Did you know that the units reported for a concentration-based equilibrium constant, Kc, depend on the balanced chemical equation? The classroom expression for H₂(g) + I₂(g) ⇌ 2HI(g) has no residual concentration units, while the usual expression for the Haber reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) gives units of dm⁶ mol⁻² when concentrations are measured in mol dm⁻³. That difference comes directly from the powers in the equilibrium expression.
For JC1 and JC2 students searching for an H2 Chemistry tutor in Hougang, Kc equilibrium constant units, A-Level Chemistry 9476 chemical equilibrium, Haber process Kc calculations, mole concept practice or Kc and Kp worked questions, the important skill is not memorising one unit. It is writing the correct balanced equation, deriving its Kc expression, then simplifying the concentration powers.
I would ask students to solve three reactions side by side. Why does one Kc have no units, another have concentration units and another have inverse concentration-squared units? If they can explain that without a reference sheet, they are reading equilibrium mathematics as a physical model rather than just pushing numbers through a calculator.
We will use carefully labelled hypothetical data, explain the important scientific distinction between textbook concentration-form Kc and rigorous dimensionless thermodynamic equilibrium constants, and finish with changed questions that can be applied across JC Chemistry chapters.
The eduKate teaching approach is to begin with the first unstable idea, rebuild it in a small tutorial, and check independent performance on unfamiliar problems. Families who are comparing three-student groups and other teaching formats should verify the available H2 Chemistry lessons, actual location and commuting time directly. This Hougang learning guide is not a claim that our physical teaching venue is in Hougang.
Hougang Chemistry learning library · SEAB A-Level H2 Chemistry 9476 syllabus · eduKate Singapore tuition enquiries
Begin with the balanced equation, not the numerical answer
For the simple reversible reaction:
aA + bB ⇌ cC + dD,
the familiar dilute-solution concentration-form equilibrium expression is written:
Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ),
provided the activities represented by these concentration terms are appropriate and pure solids or pure liquids have been omitted in the usual school convention.
The powers come from the stoichiometric coefficients of the balanced reaction. We cannot alter a coefficient for convenience and retain the same mathematical definition of Kc.
The square brackets mean equilibrium concentration, not any arbitrary initial value. Concentrations in an A-Level Kc calculation are typically given in mol dm⁻³.
So every bracket contributes a unit factor of mol dm⁻³. When we multiply and divide bracket terms, their units undergo the same index operations as the concentrations themselves.
That is the entire principle behind the different units.
The shortcut: count the relevant concentration powers
For an equilibrium expression involving appropriate dissolved or gaseous species, add the powers on the product side and subtract the powers on the reactant side.
Define the concentration-power difference:
Δn = total product concentration powers − total reactant concentration powers.
In the simplified classroom concentration convention:
Units of Kc = (mol dm⁻³)^(Δn).
If Δn = 0, the concentration units cancel.
If Δn = +1, the units are mol dm⁻³.
If Δn = −2, the units are (mol dm⁻³)⁻² = dm⁶ mol⁻².
This shortcut is useful, but it must be applied to the species that actually appear in Kc. Do not count a pure solid or pure liquid as a concentration factor merely because its coefficient appears in the balanced equation.
The full expression should always be written before simplifying units, especially if you are still learning the process.
Worked example 1: H₂ + I₂ ⇌ 2HI
Consider the hypothetical gas equilibrium:
H₂(g) + I₂(g) ⇌ 2HI(g).
Its textbook concentration expression is:
Kc = [HI]² / ([H₂][I₂]).
What happens to the units?
Numerator: (mol dm⁻³)².
Denominator: (mol dm⁻³) × (mol dm⁻³) = (mol dm⁻³)².
The numerator and denominator have the same concentration units, so they cancel.
Kc has no residual concentration units in this expression.
The power difference makes this obvious: two product concentration powers minus one plus one reactant powers gives:
Δn = 2 − (1 + 1) = 0.
