Thermodynamics

Thermodynamics

Part 1 of 5 — Enthalpy ($\Delta H$) & Hess's Law

System vs. Surroundings

• System: The reaction or process under study
• Surroundings: Everything else (including the calorimeter/solution)
• First Law of Thermodynamics: Energy is conserved: $\Delta E_{\text{universe}} = 0$; $\Delta E_{\text{sys}} = -\Delta E_{\text{surr}}$

Enthalpy ($H$)

Enthalpy change $\Delta H$ = heat transferred at constant pressure.

$\Delta H = H_{\text{products}} - H_{\text{reactants}}$

Calorimetry at constant pressure:

$q = mc\Delta T$

where $m$ = mass (g), $c$ = specific heat (J/g·°C), $\Delta T = T_{\text{final}} - T_{\text{initial}}$.

For the reaction: $\Delta H_{\text{rxn}} = -q_{\text{solution}}$ (the sign flip — if the solution warms, the reaction is exothermic)

Standard Enthalpy of Formation ($\Delta H°_f$)

$\Delta H°_f$ = enthalpy change when 1 mole of a compound is formed from its elements in their standard states.

By convention: $\Delta H°_f = 0$ for pure elements in standard states.

Hess's Law: $\Delta H°_{\text{rxn}} = \sum n \Delta H°_f(\text{products}) - \sum m \Delta H°_f(\text{reactants})$

Hess's Law (Path Independence)

Enthalpy is a state function — $\Delta H$ depends only on initial and final states, not the path.

Method: Reactions can be added/subtracted like algebraic equations to obtain a target reaction.

Rules for manipulating reactions:

1. If a reaction is reversed, flip the sign of $\Delta H$
2. If a reaction is multiplied by a factor, multiply $\Delta H$ by the same factor

Worked Example

Find $\Delta H$ for: $\text{C}(s) + \frac{1}{2}\text{O}_2(g) \to \text{CO}(g)$

Given:

• (1) $\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$; $\Delta H_1 = -393.5$ kJ
• (2) $\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \to \text{CO}_2(g)$; $\Delta H_2 = -283.0$ kJ

Reverse reaction (2): $\text{CO}_2(g) \to \text{CO}(g) + \frac{1}{2}\text{O}_2(g)$; $\Delta H = +283.0$ kJ

Add to reaction (1):

$\text{C}(s) + \text{O}_2 + \text{CO}_2 \to \text{CO}_2 + \text{CO} + \frac{1}{2}\text{O}_2$

$\text{C}(s) + \frac{1}{2}\text{O}_2(g) \to \text{CO}(g)$

$\Delta H = -393.5 + 283.0 = \mathbf{-110.5\text{ kJ}}$

Bond Enthalpy Approximation

$\Delta H_{\text{rxn}} \approx \sum (\text{bonds broken}) - \sum (\text{bonds formed})$

Breaking bonds requires energy (+); forming bonds releases energy (−). This gives an estimate; less accurate than using $\Delta H°_f$ values.

Enthalpy & Hess's Law 🎯

Key Takeaways — Part 1

• $\Delta H < 0$: exothermic (releases heat); $\Delta H > 0$: endothermic (absorbs heat)
• Calorimetry: $q = mc\Delta T$; $\Delta H_{\text{rxn}} = -q_{\text{soln}}$
• Hess's Law: $\Delta H$ is a state function; add reactions algebraically
• Reverse a reaction → flip sign of $\Delta H$; multiply by factor → multiply $\Delta H$
• Formation enthalpy: $\Delta H°_{\text{rxn}} = \sum n\Delta H°_f(\text{products}) - \sum m\Delta H°_f(\text{reactants})$
• $\Delta H°_f = 0$ for elements in standard states

Thermodynamics

Part 2 of 5 — Entropy ($\Delta S$), Spontaneity & the Laws of Thermodynamics

Entropy ($S$)

Entropy is a measure of disorder or randomness of a system.

$\Delta S = S_{\text{products}} - S_{\text{reactants}}$

Predicting the Sign of $\Delta S$

$\Delta S > 0$ (entropy increases) when:

1. Gases are produced from solids/liquids ($\Delta n_{\text{gas}} > 0$)
2. A solid dissolves to form aqueous ions
3. Temperature increases (for any substance)
4. A substance vaporizes or melts
5. Mixing occurs
6. Gas occupies larger volume

$\Delta S < 0$ (entropy decreases) when:

1. Gases are consumed ($\Delta n_{\text{gas}} < 0$, e.g., 3 mol gas → 1 mol gas)
2. A substance crystallizes/freezes
3. Complex molecules form from simpler ones (polymerization)

Laws of Thermodynamics

Zeroth Law: If A is in thermal equilibrium with B, and B with C, then A is in equilibrium with C. (Defines temperature)

First Law: Energy is conserved. $\Delta E_{\text{universe}} = 0$.

$\Delta U = q + w \quad (\text{internal energy} = \text{heat} + \text{work})$

where $w = -P\Delta V$ for expansion/compression work.

