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Thermodynamics

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Thermodynamics - Complete Interactive Lesson

Part 1: Enthalpy & Hess's Law

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.

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

$\Delta H$	Name	Energy transfer
Negative ( $\Delta H < 0$ )	Exothermic	Releases heat to surroundings
Positive ( $\Delta H > 0$ )	Endothermic

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:

If a reaction is reversed, flip the sign of $\Delta H$
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) $\text{C}(s) + \text{O}_2(g) \to \text{CO}_2(g)$ ; kJ

Reverse reaction (2): $\text{CO}_2(g) \to \text{CO}(g) + \frac{1}{2}\text{O}_2(g)$ ; 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: $q = mc\Delta T$ ;

Part 2: Entropy & Laws of Thermodynamics

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.

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

Part 3: Gibbs Free Energy, Keq & Cell Potential

Thermodynamics

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

Gibbs Free Energy ( $G$ )

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

Part 4: Phase Changes & Heating Curves

Thermodynamics

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

Phases and Phase Changes

The six phase transitions:

Transition	Direction	$\Delta H$	$\Delta S$
Fusion (melting)	solid → liquid	+ (endothermic)	+
Freezing	liquid → solid	− (exothermic)	−
Vaporization	liquid → gas	+ (endothermic)	+
Condensation

Part 5: Mixed MCAT Review

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)
✅ $Δ G = Δ H - T Δ$ : negative = spontaneous

∑mΔH°f(reactants)

\Delta H_1 = -393.5

(2)

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

;

\Delta H_2 = -283.0

\Delta H_{\text{rxn}} = -q_{\text{soln}}

Δ H_{rxn} = - q_{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

$\Delta S$	Meaning
Positive ( $\Delta S > 0$ )	Increased disorder
Negative ( $\Delta S < 0$ )	Decreased disorder

Predicting the Sign of $\Delta S$

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

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

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

Gases are consumed ( $\Delta n_{\text{gas}} < 0$ , e.g., 3 mol gas → 1 mol gas)
A substance crystallizes/freezes
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)

$\Delta H$	$\Delta S$	Spontaneous?
−	+	Always spontaneous
+	−	Never spontaneous
−	−	Depends on T (low T favors)
+	+	Depends on T (high T favors)

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

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

$\Delta G$	Process
$\Delta G < 0$	Spontaneous (thermodynamically favorable)
$\Delta G = 0$	At equilibrium
$\Delta G > 0$	Non-spontaneous (requires energy input)

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.

$\Delta G°$	$K$	Meaning
$< 0$	$K > 1$	Products favored at equilibrium
$= 0$	$K = 1$	Reactants and products roughly equal
$> 0$	$K < 1$	Reactants favored at equilibrium

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

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

For electrochemical reactions:

$\Delta G° = -nFE°_{\text{cell}}$

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

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

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

Sign of $E°_{\text{cell}}$	$\Delta G°$	Spontaneous?
$E°_{\text{cell}} > 0$	Negative	Yes
$E°_{\text{cell}} < 0$	Positive	No
$E°_{\text{cell}} = 0$	Zero	At equilibrium

$\Delta G$ 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).

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 → 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

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

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

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

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

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

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

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

Point/Line	Meaning
Triple point	All three phases coexist
Critical point	Gas and liquid become indistinguishable
Solid-liquid line	Melting points at various pressures
Liquid-gas line	Boiling points at various pressures

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.

Property	Formula	Key notes
Boiling point elevation	$\Delta T_b = K_b m i$	Higher BP than solvent
Freezing point depression	$\Delta T_f = K_f m i$	Lower FP than solvent
Osmotic pressure	$\Pi = iMRT$	$R = 0.0821$ L·atm/mol·K
Vapor pressure lowering	$\Delta P = \chi_{\text{solute}} P°_\text{solvent}$	Raoult's law

$i$ = 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: $c_{\text{liq}} = 4.18$ ; $c_{\text{steam}} = 2.01$ ; J/g·°C;
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)

Δ G = Δ H - T Δ S

✅

\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

$\Delta G = 0$ at equilibrium or phase transition temperature
$\Delta G°, K, E°_{\text{cell}}$ are all related: $- n F$

c_{\text{ice}} = 2.09

\Delta H_{\text{vap}} \gg \Delta H_{\text{fus}}

- n FE ° = Δ G ° = - RT ln K

Spontaneous at all T: when

\Delta H < 0

and

\Delta S > 0

Heating curve plateaus — never confuse with temperature changes

Phase transition:

T_{\text{transition}} = \Delta H/\Delta S

Colligative properties depend on number of particles, not identity

Thermodynamics

Predicting the Sign of $\Delta S$

Laws of Thermodynamics

Standard Molar Entropy ( $S°$ )

Spontaneity — First Look

Key Takeaways — Part 2

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

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

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

$\Delta G$ at Non-Standard Conditions

Coupled Reactions

Key Takeaways — Part 3

Heating Curves

Phase Diagrams

Colligative Properties (Review)

Key Takeaways — Part 4

Thermodynamics — Complete Topic Summary

Most Tested MCAT Thermodynamics Concepts

Thermodynamics

Thermodynamics - Complete Interactive Lesson

Part 1: Enthalpy & Hess's Law

Thermodynamics

System vs. Surroundings

Enthalpy (HHH)

Standard Enthalpy of Formation (ΔH°f\Delta H°_fΔH°f​)

Hess's Law (Path Independence)

Worked Example

Bond Enthalpy Approximation

Key Takeaways — Part 1

Part 2: Entropy & Laws of Thermodynamics

Thermodynamics

Entropy (SSS)

Part 3: Gibbs Free Energy, Keq & Cell Potential

Thermodynamics

Gibbs Free Energy (GGG)

Part 4: Phase Changes & Heating Curves

Thermodynamics

Phases and Phase Changes

Part 5: Mixed MCAT Review

Thermodynamics

High-Yield Checklist

Predicting the Sign of ΔS\Delta SΔS

Laws of Thermodynamics

Standard Molar Entropy (S°S°S°)

Spontaneity — First Look

Key Takeaways — Part 2

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

Relationship Between ΔG°\Delta G°ΔG° and KeqK_{eq}Keq​

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

ΔG\Delta GΔG at Non-Standard Conditions

Coupled Reactions

Key Takeaways — Part 3

Heating Curves

Phase Diagrams

Colligative Properties (Review)

Key Takeaways — Part 4

Thermodynamics — Complete Topic Summary

Most Tested MCAT Thermodynamics Concepts

Enthalpy ( $H$ )

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

Entropy ( $S$ )

Gibbs Free Energy ( $G$ )

Predicting the Sign of $\Delta S$

Standard Molar Entropy ( $S°$ )

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

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

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

$\Delta G$ at Non-Standard Conditions