Notice that a unitless value did not arise because HI is a special gas or because the reaction is reversible. It arose because the exponents happened to cancel for this balanced reaction.
A numerical H₂–I₂ example
Suppose the equilibrium concentrations at a stated temperature are:
- [H₂] = 0.200 mol dm⁻³.
- [I₂] = 0.200 mol dm⁻³.
- [HI] = 0.600 mol dm⁻³.
These are illustrative values for a calculation, not experimental measurements.
Substitute:
Kc = (0.600)² / [(0.200)(0.200)].
Therefore:
Kc = 0.360/0.0400 = 9.00.
Because the concentration units cancel, the value is 9.00 with no residual units in this classroom expression.
An answer of 9.00 mol dm⁻³ would attach a unit that is not supported by the algebra.
I would ask students to make the unit cancellation visible in the working at least once before using any shortcut.
Worked example 2: the Haber equilibrium
The Haber reaction is:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
The concentration expression is:
Kc = [NH₃]² / ([N₂][H₂]³).
The numerator has total concentration power 2.
The denominator has power 1 + 3 = 4.
Therefore:
Δn = 2 − 4 = −2.
The units are:
(mol dm⁻³)⁻² = dm⁶ mol⁻².
The expression has units in the familiar concentration-based calculation because the product and reactant concentration powers do not cancel.
We do not say that the Kc units are “two moles of ammonia”. The powers apply to the concentrations and their units, not merely to a count of gas molecules.
A numerical Haber-process Kc calculation
Suppose a hypothetical equilibrium mixture at a specified temperature has:
[N₂] = 0.100 mol dm⁻³, [H₂] = 0.300 mol dm⁻³ and [NH₃] = 0.0500 mol dm⁻³.
Use:
Kc = [NH₃]² / ([N₂][H₂]³).
Substitute:
Kc = (0.0500)² / [(0.100)(0.300)³].
Calculate powers before division:
(0.0500)² = 0.00250.
(0.300)³ = 0.0270.
(0.100)(0.0270) = 0.00270.
Thus:
Kc = 0.00250/0.00270 ≈ 0.926 dm⁶ mol⁻².
This is an internally consistent example built from imaginary equilibrium concentrations. Actual equilibrium values depend on temperature and the real system.
The important learning check is that both the numerical exponents and the units were processed using the same balanced reaction coefficients.
Worked example 3: one gas molecule becomes two
Consider another hypothetical equilibrium:
N₂O₄(g) ⇌ 2NO₂(g).
Its school concentration expression is:
Kc = [NO₂]²/[N₂O₄].
The product contributes concentration power 2; the reactant contributes power 1.
Therefore:
Δn = 2 − 1 = +1.
The units are:
mol dm⁻³.
If the equilibrium concentrations were [NO₂] = 0.0200 mol dm⁻³ and [N₂O₄] = 0.0100 mol dm⁻³, then:
Kc = (0.0200)²/0.0100 = 0.0400 mol dm⁻³.
This is not a contradiction with the previous example. The written reaction has a different concentration exponent balance.
Notice that the sign of Δn predicts whether the units contain a positive concentration power or its reciprocal.
What about a reaction containing pure solids?
Consider the commonly used heterogeneous equilibrium:
CaCO₃(s) ⇌ CaO(s) + CO₂(g).
For the idealised coexistence of the relevant pure solid phases, the concentration expression is:
Kc = [CO₂].
The pure solids do not appear as variable concentration terms because their thermodynamic activities are constant in the ordinary model.
Consequently:
Units of Kc = mol dm⁻³ under this classroom concentration form.
A student who counts all species in the balanced equation as though they contribute a bracket would get a misleading unit result. The right order is omit pure solids and pure liquids from the conventional expression, then count the powers that remain.
This is another reason not to use a shortcut before writing the actual chemistry.
