Back to Chemistry syllabus

Chemical Reaction Rates and Equilibrium

Distinguish reaction rates from equilibrium composition. Use reaction quotients to explain catalysts, temperature, gas compression, and inert-gas additions.

Before planning this topic, check the CSCA Exam Guide 2026 for exam dates, registration, fees, and subject requirements.

Syllabus Alignment

This study guide is organized around the publicly available CSCA syllabus for international undergraduate applicants.

Who It Is For

International students preparing for CSCA Math, Physics, Chemistry, or Chinese exams.

Related Practice

CSCA Practice · Go to questions

Past papers and worked video solutions · Timed mock exams · All subject video lessons

Practice by Topic

Jump from this tutorial to filtered practice questions for the same knowledge point.

Tutorial Content

Chemical Reaction Rates and Equilibrium

Reaction rate tells you how quickly a mixture changes. Chemical equilibrium describes the composition it approaches under fixed conditions. A reaction can reach equilibrium quickly yet produce little of the desired product, so keep these two questions separate.

1. Measuring reaction rates

For a reaction written as

the average net reaction rate over an interval $\Delta t$ can be written as

Here, square brackets denote molar concentration. These relations assume a constant reaction volume and no other reactions or material flows changing the listed concentrations. They give a positive net rate when the reaction proceeds in the forward direction and a negative one when it proceeds in reverse.

The rate of disappearance of a reactant or appearance of a product is usually reported as a positive magnitude. Divide that species' rate by its coefficient to obtain the common reaction rate. Use the same time interval and units when comparing species.

An average rate describes a finite interval. An instantaneous rate is the slope of the concentration–time curve at a particular time, with the same sign and coefficient conventions.

Example 1: Relate the rates of different species

For

the net rate of ammonia formation is $0.50\ \mathrm{mol\,L^{-1}\,min^{-1}}$. Find the rate of hydrogen consumption, assuming the conditions above.

The coefficients give

The normalized reaction rate is $0.50/2=0.25\ \mathrm{mol\,L^{-1}\,min^{-1}}$. This is different from the rate assigned to either individual species.

2. What changes the rate?

The effect of concentration depends on the reaction's rate law. Increasing a reactant concentration often increases the rate, but the exponent in a rate law is determined experimentally; it cannot generally be read from the overall balanced equation.

Temperature, catalysts, and the exposed surface area of a reacting solid can also matter. Higher temperature commonly increases rate constants. More exposed solid surface can make a heterogeneous reaction faster when access to the surface limits the process.

A catalyst provides an alternative reaction pathway with a lower activation barrier. It speeds up the approach to equilibrium but does not change the equilibrium constant or the equilibrium composition at a given temperature.

For gases, distinguish pressure from concentration. Compressing a mixture changes the reacting gases' concentrations. Adding an inert gas at fixed temperature and volume raises the total pressure without changing those concentrations in the ideal-gas model.

3. What equilibrium means

In a closed reacting system under fixed conditions, dynamic equilibrium means that the forward and reverse reaction rates are equal. The molecular reactions continue, but their opposing effects balance:

For an ongoing reversible reaction, the forward and reverse rates at equilibrium are not zero. Concentrations remain constant, but reactant and product concentrations need not be equal.

A flat concentration curve alone is not enough to prove equilibrium. For example, an open flow system can maintain a steady composition while reactions and material flows continue without being at equilibrium.

4. Equilibrium constants and the reaction quotient

For the general equation above, when the species are described by the usual ideal-gas or dilute-solution concentration model,

The concentrations in $K_c$ are equilibrium concentrations, and the exponents come from the balanced equation. Pure solids and pure liquids have unit activity and are omitted from the usual expression while those phases are present. Do not interpret this as permission to omit every liquid species: a dissolved solute still contributes. Treating a solvent's activity as $1$ is an approximation appropriate to sufficiently dilute solutions.

The thermodynamic equilibrium constant is defined using dimensionless activities. The concentration form above is the introductory approximation, so use the same concentration and standard-state convention throughout a calculation.

For a specified reaction and convention, the equilibrium constant depends on temperature. Changing the starting amounts, compressing the mixture at fixed temperature, or adding a catalyst does not change that constant. Reversing or rescaling the written reaction changes the associated expression and value of $K$.

The reaction quotient $Q_c$ has the same form but uses the current concentrations, whether or not the mixture is at equilibrium:

A large equilibrium constant indicates a product-favored equilibrium for that reaction. It does not tell you the reaction is fast or that conversion must be complete.

5. Predicting shifts in equilibrium

Le Châtelier's principle describes how a system responds to a disturbance: the ensuing reaction tends to oppose the imposed change, not necessarily undo it completely.

Adding more of a pure solid already present does not change its activity. It may change available surface area and thus the rate, but does not by itself shift the equilibrium.

For $\mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)}$, each side of the balanced equation represents two moles of gas. Isothermal compression does not shift the ideal-gas equilibrium, although the changed concentrations may affect reaction rates.

Example 2: Compress the nitrogen dioxide equilibrium

Consider the ideal-gas equilibrium

at a fixed temperature. In a conceptual two-stage calculation, halve the volume before allowing further chemical reaction.

Immediately after the volume change, both concentrations have doubled. Since

the new reaction quotient is

The reaction therefore shifts toward $\mathrm{N_2O_4}$. The $\mathrm{NO_2}$ concentration falls from its just-compressed value, but remains above its original equilibrium concentration: the shift only partly offsets the concentration increase caused by compression.

$\mathrm{NO_2}$ is brown, whereas $\mathrm{N_2O_4}$ is colorless. At the same temperature and viewing path length, the mixture would first look darker, then become lighter as equilibrium is restored, while remaining darker than before compression. Do not apply that visual prediction unchanged if compression also heats the gas or changes the optical path length.

Example 3: Add helium

Assume helium does not react and the gas mixture behaves ideally.

• At fixed temperature and volume, the reacting gases' partial pressures and concentrations do not change. There is no equilibrium shift, even though total pressure increases.
• At fixed temperature and total pressure, adding helium expands the volume. If one side of the equation has more moles of gas than the other, equilibrium shifts toward that side. If the totals are equal, there is no such shift.

The phrase “add an inert gas” is therefore not enough to determine the answer; the volume or pressure condition matters.

6. Organizing a concentration calculation

For $\mathrm{2NO_2(g)\rightleftharpoons N_2O_4(g)}$ at fixed volume, suppose the initial concentrations are $a$ and $b$. If the net reaction proceeds forward and increases $[\mathrm{N_2O_4}]$ by $x$, an initial–change–equilibrium table is

Substitute the equilibrium row into $K_c=(b+x)/(a-2x)^2$. Check the direction with $Q_c$ first, and reject any solution that produces an impossible concentration or an undefined equilibrium expression.

Further reading

• OpenStax: Chemical Reaction Rates
• OpenStax: Rate Laws
• OpenStax: Equilibrium Constants
• OpenStax: Shifting Equilibria—Le Châtelier's Principle

Content reviewed: 9 September 2026.