Chemistry Study Guide: Clausius-Clapeyron, Acids/Bases, Equilibrium, Solutions & Kinetics

Note: This guide now covers all four topics: Clausius-Clapeyron Equation, Acids/Bases, Chemical Equilibrium/Le Chatelier's, Solution Concentration, and Rate Kinetics. Fully complete based on everything sent so far.

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## PART 0: CLAUSIUS-CLAPEYRON EQUATION

### 0.1 Two-Point Form

Relates vapor pressure and temperature at two different conditions for the same substance:

$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

where R = 8.314 J/(mol·K), ΔHvap in J/mol, T in Kelvin. ΔHvap is always positive (vaporization is endothermic — it takes energy to overcome intermolecular forces and pull molecules apart into the gas phase; this doesn't release energy, so ΔHvap must be positive).

Worked Example 1: Water has VP = 24 mmHg at 25°C, ΔHvap = 40.7 kJ/mol. Find VP at 67°C.

- T1 = 298.15 K, T2 = 340.15 K

- 1/T2 − 1/T1 = 1/340.15 − 1/298.15 = −0.000415 K⁻¹

$\ln\left(\frac{P_2}{24}\right) = -\frac{-40700}{8.314}\times(-0.000415) = 2.032$

$P_2 = 24\times e^{2.032} = 24\times7.63 = 183\text{ mmHg}$

Worked Example 2: An unknown liquid has VP = 88 mmHg at 45°C and 39 mmHg at 25°C. Find ΔHvap.

- T1 = 318.15 K, T2 = 298.15 K

- ln(39/88) = ln(0.4432) = −0.8140

- 1/T2 − 1/T1 = 1/298.15 − 1/318.15 = 0.00021 K⁻¹

$-0.8140 = -\Delta H_{vap}\times\frac{0.00021}{8.314} \Rightarrow \Delta H_{vap} = 32000\text{ J/mol} = 32.0\text{ kJ/mol}$

Worked Example 3: At 20°C, VP(water) = 17.5 torr; at 60°C, VP = 149.4 torr. Find ΔHvap (R = 8.314 J/(mol·K)).

- Convert to atm: P1 = 0.0230 atm, P2 = 0.1965 atm; T1 = 293.15 K, T2 = 333.15 K

- 1/T2 − 1/T1 = −0.000409 K⁻¹

- ln(P2/P1) = ln(8.543) = 2.145

$\Delta H_{vap} = \frac{2.145}{0.0000493} = 43500\text{ J/mol} = 43.5\text{ kJ/mol}$

### 0.2 Graphical (Linear) Form

The Clausius-Clapeyron equation can be rewritten as a straight-line equation (y = mx + b) by plotting ln(P) vs. 1/T:

$\ln P = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T}\right) + \ln\beta$

- y = ln P, x = 1/T (in K⁻¹)

- slope (m) = −ΔHvap/R (always negative, since ΔHvap and R are both positive)

- y-intercept (b) = ln β (β is a constant related to the substance, not itself chemically meaningful on its own)

Data table setup: Convert given T(°C) to Kelvin (column 3), then take 1/T in K (column 4) as your x-values, and take ln(VP) (column 6) as your y-values. Plot ln(VP) vs. 1/T and fit a linear trendline.

Worked Example (Toluene data): Best-fit line: y = −4537.4x + 18.509, where x = 1/T, y = ln P

Finding ΔHvap from the slope:

$\text{slope} = -\frac{\Delta H_{vap}}{R} \Rightarrow -4537.4 = \frac{-\Delta H_{vap}}{8.314\text{ J/(mol·K)}}$

$\Delta H_{vap} = 4537.4\times8.314 = 37723\text{ J/mol} = 37.723\text{ kJ/mol}$

Finding vapor pressure at a given temperature (e.g., 25.0°C = 298.15 K):

$\ln P = -4537.4\left(\frac{1}{298.15}\right)+18.509 = 3.29$

$P = e^{3.29} = 26.84\text{ torr}$

Finding normal boiling point (where P = 760 torr, since "normal" boiling point = 1 atm = 760 torr):

$\ln(760) = -4537.4\left(\frac{1}{T}\right)+18.509$

$6.633 = -4537.4\left(\frac{1}{T}\right)+18.509$

$4537.4\left(\frac{1}{T}\right) = 18.509-6.633 = 11.877 \Rightarrow \frac{1}{T} = 0.0026167\text{ K}^{-1}$

$T = \frac{1}{0.0026167} = 382.1\text{ K} = 382.1-273.15 = 108.9°C$

### 0.3 Quick Reference

| Quantity | Formula |

|---|---|

| Two-point form | ln(P₂/P₁) = −(ΔHvap/R)(1/T₂ − 1/T₁) |

| Linear/graphing form | ln P = −(ΔHvap/R)(1/T) + ln β |

| Slope of ln P vs 1/T plot | −ΔHvap/R |

| Sign of ΔHvap | Always positive (vaporization is endothermic) |

| Normal boiling point | T where P = 760 torr (1 atm) |

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## PART 1: ACIDS AND BASES

### 1.1 Core Definitions

- Acids dissociate in water and are electrolytes.

- pH scale: 0–14, where pH < 7 = acidic, pH = 7 = neutral, pH > 7 = basic.

- Litmus test: Acids turn blue litmus red ("BAR"). Bases turn red litmus blue ("RBB").

- Acids react with metals to produce H₂ gas: Zn + 2HCl → ZnCl₂(s) + H₂(↑)

- Acids neutralize bases (like alkaline salts). Acids taste sour; bases taste bitter and feel soapy.