A comparison table of useful examples
| Balanced reaction | Textbook Kc expression | Δn | Concentration-form units |
|---|---|---|---|
| H₂ + I₂ ⇌ 2HI | [HI]²/([H₂][I₂]) | 0 | None |
| N₂ + 3H₂ ⇌ 2NH₃ | [NH₃]²/([N₂][H₂]³) | −2 | dm⁶ mol⁻² |
| N₂O₄ ⇌ 2NO₂ | [NO₂]²/[N₂O₄] | +1 | mol dm⁻³ |
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | [CO₂] | +1 | mol dm⁻³ |
| CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) | [CO₂][H₂]/([CO][H₂O]) | 0 | None |
The table is not intended as a list of five constants to memorise. In an unfamiliar question, reproduce the method and derive the units directly.
One practical improvement is to underline every state symbol before writing Kc. That prevents accidental inclusion of a pure solid and reminds you whether H₂O is a gas or a pure liquid in the particular reaction.
The H₂O(g) versus H₂O(l) trap
Water illustrates why state symbols matter. Consider the gas-phase equilibrium:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g).
Here H₂O is water vapour, which appears as a variable gas concentration in Kc. The concentration powers total 2 on each side, so the units cancel.
Now imagine a different equilibrium involving pure liquid water, H₂O(l), as a reactant. In the usual activity-based treatment, the pure liquid has constant activity and is omitted from the concentration expression.
Therefore you cannot use the rule “water is never included in Kc”. The correct rule is that pure liquid water is omitted, while water vapour may be included when it is a relevant gaseous species in the reaction.
Chemistry equations give us phase information for a reason. Ignoring (g) versus (l) can change both the equilibrium expression and its units.
What happens to Kc units when the equation is reversed?
Suppose the original reaction is:
N₂O₄(g) ⇌ 2NO₂(g).
Its concentration-form Kc has units mol dm⁻³.
Reverse the written reaction:
2NO₂(g) ⇌ N₂O₄(g).
The new expression is:
Kc,reverse = [N₂O₄]/[NO₂]².
Its concentration-power difference is now 1 − 2 = −1.
So its units are:
(mol dm⁻³)⁻¹ = dm³ mol⁻¹.
The numerical value of the reversed concentration-form Kc is the reciprocal of the original value, and the unit factor is also inverted.
For the hypothetical original Kc = 0.0400 mol dm⁻³, the reverse constant is:
Kc,reverse = 1/(0.0400 mol dm⁻³) = 25.0 dm³ mol⁻¹.
This example shows that the units change with the mathematical definition of the balanced reaction, not merely with temperature.
What happens when a reaction equation is doubled?
Again begin with N₂O₄ ⇌ 2NO₂ and Kc = [NO₂]²/[N₂O₄].
Double its coefficients:
2N₂O₄(g) ⇌ 4NO₂(g).
The new expression is:
Kc,new = [NO₂]⁴/[N₂O₄]².
That is the square of the original expression.
Therefore:
Kc,new = Kc,original².
The units must be squared too:
(mol dm⁻³)² = mol² dm⁻⁶.
If the numerical original value is 0.0400 mol dm⁻³:
Kc,new = (0.0400 mol dm⁻³)² = 0.00160 mol² dm⁻⁶.
The apparent change is an algebraic result of writing the same equilibrium using a doubled reaction. It does not mean the physical vessel suddenly attained a different composition.
Why a bigger number does not necessarily mean “more products”
Students sometimes see a large Kc value and say the products must always dominate. That statement needs careful qualification.
For a specified reaction equation, in a specified direction, at a specified temperature, a larger equilibrium constant commonly signals a more product-favoured equilibrium in the relevant activity-based sense.
But Kc values for differently written reactions cannot be compared naively. Reversing the equation inverts K, and doubling the coefficients squares it. In the concentration-based convention, the units can also differ.
So comparing the numerical Kc of N₂O₄ ⇌ 2NO₂ with the numerical Kc of N₂ + 3H₂ ⇌ 2NH₃ as if they were identical kinds of dimensionless ratios can be misleading.
A careful answer first identifies the reaction, its direction and the expression being evaluated.
