Q: What does H2 Chemistry Notes: Topic 9 - Chemical Equilibria cover? A: Master equilibrium constants, Le Chatelier shifts, and quantitative problem-solving for Core Idea 3 (Chemical Equilibria) in the 2026 H2 Chemistry syllabus.
Equilibrium mastery blends conceptual understanding with algebraic manipulation. This note covers Kc, Kp, reaction quotient reasoning, and Le Chatelier justifications tailored for the 2026 exam style.
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Status: SEAB's current H2 Chemistry (9476) syllabus PDF is labelled for 2026, and the current Chemistry Data Booklet is labelled 8873/9476/9813 for use from 2026 in non-practical papers. Core Idea 3 Topic 9 is assessed across Papers 1-3.
The core idea is simple: Equilibrium is a balance of forward and reverse rates, not a stopped reaction.
Use it as a working check: Write the correct Kc or Kp expression first. Exclude pure solids and liquids, then use Q versus K to decide direction.
Then go one layer deeper: Example: if Q<K, there are too few products for the current conditions, so the system shifts right until the ratio reaches K.
Quick revision box
What this topic tests: Kc/Kp manipulation, reaction quotient, Le Chatelier, and equilibrium calculations.
Top mistakes to avoid: Wrong equilibrium expression terms; not checking state symbols; qualitative shifts without quantitative support.
20-minute sprint plan: 5 min Kc/Kp expression recall; 10 min equilibrium math; 5 min shift + Q vs K reasoning.
1 Defining Equilibrium Constants
For reaction aA+bB⇋cC+dD:
Kc=[A]a[B]b[C]c[D]d
Exclude pure solids and liquids activity=1. For gas-phase reactions:
Kp=PAaPBbPCcPDd,Pi=partial pressure
Relationship: Kp=Kc(RT)Δn, where Δn=(c+d)−(a+b).
Use R=8.31J⋅K−1⋅mol−1 from the SEAB Chemistry Data Booklet when converting between Kc and Kp, and keep units consistent (Pa or bar) as specified in the question.
Equilibrium-expression state checkpoint
Before substituting numbers into Kc or Kp, filter each species by state symbol. This prevents a correct algebra method from being built on the wrong expression.
Species state in equation
Include in Kc?
Include in Kp?
Common trap
Aqueous, (aq)
Yes, as concentration.
No, because Kp is for gas partial pressures.
Putting aqueous ions into a gas-pressure expression.
Gas, (g)
Yes, as concentration if using Kc.
Yes, as partial pressure if using Kp
Pure liquid, (l)
No, its activity is treated as constant.
No.
Including HX2O(l)
Pure solid, (s)
No, its activity is treated as constant.
No.
Including a catalyst or precipitate in the denominator.
Worked check: for CaCOX3(s)⇌CaO(s)+COX2(g), the only term in Kp is PCO2. The solids are present in the equilibrium mixture, but they do not appear in the expression.
Misconception check: excluding a pure solid or liquid from K does not mean it is unimportant to the reaction. It means its activity is constant in the equilibrium expression.
Kp partial-pressure setup checkpoint
For gas equilibria, do not place total pressure directly into every term. Work out each gas partial pressure first.
Given information
First move
Partial-pressure link
Common trap
Moles of each gas and total pressure
Find total moles, then mole fraction for each gas.
Pi=xiPtotal
Using the same total pressure for every gas in Kp.
Equilibrium mole amounts from an ICE table
Add all gas moles at equilibrium before finding mole fractions.
xi=ni/ntotal
Using initial mole fractions after the composition has changed.
Partial pressures are already given
Substitute those partial pressures directly.
Keep all pressures in the same unit.
Mixing Pa, bar, and atm in one expression.
Question gives concentrations instead
Use Kc, or convert with the stated temperature if required.
Kp=Kc(RT)Δn
Misconception check: Kp is built from partial pressures, not from total pressure alone. Total pressure only becomes useful after it is split by mole fraction.
2 Reaction Quotient Q
Computed using same expression as K but with initial concentrations/pressures. Compare with K:
If Q<K, reaction shifts right to reach equilibrium.
If Q>K, reaction shifts left.
If Q=K, system already at equilibrium.
State this logic explicitly in answers; include direction of shift and rationale.
Q versus K decision map
Use this map before writing the ICE table, because it fixes the signs in the change row.
Comparison
Mixture has too much of
Shift needed
ICE-table sign pattern
Q<K
Reactants relative to products
Forward, towards products
Reactants decrease; products increase
Q>K
Products relative to reactants
Reverse, towards reactants
Products decrease; reactants increase
Q=K
Neither side
No net shift
No change row needed
Worked sign check: for HX2(g)+IX2(g)⇋2HI(g), suppose Q<K. The mixture needs more HI, so the change row must be
Stage
HX2
IX2
HI
Change
−x
−x
+2x
Do not choose the signs by where the larger number appears in the data table. Choose them from the shift direction, then use the balanced equation to set the coefficients.
