O_2}$ | $k[\mathrm{N_2O_5}]$ | no: coefficient 2, order 1 |
| $\mathrm{NO_2 + CO \rightarrow NO + CO_2}$ (below about 500 K) | $k[\mathrm{NO_2}]^2$ | no: CO has order 0 |
| $\mathrm{2NO + 2H_2 \rightarrow N_2 + 2H_2O}$ | $k[\mathrm{NO}]^2[\mathrm{H_2}]$ | only for NO |
| $\mathrm{2NO + O_2 \rightarrow 2NO_2}$ | $k[\mathrm{NO}]^2[\mathrm{O_2}]$ | yes, by coincidence |
The last row is the dangerous one, because it seems to confirm the shortcut. It happens to agree; the three above it show that nothing makes it agree. The iodination of propanone is the cleanest case of all: iodine is in the balanced equation, and the measured rate equation, $k[\mathrm{CH_3COCH_3}][\mathrm{H^+}]$, has no iodine in it but does contain hydrogen ions, which are not consumed at all — they are a catalyst. A rate equation can leave out a reactant and include something that is not a reactant.
A rate is always in mol/(L s). Concentrations are in mol/L. The rate constant must have whatever units make the two sides of the rate equation agree, so its units depend on the overall order.
| Overall order | Rate equation | Units of $k$ |
|---|---|---|
| 0 | $k$ | mol/(L s) |
| 1 | $k[\mathrm{A}]$ | 1/s |
| 2 | $k[\mathrm{A}][\mathrm{B}]$ or $k[\mathrm{A}]^2$ | L/(mol s) |
| 3 | $k[\mathrm{A}]^2[\mathrm{B}]$ | L²/(mol² s) |
Each extra order means dividing by another mol/L, which puts another L/mol into the units. The pattern is worth working out rather than memorising, because a units check is the quickest way to catch a rate equation with a wrong order in it.
Copying the orders from the balanced equation. The coefficients count particles in the whole change; the orders describe the slowest step, and they have to be measured. Sometimes the two agree and most of the time they do not.
Thinking a zero-order substance does not react. It reacts completely, in the ratio the balanced equation gives. Its concentration simply does not control the rate.
Adding instead of multiplying. The rate equation is a product. Doubling a concentration of order two multiplies the rate by four; it does not add two to anything.
Treating $k$ as fixed for all conditions. It is constant for one reaction at one temperature. Changing the temperature, or adding a catalyst, changes $k$; changing a concentration never does.
For $\mathrm{2NO + 2H_2 \rightarrow N_2 + 2H_2O}$ at a certain temperature, $\mathrm{rate} = k[\mathrm{NO}]^2[\mathrm{H_2}]$ with $k = 400$ L²/(mol² s). A mixture has $[\mathrm{NO}] = 0.02$ mol/L and $[\mathrm{H_2}] = 0.05$ mol/L.
The rate equation, its constant and the two concentrations are all that is needed.
Raise each to its order: $0.02^2 = 0.0004$, and $0.05^1 = 0.05$.
Only NO is squared; the power belongs to its own bracket.
Rate $= 400 \times 0.0004 \times 0.05 = 0.008$ mol/(L s).
Multiply the constant by both results; the units come out as a rate.
The same reaction runs at 0.008 mol/(L s). The NO concentration is tripled and the hydrogen concentration is halved.
Two changes, so two factors.
NO is second order: a factor of 3 becomes $3^2 = 9$. Hydrogen is first order: halving it halves the rate.
Each factor is raised to its own order.
New rate $= 0.008 \times 9 \div 2 = 0.036$ mol/(L s).
The squared factor dominates, so the rate rises even though one concentration fell.
A is first order, so doubling it doubles the rate: a factor of 2.
The order in A is one.
B is second order, so doubling it gives a factor of $2^2 = 4$, and together the factor is $2 \times 4 = \ldots$
…8, so the new rate is $0.05 \times 8 = 0.4$ mol/(L s).
For a reaction between A and B at a fixed temperature, the rate equation is $\mathrm{rate} = k[\mathrm{A}]^{1}[\mathrm{B}]^{0}$ with $k = 6$ in the units that suit it. In a mixture with $[\mathrm{A}] = 0.1$ mol/L and $[\mathrm{B}] = 0.3$ mol/L, what is the rate, in mol/(L s)?
Answer: mol/(L s)
In each row one concentration in a rate equation is multiplied by the factor shown and everything else is held the same. By what factor is the rate multiplied?
| order in that substance | concentration multiplied by | rate multiplied by | |
|---|---|---|---|
| first row | 0 | 2 | |
| second row | 1 | 2 | |
| third row | 2 | 2 | |
| fourth row | 2 | 3 |
The balanced equation for a reaction is $\mathrm{2A + 2B \rightarrow products}$. Experiments show the reaction is order $1$ in A and order $0$ in B. Write the rate equation, using $x$ for $[\mathrm{A}]$, $y$ for $[\mathrm{B}]$ and $k$ for the rate constant.
Answer:
Propanone reacts with iodine in acid: $\mathrm{CH_3COCH_3 + I_2 \rightarrow CH_3COCH_2I + HI}$. The measured rate equation is $\mathrm{rate} = k[\mathrm{CH_3COCH_3}][\mathrm{H^+}]$. The iodine concentration alone is multiplied by $2$. What happens to the rate?
The rate is always measured in mol/(L s). Match each rate equation to the units its rate constant must have.
| mol/(L s) | 1/s | L/(mol s) | L²/(mol² s) | |
|---|---|---|---|---|
| rate = k | ||||
| rate = k[A] | ||||
| rate = k[A][B] | ||||
| rate = k[A]²[B] |
A reaction has the rate equation $\mathrm{rate} = k[\mathrm{A}]^2[\mathrm{B}]$. Put the three mixtures in order of rate, fastest first.
Number the steps in order (write the number in the box):
In polluted air, nitrogen monoxide destroys ozone: $\mathrm{NO + O_3 \rightarrow NO_2 + O_2}$, with the measured rate equation $\mathrm{rate} = k[\mathrm{NO}][\mathrm{O_3}]$. In a test chamber, $[\mathrm{NO}] = 0.09$ mol/L and $[\mathrm{O_3}] = 0.08$ mol/L, and the rate is $0.0216$ mol/(L s). What is the rate constant, in L/(mol s)?
Answer: L/(mol s)
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A reaction has the rate equation $\mathrm{rate} = k[\mathrm{X}]^2[\mathrm{Y}]$, and a mixture reacts at $0.03$ mol/(L s). A new mixture has $[\mathrm{X}]$ multiplied by $3$ and $[\mathrm{Y}]$ divided by $2$, at the same temperature. What is its rate, in mol/(L s)?
Answer: mol/(L s)
You can use a rate equation to calculate a rate and to predict the effect of a change. Say out loud what happens to the rate of a reaction that is second order in A when the concentration of A is tripled, and why iodine can react with propanone without appearing in its rate equation. Next: finding the orders yourself from a table of experiments.
10. Your turn: $\mathrm{rate} = k[\mathrm{A}][\mathrm{B}]^2$, and the rate is 0.05 mol/(L s). A is doubled and B is doubled. What is the new rate?, step 3