Important scientific distinction: rigorous K versus textbook Kc
You may encounter a Chemistry reference stating that an equilibrium constant is dimensionless. You may also encounter a school calculation giving Kc in dm⁶ mol⁻².
Both statements can make sense under different conventions.
In rigorous thermodynamics, equilibrium constants are formulated from dimensionless activities. A gas activity, for example, can be expressed relative to a standard pressure under an appropriate ideal-gas approximation; a solute activity can be represented relative to a standard concentration with corrections for non-ideality.
When each activity is dimensionless, the thermodynamic equilibrium constant formed from those activities is dimensionless too.
In familiar A-Level concentration-form calculations, Kc is often written directly as a ratio of concentration powers without explicitly dividing each concentration by a standard concentration. The resulting numerical expression can then have residual mol dm⁻³ units.
The practical school rule is: follow the definition, supplied data and unit convention in the question. Do not remove legitimate concentration-form units merely because an advanced source defines a dimensionless thermodynamic constant.
Understanding the distinction is useful for later university chemistry, but it should not make straightforward JC questions harder than necessary.
What about Kp?
Kp describes a gas equilibrium in terms of partial pressures rather than equilibrium molar concentrations.
Its pressure-form units can similarly depend on the sum of powers in the expression. If a question uses pressures in pascals, the corresponding pressure powers differ from a question using bars.
For the Haber process:
Kp = p(NH₃)²/[p(N₂)p(H₂)³] in the familiar pressure form.
The net pressure power is 2 − (1 + 3) = −2.
At H2 level, the syllabus expects students to deduce appropriate Kp expressions and work with partial-pressure data. Do not assume that deriving a general numerical conversion equation between Kp and Kc is compulsory; the revised 9476 syllabus explicitly states that treatment of the general relationship between Kp and Kc is not required.
That does not prevent students from using the stated ideal-gas and partial-pressure ideas where a given question supports them. It means study time should follow the actual published learning outcomes.
Why Kc units matter when checking calculations
Unit analysis is a useful error detector.
Imagine a student calculates the Haber Kc as:
[NH₃]/([N₂][H₂]),
omitting the required coefficient-derived powers. Their units would give an entirely different concentration exponent from the correct expression.
The error began in the chemical equation-to-expression step, not the arithmetic.
Similarly, if the correct expression should be dimensionless but the student reports mol dm⁻³, they may have forgotten to square a concentration or incorrectly inserted another species.
Units cannot tell you whether every numerical value is right, but they can flag inconsistencies early.
An exam repair table
| Student statement | First mistake | Repair question |
|---|---|---|
| Kc always has no units | Confused dimensionless thermodynamic K with textbook concentration-form units | What are the concentration powers in the supplied expression? |
| Kc always has mol dm⁻³ units | Assumed one concentration remains in every case | Do numerator and denominator powers cancel? |
| Haber Kc has mol dm⁻³ units | Miscounted total exponents | What is 2 − (1 + 3)? |
| Pure CaCO₃ belongs in the denominator | Ignored constant pure-solid activity | Which species are variable gases or solutes? |
| H₂O is always omitted | Ignored the water state symbol | Is the water (l) or (g)? |
| Reversing an equation keeps the same Kc units | Forgot reciprocal changes both value and unit factor | What happens when numerator and denominator swap? |
| Double equation, double Kc | Ignored exponent rules | Does the whole original expression become squared? |
Seven short changed questions
- Derive the concentration-form units of Kc for H₂ + I₂ ⇌ 2HI.
- Derive the units for N₂ + 3H₂ ⇌ 2NH₃.
- Derive the units for N₂O₄ ⇌ 2NO₂.
- At the same temperature, reverse N₂O₄ ⇌ 2NO₂. What happens to the numerical value and its units?
- Write the Kc expression for CaCO₃(s) ⇌ CaO(s) + CO₂(g) and state its concentration-form unit.
- Why does H₂O(g) appear in Kc for CO + H₂O ⇌ CO₂ + H₂ while pure H₂O(l) is generally omitted?
