Spontaneity, the equilibrium constant, and the unit trap — keep this beside you.
Rearranged: ln K = −ΔG° / (R·T) → K = e(−ΔG°/RT)
R = 8.31 J K⁻¹ mol⁻¹ • T in kelvin • °C + 273
| ΔG° | Reaction | ln K | K | Favours |
|---|---|---|---|---|
| < 0 | spontaneous → | + | K > 1 | products |
| = 0 | at equilibrium | 0 | K = 1 | balanced |
| > 0 | non-spont. → | − | K < 1 | reactants |
The Monday mistake: a negative ΔG° means K > 1, not K < 1. Negative = downhill = products win.
ΔH° is in kJ. ΔS° is in J. They don't mix.
ΔS° (J) — ÷ 1000 → ΔS° (kJ) before subtracting.
R is in J, so ΔG° comes out in J mol⁻¹.
Answer (J) — ÷ 1000 → Answer (kJ mol⁻¹).
If R = 8.31, your raw answer is in JOULES. Always ÷ 1000 for kJ.
ΔH° = −92.6 kJ mol⁻¹ ΔS° = −199 J K⁻¹ mol⁻¹ΔG° = (−92.6) − (298 × −0.199)ΔG° = −92.6 + 59.3 = −33.3 kJ mol⁻¹Golden rule: add a stress and the system shifts to oppose / undo that stress.
Add a species → shift away from it.
Remove a species → shift toward it.
Kc does NOT change.
↑ pressure → shift to the side with fewer moles of gas.
↓ pressure → shift to more moles.
Count gas moles only. Kc does NOT change.
↑ T → shift in the endothermic direction.
↓ T → shift in the exothermic direction.
Temperature is the only change that moves Kc.
Speeds up forward and reverse equally. Reaches equilibrium faster but does not shift position and does not change K.
| Compare | Meaning | Shift |
|---|---|---|
| Q < K | too few products | → forward (right) |
| Q = K | at equilibrium | no shift |
| Q > K | too many products | ← reverse (left) |
For each question you lost a mark on: find your error type below, do the Fix-It move in red pen, then write one sentence saying what you now know.
Laptops stay shut until your teacher has seen your red-pen fixes and given you the go. The quiz proves the fix worked — it is not where the fixing happens.
On “3 — 2 — 1 — Show Me”, boards up together. If your units aren't boxed, the answer isn't finished.
Q. A system is at equilibrium. The temperature is increased. Kc increases. Is the forward reaction exothermic or endothermic?
One word is a 50/50 guess. Fill the logic chain — that's the IB mark.
Leave your board face-up on the desk as you leave.
While the HL group works through ΔG°, prove each rule below on your whiteboard. Same “Show Me” command, same layout — you hold up Kc manipulations.
| Do this to the equation… | …and this happens to Kc |
|---|---|
| Reverse the reaction | Kc → 1 / Kc |
| Multiply all coefficients by n | Kc → (Kc)n |
| Halve all coefficients | Kc → √Kc (Kc½) |
| Add two reactions together | Kc → K1 × K2 |
For N₂ + 3H₂ ⇌ 2NH₃, Kc = 50. Write Kc for the reverse reaction.
Answer on board: 1/50 = 0.02.
For the same forward reaction with all coefficients halved (½N₂ + 1½H₂ ⇌ NH₃), write Kc.
Answer on board: √50 ≈ 7.07.
• Changing concentration or pressure shifts the position but never changes Kc.
• ↑ pressure shifts toward fewer moles of gas — for N₂ + 3H₂ ⇌ 2NH₃ that's 4 mol → 2 mol, so toward NH₃.
• Pure solids and liquids are left out of the Kc expression.
Cut along the lines. For each STRESS card, place it next to the correct SHIFT. Turn & talk: say the rule out loud before you commit.
Self-check (fold under before sorting): Add reactant → right · Remove product → right · ↑P → fewer gas moles · ↑T → endothermic · ↓T → exothermic · Catalyst → no shift.