### 1.2 Three Acid-Base Theories

1) Arrhenius Theory

- Acid = produces H⁺ ion in water

- Base = produces OH⁻ ion in water

- Needs water as solvent (aqueous phase) — can't be an acid/base "by itself"

- HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻ (Arrhenius acid; H₃O⁺ = hydronium ion)

- NaOH(aq) + H₂O(l) → Na⁺(aq) + OH⁻ (Arrhenius base)

2) Brønsted-Lowry (B/L) Theory

- B/L Acid = proton donor (anything that can give a proton, H⁺)

- B/L Base = proton acceptor/receiver

- Conjugate pairs: the acid that donates a proton becomes its conjugate base (C.B); the base that accepts a proton becomes its conjugate acid (C.A).

  - C·A = gained a proton

  - C·B = lost a proton

- Example: HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)

  - HCl = B/L acid (donates proton) → becomes C.B (Cl⁻)

  - H₂O = B/L base (accepts proton) → becomes C.A (H₃O⁺)

- Water is amphoteric — it can act as either an acid or a base depending on what it reacts with.

3) Lewis Acid-Base Theory

- Lewis Acid = electron pair acceptor (needs room for more electrons — incomplete octet, e.g., BF₃, AlCl₃)

- Lewis Base = electron pair donor (has lone pairs ready to donate, e.g., NH₃)

- Forms a coordinate covalent bond (both electrons come from the same atom)

- Example: NH₃ + BF₃ → NH₃–BF₃ (N donates lone pair; B accepts it)

### 1.3 Acid Strength & Ka (Acid Dissociation Constant)

$K_a = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$

- Strong acid: Ka >> 1 → reaction goes essentially all the way forward (→), fully ionizes

- Weak acid: Ka << 1 → equilibrium favors reactants (weak acid HAs a Ka value; must use ICE tables)

- Smaller Ka = weaker acid. (Larger Ka = stronger acid)

- If the conjugate base is stable (full outer shell / low reactivity), the acid is stronger.

Ion Product Constant of Water (Kw):

$K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0\times10^{-14} \text{ (at 25°C)}$

$K_w = K_a \times K_b$

In pure water: [H₃O⁺] = [OH⁻] = 1.0×10⁻⁷ M

Solution type rules:

| Solution | Relationship |

|---|---|

| Acidic | [H₃O⁺] > [OH⁻] |

| Basic | [H₃O⁺] < [OH⁻] |

| Neutral | [H₃O⁺] = [OH⁻] |

### 1.4 pH / pOH Formulas

$pH = -\log[\text{H}_3\text{O}^+] \qquad pOH = -\log[\text{OH}^-]$

$pH + pOH = 14$

$[\text{H}_3\text{O}^+] = 10^{-pH} \qquad [\text{OH}^-] = 10^{-pOH}$

Each whole pH unit = a factor of 10 in strength (pH 3 is 10× weaker acid than pH 2; pH 4 is 100× weaker than pH 2).

### 1.5 Weak Acid Equilibrium (ICE Tables)

General equation for any weak acid:

$HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$

Worked Example: Find pH of 0.200 M HNO₂ solution, Ka = 4.6×10⁻⁴ @ 25°C

| | [HNO₂] | [H₃O⁺] | [NO₂⁻] |

|---|---|---|---|

| I | 0.200 | 0 | 0 |

| C | −x | +x | +x |

| E | 0.200−x | x | x |

$K_a = \frac{x \cdot x}{0.200 - x} \approx \frac{x^2}{0.200}\text{ (assume x is small, ignore in denominator)}$

$4.6\times10^{-4} = \frac{x^2}{0.200} \Rightarrow x = 9.6\times10^{-3}\text{ M}$

Check validity (5% rule):

\frac{9.6\times10^{-3}}{0.200}\times100 = 4.8\% \quad (4.8\% < 5\% \Rightarrow \text{approximation valid})

$pH = -\log(9.6\times10^{-3}) = 2.02$

% Ionization formula:

$\%\text{ Ionization} = \frac{[\text{Ionized acid}]}{[\text{Initial}]}\times100$

### 1.6 Strong Acid/Base — Complete Ionization

Strong acids/bases ionize 100% — no ICE table needed, no Ka.

$HCl(aq) + H_2O(l) \rightarrow H_3O^+ + Cl^- \quad (1.0\text{ M} \rightarrow 1.0\text{ M})$

For strong bases like NaOH or Ca(OH)₂, [OH⁻] comes directly from stoichiometry:

- NaOH: [OH⁻] = concentration given (1:1 ratio)

- Ca(OH)₂: [OH⁻] = 2 × concentration given (2 OH⁻ per formula unit)

### 1.7 Periodic Trends in Acid Strength

- Across a period (left → right): acidity increases (e.g., NH₃ < H₂O < HF) — driven by increasing electronegativity, which pulls harder on the H, weakening the H–Y bond.

- Down a group: acidity increases (e.g., HF < HCl < HBr < HI) — driven by increasing atomic size, which makes the H–Y bond longer/weaker so the proton comes off more easily.

- Acid strength = ease with which the proton (H⁺) can come out.

