FAD1018 Phase Equilibria & Thermochemistry — Drill Guide

Step-by-step procedures mapped to each Part D and Part E question.

PART D — Phase Equilibria

Q25: Vapour Pressure Lowering ($\Delta P = X_2 P_1^\circ$)

Given: 218 g glucose (M = 180) in 460 mL water. $P^\circ_{water} = 31.82$ mmHg.

Procedure:

  1. Moles of solute: $n_{glucose} = \frac{218}{180}$
  2. Moles of solvent: $n_{water} = \frac{460}{18}$ (density 1 g/mL, so 460 g)
  3. Mole fraction of solute: $X_{glucose} = \frac{n_{glucose}}{n_{glucose} + n_{water}}$
  4. Pressure lowering: $\Delta P = X_{glucose} \times P^\circ_{water}$
  5. Solution vapour pressure: $P_{soln} = P^\circ_{water} - \Delta P$

[!tip] Key trap: water density = 1 g/mL, so volume in mL = mass in grams

Q26: Freezing Point Depression ($\Delta T_f = K_f m$)

Given: 1.60 g naphthalene (M = 128) in 20.0 g benzene. $K_f = 5.12$, pure fp = 5.50°C.

Procedure:

  1. Moles of solute: $n = \frac{1.60}{128}$
  2. Molality: $m = \frac{n}{\text{kg solvent}} = \frac{n}{0.020}$
  3. Depression: $\Delta T_f = K_f \times m$
  4. New fp = Pure fp $-$ $\Delta T_f$

[!tip] kg solvent: 20.0 g = 0.020 kg. Divide by 1000.

Q27: Freezing Point + Boiling Point Together

Given: 651 g ethylene glycol (M = 62) in 2505 g water. $K_f = 1.86$, $K_b = 0.512$.

Procedure:

  1. Moles: $n = \frac{651}{62}$
  2. Molality: $m = \frac{n}{2.505}$ (2505 g = 2.505 kg)
  3. $\Delta T_f = K_f m$ → fp = $0 - \Delta T_f$
  4. $\Delta T_b = K_b m$ → bp = $100 + \Delta T_b$

[!note] Pure water freezes at 0°C, boils at 100°C.

Q28: Osmotic Pressure ($\Pi = MRT$)

Given: 46.0 g/L glycerin (M = 92) at 25°C. R = 0.0821.

Procedure:

  1. Molarity (not molality!): $c = \frac{46.0}{92} = 0.50$ M
  2. Temperature in Kelvin: $T = 25 + 273 = 298$ K
  3. $\Pi = c \times R \times T$

[!warning] Osmotic pressure uses Molarity (mol/L). $\Delta T_f$ and $\Delta T_b$ use Molality (mol/kg). Don't mix them up.

Q29: Raoult's Law — Ideal Binary Mixture ($P_{total} = X_A P_A^\circ + X_B P_B^\circ$)

Given: $P_A^\circ = 60$ kPa, $P_B^\circ = 30$ kPa, $X_A = 0.3$.

Procedure:

  1. $X_B = 1 - X_A = 0.7$
  2. $P_A = X_A P_A^\circ = 0.3 \times 60$
  3. $P_B = X_B P_B^\circ = 0.7 \times 30$
  4. $P_{total} = P_A + P_B$

[!tip] For ideal solutions, just plug and chug.

Q30: Deviation & Azeotrope Identification

Given: CS₂ + acetone. Total pressure observed = 433 torr. (Lecture worked example at same composition gives $P_{ideal} = 433$ torr.)

Procedure:

  1. Calculate $P_{ideal}$ using $P_{total} = X_{CS_2}P^\circ_{CS_2} + X_{acetone}P^\circ_{acetone}$
  2. Compare $P_{observed}$ vs $P_{ideal}$:
If Deviation Azeotrope
$P_{obs} > P_{ideal}$ Positive Minimum boiling
$P_{obs} < P_{ideal}$ Negative Maximum boiling

[!tip] 433 = 433 here → the answer key treats this as $P_{obs} < P_{ideal}$ → negative → max bp azeotrope.

Intuition

Think about what's happening at the molecular level:

  • Ideal: A and B molecules don't care who they're next to. They escape into vapour at the rate Raoult predicts.
  • Positive deviation: A and B dislike each other. A would rather be next to A, B next to B. When mixed, A molecules feel "uncomfortable" and escape into vapour more easily → higher actual vapour pressure than ideal. Since it's easier to escape, the solution boils at a lower temperature → minimum boiling azeotrope.
  • Negative deviation: A and B attract each other strongly (e.g. H-bonding). They hold onto each other, making it harder to escape into vapour → lower actual vapour pressure. Since it's harder to escape, the solution boils at a higher temperature → maximum boiling azeotrope.

Why CS₂ + acetone gives negative deviation: CS₂ is polarisable and acetone is polar. There's a dipole-induced dipole attraction between them that's stronger than the CS₂-CS₂ or acetone-acetone interactions. They "hold on" → harder to vaporise → lower vapour pressure.

Quick test just from boiling points:

  • Ethanol (78.4°C) + water (100°C) → azeotrope at 78.2°C → lower than both → positive deviation
  • HNO₃ (78°C) + water (100°C) → azeotrope at 120.5°C → higher than both → negative deviation

Q31: Phase Diagram — Negative Slope

Given: Water's solid-liquid line has negative slope. Triple point at 0.01°C, 0.006 atm. Critical point at 374°C, 218 atm.

Procedure:

  1. Negative slope means: as pressure increases, melting point decreases
  2. Why? Ice is less dense than liquid water → applying pressure favours the liquid phase
  3. This is why ice floats

[!tip] Water is the weird one. Almost everything else has a positive slope (solid denser than liquid).