Second Law: The entropy of the universe always increases for spontaneous processes.

$\Delta S_{\text{universe}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0 \quad (\text{spontaneous})$

Third Law: The entropy of a perfect crystal at absolute zero (0 K) is exactly zero. This establishes an absolute reference for entropy measurements.

Standard Molar Entropy ($S°$)

$S° > 0$ for all pure substances (third law). More complex molecules have higher $S°$ values.

$\Delta S°_{\text{rxn}} = \sum n S°(\text{products}) - \sum m S°(\text{reactants})$

Spontaneity — First Look

A process is spontaneous if it can proceed without continuous external input of energy.

Caution: Spontaneous does NOT mean fast (kinetics is separate from thermodynamics).

• Exothermic processes ($\Delta H < 0$) tend to be spontaneous — enthalpy drives spontaneity
• Disorder increasing ($\Delta S > 0$) tends to be spontaneous — entropy drives spontaneity
• But both factors matter — see Gibbs free energy (Part 3)

Entropy & Laws of Thermodynamics 🎯

Key Takeaways — Part 2

• Entropy increases when gases form, substances dissolve, temperature rises, or mixing occurs
• Entropy decreases when gases condense, substances crystallize, or complexity increases
• Second Law: $\Delta S_{\text{universe}} > 0$ for all spontaneous processes
• Third Law: $S = 0$ for perfect crystal at 0 K; all $S° > 0$ at room temperature
• Spontaneous ≠ fast; thermodynamics says nothing about how quickly a reaction occurs
• $\Delta H < 0$ favors spontaneity; $\Delta S > 0$ favors spontaneity; temperature controls the trade-off

Thermodynamics

Part 3 of 5 — Gibbs Free Energy, $K_{eq}$ & Cell Potential

Gibbs Free Energy ($G$)

$\Delta G = \Delta H - T\Delta S$

$\Delta G$ combines both enthalpy and entropy to predict spontaneity:

Temperature governs which effect wins when $\Delta H$ and $\Delta S$ have the same sign.

Standard Gibbs Free Energy ($\Delta G°$)

$\Delta G° = \Delta H° - T\Delta S°$

Also calculable from formation values: $\Delta G° = \sum n\Delta G°_f(\text{products}) - \sum m\Delta G°_f(\text{reactants})$

Relationship Between $\Delta G°$ and $K_{eq}$

$\Delta G° = -RT\ln K$

where $R = 8.314$ J/mol·K, $T$ in Kelvin.

Large negative $\Delta G°$ → large $K$ → reaction runs nearly to completion.

Relationship to Cell Potential ($E° _{\text{cell}}$)

For electrochemical reactions:

$\Delta G° = -nFE°_{\text{cell}}$

where:

• $n$ = moles of electrons transferred
• $F = 96{,}485$ C/mol (Faraday constant)
• $E°_{\text{cell}}$ measured in volts (V)

Connection between $\Delta G°$, $K$, and $E°_{\text{cell}}$:

$\Delta G° = -RT\ln K = -nFE°_{\text{cell}}$

$\Delta G$ at Non-Standard Conditions

When the system is not at standard conditions:

$\Delta G = \Delta G° + RT\ln Q$

At equilibrium, $\Delta G = 0$ and $Q = K$, which yields $\Delta G° = -RT\ln K$ (as above).

Coupled Reactions

In biochemistry, non-spontaneous reactions are driven by coupling with spontaneous ones:

Example: Glucose phosphorylation ($\Delta G° = +13.8$ kJ) is coupled with ATP hydrolysis ($\Delta G° = -30.5$ kJ):

$\Delta G°_{\text{total}} = 13.8 + (-30.5) = -16.7\text{ kJ} \quad \text{(spontaneous)}$

This is how cells use ATP to drive energetically unfavorable biochemical reactions.

Gibbs Free Energy, $K_{eq}$ & $E°_{\text{cell}}$ 🎯

Key Takeaways — Part 3

• $\Delta G = \Delta H - T\Delta S$: negative $\Delta G$ = spontaneous
• $\Delta G° = -RT\ln K$: negative $\Delta G°$ → $K > 1$ (products favored)
• $\Delta G° = -nFE°_{\text{cell}}$: positive $E°_{\text{cell}}$ → negative $\Delta G°$ → spontaneous galvanic cell
• At equilibrium: $\Delta G = 0$, $Q = K$
• Non-standard conditions: $\Delta G = \Delta G° + RT\ln Q$
• Coupled reactions: add $\Delta G°$ values; ATP hydrolysis drives non-spontaneous biochemical reactions
• The $\Delta G$, $K$, and $E°_{\text{cell}}$ are all interconnected — if you know one, you can find the others

Thermodynamics

Part 4 of 5 — Phase Changes, Heating Curves & Colligative Properties

Phases and Phase Changes

The six phase transitions:

At the phase transition temperature (e.g., boiling point), $\Delta G = 0$:

$\Delta G = \Delta H - T\Delta S = 0 \Rightarrow T_{\text{transition}} = \frac{\Delta H}{\Delta S}$

Heating Curves

A heating curve shows temperature vs. heat added for a substance.