Factors affecting proton release (oxyacids, Y–O–H structure):

1. Electronegativity of Y: more electronegative Y → weak pull on O–H → stronger acid (Y pulls electron density away from O–H bond, weakening it)

2. Number of oxygens attached to Y: more oxygens = stronger acid ("greater # of oxygen = stronger acid")

   - Example: HClO < HClO₂ < HClO₃ < HClO₄ (increasing O → increasing acid strength)

### 1.8 Acid/Base Structural Classifications

- Oxyacid: has structure Y–O–H; H⁺ (acidic proton) attached to an oxygen (e.g., HNO₃, H₂SO₄)

- Binary acid: H + monoatomic ion, no oxygen (e.g., HCl, HBr)

- Monoprotic: 1 acidic proton (HCl, HNO₃)

- Diprotic: 2 acidic protons (H₂SO₄, H₂CO₃)

- Triprotic: 3 acidic protons (H₃PO₄) — "ready to give up 3 protons"

### 1.9 Key Practice Problems (with answers, worked in your notes)

1. Conjugate base of H₂PO₄⁻? → Answer: HPO₄²⁻ (loses one H⁺)

2. NOT a conjugate acid-base pair? → check each pair donates/accepts exactly 1 proton

3. Stronger acid → weaker conjugate base (inverse relationship)

4. Increasing acid strength: HClO₂ < HClO₄ ... actually ranked by O count: HClO < HClO₂ < HClO₃ < HClO₄

5. Highest pH among 0.10 M acids (compare Ka): smallest Ka = weakest acid = highest pH → HCN (Ka = 4.9×10⁻¹⁰)

6. Neutral solution: [H₃O⁺] = [OH⁻] (not [H₂O] = [H₃O⁺])

7. [OH⁻] in pure water at 30°C, Kw = 1.47×10⁻¹⁴:

   $[\text{OH}^-] = \sqrt{K_w} = \sqrt{1.47\times10^{-14}} = 1.21\times10^{-7}\text{ M}$

8. pH of pure water at 40°C, Kw = 2.92×10⁻¹⁴:

   $[\text{H}^+] = \sqrt{2.92\times10^{-14}} = 1.7\times10^{-7} \Rightarrow pH = 6.769$

9. Kw of pure water at 50°C if pH = 6.630:

   $[\text{H}^+] = 10^{-6.630} = 2.34\times10^{-7} \Rightarrow K_w = (2.34\times10^{-7})^2 = 5.5\times10^{-14}$

10. [H₃O⁺] in solution with 5.5×10⁻⁵ M OH⁻ @ 25°C:

    [\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0\times10^{-14}}{5.5\times10^{-5}} = 1.8\times10^{-10}\text{ M (basic, since H}_3\text{O}^+ < \text{OH}^-\text{)}

11–16. Straightforward pH/pOH/[H₃O⁺]/[OH⁻] conversions using the four core formulas above.

17. pH of 0.023 M HNO₃ (strong acid): pH = −log(0.023) = 1.64

18. Strongest acid = largest Ka → HClO₂ (Ka = 1.1×10⁻²)

19. Weakest acid = smallest Ka → HIO (Ka = 2.3×10⁻¹¹)

20. % ionization of 0.337 M HF (Ka = 3.5×10⁻⁴):

    $x = \sqrt{3.5\times10^{-4}\times0.337} = 0.011 \Rightarrow \%I = \frac{0.011}{0.337}\times100 = 2.26\%$

21. pOH of 0.235 M NaOH: pOH = −log(0.235) = 0.63

22. [OH⁻] in 0.169 M Ca(OH)₂: [OH⁻] = 2 × 0.169 = 0.338 M

23. Base in baking soda: NaHCO₃ (sodium bicarbonate)

24. Weak diprotic acid: H₂CO₃

25. Weak acid example: HCO₂H (formic acid)

26. Strongest conjugate base comes from the weakest acid (HCN → CN⁻ is the strongest C.B. here)

29. pOH when pH = 9.85: pOH = 14 − 9.85 = 4.15

30–35. More Ka/Kb comparisons — always: larger Ka/Kb = stronger acid/base; smaller = weaker.

$K_b(\text{CN}^-) = \frac{K_w}{K_a(\text{HCN})} = \frac{1.0\times10^{-14}}{4.9\times10^{-10}} = 2.0\times10^{-5}$

34. Lewis acid identification: BCl₃ (electron-deficient, accepts electron pairs)

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## PART 2: CHEMICAL EQUILIBRIUM & LE CHATELIER'S PRINCIPLE

### 2.1 Core Definitions

- Kinetics = rate of reaction

- Equilibrium = thermodynamics/energy exchanges of the reaction (whether the reaction is moving right or left)

- At equilibrium, the reaction is a continuous cycle from reactants to products and back (not "stopped")

### 2.2 Equilibrium Constant, K

For a generic reaction: aA + bB ⇌ cC + dD

$K = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} = \frac{[\text{products}]}{[\text{reactants}]}$

This relationship (balanced equation ↔ equilibrium expression) is called the Law of Mass Action.

- Large K = reaction favors products (shifted right)

- Small K = reaction favors reactants (shifted left)

Examples from your notes:

- 2N₂O₅(g) ⇌ 4NO₂(g) + O₂(g): K = [NO₂]⁴[O₂] / [N₂O₅]²

- CH₃OH(g) ⇌ CO(g) + 2H₂(g): K = [CO][H₂]² / [CH₃OH]

- C₃H₈(g) + 5O₂(g) ⇌ 3CO₂(g) + 4H₂O(g): Kc = [CO₂]³[H₂O]⁴ / [C₃H₈][O₂]⁵

Homogeneous equilibria: all species in the same phase (usually all gases)

Heterogeneous equilibria: species in different phases — pure solids/liquids are omitted from K expression (their "concentration" is constant)

- Example: 2CO(g) ⇌ CO₂(g) + C(s) → Kc = [CO₂] / [CO]² (solid C omitted)

- Example: CO₂(g) + H₂O(l) ⇌ H⁺(aq) + HCO₃⁻(aq) → Kc = [H⁺][HCO₃⁻] / [CO₂] (liquid water omitted)

