Year 12

AQA A level Chemistry Equilibria Notes

Equilibria

— Chemistry · AS Level (7405) · Le Chatelier, Kc & Kp —
★ Exam Tip: Equilibria appears in both MCQs and structured questions. Learn: Le Chatelier answers (temperature → enthalpy sign, pressure → moles of gas), rate vs yield, and how to set up an I-C-E table for Kc and Kp with units. A catalyst never changes yield or K.

1. Reversible Reactions & Dynamic Equilibrium

Reversible reaction: a reaction that can proceed both forwards and backwards as per the conditions applied (shown by ⇌). Equilibria only apply to reversible reactions.

e.g. H₂ + I₂ ⇌ 2HI is reversible — it never goes to completion, hence the ⇌ sign.

Dynamic equilibrium: established when the rate of the forward reaction equals the rate of the backward reaction.

Two conditions:

  • The reaction must be reversible.
  • The system must be closed → energy can be exchanged in and out of the system, but matter cannot.
ratetimeforward rate ↓backward rate ↑ratef = ratebequilibrium
reactants only at start: forward fast, backward zero; they meet at equilibrium and stay equal (both ≠ 0 — "dynamic")

At equilibrium nothing looks like it changes (constant concentrations) but both reactions are still going on at the same rate.

2. Le Chatelier's Principle

If a change is made to a system at equilibrium, the equilibrium shifts to oppose that change.
  • Equilibrium shifts to the right ⇒ concentration of products increases.
  • Equilibrium shifts to the left ⇒ concentration of reactants increases.
# Change Equilibrium shifts Result
1 [reactants] ↑ to the right products ↑, reactants ↓
2 [reactants] ↓ to the left reactants ↑, products ↓
3 [products] ↑ to the left reactants ↑, products ↓
4 [products] ↓ to the right products ↑, reactants ↓
5 temperature ↑ the endothermic reaction proceeds heat absorbed
6 temperature ↓ the exothermic reaction proceeds heat released
7 pressure ↓ side with more moles of gas ↑ no. of particles → ↑ collisions → ↑ pressure
8 pressure ↑ side with fewer moles of gas ↓ no. of particles → ↓ collisions → ↓ pressure
Board correction: in row 2 the board line reads "reactants increase, products increase" — the second part should be products decrease (as in row 3).
Rate vs yield (very important): Yield depends on the position of equilibrium. Rate depends on temperature (etc.). Raising T increases the rate of both forward and backward reactions regardless of which is exothermic — but it moves the position towards the endothermic side, so yield of an exothermic forward reaction decreases.
Watch: equal moles of gas on both sides (e.g. H₂ + I₂ ⇌ 2HI) ⇒ changing pressure has no effect on the position. And if a reaction is not reversible (goes to completion) Le Chatelier does not apply — changing conditions shifts nothing.

3. Choosing Conditions — Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH = −92 kJ mol⁻¹
4 mol gas → 2 mol gas: forward reaction exothermic (and fewer moles); backward reaction endothermic
N₂ + 3H₂feed incompress200 atmreactor: iron450 °Ccool: NH₃condenses (l)unreacted N₂ + H₂ recycled back
NH₃ has the higher b.p. (H-bonding) so it condenses first and is removed → equilibrium keeps shifting right
① Temperature

Use a lower temperature as the forward reaction is exothermic → we use 450 °C.
→ Compromise between yield and rate. Any lower: yield is good but the rate of reaction is too slow.

② Pressure

Use a high pressure as there are fewer moles of gas on the product side → we use 200 atm.
→ Going higher increases the cost of maintenance and machinery (vessels/pipes, safety, money) — good yield, good rate at 200 atm.

③ Remove NH₃ continuously

Pump/extract out ammonia → equilibrium shifts to the right → more ammonia produced. Easy to separate as ammonia condenses before H₂ and N₂ because it has a higher m.p. and b.p. (hydrogen bonding).

④ Feed N₂ + H₂ continuously

Pump in hydrogen and nitrogen continuously → equilibrium shifts to the right → more ammonia produced. (So true dynamic equilibrium is never really established in the plant.)

⑤ Iron catalyst

Activation energy decreases → rate of reaction increases. Yield does NOT increase.

