⚡ Gibbs Free Energy and Spontaneity

Part 1 of 7 — ΔG = ΔH − TΔS

Gibbs Free Energy ($G$) is the single most important thermodynamic quantity for chemists. It combines enthalpy and entropy into one number that tells you whether a reaction is spontaneous — without needing to calculate the entropy of the surroundings.

Defining Gibbs Free Energy

$G = H - TS$

The change in Gibbs free energy at constant temperature:

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

Where Does This Come From?

Recall: $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}$

And: $\Delta S_{\text{surroundings}} = -\Delta H_{\text{system}}/T$

So: $\Delta S_{\text{universe}} = \Delta S_{\text{sys}} - \Delta H_{\text{sys}}/T$

Multiply by $-T$:

$-T\Delta S_{\text{universe}} = \Delta H_{\text{sys}} - T\Delta S_{\text{sys}} = \Delta G$

Since $\Delta S_{\text{universe}} > 0$ for spontaneous processes:

$\Delta G < 0 \quad \text{(spontaneous)}$

The Spontaneity Criterion

Why Gibbs Free Energy Is So Useful

• It accounts for both enthalpy and entropy
• It is a property of the system only — no need to calculate $\Delta S_{\text{surroundings}}$
• It connects directly to equilibrium and electrochemistry

What "Free" Means

"Free energy" is the maximum amount of energy available to do useful work (non-$PV$ work) in a reaction.

$w_{\text{max}} = \Delta G$

If $\Delta G = -100$ kJ, the reaction can do at most 100 kJ of useful work.

Temperature and Spontaneity

From $\Delta G = \Delta H - T\Delta S$, we see that temperature affects spontaneity through the $T\Delta S$ term:

• At low temperatures: $\Delta H$ dominates ($T\Delta S$ is small)
• At high temperatures: $T\Delta S$ dominates ($T\Delta S$ is large)

The Crossover Temperature

When $\Delta G = 0$ (equilibrium):

$T = \frac{\Delta H}{\Delta S}$

This is the temperature at which the reaction switches between spontaneous and nonspontaneous.

Example

For ice melting: $\Delta H = +6.01$ kJ/mol, $\Delta S = +22.0$ J/(mol·K)

$T = \frac{6010}{22.0} = 273 \text{ K} = 0°\text{C}$

Above 273 K: melting is spontaneous ($\Delta G < 0$). Below 273 K: freezing is spontaneous.

Gibbs Free Energy Concept Quiz 🎯

Gibbs Free Energy Calculations 🧮

1. $\Delta H = -100$ kJ, $\Delta S = +50$ J/K, $T = 298$ K. Calculate $\Delta G$ in kJ. (to 3 significant figures)

2. $\Delta H = +200$ kJ, $\Delta S = +500$ J/K, $T = 500$ K. Calculate $\Delta G$ in kJ.

3. A reaction has $\Delta H = +30$ kJ and $\Delta S = +100$ J/K. At what temperature (in K) is $\Delta G = 0$?

Gibbs Free Energy Concepts 🔽

Exit Quiz — Gibbs Free Energy ✅

🔀 Four ΔH/ΔS Combinations

Part 2 of 7 — Always, Never, or Temperature-Dependent

The equation $\Delta G = \Delta H - T\Delta S$ has four possible sign combinations for $\Delta H$ and $\Delta S$. Understanding these four cases is essential for AP Chemistry and lets you quickly assess spontaneity.