### 2.3 Manipulating Equilibrium Expressions

1. Reversing the reaction → invert K:

   $K_{reverse} = \frac{1}{K_{forward}}$

2. Multiplying coefficients by a factor n → raise K to that power:

   $K' = K^n$

### 2.4 Kc vs. Kp (Gas-Phase Equilibria)

$K_p = P_{CO_2}^3 \cdot P_{H_2O}^4 / (P_{C_3H_8}\cdot P_{O_2}^5) \text{ — same idea, using partial pressures}$

Relationship between Kp and Kc:

$K_p = K_c(RT)^{\Delta n} \qquad K_c = \frac{K_p}{(RT)^{\Delta n}}$

where:

- R = 0.08206 L·atm/(mol·K)

- T = temperature in Kelvin

- Δn = (sum of gas coefficients of products) − (sum of gas coefficients of reactants)

Special cases:

- If Δn = negative → Kc > Kp always

- If Δn = 0 → Kp = Kc (since (RT)⁰ = 1)

Worked Example: 2NO(g) + O₂(g) ⇌ 2NO₂(g), Kp = 2.2×10¹² @ 25°C. Find Kc.

- Δn = 2 − (2+1) = −1

- $K_c = K_p(RT)^{-\Delta n} = (2.2\times10^{12})\times[(0.08206)(298)]^{1} = 5.4\times10^{13}$

Worked Example: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Kc = 8.0×10²⁵ @ 27°C (300.15 K). Find Kp.

- Δn = 2 − 3 = −1

- $K_p = K_c(RT)^{\Delta n} = (8.0\times10^{25})\times(0.08206\times300.15)^{-1} = 3.25\times10^{24}$

### 2.5 Le Chatelier's Principle

Statement: When a chemical system at equilibrium is stressed/disturbed, the system shifts in the direction that minimizes the disturbance. (System moves away from the stress.)

Stressor 1 — Concentration

| Change | Shift |

|---|---|

| Add reactants | Shifts right (toward products) |

| Add products | Shifts left (toward reactants) |

| Remove products | Shifts right (toward products) |

| Remove reactants | Shifts left (toward reactants) |

Example: CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(g)

- Add CH₄ → shifts right

- Remove O₂ → shifts left

Stressor 2 — Temperature

Heat can be treated as a "reactant" or "product" depending on the reaction type:

- Exothermic (ΔH negative): heat is released → treat heat as a product

  - Add heat → shifts left

  - Remove heat → shifts right

- Endothermic (ΔH positive): heat is absorbed → treat heat as a reactant

  - Add heat → shifts right

  - Remove heat → shifts left

Example: CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(g) + Heat (ΔH = −38.7 kJ/mol, exothermic)

Stressor 3 — Pressure / Volume

Gas law relationships: V ∝ n, P ∝ 1/V, P ∝ n

| Change | Effect |

|---|---|

| Add pressure (or decrease volume) | Shifts toward the side with fewer moles of gas |

| Remove pressure (or increase volume) | Shifts toward the side with more moles of gas |

| Equal moles gas on both sides | No shift (pressure/volume changes don't affect equilibrium) |

Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (4 mol reactants → 2 mol products)

- Add pressure → shifts right (toward fewer moles, products)

- Remove pressure → shifts left (toward reactants, more moles)

Example with no shift: C(s) + S₂(g) ⇌ CS₂(g) (1 mol gas reactant, 1 mol gas product → pressure change has no effect)

### 2.6 Practice Problems (Le Chatelier)

Given N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (Haber process):

- Increasing volume decreases pressure (P ∝ 1/V)

- Reactant side has larger # mols → shifts left

- Product side has larger # mols → shifts right

- Equal # mols on both sides → no change/neutral

### 2.7 Combining/Manipulating Equilibrium Constants

Rule 1 — Reversing a reaction: invert K

$K_{reverse} = \frac{1}{K_{forward}}$

Rule 2 — Multiplying coefficients by a factor n: raise K to that power

$K' = K^n$

Rule 3 — Adding two or more individual equations to get an overall equation: multiply the corresponding K's together

$K_{overall} = K_1 \cdot K_2$

Worked Example: Given

- N₂(g) + 3H₂(g) ⇌ 2NH₃(g), K = 3.7×10⁸

- Find K for: NH₃(g) ⇌ ½N₂(g) + 3/2H₂(g)

Step 1 — reverse the given equation: 2NH₃(g) ⇌ N₂(g) + 3H₂(g), K₁ = 1/(3.7×10⁸)

Step 2 — multiply by ½ (divide all coefficients by 2), raise K to the ½ power:

$K_2 = \left[\frac{1}{3.7\times10^8}\right]^{1/2} = 2.7\times10^{-9}$

Worked Example (3-step, combining two reactions): Given

- A ⇌ 2B, K₁ = 1.6×10⁴

- 2B ⇌ Z, K₂ = 1.25×10⁻¹

- Find K for: 3Z ⇌ 3A

Step 1 — add eq. 1 and 2: A ⇌ Z, K₃ = K₁·K₂ = (1.6×10⁴)(1.25×10⁻¹)

Step 2 — reverse eq. 3: Z ⇌ A, K₄ = 1/K₃

Step 3 — multiply eq. 4 by factor of 3: 3Z ⇌ 3A

$K_5 = \left[\frac{1}{(1.6\times10^4)(1.25\times10^{-1})}\right]^3 = 1.25\times10^{-10}$

Simple Kc calculation example: H₂(g) + I₂(g) ⇌ 2HI(g); [H₂] = 0.11 M, [I₂] = 0.11 M, [HI] = 0.78 M

$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.78)^2}{(0.11)(0.11)} = 50$

### 2.8 Reaction Quotient, Q

Used to determine which direction the equilibrium will shift when the system is not yet at equilibrium (reactants and products both present but not at their equilibrium values).