Never write: "iron increases the yield of ammonia". It only makes ammonia faster. (The iron is used as a honeycomb bed for a large surface area.) Likewise for any catalyst.
Business reasoning: a factory wants to spend little and make a lot — so accept a compromise temperature/pressure rather than the "ideal" values (feasibility / cost of high-pressure equipment and hazards).

4. Choosing Conditions — Contact Process

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)   ΔH = −196 kJ mol⁻¹
3 mol gas → 2 mol gas: forward reaction exothermic; backward endothermic. (Makes H₂SO₄.)
① Temperature

Lower T favours the exothermic forward reaction → we use 450 °C: compromise between yield and rate; any lower, yield is good but rate is too slow.

② Pressure

High pressure would favour fewer moles of gas (right) but we still use only 1–2 atm: we already get 99.5 % yield at this pressure, and anything higher is useless as it only increases costs.

③ Separation

Not really necessary as we have 99.5 % yield already.

④ V₂O₅ catalyst

Activation energy decreases → rate increases (yield does not increase).

Haber Contact
Reaction N₂ + 3H₂ ⇌ 2NH₃ 2SO₂ + O₂ ⇌ 2SO₃
ΔH −92 −196
Temperature 450 °C 450 °C
Pressure 200 atm 1–2 atm
Catalyst iron V₂O₅
Extra remove NH₃ (condense), feed reactants no separation needed, 99.5 % yield

5. Past-Paper Questions — Le Chatelier

Q10 (V²⁺)MCQ

V²⁺ + H₂O ⇌ V³⁺ + ½H₂ + OH⁻. What would favour V²⁺?

Need a shift to the left → add OH⁻ (or remove a right-hand species is wrong).

A ✗ adding an acid removes OH⁻ → shifts rightB ✗ removing V³⁺ → shifts rightC ✗ H₂ escaping → shifts rightD ✓ more alkaline = more OH⁻ → shifts left

Answer: D

Q10 (methyl ethanoate)MCQ

CH₃CO₂CH₃ + H₂O ⇌ CH₃CO₂H + CH₃OH; hydrolysis slow in neutral solution but measurable in acid. Function of HCl?

HCl does not alter either side of the equilibrium — it only speeds it up: catalytic action (H⁺ catalyst for ester hydrolysis).

Answer: C

Q19 (pool water)MCQ

OCl⁻(aq) + H₂O ⇌ OH⁻(aq) + HOCl(aq). UV breaks down OCl⁻ → Cl⁻ + ½O₂ (this second reaction is not reversible). Highest [HOCl]?

Only the first is an equilibrium. Remove OH⁻ → shifts right → more HOCl: acidify (H⁺ + OH⁻ → H₂O).

B, D ✗ act on the complete (non-reversible) UV reaction, so no shift   C ✗ extra OH⁻ shifts LEFT

Answer: A

Q18 (Haber, T ↑)MCQ

N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92. Temperature increased — which is correct?

Rate depends on temperature: both forward and backward rates increase, whatever the equilibrium position does.

Answer: A

Q20 (N₂O₄ ⇌ 2NO₂)MCQ · ΔH = +57

T increased at constant pressure — volume of the mixture?

Forward endothermic → T↑ shifts right (absorbs heat) → 2 mol NO₂ from 1 mol N₂O₄ → more gas → volume ↑. Plus thermal expansion of gases. Both reasons.

A ✗ ignores thermal expansion   C, D ✗ shift is to the right, not left

Answer: B

Q32 (H₂ + I₂ ⇌ 2HI)MCQ · ΔH = +53

Which change makes the purple mixture paler (less I₂)?

Need shift right (uses up I₂). Forward is endothermic ⇒ increase T. Pressure: 2 mol gas each side ⇒ no effect.

A, C ✗ pressure has no effect   B ✗ lower T shifts left (more I₂)

Answer: D

% Zpressurelow Thigh T
Q22: same P, lower T gives higher yield ⇒ forward exothermic; higher P ⇒ higher yield ⇒ fewer moles of gas
Q22MCQ · yield graphs of X + Y ⇌ Z

Percentage yield of Z rises with pressure and is higher at lower temperature. Which statement is correct?