The Four Cases

Case 1: ΔH < 0, ΔS > 0 — Always Spontaneous ✅

$\Delta G = (\text{negative}) - T(\text{positive}) = \text{always negative}$

• Both terms favor spontaneity
• Spontaneous at all temperatures
• Example: combustion of hydrocarbons

Case 2: ΔH > 0, ΔS < 0 — Never Spontaneous ❌

$\Delta G = (\text{positive}) - T(\text{negative}) = \text{always positive}$

• Both terms oppose spontaneity
• Never spontaneous (the reverse reaction is always spontaneous)
• Example: the reverse of combustion

Case 3: ΔH < 0, ΔS < 0 — Spontaneous at Low T 🥶

$\Delta G = (\text{negative}) - T(\text{negative})$

• Exothermic but entropy-decreasing
• At low T: $|\Delta H| > |T\Delta S|$ → $\Delta G < 0$
• At high T: $|T\Delta S| > |\Delta H|$ → $\Delta G > 0$
• Example: freezing of water

Case 4: ΔH > 0, ΔS > 0 — Spontaneous at High T 🔥

$\Delta G = (\text{positive}) - T(\text{positive})$

• Endothermic but entropy-increasing
• At high T: $|T\Delta S| > |\Delta H|$ → $\Delta G < 0$
• At low T: $|\Delta H| > |T\Delta S|$ → $\Delta G > 0$
• Example: melting of ice, vaporization

Summary Table

The Crossover Temperature

For Cases 3 and 4, the temperature where $\Delta G = 0$:

$T_{\text{crossover}} = \frac{\Delta H}{\Delta S}$

(Both $\Delta H$ and $\Delta S$ must have the same sign for this temperature to be positive and physically meaningful.)

Real-World Examples

Case 1 (Always Spontaneous): Combustion

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$

• $\Delta H < 0$ (releases heat)
• $\Delta S > 0$ ($\Delta n_{\text{gas}} = 3 - 3 = 0$, but products are more complex — actually $\Delta S$ can be slightly negative for this specific reaction at standard conditions)

Case 3 (Low T): Freezing Water

$\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(s)$

• $\Delta H < 0$ (releases heat — exothermic)
• $\Delta S < 0$ (liquid → solid, more ordered)
• Spontaneous only below 273 K

Case 4 (High T): Melting Ice

$\text{H}_2\text{O}(s) \rightarrow \text{H}_2\text{O}(l)$

• $\Delta H > 0$ (absorbs heat — endothermic)
• $\Delta S > 0$ (solid → liquid, more disordered)
• Spontaneous only above 273 K

Four Cases Quiz 🎯

Classify the Reaction 🧮

For each combination, type "always", "never", "low T", or "high T" for when the reaction is spontaneous:

1. $\Delta H < 0$, $\Delta S > 0$

2. $\Delta H > 0$, $\Delta S > 0$

3. $\Delta H > 0$, $\Delta S < 0$

Spontaneity and Temperature 🔽

Exit Quiz — Four Cases ✅

🏗️ Standard Free Energy of Formation

Part 3 of 7 — Calculating ΔG° from Tables

Just as we used $\Delta H°_f$ to calculate $\Delta H°_{\text{rxn}}$, we can use standard free energies of formation ($\Delta G°_f$) to calculate $\Delta G°_{\text{rxn}}$. The formula is identical in structure.

Standard Free Energy of Formation ($\Delta G°_f$)

The free energy change when one mole of a compound is formed from its elements in their standard states at standard conditions.

The Master Equation

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

Key Rule

$\Delta G°_f = 0 \text{ for all elements in their standard states}$

(Same convention as $\Delta H°_f$)

Sample Values

Worked Example

Calculate $\Delta G°$ for: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

$\Delta G° = [(-394.4) + 2(-237.1)] - [(-50.7) + 2(0)]$ $= [-394.4 - 474.2] - [-50.7]$ $= -868.6 + 50.7 = -817.9 \text{ kJ}$

The large negative $\Delta G°$ confirms that combustion of methane is very spontaneous.

Two Methods to Calculate ΔG°

1. From $\Delta G°_f$ values (this method) — direct lookup
2. From $\Delta H°$ and $\Delta S°$: $\Delta G° = \Delta H° - T\Delta S°$

Both methods give the same answer at 25°C.

Standard Free Energy Quiz 🎯

ΔG° Calculations 🧮

Use these $\Delta G°_f$ values (kJ/mol): $\text{CO}_2(g) = -394.4$, $\text{H}_2\text{O}(l) = -237.1$, $\text{C}_2\text{H}_6(g) = -32.0$, $\text{O}_2(g) = 0$

1. Calculate $\Delta G°$ for: $\text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$ (in kJ, to 1 decimal)

2. Is this reaction spontaneous under standard conditions? (type "yes" or "no")

Round all answers to 3 significant figures.