$Q_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \qquad Q_p = \frac{[P_C]^c[P_D]^d}{[P_A]^a[P_B]^b}$

Comparison rules (memorize):

| Condition | Meaning |

|---|---|

| Q > K | Shifts to reverse direction (toward reactants) |

| Q < K | Shifts forward direction (toward products) |

| Q = K | System is already at equilibrium |

| Q = ∞ | Shifts to reverse direction ([reactant] = 0) |

| Q = 0 | Shifts to forward direction ([product] = 0) |

Worked Example: CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g), Kc = 3.59

[CH₄] = 1.15, [H₂S] = 1.20, [CS₂] = 1.51, [H₂] = 1.08. Find Qc and predict direction.

$Q_c = \frac{[1.51]^1[1.08]^4}{[1.15]^1[1.20]^2} = 2.57$

Since Qc = 2.57 < Kc = 3.59 → shifts forward (toward products)

### 2.9 Calculating Equilibrium Concentrations from K (ICE Tables)

Case A — K and all but one equilibrium concentration are known (no ICE table needed, solve directly):

Example: 2COF₂(g) ⇌ CO₂(g) + CF₄(g), Kc = 2.00 @ 1000°C; [COF₂] = 0.255 M, [CF₄] = 0.118 M. Find [CO₂].

$[\text{CO}_2] = \frac{K_c\cdot[\text{COF}_2]^2}{[\text{CF}_4]} = \frac{2.0\times(0.255)^2}{0.118} = 1.10\text{ M}$

Case B — Initial concentration and one equilibrium concentration are given (build an ICE table; when product's initial isn't stated, it's assumed to be zero):

Example: A(g) ⇌ 2B(g); [A]initial = 1.0 M, [A]eq = 0.75 M. Find Kc.

| | [A] | [B] |

|---|---|---|

| I | 1.0 | 0 |

| C | −0.25 | +2×0.25 |

| E | 0.75 | 0.50 |

$K_c = \frac{[\text{B}]^2}{[\text{A}]} = \frac{(0.50)^2}{0.75} = 0.33$

Case C — Only initial concentrations and K are given (must solve for x algebraically, often using the small-x/5% approximation like in weak-acid problems):

Example: 2H₂S(g) ⇌ 2H₂(g) + S₂(g), Kc = 1.67×10⁻⁷ @ 800°C; [H₂S]initial = 0.025 M. Find equilibrium concentrations.

| | [H₂S] | [H₂] | [S₂] |

|---|---|---|---|

| I | 0.025 | 0 | 0 |

| C | −2x | +2x | +x |

| E | 0.025−2x | 2x | x |

$K_c = \frac{(2x)^2(x)}{(0.025-2x)^2} \approx \frac{4x^3}{(0.025)^2}\text{ (approximate: ignore 2x in denominator)}$

$1.67\times10^{-7} = \frac{4x^3}{(0.025)^2} \Rightarrow x = 2.97\times10^{-4}$

Validity check (5% rule): $\frac{2.97\times10^{-4}}{0.025}\times100 = 1.19\%$ → since 1.19% < 5%, the approximation is valid.

Results: [H₂] = 5.97×10⁻⁴ M, [S₂] = 2.97×10⁻⁴ M, [H₂S] = 0.024 M

Example with a solid (heterogeneous): CuS(s) + O₂(g) ⇌ Cu(s) + SO₂(g), Kc = 1.5, [O₂]initial = 2.9 M. Find [O₂]eq.

Since CuS and Cu are solids, they're omitted from K:

$K_c = \frac{[\text{SO}_2]}{[\text{O}_2]}$

| | [O₂] | [SO₂] |

|---|---|---|

| I | 2.9 | 0 |

| C | −x | +x |

| E | 2.9−x | x |

$1.5 = \frac{x}{2.9-x} \Rightarrow x = 1.74 = [\text{SO}_2]_{eq}$

$[\text{O}_2]_{eq} = 2.9 - 1.74 = 1.16\text{ M}$

### 2.10 Practice Problems (Equilibrium Worksheet)

1. Missing K from reversed/scaled reaction: H₂(g)+Br₂(g)⇌2HBr(g), Kc₁=3.8×10⁴. Find Kc for 4HBr(g)⇌2H₂(g)+2Br₂(g) (this is the reverse, times 2):

   $K_2 = \left(\frac{1}{3.8\times10^4}\right)^2 = 6.9\times10^{-10}$

2. 2COF₂(g)⇌CO₂(g)+CF₄(g), Kc=2.2×10⁶. Find Kc for the reverse reaction doubled: 2CO₂+2CF₄⇌4COF₂

   $K_{new} = \left(\frac{1}{2.2\times10^6}\right)^2 = 2.1\times10^{-13}$

3. Find Kp from equilibrium partial pressures: 2CO(g)+O₂(g)⇌2CO₂(g); P(CO)=6.8×10⁻¹¹ atm, P(O₂)=1.3×10⁻³ atm, P(CO₂)=0.041 atm

   $K_p = \frac{[0.041]^2}{[6.8\times10^{-11}]^2[1.3\times10^{-3}]} = 2.8\times10^{20}$

4. Find Kp from Kc: H₂(g)+I₂(g)⇌2HI(g), Kc=6.2×10² @ 25°C. Δn = 2−(1+1) = 0, so Kp = Kc = 6.2×10² (when Δn=0, Kp=Kc regardless of RT).