- If Kc = 0.250 mol dm⁻³ for A ⇌ 2B in a simplified ideal-gas model, what is Kc for 2A ⇌ 4B?
Answers
1. No residual units. 2. (mol dm⁻³)⁻² = dm⁶ mol⁻². 3. mol dm⁻³. 4. Reciprocal numerical value and reciprocal unit factor, dm³ mol⁻¹. 5. Kc = [CO₂], with mol dm⁻³ under the classroom convention. 6. Water vapour is a variable gas-phase species while pure liquid water has constant activity in the model. 7. (0.250 mol dm⁻³)² = 0.0625 mol² dm⁻⁶.
A student who can explain Question 7 correctly has learnt more than a list of Kc units. They can transform both a chemical equation and its associated mathematical quantity consistently.
A weekly JC study routine
- Session 1: balance five reversible equations and mark every species’ phase.
- Session 2: write Kc expressions without any numbers.
- Session 3: assign concentration powers, simplify their units and predict whether they cancel.
- Session 4: calculate a Kc value from equilibrium concentrations, including the correct units.
- Session 5: reverse and double one equation, then transform Kc and its units.
- Session 6: compare a homogeneous gas equilibrium with a heterogeneous system containing pure solids.
- Session 7: solve a mixed Kc/Kp question without chapter notes, being precise about the supplied definitions and units.
For JC1, these steps strengthen algebra and stoichiometry. For JC2, they prevent avoidable mistakes in equilibrium, solubility and integrated problem-solving where the question may hide the formula within a longer story.
What parents choosing H2 Chemistry support around Hougang can ask
Bring a marked equilibrium calculation and ask the tutor whether the student’s error arose from the balanced equation, state symbols, choice of equilibrium rather than initial concentrations, exponent algebra, or concentration units.
These problems require different corrections. A useful three-student tutorial or one-to-one lesson should end with a changed reaction whose Kc expression and unit the student must deduce independently.
Confirm current H2 syllabus familiarity, lesson availability, actual venue and travel requirements directly; a Hougang-search-focused article does not establish that a physical tuition class is situated inside Hougang.
Why this matters beyond examinations
Scientists use dimensional analysis to check models in engineering, physical chemistry and materials research. When a numerical result acquires an unexpected unit, it may reveal that a concentration, pressure or stoichiometric exponent has been omitted.
At university, the distinction between a concentration-form expression and a properly normalised activity-based thermodynamic equilibrium constant becomes more important. JC learners need not master every advanced convention now, but they can develop the transferable habit of checking what quantity is defined, how it is calculated and which units follow from the definition.
Frequently asked questions
Does Kc always have units?
In school concentration-form expressions, not always. Its reported units depend on the net concentration powers for the balanced reaction.
Why is Haber-process Kc written with dm⁶ mol⁻²?
The product concentration power totals 2 and the reactant powers total 4, giving (mol dm⁻³)⁻².
Can Kc be dimensionless?
Yes. The units cancel in some textbook concentration expressions. Rigorous thermodynamic equilibrium constants are separately defined from dimensionless activities.
Do solids and liquids change the Kc units?
Pure solids and pure liquids are omitted from the usual activity-derived equilibrium expression, so they do not contribute variable concentration powers.
What happens to Kc units when I reverse the equation?
The units invert along with the equilibrium constant expression, under a consistent concentration convention.
Is the Kp–Kc conversion required for H2 Chemistry 9476?
The revised syllabus covers Kc and Kp expressions and calculations but says that treatment of the general relationship between Kp and Kc is not required.
Continue learning
Why reversing and doubling equilibria change Kc · Why pure solids are omitted from Kc · Why Kp uses partial pressures · Why Qc is not Kc until equilibrium · Hougang Chemistry Tutor Library · eduKate Singapore consultations.
All numerical equilibrium compositions are illustrative. Use the question’s given temperature, units, equilibrium definitions and examination-year syllabus conventions in graded work.