Comparing the Three Formation Quantities

Common AP Mistake

Students often confuse these three quantities. Remember:

• $\Delta H°_f$ and $\Delta G°_f$ are zero for elements in standard states
• $S°$ is NOT zero — it is always positive at $T > 0$ K

Formation Free Energy Concepts 🔽

Exit Quiz — Standard Free Energy ✅

⚖️ ΔG and Equilibrium

Part 4 of 7 — ΔG° = −RT ln K

One of the most powerful relationships in all of chemistry connects Gibbs free energy to the equilibrium constant. This equation bridges thermodynamics and equilibrium — two pillars of AP Chemistry.

The Key Equation

$\Delta G° = -RT\ln K$

What This Equation Tells Us

Important Nuance

$\Delta G° < 0$ does NOT mean the reaction goes to completion. It means $K > 1$, so products are favored, but reactants are still present at equilibrium.

Solving for K from ΔG°

Rearranging: $K = e^{-\Delta G°/(RT)}$

Worked Example

Find $K$ at 25°C for a reaction with $\Delta G° = -5.40$ kJ/mol.

$K = e^{-\Delta G°/(RT)} = e^{-(-5400)/(8.314 \times 298)}$

$K = e^{5400/2477.6} = e^{2.180} = 8.85$

Solving for ΔG° from K

If $K = 1.0 \times 10^{10}$ at 298 K:

$\Delta G° = -RT\ln K = -(8.314)(298)\ln(1.0 \times 10^{10})$

$\Delta G° = -(2477.6)(23.03) = -57{,}050 \text{ J} = -57.1 \text{ kJ}$

Converting Between ln and log

$\ln K = 2.303 \log K$

So: $\Delta G° = -2.303 RT \log K$

ΔG° and K Concept Quiz 🎯

ΔG° and K Calculations 🧮

Use $R = 8.314$ J/(mol·K), $T = 298$ K

1. If $\Delta G° = -17.1$ kJ/mol, what is $K$? (round to nearest whole number)

2. If $K = 1.0 \times 10^{5}$ at 298 K, what is $\Delta G°$? (in kJ/mol, to 1 decimal)

3. If $\Delta G° = +10.0$ kJ/mol, is $K$ greater than or less than 1? (type "greater" or "less")

ΔG° and Equilibrium 🔽

Exit Quiz — ΔG° and K ✅

📊 Non-Standard Conditions — ΔG = ΔG° + RT ln Q

Part 5 of 7 — Real-World Free Energy

$\Delta G°$ tells us about equilibrium, but real reactions rarely start at standard conditions. To determine spontaneity at any concentration, pressure, or composition, we need $\Delta G$ (without the °), which uses the reaction quotient $Q$.

The Non-Standard Free Energy Equation

$\Delta G = \Delta G° + RT\ln Q$

Recall: Q vs K

• $Q$ = reaction quotient (calculated from current concentrations)
• $K$ = equilibrium constant (concentrations at equilibrium)

$Q = \frac{[\text{products}]^n}{[\text{reactants}]^m}$ (same form as K, but not at equilibrium)

Interpreting ΔG, Q, and K

Key Insight

At equilibrium, $Q = K$ and $\Delta G = 0$:

$0 = \Delta G° + RT\ln K$ $\Delta G° = -RT\ln K$

This is how we derived the $\Delta G°$–$K$ relationship!

The Big Picture

• $\Delta G°$ tells you WHERE equilibrium lies (the value of $K$)
• $\Delta G$ tells you WHICH DIRECTION the reaction will go from current conditions
• A reaction with $\Delta G° > 0$ can still proceed forward if $Q$ is small enough

Worked Example

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

$\Delta G° = -33.0$ kJ/mol at 298 K

Calculate $\Delta G$ when $P_{\text{N}_2} = 1.0$ atm, $P_{\text{H}_2} = 3.0$ atm, $P_{\text{NH}_3} = 0.50$ atm.