3 Le Chatelier's Principle
When systems at equilibrium are disturbed, they shift to counteract the change. Provide particle-level reasoning:
Pressure/volume changes: for gases, stress number of moles Δn.
Temperature: consider exothermic/endothermic direction using ΔH.
Catalyst: no effect on position; only speeds reaching equilibrium.
Include quantitative context when data available (e.g. new equilibrium constant values).
Le Chatelier disturbance checkpoint
Before writing the explanation, sort the disturbance into the right lane. This keeps the answer from saying K changes when only the equilibrium position changes.
Disturbance
First question to ask
What changes immediately
What to say about K
Add or remove a reactant or product
Which side has been made too large in Q?
Concentration or partial pressure of that species
K is unchanged at the same temperature.
Compress or expand a gaseous mixture
Which side has fewer gas moles?
Total pressure and partial pressures
K is unchanged at the same temperature.
Change temperature
Is the forward reaction exothermic or endothermic?
The value of K and the equilibrium composition
K changes because temperature changes.
Add a catalyst
Is the system already at equilibrium?
Rate of forward and reverse reactions
K is unchanged; position is unchanged.
Fast check: if your sentence says "the system shifts to increase K", rewrite it. At fixed temperature, the system shifts so that Q moves back towards the existing K.
Temperature and K direction checkpoint
Temperature is the disturbance that changes the equilibrium constant. Treat heat like a reagent first, then translate the shift into how K changes.
Forward reaction type
Temperature change
Shift direction
What happens to K
Exothermic forward reaction
Increase temperature
Reverse, towards reactants
K decreases because the product-to-reactant ratio at the new equilibrium is smaller.
Exothermic forward reaction
Decrease temperature
Forward, towards products
K increases because the product-to-reactant ratio at the new equilibrium is larger.
Endothermic forward reaction
Increase temperature
Forward, towards products
K increases because the product-to-reactant ratio at the new equilibrium is larger.
Endothermic forward reaction
Decrease temperature
Reverse, towards reactants
K decreases because the product-to-reactant ratio at the new equilibrium is smaller.
Worked check: for NX2(g)+3HX2(g)⇋2NHX3(g), the forward reaction is exothermic. Raising temperature favours the reverse direction, so the equilibrium mixture contains proportionally less NHX3 and the value of K is smaller at the higher temperature.
Misconception check: concentration, pressure, and catalyst changes can shift position or change rate at fixed temperature, but they do not change K. Temperature changes K because the forward and reverse directions are affected differently by heat.
4 ICE Table Method
Use Initial-Change-Equilibrium tables for calculations. Example for NX2OX4(g)⇋2NOX2(g):
Stage
NX2OX4
NOX2
Initial
C
0
Change
−x
+2x
Equilibrium
C−x
2x
Plug into Kc=C−x(2x)2. Solve for x (often via quadratic). Check reasonableness (concentration cannot be negative).
ICE root sanity checkpoint
After solving for x, check the answer against the physical limits in the ICE table before accepting it.
Situation after solving
What to check
What to do in the answer
Common trap
Two algebraic roots appear
Substitute each root into every equilibrium row.
Reject any root that gives a negative amount, concentration, or partial pressure.
Keeping the larger root just because it came from the calculator first.
The shift consumes a limiting reactant
Compare x with the smallest allowed change from the reactant row.
State the allowed range, then choose only a root inside it.
Letting 0.200−2x become negative in a forward shift.
A small-x approximation is used
Check the percentage change from the initial value.
Keep the approximation only if the change is negligible for the required precision.
Assuming small K or large K always makes every change negligible.
Final value looks plausible but units are mixed
Check whether the table used concentration or partial pressure throughout.
Keep Kc tables in concentration terms and Kp tables in partial-pressure terms.
Solving a hybrid table with both mol⋅L−1
Worked check: in a forward shift with an equilibrium row [SOX2]=0.200−2x, the root must satisfy 0.200−2x>0, so x<0.100. A calculator root of x=0.132 is rejected before any final concentrations are reported.
5 Worked Example
Question:2SOX2(g)+OX2(g)⇋2SOX3(g) has Kc=280 at 700K. If [SO2]=[O2]=[SO3]=0.200mol⋅L−1 initially, determine the equilibrium concentrations.
Solution:
Compute Q: Result:Q=(0.200)2(0.200)(0.200)2=0.008000.0400=5.00 Since Q<Kc, position shifts to right.