A ✗ lower T of an exothermic reaction increases KcB ✗ lower T decreases the rateC ✗ pressure never changes KcD ✓ the reaction is exothermic in the forward direction

Answer: D

% NH₃pressure400 °C500 °C
Q23 (graph A): both rise with P; the 400 °C curve is higher (exothermic ⇒ lower T = more NH₃)
Q23 (Haber graphs)MCQ

% NH₃ at equilibrium vs pressure for 400 °C and 500 °C.

Yield ↑ with pressure (4 → 2 mol) ⇒ curves rise. Exothermic ⇒ lower T (400 °C) higher yield ⇒ 400 °C curve above 500 °C.

B ✗ labels swapped   C, D ✗ yield decreasing with pressure

Answer: A

Q46 (2NO₂ ⇌ N₂O₄)MCQ · ΔH = −57

Conditions for greatest % of N₂O₄?

Right side has fewer moles ⇒ high pressure; forward exothermic ⇒ low temperature.

Answer: B (high P, low T)

Methanol · CO + 2H₂ ⇌ CH₃OHStructured Q2 · ΔH = −91 kJ mol⁻¹

(a)(i) State Le Chatelier's principle. If a change is made to a system at equilibrium, the equilibrium shifts to oppose that change. [2]

(ii) Conditions for higher yield, with explanation:

  • Temperature: lower. The forward reaction is exothermic, so it is favoured by lower T (it releases heat, opposing the fall in T).
  • Pressure: higher. There are fewer moles of gas on the product side (3 → 1), so higher pressure shifts right, increasing methanol yield and so decreasing the pressure.

6. The Equilibrium Constant Kc

For a reversible reaction wA + xB ⇌ yC + zD we define:

Kc = [C]yeqm [D]zeqm[A]weqm [B]xeqm
Where: [A]eqm = concentration of A in mol dm⁻³ at equilibrium | powers = the balancing numbers | products on top, reactants below

Writing Kc and deducing its units

(a) H₂ + I₂ ⇌ 2HI
Kc = [HI]²[H₂][I₂] ⇒ units (mol dm⁻³)²(mol dm⁻³)(mol dm⁻³) = no units
(b) 2H₂S ⇌ 2H₂ + S₂
Kc = [H₂]²[S₂][H₂S]² ⇒ units (mol dm⁻³)² × (mol dm⁻³)(mol dm⁻³)² = mol dm⁻³
(c) 2SO₂ + O₂ ⇌ 2SO₃
Kc = [SO₃]²[SO₂]²[O₂] ⇒ units (mol dm⁻³)²(mol dm⁻³)²(mol dm⁻³) = dm³ mol⁻¹
Units trick: count concentration terms on top vs bottom. Equal ⇒ no units; extra on top ⇒ mol dm⁻³ (to that power); extra on the bottom ⇒ dm³ mol⁻¹ (to that power).
  • A higher Kc usually means a high concentration of products, so equilibrium lies in favour of the right-hand side.
  • Solids have no concentration — never include them in Kc (liquid solutions and gases are included; a solid is left out completely). e.g. A(s) + 2B(g) ⇌ C(g): Kc = [C]/[B]².

Kc belongs to one version of the equation

Take NO + ½O₂ ⇌ NO₂ (Kc1),   2NO + O₂ ⇌ 2NO₂ (Kc2),   2NO₂ ⇌ 2NO + O₂ (Kc3):

Kc1 Kc2 Kc3
[NO₂][NO][O₂]½ [NO₂]²[NO]²[O₂] [NO]²[O₂][NO₂]²
RelationsKc1 = √Kc2 = √(1/Kc3)
Kc2 = 1/Kc3 = (Kc1)²
Kc3 = 1/Kc2 = 1/(Kc1)²
Rule: double the equation ⇒ K squared; halve ⇒ square root; reverse ⇒ 1/K.

7. Calculating Kc (I-C-E tables)

Method: Initial → Change (use the mole ratio; reactants −, products +) → Equilibrium (moles). Divide moles by volume to get concentrations, then substitute in Kc. If moles on both sides are equal the volume cancels.
H₂ + I₂ ⇌ 2HI

2.5 mol H₂ and 2.5 mol I₂ in a 4 dm³ vessel; equilibrium mixture contains 4 mol HI. Find Kc.

H₂ I₂ 2HI
Initial 2.5 2.5 0
Change −2 −2 +4
Equilibrium 0.5 0.5 4
Kc = [HI]² / ([H₂][I₂])= (4/4)² / ((0.5/4) × (0.5/4))Kc = 64   no units
X₂ + Y₂ ⇌ 2XY

0.5 mol X₂ and 0.5 mol Y₂ react; 0.6 mol XY at equilibrium. Find Kc.

