Chemical kinetics: reaction rate and worked exercises

It is the end of September 2026, technical institute (an Italian technical secondary school), Chemistry, Materials and Biotechnology specialisation: in Year 4, chemistry opens the year with chemical kinetics, even before reaching equilibrium — it is the chapter that provides the tools (reaction rate, order, rate constant) on which the whole rest of the block is built. The first test typically falls between the end of October and November, so between four and eight weeks from now. The same kinetics, with a very similar approach, is also covered in Year 4 at the science high school (liceo scientifico), shortly before chemical equilibrium and Le Châtelier (I talk about it separately in Chemical equilibrium and Le Châtelier: Kc made clear): the method you see here holds, with the same formulas, in both tracks.

Before we start, a prerequisite that is often taken for granted by whoever writes the textbook but not by whoever reads it: all of kinetics works with concentrations, so if molarity is not yet secure it is worth revising it — I talk about it in Solution concentration: molarity and dilutions.

What the reaction rate is

The reaction rate measures how quickly the concentration of a reactant or a product changes over time. For a generic reaction

the average rate with respect to one species, over a time interval $\Delta t$, is the change in its concentration divided by the time taken:

Two details that explain this formula, which looks more complicated than necessary:

• The minus sign in front of the reactants is there because their concentration decreases over time ($\Delta [A]$ is negative), and the reaction rate by convention is always a positive number.
• You divide by the stoichiometric coefficient ($a$, $b$, $c$, $d$) because different species in the same reaction are consumed or formed at different rates, proportional to their coefficients: by dividing, you get a single reaction rate, the same whichever species you are monitoring.

The rate law: careful, the exponents are not the coefficients

For many reactions, the rate can be written as a function of the concentrations of the reactants through the rate law (or rate equation):

where $k$ is the rate constant (it depends on the reaction and on the temperature, not on the concentrations) and $m$, $n$ are the partial orders of reaction with respect to $A$ and $B$. Their sum $m+n$ is the overall order of the reaction.

This is where the most frequent mistake of the whole topic hides: $m$ and $n$ are not, in general, equal to the stoichiometric coefficients $a$ and $b$. Unlike the equilibrium constant $K_c$, where the exponents are always the coefficients of the balanced reaction, in the rate law the orders of reaction are determined only experimentally, with laboratory data — they cannot be read off the balanced equation. A reaction can have coefficient 2 on a reactant and be first order with respect to that reactant, if that is how the experimental data behave.

The method of initial rates

The standard way to find the orders of reaction from a table of experimental data is the method of initial rates: you compare two experiments in which the concentration of only one reactant changes, keeping the other fixed, and you observe how the rate changes.

Procedure, step by step:

1. Find in the table two rows in which the concentration of $A$ changes and that of $B$ stays constant.
2. Calculate the ratio between the two concentrations of $A$ and the ratio between the two corresponding rates.
3. The order $m$ with respect to $A$ is the exponent that satisfies: (ratio of concentrations)${}^m$ = (ratio of rates).
4. Repeat the same comparison with two rows in which $B$ changes and $A$ stays constant, to find $n$.
5. Once $m$ and $n$ are known, find $k$ by substituting the values of any row into the rate law.

The key point is always the same: isolate one variable at a time, exactly as you do in physics when looking for the relationship between two quantities while holding all the others fixed.

The units of $k$ change with the order

A detail that the textbook often leaves implicit, but which is asked in tests to check whether you have understood the formula: the unit of the rate constant $k$ depends on the overall order of the reaction, because $v$ is always measured in $\text{mol}/(\text{L}\cdot \text{s})$ while $[A{]}^m[B{]}^n$ changes unit with the exponent.

A good final check on any exercise: if you find a value of $k$ with a unit different from the one expected for the order you have calculated, you have almost certainly got an exponent wrong somewhere.

The factors that affect the reaction rate

Four factors, all explainable with the same underlying theory (collision theory, seen below):

• Concentration of the reactants: more molecules in the same volume, more possible collisions per unit of time, so a higher rate.
• Temperature: by raising it, the molecules move faster and with more average kinetic energy, so they collide more often and with more effective collisions.
• Contact surface: for solid reactants, breaking the solid into smaller pieces increases the area exposed to the rest of the mixture, so it increases the frequency of useful collisions.
• Catalyst: a substance that speeds up the reaction without being consumed by it, by providing an alternative reaction pathway with a lower activation energy (see below).

Collision theory

Collision theory explains, at the microscopic level, why these factors work: a reaction happens when two reactant molecules collide with two conditions met together:

1. Correct orientation: the molecules must collide the right way round so that the bonds can break and re-form in the way the reaction requires.
2. Sufficient energy: the collision must have kinetic energy at least equal to the activation energy $E_a$, the minimum energy threshold for the collision to actually lead to the reaction, instead of being an “elastic” collision with no consequences.

Not all collisions are effective: most collisions between molecules in a reacting mixture produce nothing, because one condition or the other is missing. This is why raising the concentration or the temperature, which respectively increase the total number of collisions and the fraction of collisions with sufficient energy, speeds up the reaction.