5. Find [SO₃] given Kc and other equilibrium concentrations: 2SO₂(g)+O₂(g)⇌2SO₃(g), Kc=1.7×10⁸; [SO₂]eq=0.0034 M, [O₂]eq=0.0018 M

   $[\text{SO}_3]^2 = K_c[\text{SO}_2]^2[\text{O}_2] = (1.7\times10^8)(0.0034)^2(0.0018) \Rightarrow [\text{SO}_3] = 1.9\text{ M}$

6. ICE table with a heterogeneous reaction (see 2.9 CuS example above): [O₂]eq = 1.16 M

7. Find Kc and Kp from initial + equilibrium concentrations: CO(g)+2H₂(g)⇌CH₃OH(g) @ 980°C; [CO]initial=0.27 M, [H₂]initial=0.49 M, [CH₃OH]eq=0.11 M

   - ICE: [CO]eq = 0.27−0.11 = 0.16 M; [H₂]eq = 0.49−2(0.11) = 0.27 M

   $K_c = \frac{0.11}{(0.16)(0.27)^2} = 9.43$

   $K_p = K_c(RT)^{\Delta n} = 9.43\times(0.08206\times1253)^{-2} = 8.9\times10^{-4}$

8. Reaction quotient direction: N₂O₄(g)⇌2NO₂(g), Kc=5.85×10⁻³; [NO₂]=0.0255 M, [N₂O₄]=0.0331 M

   $Q_c = \frac{(0.0255)^2}{0.0331} = 0.0196$

   Since Qc (0.0196) > Kc (5.85×10⁻³) → too much product relative to equilibrium → shifts in reverse (left), converting product back to reactant until equilibrium is reached.

---

## PART 3: SOLUTION CONCENTRATION

### 3.1 Concentration Unit Definitions

$\text{Molarity } (M) = \frac{\text{mol solute}}{\text{L solution}} \qquad \text{Molality } (m) = \frac{\text{mol solute}}{\text{kg solvent}}$

$\text{Mass \%} = \frac{\text{mass solute}}{\text{mass solution}}\times100 \qquad \text{Mole fraction } (X) = \frac{\text{mol component}}{\text{total mol all components}}$

Key distinction: molarity uses volume of solution; molality uses mass of solvent only. Mole fraction and mass % always use the whole solution (solute + solvent).

Worked Examples:

- Molality of 27.8 g LiI in 500.0 mL water: mol LiI ÷ kg water = 0.415 m

- Mass of NH₃ needed for 0.250 m in 475 g methanol: 0.250 mol/kg × 0.475 kg = 0.119 mol → 2.02 g

- Moles of KF in 0.175 m solution with 347 g water: 0.175 mol/kg × 0.347 kg = 0.061 mol

- Molality of 0.500 mol CaF₂ in 11.5 mol H₂O: convert mol H₂O to kg (11.5 × 18.02 g/mol = 207 g = 0.207 kg) → 0.500/0.207 = 2.41 mol/kg

- Mass % of 98.6 g NaCl in 875 mL solution (density 1.06 g/mL): mass solution = 875×1.06 = 927.5 g → (98.6/927.5)×100 = 10.6%

- Mass of CuCl₂ in 75.85 g of a 22.4% solution: 75.85 × 0.224 = 17.0 g

- Mole fraction of total ions from 0.400 mol MgCl₂ in 850.0 g water: MgCl₂ → Mg²⁺ + 2Cl⁻ = 3 mol ions total; mol H₂O = 850.0/18.02 = 47.2 mol → X(ions) = 1.2/(1.2+47.2) = 0.027

  (Note: for ionic compounds, multiply moles of compound by the number of ions it dissociates into before computing mole fraction or total particle count — this matters for colligative properties too, see the van't Hoff factor below.)

- Molarity of 0.0433 m LiF solution (density 1.10 g/mL): Answer 0.0476 M (convert via assuming 1 kg solvent, add solute mass, use density to get volume, then mol/L)

- Molarity of a solution: 27.8 g LiI dissolved to 500 mL — reminder: molarity always needs the final solution volume, not solvent volume.

### 3.2 Colligative Properties

Colligative properties depend on the number of dissolved particles, not their identity. For ionic compounds, multiply concentration by the van't Hoff factor (i) = number of ions produced per formula unit (e.g., NaCl → i=2, MgCl₂ → i=3, nonionic/molecular compounds like sucrose or glycerin → i=1).

Freezing Point Depression:

$\Delta T_f = K_f \cdot m \cdot i$

Boiling Point Elevation:

$\Delta T_b = K_b \cdot m \cdot i$

For water: Kf = 1.86°C/m, Kb = 0.512°C/m

Worked Example: Freezing point depression of 30.7 g glycerin (C₃H₈O₃, 92.09 g/mol, nonionic so i=1) in 376 mL water.

- mol glycerin = 30.7/92.09 = 0.333 mol; kg water = 0.376 kg

- m = 0.333/0.376 = 0.887 mol/kg

$\Delta T_f = (1.86)(0.887)(1) = 1.65°C$

Worked Example: Boiling point elevation of an aqueous sucrose solution is 0.39°C. Find mass of sucrose (342.30 g/mol) in 500.0 g water.

$m = \frac{\Delta T_b}{K_b} = \frac{0.39}{0.512} = 0.762\text{ mol/kg}$

mol sucrose = 0.762 × 0.500 kg = 0.381 mol → mass = 0.381 × 342.30 = 130.0 g

### 3.3 Vapor Pressure Lowering (Raoult's Law)

$P_{solution} = X_{solvent}\cdot P^\circ_{solvent}$

Worked Example: Vapor pressure at 55°C of a solution with 34.2 g NaCl in 375 mL water (P° = 118.1 torr).

