⚖️ Concept Overview
10.1–10.4 — Collision Theory & Reaction Rates
Collision Theory: For a reaction to occur, particles must collide with (1) correct orientation AND (2) sufficient energy ≥ activation energy (Eₐ).
- Reaction rate = change in concentration / change in time (mol/L·s)
- Rate stoichiometry: if A disappears twice as fast as B appears, use mole ratios from the equation
Factors that increase reaction rate:
- ↑ Temperature → more KE → more collisions with sufficient energy
- ↑ Concentration → more particles → more frequent collisions
- ↑ Surface area → more exposed particles → more collisions possible
- ↑ Pressure (gases) → particles closer together → more collisions
- Catalyst → lowers activation energy → more collisions succeed
- Simpler molecules / lower molecular complexity → easier to orient correctly
10.4 — Potential Energy (PE) Diagrams
- Reactants: starting energy level on left
- Activated complex: peak (highest point) — unstable transition state
- Products: ending energy level on right
- Activation energy (Eₐ): energy from reactants to peak (always positive)
- ΔH: difference between products and reactants energy
- Exothermic: products lower than reactants (ΔH < 0) → Eₐ reverse > Eₐ forward
- Endothermic: products higher than reactants (ΔH > 0) → Eₐ reverse < Eₐ forward
- Catalyst: lowers Eₐ for BOTH forward and reverse — does NOT change ΔH or product/reactant energy levels
10.5–10.11 — Equilibrium & Keq
- Completion reaction: goes to completion; single arrow (→). Products form until reactant is used up.
- Reversible reaction: double arrow (⇌); reactants and products both present at equilibrium.
- Dynamic equilibrium: forward and reverse rates are EQUAL. Concentrations are constant (not equal). Both reactions still occur.
- Only include aqueous (aq) and gas (g) species — pure solids (s) and liquids (l) are excluded
- Keq > 1: products-favored (more products than reactants at equilibrium)
- Keq < 1: reactants-favored (more reactants than products at equilibrium)
10.12 — Le Chatelier's Principle
Le Chatelier's Principle: When a system at equilibrium is stressed, it shifts to counteract the stress and re-establish equilibrium.
| Stress Applied | Direction of Shift |
|---|---|
| Add reactant | → Right (makes more products) |
| Remove reactant | ← Left (makes more reactants) |
| Add product | ← Left (makes more reactants) |
| Remove product | → Right (makes more products) |
| Increase pressure / decrease volume | Toward fewer moles of gas |
| Decrease pressure / increase volume | Toward more moles of gas |
| Increase temperature | Endothermic direction (treat heat as reactant/product) |
| Decrease temperature | Exothermic direction |
| Add catalyst | No shift — just reaches equilibrium faster |
10.12 — ICE Charts
ICE = Initial, Change, Equilibrium. Used to find equilibrium concentrations or Keq.
I: 0.500 / 0.500 / 0
C: −x / −x / +2x
E: 0.500−x / 0.500−x / 2x
- Change row signs follow stoichiometry: reactants decrease (−), products increase (+)
- Plug equilibrium row into Keq expression and solve for x
🃏 Flashcards
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⚡ Quick Review
- Collision theory: correct orientation + energy ≥ Eₐ → successful reaction
- Factors ↑ rate: ↑T, ↑concentration, ↑surface area, ↑pressure (gases), catalyst
- Catalyst: lowers Eₐ only — no change to ΔH or Keq
- PE diagram: peak = activated complex; Eₐ = reactants to peak; ΔH = reactants to products
- Keq expression: [products]^coeff / [reactants]^coeff (only aq and g species)
- Keq > 1: products-favored; Keq < 1: reactants-favored
- Le Chatelier: add → shift away from what you added; remove → shift toward
- Pressure ↑ → shift toward fewer gas moles
- Temperature ↑ → shift endothermic direction (changes Keq)
- Catalyst → no shift, just faster equilibrium
- ICE chart: I + C = E; Change uses stoichiometric ratios; plug E into Keq
- Dynamic equilibrium: rates equal, concentrations constant (not necessarily equal)
📝 Practice Quiz
20 multiple choice + 3 short answer.