Activation energy and the energy diagram

The energy diagram of a reaction shows the energy of the system along the horizontal axis called the “reaction coordinate”, from the reactants to the products. The typical profile:

• it starts from the energy level of the reactants;
• it rises to a maximum, called the transition state (or activated complex): an unstable, very short-lived arrangement in which the old bonds are breaking and the new ones are forming;
• it falls to the energy level of the products.

The activation energy $E_a$ is the difference between the energy of the transition state and that of the reactants: it is the “barrier” the reaction has to overcome, regardless of whether the overall reaction releases energy (exothermic) or absorbs it (endothermic). An exothermic reaction has products at a lower energy than the reactants, but can still have a high activation energy — this is the reason why, for example, wood does not burn by itself at room temperature even though combustion is strongly exothermic: you need a trigger that supplies the energy to overcome $E_a$.

The catalyst lowers $E_a$, it does not shift the equilibrium

A very common mistake, which comes from confusing kinetics with equilibrium: a catalyst lowers the activation energy (by providing an alternative reaction pathway, with a transition state at a lower energy), and so speeds up both the forward and the reverse reaction to the same extent. But it does not change the energies of reactants and products, so it does not shift the position of equilibrium and does not change $K_c$: it only makes the equilibrium get reached faster, without changing where it lies. Kinetics and equilibrium answer different questions — how fast, and where — and a catalyst acts only on the first.

A note on the Arrhenius equation

The relationship between the rate constant $k$, the temperature $T$ and the activation energy $E_a$ is given by the Arrhenius equation:

where $A$ is a frequency factor (linked to the number of collisions with the correct orientation) and $R$ is the gas constant. All you need for the Year 4 test is to recognise it: it says, in one formula, that $k$ grows with temperature and decreases as $E_a$ increases — consistent with everything we have seen above.

Typical mistakes

Three worked exercises

1. Finding the orders and the rate constant with the method of initial rates. For the reaction $A+B\to C$ these data were collected:

Find the order with respect to $A$, to $B$, the overall order and the constant $k$.

I compare experiments 1 and 2, where $[B]$ stays constant and $[A]$ doubles:

I compare experiments 2 and 3, where $[A]$ stays constant and $[B]$ doubles:

The rate law is therefore $v=k[A{]}^2[B{]}^1$, overall order $m+n=3$. I find $k$ by substituting the data of experiment 1:

The unit consistent with an overall order of 3 is ${\text{L}}^2/({\text{mol}}^2\cdot \text{s})$: a good check that the calculation of the orders is right.

2. Effect of temperature on the rate. A reaction at $20{}^{\circ }\text{C}$ has an initial rate $v_1=4.0\times 10^{-4}\text{mol}/(\text{L}\cdot \text{s})$. When the temperature is raised, the fraction of molecules with enough energy to overcome $E_a$ grows, and the rate observed at $40{}^{\circ }\text{C}$ becomes $v_2=1.6\times 10^{-3}\text{mol}/(\text{L}\cdot \text{s})$, with the same concentrations. By how many times has the rate increased, and what changed in the rate law to produce this increase?

The rate has increased by 4 times. The concentrations are the same in both cases, so the only term of the rate law $v=k[A{]}^m[B{]}^n$ that can have changed is $k$: it is the rate constant that has grown, not the orders of reaction (which depend on the mechanism, not on the temperature) nor the concentrations. This is exactly what the Arrhenius equation describes: at a higher temperature, a larger fraction of molecules has enough kinetic energy to overcome $E_a$ in collisions, so $k$ increases.

3. Catalyst and activation energy. An uncatalysed reaction has $E_a=120\text{kJ/mol}$. In the presence of a catalyst, the activation energy drops to $E_{a,\text{cat}}=75\text{kJ/mol}$. The reaction is exothermic, with $\Delta H=-50\text{kJ/mol}$. Does the catalyst change $\Delta H$ or the position of equilibrium? By how much has the energy barrier decreased?

The barrier has decreased by

The catalyst does not change $\Delta H$: the energy of the reactants and that of the products stay the same ($\Delta H=-50\text{kJ/mol}$ stays unchanged), because the catalyst acts only on the transition state, lowering its energy and so the barrier that the forward and reverse reactions have to overcome. As a consequence it does not shift the position of equilibrium: forward and reverse rates increase in the same proportion, and equilibrium is reached faster but at the same point.

There is no chemistry practice test, for now

For this topic there is not yet a dedicated practice test on the site: you will still find all those available for the other subjects at /simulazioni/liceo/.

The chapter that follows kinetics directly, in many Year 4 programmes, is chemical equilibrium: I talk about it in Equilibrio chimico e Le Châtelier in quarta. For the rest of the science programme, you will find everything at /materie/scienze/. For the sign of $\Delta H$ and the heat of reaction that appear in the energy diagram, the reference is Termochimica: entalpia e legge di Hess con esercizi.

If kinetics is creating more doubts than certainties before the test, the first introductory call is free and lasts half an hour.