- mol NaCl = 34.2/58.44 = 0.585 mol; since NaCl → 2 ions, effective mol solute = 0.585×2 = 1.17 mol

- mol water = 375 g/18.02 = 20.8 mol

- X(water) = 20.8/(20.8+1.17) = 0.947

$P = 0.947 \times 118.1 = 112\text{ torr}$

### 3.4 Osmotic Pressure

$\pi = MRT \quad (\text{times } i \text{ for ionic solutes})$

where R = 0.08206 L·atm/(mol·K), T in Kelvin, M = molarity.

Worked Example: A compound with molar mass 598 g/mol; 35.8 mg dissolved in 175 mL solution at 25°C. Find osmotic pressure.

- mol = 0.0358 g / 598 g/mol = 5.99×10⁻⁵ mol

- M = 5.99×10⁻⁵ mol / 0.175 L = 3.42×10⁻⁴ M

$\pi = (3.42\times10^{-4})(0.08206)(298) = 8.37\times10^{-3}\text{ atm} = 6.36\text{ torr}$

Comparing osmotic pressure across solutions: since π ∝ M×i, rank by (concentration × van't Hoff factor). Example: comparing 0.011 M sucrose (i=1), 0.00095 M glucose (i=1), 0.0060 M glycerin (i=1) — since all are nonionic (i=1 for all), just rank by concentration: sucrose > glycerin > glucose for decreasing osmotic pressure.

### 3.5 Quick Practice Reference

| Problem type | Key formula |

|---|---|

| Molality | m = mol solute / kg solvent |

| Molarity | M = mol solute / L solution |

| Mass % | (mass solute/mass solution)×100 |

| Mole fraction | mol component / total mol |

| ΔTf | Kf·m·i |

| ΔTb | Kb·m·i |

| Vapor pressure | X(solvent)×P°(solvent) |

| Osmotic pressure | π = MRTi |

---

## PART 4: RATE KINETICS

### 4.1 Rate of Reaction & Stoichiometry

For a reaction aA + bB → cC + dD, the rate of disappearance/appearance of each species relates through stoichiometric coefficients:

$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$

General conversion:

$\frac{\Delta[\text{want}]}{\Delta t} = \frac{\text{coefficient of want}}{\text{coefficient of have}}\times\frac{\Delta[\text{have}]}{\Delta t}$

Worked Example: 2HBr(g) → H₂(g) + Br₂(g); rate of disappearance of HBr = 0.301 M/s. Find rate of appearance of Br₂.

$\text{Rate} = \frac{1}{2}\times0.301 = 0.151\text{ M/s}$

Worked Example (reverse direction): 2O₃(g) → 3O₂(g); rate of formation of O₂ = 7.78×10⁻¹ M/s. Find rate of loss of O₃.

$\frac{\Delta[\text{O}_3]}{\Delta t} = \frac{2}{3}\times7.78\times10^{-1} = 0.519\text{ M/s}$

### 4.2 Rate Laws and Reaction Order

$\text{Rate} = k[A]^m[B]^n$

- m, n are the individual reaction orders (determined experimentally, NOT from stoichiometric coefficients)

- Overall order = m + n (sum of all exponents)

- Doubling [Y] when Y has exponent p multiplies the rate by 2^p

- Comparing two experiments (holding one reactant constant) isolates the order in the other reactant — divide rate expressions to solve for exponents

Worked Example: Rate = k[X][Y]², doubling [Y] → rate increases by 2² = factor of 4

Worked Example: Rate = k[X]⁴[Y]³, doubling [Y] → rate increases by 2³ = factor of 8

Units of k depend on overall reaction order (n):

$k\text{ units} = M^{1-n}\cdot s^{-1}$

| Overall order | Units of k |

|---|---|

| 0 | M/s |

| 1 | s⁻¹ |

| 2 | 1/(M·s) or M⁻¹s⁻¹ |

| 3 | 1/(M²·s) or M⁻²s⁻¹ |

| 3/2 | M⁻¹ᐟ²s⁻¹ |

Worked Example: Rate = k[X][Y]³ → order = 1+3 = 4th order → units of k = 1/(M³·s)

Worked Example: Rate = k[X][Y]^(1/2) → order = 1.5 → units of k = M⁻¹ᐟ²s⁻¹

### 4.3 Determining Rate Law from Initial Rates (Method of Initial Rates)

Given a table of [reactant]ᵢ vs. initial rate for several trials, compare pairs of experiments where only one concentration changes, and see how the rate changes (the ratio of rates equals the ratio of concentrations raised to that reactant's order).

Worked Example: 2N₂O₅(g) → 4NO₂(g) + O₂(g)

| [N₂O₅]ᵢ (M) | Initial Rate (M⁻¹s⁻¹) |

|---|---|

| 0.093 | 4.84×10⁻⁴ |

| 0.084 | 4.37×10⁻⁴ |

| 0.224 | 1.16×10⁻³ |

Comparing any two trials, the rate ratio ≈ concentration ratio (to the first power) → reaction is first order in N₂O₅:

$\text{Rate} = k[\text{N}_2\text{O}_5]^1$

(This is the general workflow for the potassium permanganate/oxalic acid rate law problem too — set up a table of initial concentrations and initial rates for each reactant, hold one concentration fixed between two trials, and solve for the order in the other reactant from how the rate changes.)