X₂ Y₂ 2XY
I 0.5 0.5 0
C −0.3 −0.3 +0.6
E 0.2 0.2 0.6
Kc = (0.6/V)² / ((0.2/V)(0.2/V))V cancels (2 mol gas each side)Kc = 9   no units

Past-paper: Kc

Q1 — HI dissociationStructured

2HI ⇌ H₂ + I₂.

(a) Kc = [H₂]eqm[I₂]eqm[HI]²eqm

(b) At 120 °C: [HI] = 1.47, [H₂] = [I₂] = 0.274 mol dm⁻³.

Kc = [0.274][0.274] / [1.47]²Kc = 0.035 no units

(c) Why harder to find Kc at room temperature? Iodine is a solid at room temperature, so [I₂(g)] may be too small to measure; also the rate may be too slow, taking too long to reach equilibrium. [2]

Q1(b) — esterificationStructured

CH₃CO₂H + C₂H₅OH ⇌ CH₃CO₂C₂H₅ + H₂O

(i) Dynamic equilibrium: established when the rate of the forward reaction equals the rate of the backward reaction in a reversible reaction.

(ii) Kc = [CH₃CO₂C₂H₅][H₂O][CH₃CO₂H][C₂H₅OH]

(c) 6.0 g ethanoic acid + 6.0 g ethanol + 4.4 g ethyl ethanoate; at equilibrium 0.040 mol acid.

moles: acid 6.0/60 = 0.1; ethanol 6.0/46 = 0.13; ester 4.4/88 = 0.05; water 0
acid alcohol ester H₂O
initially 0.1 0.13 0.05 0
change −0.06 −0.06 +0.06 +0.06
equilibrium 0.040 0.07 0.11 0.06
Kc = (0.11 × 0.06) / (0.07 × 0.04)Kc = 2.4

(iii) Kc has no units because the units cancel out (equal number of concentration terms top and bottom).

Q3 — related equilibriaMCQ

I: 2X₂ + Y₂ ⇌ 2X₂Y, Kc = 2. II: X₂Y ⇌ X₂ + ½Y₂. Kc for II?

II = reverse of I and halvedKc(II) = 1/√21/√2

B ✗ reversed only ½ (no square root)   C ✗ reversed only (1/K squared)   D ✗ K cannot be negative

Answer: A

Q30 — Ag⁺ + Fe²⁺ ⇌ Ag(s) + Fe³⁺MCQ

1.0 mol AgNO₃ + 1.0 mol FeSO₄ in 1.00 dm³; 0.44 mol Ag⁺ at equilibrium.

Solid Ag is not included (left out of Kc).

Ag⁺ Fe²⁺ Ag(s) Fe³⁺
I 1 1 0 0
C −0.56 −0.56 +0.56 +0.56
E 0.44 0.44 — 0.56
Kc = [Fe³⁺] / ([Ag⁺][Fe²⁺])= 0.56 / (0.44 × 0.44)Kc = 2.89

Answer: D

Q48 — Haber, find [NH₃]MCQ

500 K, vessel 6.0 dm³; equilibrium 7.2 mol N₂, 12.0 mol H₂; Kc = 6.0 × 10⁻² at 500 K. [NH₃]?

Kc = [NH₃]² / ([N₂][H₂]³)[N₂] = 7.2/6 = 1.2; [H₂] = 12/6 = 26 × 10⁻² = x² / (1.2 × 2³)x² = 0.576x = 0.76 mol dm⁻³

Answer: B

conc.timeX: 0.5 → 0.25Y: 0 → 0.50.250.5
Q5: [X] falls to 0.25, [Y] rises to 0.5 (2 Y for each X), both level off ⇒ graph C, Kc = 1
Q5 — dimer X ⇌ 2YMCQ

X(aq) ⇌ 2Y(aq); [X] starts 0.50 mol dm⁻³ and falls to 0.25 at equilibrium.

X 2Y
I 0.5 0
C −0.25 +0.5
E 0.25 0.5
Kc = [0.5]² / 0.25Kc = 1 mol dm⁻³

Answer: C

8. The Equilibrium Constant Kp

For gases it is easier to work with pressures ⇒ Kp.