Step 1: Calculate Q

$Q = \frac{(P_{\text{NH}_3})^2}{(P_{\text{N}_2})(P_{\text{H}_2})^3} = \frac{(0.50)^2}{(1.0)(3.0)^3} = \frac{0.25}{27} = 0.00926$

Step 2: Calculate ΔG

$\Delta G = \Delta G° + RT\ln Q$ $= -33{,}000 + (8.314)(298)\ln(0.00926)$ $= -33{,}000 + (2478)(-4.682)$ $= -33{,}000 + (-11{,}602)$ $= -44{,}602 \text{ J} = -44.6 \text{ kJ}$

Since $\Delta G < 0$ and $Q < K$, the forward reaction is spontaneous — more $\text{NH}_3$ will form.

Non-Standard ΔG Quiz 🎯

Non-Standard ΔG Calculations 🧮

For a reaction with $\Delta G° = -10.0$ kJ/mol at $T = 298$ K:

1. What is $\Delta G$ when $Q = 1$? (in kJ/mol)

2. What is $\Delta G$ when $Q = K$ (at equilibrium)? (in kJ/mol)

3. If $Q > K$, is $\Delta G$ positive or negative? (type "positive" or "negative")

Round all answers to 3 significant figures.

Q, K, and ΔG 🔽

Exit Quiz — Non-Standard ΔG ✅

🛠️ Problem-Solving Workshop — Gibbs Free Energy

Part 6 of 7 — Practice and Integration

This workshop brings together all the Gibbs free energy tools: calculating ΔG from ΔH and ΔS, using formation values, the four sign cases, ΔG°-K relationships, and non-standard conditions.

Problem-Solving Flowchart

What Are You Given? → What Method to Use?

Common Unit Traps

Mixed ΔG Problems 🎯

Multi-Step Calculation Workshop 🧮

1. A reaction has $\Delta H° = +50$ kJ and $\Delta S° = +150$ J/K. What is $\Delta G°$ at 400 K? (in kJ)

2. For the reaction in (1), what is $K$ at 400 K? (round to nearest whole number; use $e^{3.01} \approx 20$)

3. A reaction has $\Delta G° = -20$ kJ/mol. What is $K$ at 298 K? (round to nearest thousand; use $e^{8.07} \approx 3200$)

Problem Strategy Selection 🔽

Exit Quiz — Problem-Solving Workshop ✅

🎯 Synthesis & AP Review — Gibbs Free Energy

Part 7 of 7 — Mastering the Connections

Gibbs free energy ties together enthalpy, entropy, equilibrium, and electrochemistry. This final review ensures you can navigate all the relationships and solve any AP-level problem.

The Web of Thermodynamic Equations

Core Equations

The Four Sign Cases

Critical Relationships

• $\Delta G° < 0 \Leftrightarrow K > 1$ (products favored)
• $\Delta G° = 0 \Leftrightarrow K = 1$
• $\Delta G° > 0 \Leftrightarrow K < 1$ (reactants favored)
• $\Delta G < 0$: forward reaction proceeds
• $\Delta G = 0$: at equilibrium
• $\Delta G > 0$: reverse reaction proceeds

Comprehensive AP Review 🎯

Integration Problems 🧮

1. $\Delta H° = -180$ kJ, $\Delta S° = -250$ J/K. What is the crossover temperature? (in K)

2. At 298 K, $\Delta G° = -57.1$ kJ/mol. What is $K$? (use $e^{23.0} \approx 10^{10}$; express as a power of 10)

3. A reaction has $\Delta G° = +5.0$ kJ/mol. At what value of $Q$ does $\Delta G = 0$ at 298 K? (i.e., what is $K$? Round to nearest tenth; use $e^{-2.02} \approx 0.1$)

Final Concept Review 🔽

Final Exit Quiz — Gibbs Free Energy Mastery ✅