### 4.4 Integrated Rate Laws & Half-Life

First-order reactions:

$\ln\frac{[A]_0}{[A]} = kt \qquad t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$

Key property: first-order half-life is independent of concentration — it's the same no matter how much you start with.

Worked Example: [CH₃NC] decays with k = 9.45×10⁻⁵ s⁻¹. Find half-life.

$t_{1/2} = \frac{0.693}{9.45\times10^{-5}} = 7.33\times10^3\text{ s}$

Worked Example: How many half-lives for concentration to drop to 25% of original?

$(1/2)^n = 0.25 \Rightarrow n = 2$

Worked Example: First-order reaction, k = 0.76 s⁻¹. Time to decrease to 12% of initial concentration?

$\ln\left(\frac{1}{0.12}\right) = kt \Rightarrow 2.12 = (0.76)(t) \Rightarrow t = 2.8\text{ s}$

Worked Example: First-order reaction takes 24 minutes to decrease to 25% of initial value. Find k.

$\ln\left(\frac{1}{0.25}\right) = k(24\times60\text{ s}) \Rightarrow 1.386 = k(1440) \Rightarrow k = 9.6\times10^{-4}\text{ s}^{-1}$

Worked Example: SO₂Cl₂ first-order, k = 2.20×10⁻⁵ s⁻¹ @ 593 K. What % remains after 6.00 hours (21600 s)?

$\ln\frac{[A]_0}{[A]} = (2.20\times10^{-5})(21600) = 0.475$

$\frac{[A]}{[A]_0} = e^{-0.475} = 0.622 \Rightarrow \textbf{62.2\%}$

Second-order reactions:

$\frac{1}{[A]} - \frac{1}{[A]_0} = kt \qquad t_{1/2} = \frac{1}{k[A]_0}$

Key property: second-order half-life DOES depend on initial concentration (unlike first order).

Worked Example: Second-order reaction, k = 0.54 M⁻¹s⁻¹, [A]₀ = 0.33 M. Find half-life.

$t_{1/2} = \frac{1}{(0.54)(0.33)} = 5.6\text{ s}$

### 4.5 Arrhenius Equation — Two-Point Form

Used to find the rate constant at a different temperature, or to find activation energy, given two (k, T) data points.

$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

where R = 8.314 J/(mol·K), Ea in J/mol, T in Kelvin.

Worked Example: Ea = 126 kJ/mol, k at 30°C (303 K) = 9.8×10⁻³ min⁻¹. Find k at 35°C (308 K).

$\ln\left(\frac{k_2}{9.8\times10^{-3}}\right) = -\frac{126000}{8.314}\left(\frac{1}{308}-\frac{1}{303}\right)$

Solving gives $k_2 = 2.2\times10^{-2}\text{ min}^{-1}$

Worked Example: k = 1.35×10² s⁻¹ @ 25.0°C (298 K), Ea = 55.5 kJ/mol. Find k at 95.0°C (368 K).

$\ln\left(\frac{k_2}{135}\right) = -\frac{55500}{8.314}\left(\frac{1}{368}-\frac{1}{298}\right) \Rightarrow k_2 = 9.56\times10^3\text{ s}^{-1}$

Worked Example: Same k and T₁ as above, Ea = 85.6 kJ/mol. Find k at 75.0°C (348 K).

$k_2 = 1.92\times10^4\text{ s}^{-1}$

### 4.6 Quick Practice Reference

| Order | Integrated law | Half-life | k units |

|---|---|---|---|

| 0 | [A] = [A]₀ − kt | [A]₀/2k | M/s |

| 1 | ln[A] = ln[A]₀ − kt | ln2/k (constant) | s⁻¹ |

| 2 | 1/[A] = 1/[A]₀ + kt | 1/(k[A]₀) | M⁻¹s⁻¹ |

---

## Quick-Reference Formula Sheet

| Concept | Formula |

|---|---|

| pH | pH = −log[H₃O⁺] |

| pOH | pOH = −log[OH⁻] |

| pH + pOH | = 14 |

| [H₃O⁺] from pH | = 10⁻ᵖᴴ |

| [OH⁻] from pOH | = 10⁻ᵖᴼᴴ |

| Kw | = [H₃O⁺][OH⁻] = 1.0×10⁻¹⁴ (@25°C) |

| Kw relation | Kw = Ka × Kb |

| Ka | = [H₃O⁺][A⁻] / [HA] |

| % Ionization | = ([ionized]/[initial]) × 100% |

| Kp–Kc relation | Kp = Kc(RT)^Δn |

| Δn | = Σ(mol gas products) − Σ(mol gas reactants) |

| K (equilibrium) | = [products]/[reactants] (coefficients as exponents) |

| Reverse a reaction | K_reverse = 1/K_forward |

| Scale coefficients by n | K' = K^n |

| Add reactions | K_overall = K1 × K2 |

| Q vs K | Q>K → shifts reverse; Q<K → shifts forward; Q=K → at equilibrium |

| Molality | mol solute / kg solvent |

| ΔTf, ΔTb | Kf·m·i , Kb·m·i |

| Vapor pressure | X(solvent) × P°(solvent) |

| Osmotic pressure | π = MRTi |

| First-order integrated law | ln[A]₀/[A] = kt; t½ = 0.693/k |

| Second-order integrated law | 1/[A] − 1/[A]₀ = kt; t½ = 1/(k[A]₀) |

| Arrhenius 2-point | ln(k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁) |

| Clausius-Clapeyron (2-point) | ln(P₂/P₁) = −(ΔHvap/R)(1/T₂ − 1/T₁) |

| Clausius-Clapeyron (graph form) | ln P = −(ΔHvap/R)(1/T) + ln β; slope = −ΔHvap/R |