DefinitionsMole fraction of gas A = moles of gas Atotal moles in mixture
Partial pressure of A = mole fraction of A × total pressure
For wA + xB ⇌ yC + zD:  Kp = (pC)yeqm (pD)zeqm(pA)weqm (pB)xeqm
Note: the board's general expression has slightly mismatched powers; the correct pattern is: power = the coefficient of that gas in the balanced equation.
Where: (pA)eqm = partial pressure of gas A at equilibrium (Pa or atm). Sum of partial pressures = total pressure.
  • Don't include solids or liquids in Kp.
  • The value of an equilibrium constant remains constant if temperature is constant.
  • Exothermic: T ↑ ⇒ Kc and Kp ↓  |  Endothermic: T ↑ ⇒ Kc and Kp ↑.
  • Changing pressure or using a catalyst has NO effect on Kc and Kp.
Ktemperatureexo: K ↓ as T ↑endo: K ↑ as T ↑
only temperature changes K; K never depends on pressure or catalyst
Why? Higher T shifts the position towards the endothermic side. For an exothermic forward reaction that means fewer products ⇒ smaller K.

9. Calculating Kp

PCl₅ ⇌ PCl₃ + Cl₂

Total pressure 1 × 10⁶ Pa, 550 °C. At equilibrium: 0.3 mol PCl₅, 0.7 mol PCl₃, 0.7 mol Cl₂. Find Kp.

total moles = 0.3 + 0.7 + 0.7 = 1.7mole fractions: 0.3/1.7 ; 0.7/1.7 ; 0.7/1.7p = mole fraction × 1 × 10⁶ PaKp = p(PCl₃) × p(Cl₂)p(PCl₅)= (0.7/1.7 × 10⁶)² / (0.3/1.7 × 10⁶)   units Pa²/Pa = PaKp = 9.6 × 10⁵ Pa
Board correction: the board writes "0.17" as the total moles; it should be 1.7 (0.3 + 0.7 + 0.7). The answer 9.6 × 10⁵ Pa is right.
N₂O₄ ⇌ 2NO₂ (350 K)Structured (b)

2.00 mol N₂O₄ sealed at 350 K. At equilibrium total P = 140 kPa and 1.84 mol N₂O₄ present. (Equation: 2NO₂ ⇌ N₂O₄.)

(i) Kp = p(N₂O₄)eqmp(NO₂)²eqm

2NO₂ N₂O₄
I 0 2
C +0.32 −0.16
E 0.32 1.84

(ii) n(NO₂) = 0.32 mol.

(iii)total = 0.32 + 1.84 = 2.16 molx(NO₂) = 0.32/2.16 = 4/27x(N₂O₄) = 1.84/2.16 = 23/27
(iv)p(NO₂) = 4/27 × 140 × 10³ = 20 740 ≈ 21 000 Pap(N₂O₄) = 23/27 × 140 × 10³ = 119 259 ≈ 120 000 Pa
(v)Kp = 120 000 / (21 000)²Kp = 2.72 × 10⁻⁴ Pa⁻¹units = Pa / Pa² = Pa⁻¹
Note: using the unrounded pressures (119 259 and 20 740) gives 2.77 × 10⁻⁴ Pa⁻¹ — rounding p early changes the 3rd significant figure, so keep full digits until the end.
Contact process Q2Structured

(a) Roasting pyrites: 4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)  (the board wrote it unbalanced: FeS₂ + O₂ → Fe₂O₃ + SO₂)

(d)(i) Higher temperature: advantage — rates of the forward and backward reactions become fast, so equilibrium is reached faster. Disadvantage — the forward reaction is exothermic, so a higher T shifts equilibrium to the left and decreases the yield of SO₃. [4]

(ii) Kp = (pSO₃)²eqm(pSO₂)²eqm (pO₂)eqm

(iii) 2.00 mol SO₂ + 2.00 mol O₂; at equilibrium P = 2.00 × 10⁵ Pa and 1.80 mol SO₃.