ICE table with change x:
Stage
SOX2
OX2
SOX3
Initial
0.200
0.200
0.200
Change
−2x
−x
+2x
Equilibrium
0.200−2x
0.200−x
0.200+2x
Substitute into Kc:
280=(0.200−2x)2(0.200−x)(0.200+2x)2
Rearrange and solve for x. The cubic simplifies to 280(0.200−2x)2(0.200−x)−(0.200+2x)2=0. Solving (by iteration) gives x=0.0714 (3 s.f.).
Equilibrium concentrations:
[SOX2]=0.200−2x=0.0571mol⋅L−1
[OX2]=0.200−x=0.1286mol⋅L−1
[SOX3]=0.200+2x=0.3429mol⋅L−1
Check:
(0.0571)2(0.1286)(0.3429)2≈2.80×102
The large Kc value drives equilibrium heavily towards SOX3, leaving only modest amounts of SOX2.
(When calculators with equation solvers are unavailable, apply successive substitution or quadratic rearrangement; show the method used to secure reasoning marks.)
6 Industrial Applications
6.1 Haber Process NX2+3HX2⇋2NHX3
Exothermic ΔH<0; lower temperature favours NH3 but reduces rate.
High pressure favours fewer moles (forward reaction).
Use iron catalyst with promoters to increase rate.
Because NHX3 is the desired product, process design emphasises efficient downstream removal to keep shifting the equilibrium mixture toward products.
6.2 Contact Process SOX2+21OX2⇋SOX3
Exothermic; uses 450∘C and VX2OX5 catalyst.
Excess oxygen drives forward reaction.
Remove SO3 as formed to shift equilibrium (Le Chatelier).
Sulfur dioxide: oxidised reactant in the Contact-process equilibrium.
Sulfur trioxide: product species removed to increase conversion.
Treat this as an equilibrium-engineering pattern: control temperature, use catalyst for rate, and remove product species to maximise yield.
7 Common Misconceptions
Assuming catalysts change K; they do not.
Forgetting temperature is the only condition affecting K.
Failing to square concentrations when stoichiometric coefficient > 1.
Using Kc with partial pressures without converting.
8 Quick Drills
For PClX5(g)⇋PClX3(g)+ClX2(g), given Kp=0.012 at 500K and initial PClX5 pressure 1.00bar, calculate equilibrium partial pressures.
Explain, using Le Chatelier's principle, how increasing temperature affects equilibrium yield of methanol in CO+2HX2⇋CHX3OH
A mixture contains 0.300molNX2 and 0.300molHX2
Common exam mistakes
Mistake: Writing "the equilibrium shifts to increase Kc" or "to decrease Kc" - Kc is fixed at a given temperature and never shifts; only the position of equilibrium shifts in response to concentration or pressure changes.
Mistake: Including pure solids or pure liquids in the equilibrium expression - their activities are defined as 1 and must be omitted from Kc and Kp.
Mistake: Forgetting to raise concentrations to the power of the stoichiometric coefficient - for 2NOX2, the term is [NOX2]2
Mistake: Assuming a catalyst shifts the equilibrium position - catalysts speed up attainment of equilibrium but do not change Kc, Kp, or the equilibrium concentrations.
Mistake: Mixing Kc (concentration) with Kp (partial pressure) in the same expression - they are distinct constants related by Kp=Kc(RT)Δn
Mistake: Stating "increasing pressure increases yield" without checking Δn - pressure only shifts equilibrium if there is a net change in moles of gas (Δn=0).
Mistake: Stopping an ICE table calculation without checking that equilibrium concentrations are all positive - a negative concentration signals an arithmetic error.
Frequently asked questions
What is the only condition that changes the value of Kc? Temperature is the only factor that alters Kc. Changing concentration, pressure, or adding a catalyst shifts the position of equilibrium but leaves Kc unchanged.
When should I use Kp instead of Kc? Use Kp when the question gives or asks for partial pressures of gases. If concentrations in mol⋅L−1 are given, use Kc. You can interconvert using Kp=Kc(RT)Δn with R=8.31J⋅K−1⋅mol−1 from the data booklet.
How do I decide which direction the reaction shifts? Calculate the reaction quotient Q using the current concentrations or pressures. If Q<K, the reaction shifts to the right (towards products); if Q>K, it shifts to the left (towards reactants).
How do compromise industrial conditions relate to equilibrium principles? Industrial processes like the Haber process use moderate temperatures (around 450∘C) and high pressures (around 200atm) because the optimal equilibrium conditions (low temperature for high yield) would be too slow without a catalyst. The operating conditions are a practical trade-off between yield, rate, and cost.
Struggling with Chemical Equilibria? Our H2 Chemistry tuition programme covers this topic with structured practice, Paper 4 practical drills, and worked exam solutions.
Use these frameworks to articulate equilibrium reasoning with confidence.