Key features:

1. Slanted region: Temperature rises; $q = mc\Delta T$; specific heat governs slope
2. Flat region (plateau): Phase change at constant T; $q = n\Delta H_{\text{fus}}$ or $q = n\Delta H_{\text{vap}}$
3. Steeper slope = lower specific heat; gentle slope = higher specific heat

Specific heats to know (MCAT):

• Water (liquid): $4.18$ J/g·°C (highest of common substances)
• Steam: $2.01$ J/g·°C
• Ice: $2.09$ J/g·°C
• $\Delta H_{\text{fus}}$(water) = $6.01$ kJ/mol
• $\Delta H_{\text{vap}}$(water) = $40.7$ kJ/mol

Example calculation: Heat to convert 18.0 g (1.00 mol) of ice at −10°C to steam at 110°C:

1. Heat ice: $18.0 \times 2.09 \times 10 = 376.2$ J
2. Melt ice: $1.00 \times 6010 = 6010$ J
3. Heat water: $18.0 \times 4.18 \times 100 = 7524$ J
4. Boil water: $1.00 \times 40{,}700 = 40{,}700$ J
5. Heat steam: $18.0 \times 2.01 \times 10 = 361.8$ J

Total = $376.2 + 6010 + 7524 + 40{,}700 + 361.8 \approx \mathbf{54{,}972\text{ J} \approx 55.0\text{ kJ}}$

Phase Diagrams

A phase diagram plots pressure vs. temperature; shows stable phase regions divided by boundary curves.

Water is unusual: Solid-liquid line has negative slope (ice is less dense than water; increasing pressure melts ice).

Colligative Properties (Review)

Colligative properties depend on the number of solute particles, not their identity.

$i$ = van 't Hoff factor (particles per formula unit); $m$ = molality (mol solute / kg solvent).

Phase Changes & Heating Curves 🎯

Key Takeaways — Part 4

• Endothermic phase changes: melting, vaporization, sublimation (require heat input)
• Exothermic phase changes: freezing, condensation, deposition (release heat)
• Phase transition temperature: $T = \Delta H / \Delta S$ (where $\Delta G = 0$)
• Heating curve plateaus = phase transitions at constant T; sloped regions = $q = mc\Delta T$
• Water: $c_{\text{liq}} = 4.18$; $c_{\text{steam}} = 2.01$; $c_{\text{ice}} = 2.09$ J/g·°C; $\Delta H_{\text{vap}} \gg \Delta H_{\text{fus}}$
• Phase diagram: triple point (all 3 coexist); critical point (gas/liquid no distinction)
• Water's negative solid-liquid slope = unique (ice less dense than liquid water)

Thermodynamics

Part 5 of 5 — Mixed MCAT Review

High-Yield Checklist

• ✅ $\Delta H$: exothermic (−) releases heat; endothermic (+) absorbs heat
• ✅ Hess's Law: $\Delta H$ is a state function; add reactions algebraically
• ✅ $\Delta S > 0$: more disorder (gas formed, dissolution, mixing, heating)
• ✅ $\Delta G = \Delta H - T\Delta S$: negative = spontaneous
• ✅ $\Delta G° = -RT\ln K = -nFE°_{\text{cell}}$
• ✅ Phase transitions: exothermic going toward solid; endothermic going toward gas
• ✅ Heating curve: plateaus at phase transitions ($q = n\Delta H$); slopes = $q = mc\Delta T$
• ✅ Colligative properties: $\Delta T_b = K_b m i$; $\Delta T_f = K_f m i$; $\Pi = iMRT$

Mixed Thermodynamics — MCAT-Style Questions 🎯

Thermodynamics — Complete Topic Summary

Part 1: Enthalpy, exothermic/endothermic, calorimetry ($q = mc\Delta T$), Hess's Law, standard enthalpies of formation.

Part 2: Entropy, predicting $\text{sign}(\Delta S)$, First/Second/Third Laws, entropy at phase transitions, spontaneity matrix.

Part 3: Gibbs free energy ($\Delta G = \Delta H - T\Delta S$), relationship to $K_{eq}$ ($\Delta G° = -RT\ln K$), relationship to cell potential ($\Delta G° = -nFE°$), coupled reactions.

Part 4: Phase changes (endo/exothermic), heating curves, specific heats of water, phase diagrams, triple point, critical point, colligative properties review.

Part 5: Integrated MCAT practice.

Most Tested MCAT Thermodynamics Concepts

1. $\Delta G = 0$ at equilibrium or phase transition temperature
2. $\Delta G°, K, E°_{\text{cell}}$ are all related: $-nFE° = \Delta G° = -RT\ln K$
3. Spontaneous at all T: when $\Delta H < 0$ and $\Delta S > 0$
4. Heating curve plateaus — never confuse with temperature changes
5. Phase transition: $T_{\text{transition}} = \Delta H/\Delta S$
6. Colligative properties depend on number of particles, not identity