2SO₂ O₂ 2SO₃
I 2 2 0
C −1.8 −0.9 +1.8
E 0.2 1.1 1.8
total = 0.2 + 1.1 + 1.8 = 3.1 molp(SO₂) = 0.2/3.1 × 2 × 10⁵; p(O₂) = 1.1/3.1 × 2 × 10⁵; p(SO₃) = 1.8/3.1 × 2 × 10⁵Kp = (1.8/3.1 × 2×10⁵)² / ((0.2/3.1 × 2×10⁵)² (1.1/3.1 × 2×10⁵))Kp = 1.14 × 10⁻³ Pa⁻¹

Note the question uses 400 °C for this step; the board's general conditions list 450 °C.

Haber Q1 — bond energy & KpStructured

N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹; N≡N = 944, H–H = 436.

(a)(i) Bond energy: the minimum energy required to break 1 mole of a bond in gaseous molecules.

(ii) N–H bond energyΔH = E(reactants) − E(products)−92 = 944 + 3(436) − 6(N–H)6(N–H) = 2344N–H = 390.67 ≈ 391 kJ mol⁻¹

(c) At 2.00 × 10⁷ Pa: 1.00 mol N₂ + 3.00 mol H₂; equilibrium has 0.300 mol NH₃.

N₂ 3H₂ 2NH₃
I 1 3 0
C −0.15 −0.45 +0.3
E 0.85 2.55 0.3
(i) N₂ = 0.85 mol; H₂ = 2.55 mol(ii) total = 0.85 + 2.55 + 0.3 = 3.70 molp(NH₃) = 0.3/3.7 × 2 × 10⁷p(NH₃) = 1.62 × 10⁶ Pa

(d) Partial pressures: N₂ 2.20 × 10⁶; H₂ 9.62 × 10⁵; NH₃ 1.40 × 10⁴ Pa. (i) Kp = (pNH₃)²(pN₂)(pH₂)³

Kp = (1.4 × 10⁴)² / ((2.2 × 10⁶)(9.62 × 10⁵)³)Kp = 1.00 × 10⁻¹⁶ Pa⁻²units = Pa² / (Pa × Pa³) = Pa⁻²

(iii) Same T, smaller container: pressure increases (P ∝ 1/V, T constant) ⇒ equilibrium shifts to fewer moles ⇒ yield of NH₃ increases; Kp — no effect (only T changes K).

Kp in terms of x and P (HI)MCQ

2HI ⇌ H₂ + I₂: start with b mol HI, x mol dissociate; total pressure P.

2HI H₂ I₂
I b 0 0
C −x +x/2 +x/2
E b−x x/2 x/2
total = (b − x) + x/2 + x/2 = bKp = (x/2b × P)² / ((b−x)/b × P)²P and b cancelKp = x² / 4(b − x)²

Answer: D

Q16 — N₂ + O₂ ⇌ 2NOMCQ · ΔH = +180

Which factors change Kp?

Only temperature. Pressure ✗, catalyst ✗ (they change rate/position, not K).

Answer: D (T ✓, pressure ✗, catalyst ✗)

10. Quick Recap — Full Chapter Cheat-Sheet

✓ Concepts
  • Dynamic equilibrium: rate forward = rate backward; needs a reversible reaction in a closed system (energy in/out, no matter).
  • Le Chatelier: the equilibrium shifts to oppose the change. Conc ↑ reactants → right; conc ↑ products → left. T ↑ → endothermic direction; T ↓ → exothermic direction. P ↑ → fewer gas moles; P ↓ → more gas moles.
  • Catalyst: ↓ Ea, ↑ rate, no change in yield, position or K. T ↑ always ↑ both rates.
  • Haber: 450 °C, 200 atm, iron, remove NH₃ (condenses, H-bonding) & recycle. Contact: 450 °C, 1–2 atm, V₂O₅, 99.5 % yield.
✓ Kc and Kp
  • Kc = [products]coef / [reactants]coef (mol dm⁻³, at eqm; omit solids). Kp = same with partial pressures (gases only).
  • Partial pressure = mole fraction × total pressure; mole fraction = moles A / total moles.
  • Units: work out by cancelling; count terms top vs bottom.
  • Reverse ⇒ 1/K; multiply equation by n ⇒ Kn; halve ⇒ √K.
  • K changes with temperature only: exo T↑ ⇒ K↓; endo T↑ ⇒ K↑. Pressure/catalyst: no effect. Large K ⇒ equilibrium on the right.
  • Always use an I-C-E table; if moles of gas are equal on both sides, volume cancels in Kc.