1. According to Collision Theory, which TWO conditions must be met for a successful reaction?
2. Which factor does NOT directly increase the rate of a chemical reaction?
3. On a PE diagram for an exothermic reaction, which is true?
4. At chemical equilibrium:
5. For: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the correct Keq expression is:
6. A reaction has Keq = 0.0001. At equilibrium, the reaction mixture is:
7. For: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ. If more H₂ is added, the equilibrium will:
8. For: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ. If pressure is increased (volume decreased), the equilibrium will:
9. For: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ. If temperature is increased, what happens to Keq and the equilibrium position?
10. Adding a catalyst to an equilibrium system:
11. For the reaction: H₂(g) + I₂(g) ⇌ 2HI(g). Initial concentrations: [H₂] = 0.500 M, [I₂] = 0.500 M, [HI] = 0. At equilibrium, [HI] = 0.786 M. What is [H₂] at equilibrium?
12. Using the data from Q11 ([H₂]_eq = 0.107 M, [I₂]_eq = 0.107 M, [HI]_eq = 0.786 M), calculate Keq.
13. For: CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g). If volume is increased (pressure decreases), equilibrium shifts:
14. The Haber process uses: N₂ + 3H₂ ⇌ 2NH₃ ΔH = −92 kJ. To maximize NH₃ production, industrial conditions should include:
15. For: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). The correct Keq expression is:
16. For: 2NO₂(g) ⇌ 2NO(g) + O₂(g). If some O₂ is removed from the equilibrium system, the equilibrium will:
17. Which statement correctly describes a catalyst's effect on activation energy?
18. On a concentration-vs-time graph for a reaction reaching equilibrium, what happens to reactant and product concentrations after equilibrium is reached?
19. For 2A → B, if A disappears at a rate of 0.40 mol/L·s, at what rate does B appear?
20. For the equilibrium: CaCO₃(s) ⇌ CaO(s) + CO₂(g). The Keq expression is:
Short Answer Questions
SA1. For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ. (a) Write the Keq expression. (b) Predict the direction of shift for each stress: (i) add more N₂, (ii) increase pressure, (iii) increase temperature, (iv) add a catalyst. Explain each.
(b)(i) Add N₂ → shift right (away from what was added; makes more NH₃)
(ii) Increase pressure → shift right (left side has 4 mol gas, right has 2 mol gas; shift toward fewer moles)
(iii) Increase temperature → shift left (reaction is exothermic; treat heat as product; adding heat shifts left, endothermic direction; Keq decreases)
(iv) Add catalyst → no shift (catalyst lowers Eₐ for both directions equally; reaches equilibrium faster but position is unchanged)
SA2. Set up and solve an ICE chart for: H₂(g) + I₂(g) ⇌ 2HI(g). Initial: [H₂] = 0.500 M, [I₂] = 0.500 M, [HI] = 0. At equilibrium, [HI] = 0.786 M. (a) Complete the ICE chart. (b) Find equilibrium concentrations of H₂ and I₂. (c) Calculate Keq.
H₂ I₂ 2HI
I: 0.500 0.500 0
C: −x −x +2x
E: 0.500−x 0.500−x 2x
2x = 0.786 → x = 0.393
(b) [H₂]_eq = 0.500 − 0.393 = 0.107 M; [I₂]_eq = 0.107 M
(c) Keq = [HI]²/([H₂][I₂]) = (0.786)²/(0.107)(0.107) = 0.618/0.01145 = 54.0
(Keq > 1: products-favored)
SA3. Describe and sketch a potential energy diagram for an exothermic reaction. Label: (a) reactants, (b) products, (c) activated complex, (d) Eₐ forward, (e) Eₐ reverse, (f) ΔH. Then explain why Eₐ(forward) < Eₐ(reverse) for an exothermic reaction.
(a-f) Labels: Reactants (left), Products (right, lower than reactants), Activated Complex (top of hill), Eₐ forward = distance from reactants up to peak (smaller), Eₐ reverse = distance from products up to peak (larger), ΔH = difference between reactants and products (negative, shown as downward arrow).
Explanation: In an exothermic reaction, products are at lower energy than reactants. The activated complex (peak) is the same height for both forward and reverse reactions. Since products are lower, the reverse reaction must climb a LARGER energy hill (products → peak) than the forward reaction (reactants → peak). Therefore Eₐ(forward) < Eₐ(reverse) for exothermic reactions.