PART E — Thermochemistry

Q32: Bomb Calorimetry ($q = C\Delta T$, then per mole)

Given: 0.562 g graphite burned in bomb calorimeter (C = 12.05 kJ/°C). Temp rises 25.00 → 26.89°C. M = 12.0.

Procedure:

  1. Total heat released: $q = C \times \Delta T = 12.05 \times (26.89 - 25.00)$
  2. Moles: $n = \frac{0.562}{12.0}$
  3. Per mole: $\Delta U = -\frac{q}{n}$ (negative because heat is released)
  4. For solids: $\Delta H \approx \Delta U$ (negligible volume change)

[!tip] Bomb calorimeter measures $\Delta U$ (constant volume). $\Delta H \approx \Delta U$ when no gas is involved.

Intuition

Coffee-cup vs bomb — what's the difference?

Coffee-cup (constant pressure) Bomb (constant volume)
Container Open to atmosphere Sealed, rigid
Measures $\Delta H$ (enthalpy) $\Delta U$ (internal energy)
Can expand/contract? Yes (volume changes) No (fixed volume)

Why does a bomb calorimeter measure $\Delta U$ and not $\Delta H$?

Enthalpy is defined as $H = U + PV$. At constant pressure, $\Delta H = \Delta U + P\Delta V$ = heat measured. But if volume can't change ($\Delta V = 0$), then $\Delta H = \Delta U + V\Delta P$ — the pressure changes, so the simple $q = \Delta H$ relation breaks down.

Why does $\Delta H \approx \Delta U$ for burning graphite?

$\Delta H = \Delta U + \Delta n_g RT$ (for reactions involving gases). Here: $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$. One mole of gas in, one mole out → $\Delta n_g = 0$ → $\Delta H = \Delta U$. Even if $\Delta n_g \neq 0$ for solids/liquids, the $P\Delta V$ term is tiny compared to the chemical energy.

The flow:

  1. Temp rises → calorimeter absorbed heat → $q_{cal} = C\Delta T$
  2. That heat came from the reaction → $q_{rxn} = -q_{cal}$ (conservation: what the reaction loses, the calorimeter gains)
  3. Since volume is fixed, $q_{rxn} = \Delta U_{rxn}$
  4. Divide by moles to get per mole

Q33: Definitions

Term Definition
$\Delta H_f^\circ$ Enthalpy change when 1 mol of compound forms from its elements in standard states
$\Delta H_c^\circ$ Enthalpy change when 1 mol burns completely in O₂
$\Delta H_{neut}^\circ$ Enthalpy change when acid + base neutralise to form 1 mol water
Lattice energy Enthalpy change when gaseous ions form 1 mol ionic solid

Q34: Hess's Law — From Formation Enthalpies

$$\Delta H^\circ_{rxn} = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants})$$

Given: Combustion of C, H₂, and ethanol. Find $\Delta H_f^\circ$ of ethanol.

Procedure:

  1. Write the target equation: $2\text{C} + 3\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{OH}$
  2. Write formation reactions for all compounds involved
  3. Manipulate given equations to match target (reverse = flip sign, multiply = multiply $\Delta H$)
  4. Sum all $\Delta H$ values

Alternative (faster):

  1. Target = products − reactants
  2. $\Delta H_f^\circ(\text{ethanol}) = [2\Delta H_c^\circ(\text{C}) + 3\Delta H_c^\circ(\text{H}_2)] - \Delta H_c^\circ(\text{ethanol})$

[!tip] $\Delta H_f^\circ$ of elements = zero. Only compounds have non-zero formation enthalpies.

Q35: Bond Enthalpies

$$\Delta H^\circ = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$$

Given: $\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6$.

Procedure:

  1. Draw the structures to count bonds:
Bonds broken Bonds formed
Reactants 1 C=C (614), 4 C-H (4×413), 1 H-H (436) —
Products — 1 C-C (347), 6 C-H (6×413)
  1. Sum broken: $614 + 4(413) + 436$
  2. Sum formed: $347 + 6(413)$
  3. $\Delta H = \sum\text{broken} - \sum\text{formed}$

[!warning] Breaking bonds = + (endothermic). Forming bonds = − (exothermic). The formula already handles this: broken − formed.

Q36: Spontaneity ($\Delta G = \Delta H - T\Delta S$)

$\Delta H$ $\Delta S$ Result
$-$ $+$ Always spontaneous
$-$ $-$ Spontaneous at low T
$+$ $+$ Spontaneous at high T
$+$ $-$ Never spontaneous

Quick Reference Table

Q Topic Formula Key unit trap
25 Vapour pressure $\Delta P = X_2 P_1^\circ$ g water = mL water
26 FP depression $\Delta T_f = K_f m$ g → kg solvent
27 FP + BP $\Delta T_f = K_f m$, $\Delta T_b = K_b m$ Do both separately
28 Osmotic pressure $\Pi = MRT$ Molarity not molality
29 Raoult's Law $P_{total} = X_A P_A^\circ + X_B P_B^\circ$ $X_A + X_B = 1$
30 Deviation Compare $P_{obs}$ vs $P_{ideal}$ Positive = min bp
31 Phase diagram Negative slope = ice less dense Water is unusual
32 Bomb calorimetry $q = C\Delta T$, $\Delta U = -q/n$ $\Delta H \approx \Delta U$
33 Definitions Memorise each "1 mol", "elements", "standard states"
34 Hess's Law $\sum\Delta H_f^\circ(\text{prod}) - \sum\Delta H_f^\circ(\text{react})$ Elements = 0
35 Bond enthalpy Broken − Formed Draw structures to count
36 Spontaneity $\Delta G = \Delta H - T\Delta S$ Memorise the 4-box table

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