Rate of reaction, rate equations, orders and rate constants

This is A Level (IA2) content, examined on the Unit 4 paper, WCH14, and AS knowledge from Unit 2 Topic 9A (collision theory, the Maxwell-Boltzmann distribution, rate = 1/time and rates from tangents) is assumed throughout.

The practical side of this topic, the core practicals CP9a, CP9b and CP10, is also examined on the Unit 6 practical paper, WCH16.

Kinetics is the largest physical-chemistry topic on WCH14: about a sixth of the marks across the 2024-2025 papers.

At AS you explained rates with collisions: more particles in a volume, more collisions per second, a faster reaction.

That gives the direction of a change but not its size.

If you double the concentration of a reactant, does the rate double, quadruple or stay the same?

Collision theory cannot say, and the answer differs between reactants, even in one reaction.

The rate equation is the tool that answers it.

It is a short equation, found by experiment, that tells you exactly how the rate depends on the concentration of each substance.

Once you have it you can predict the rate under any conditions, and, as later sections show, it gives you evidence about the steps a reaction goes through.

The rate of a reaction is how quickly a reactant is used up or a product is made.

In solution it is measured as a change in concentration per unit time, so its unit is mol dm⁻³ s⁻¹ (or mol dm⁻³ min⁻¹ if time is in minutes).

On a concentration-time graph the rate at any instant is the gradient of the tangent at that time; for a reactant the gradient is negative, and the rate is its size.

The figure below shows a reaction X → 2Y.

Two Y form for every X used, so [Y] rises twice as fast as [X] falls.

A rate therefore has to say which substance it refers to: "the rate of formation of Y" is twice "the rate of loss of X" here.

Both curves get flatter with time.

As X is used up there are fewer X particles per dm³, fewer collisions per second, and the rate falls.

That is why the rate equation uses concentrations at the moment the rate is measured.

The general form to learn is:

where m and n are 0, 1 or 2.

Here [A] and [B] are concentrations in mol dm⁻³, $m$ is the order with respect to A, $n$ the order with respect to B, and $m + n$ is the overall order of the reaction.

The constant $k$ is the rate constant.

What each order means for the rate:

The general rule: if [A] is multiplied by a factor $f$, the rate is multiplied by $f^m$.

When two concentrations change together, multiply the two factors.

Why orders cannot be read from the equation. It is tempting to take the orders from the balanced equation, because the coefficients look like "how many of each particle collide".

They do not.

Most reactions happen in several steps, and only the slow step and the steps before it control the rate (the sections on mechanisms build this up).

For example, the reaction $2\text{NO(g)} + 2\text{H}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O(g)}$ has the experimentally found rate equation rate =$k[\text{NO}]^2[\text{H}_2]$: second order in NO but only first order in H₂, although the coefficient of H₂ is 2.

Orders come only from experiments.

At a fixed temperature $k$ is fixed: changing a concentration changes the rate, not $k$.

A large $k$ means a fast reaction.

$k$ rises steeply with temperature (see the Arrhenius section), which is how temperature enters the rate equation even though no temperature term appears in it.

The units of $k$ depend on the overall order, so derive them every time; never learn one unit for all.

Rearrange the rate equation for $k$ and put in units:

Each extra order adds one more "dm³ mol⁻¹".

"mol⁻¹ dm³ s⁻¹" is equally correct.

Meaning of the answer. A rate constant in s⁻¹ tells you at once that the reaction is first order overall: the units carry the order, so a $k$ without units is incomplete.

Later sections show how rate data are collected, how orders are found from graphs and tables, how rate equations point to mechanisms (including the SN1 and SN2 evidence), and how $k$ changes with temperature to give the activation energy.

In Unit 5, transition-metal catalysts change the steps, and so the rate equation.

Learn each definition word for word: an order is marked on "power", "concentration" and "rate equation".

1. A reaction has the rate equation rate = $k[\text{A}]^2$. What are the units of $k$? (1)

• A mol dm⁻³ s⁻¹
• B s⁻¹
• C dm³ mol⁻¹ s⁻¹
• D dm⁶ mol⁻² s⁻¹ Total [1]

2. For a reaction, rate = $k[\text{A}][\text{B}]^2$. The concentration of A is tripled and the concentration of B is halved, at constant temperature. By what factor does the rate change? (1)

• A × 0.75
• B × 1.5
• C × 2.25
• D × 4.5 Total [1]

3. A student says that for the reaction 2X + Y → Z the rate equation must be rate = $k[\text{X}]^2[\text{Y}]$, because two X and one Y react. Explain why the student may be wrong. (2) Total [2]

4. Complete the table. (4)

Total [4]

5. Peroxodisulfate ions oxidise iodide ions: $\text{S}_2\text{O}_8^{2-}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{SO}_4^{2-}\text{(aq)} + \text{I}_2\text{(aq)}$. The rate equation is rate =$k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$, where the rate is the rate of formation of iodine.
(a) State the order with respect to iodide ions and the overall order. (2)
(b) When $[\text{S}_2\text{O}_8^{2-}]$ = 0.040 mol dm⁻³ and $[\text{I}^-]$ = 0.080 mol dm⁻³, iodine forms at $1.2 \times 10^{-5}$ mol dm⁻³ s⁻¹. Calculate the value of $k$, with its units. (3)
(c) State the rate at which iodide ions are used up at this moment. (1)
(d) Explain why the rate of formation of iodine falls as the reaction goes on. (2) Total [8]

6. A first-order reaction, rate = $k[\text{A}]$, has a rate of $4.5 \times 10^{-2}$ mol dm⁻³ min⁻¹ when [A] = 0.30 mol dm⁻³. Calculate $k$ in s⁻¹. (2) Total [2]