The Hook: The Ultimate Decision Maker
Imagine you’re the coach of a cricket team:
- Batting strength (ΔH): How many runs you can score (energy factor)
- Fielding discipline (ΔS): How organized or chaotic your field placement is (entropy factor)
- Match outcome (ΔG): Win or lose (spontaneity!)
The winning formula: You don’t just need good batting OR good fielding - you need the RIGHT COMBINATION at the RIGHT TEMPERATURE of the match!
Three real-life mysteries:
Ice cubes in your drink (25°C): Melting is endothermic (ΔH > 0, unfavorable), yet happens spontaneously. Why?
Rusting of iron: Happens without any external energy input. What drives this?
Charging your phone: Requires external energy. Why doesn’t the battery charge itself?
The answer to all three lies in Gibbs Free Energy (G) - the ultimate spontaneity predictor that combines enthalpy and entropy!
Movie connection (Oppenheimer): Just like atomic reactions release energy spontaneously when ΔG < 0, the same principle determines which nuclear processes can happen!
The Core Concept
What is Gibbs Free Energy?
Gibbs Free Energy (G) is the energy available to do useful work at constant temperature and pressure.
$$\boxed{G = H - TS}$$where:
- H = Enthalpy (heat content)
- T = Absolute temperature (Kelvin)
- S = Entropy (disorder)
For a process:
$$\boxed{\Delta G = \Delta H - T\Delta S}$$In simple terms: Gibbs energy tells you whether a reaction will happen spontaneously by balancing the enthalpy drive (release energy) against the entropy drive (increase disorder).
Why “Free” Energy?
- It’s the energy “free” (available) to do useful work
- Maximum work obtainable from a process at constant T and P
- $w_{max} = -\Delta G$
The Golden Rule of Spontaneity
At constant temperature and pressure:
$$\Delta G < 0 \quad \Rightarrow \quad \text{Spontaneous (forward direction)}$$ $$\Delta G = 0 \quad \Rightarrow \quad \text{Equilibrium (no net change)}$$ $$\Delta G > 0 \quad \Rightarrow \quad \text{Non-spontaneous (reverse direction favored)}$$This is THE most important criterion in all of thermodynamics!
Why? Because ΔG combines both enthalpy AND entropy into one number:
$$\Delta G = \Delta H - T\Delta S = -T\Delta S_{universe}$$When $\Delta G < 0$, this means $\Delta S_{universe} > 0$ (Second Law satisfied!)
Connection to Entropy
Remember from Entropy:
$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$$At constant T and P:
$$\Delta S_{surroundings} = -\frac{\Delta H_{system}}{T}$$Therefore:
$$\Delta S_{universe} = \Delta S - \frac{\Delta H}{T}$$Multiply by -T:
$$-T\Delta S_{universe} = \Delta H - T\Delta S = \Delta G$$Beautiful result:
$$\boxed{\Delta G = -T\Delta S_{universe}}$$Interpretation:
- $\Delta G < 0$ ⟺ $\Delta S_{universe} > 0$ ⟹ Spontaneous
- $\Delta G = 0$ ⟺ $\Delta S_{universe} = 0$ ⟹ Equilibrium
- $\Delta G > 0$ ⟺ $\Delta S_{universe} < 0$ ⟹ Non-spontaneous
So instead of calculating ΔS_system AND ΔS_surroundings, we just calculate ΔG!
The Four Cases: ΔH vs ΔS Competition
Case 1: ΔH < 0, ΔS > 0 (Always Spontaneous)
$$\Delta G = \Delta H - T\Delta S = \text{(negative)} - T \times \text{(positive)} = \text{negative at all T}$$Examples:
Combustion: $\text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}$
- ΔH < 0 (exothermic, releases heat)
- ΔS > 0 (more gas molecules from breaking liquid water if products are gas)
- Result: Burns spontaneously at all temperatures!
Dissolution of NH₄NO₃ in water:
- ΔH < 0 (lattice energy < hydration energy for this case)
- ΔS > 0 (solid → dispersed ions)
- Result: Dissolves spontaneously
Intuition: Both enthalpy AND entropy favor the reaction - double win!
Case 2: ΔH > 0, ΔS < 0 (Never Spontaneous)
$$\Delta G = \Delta H - T\Delta S = \text{(positive)} - T \times \text{(negative)} = \text{positive at all T}$$Examples:
Reverse of combustion: Making methane from CO₂ and H₂O
- ΔH > 0 (endothermic)
- ΔS < 0 (disorder decreases)
- Result: Never happens spontaneously!
Freezing water at 50°C:
- ΔH > 0 (need to remove heat, but that means system needs to absorb heat to freeze - wait, freezing is exothermic!)
Let me correct this:
- Water freezing at 50°C:
- Actually ΔH < 0 for freezing (exothermic)
- But ΔS < 0 (liquid → solid)
- At high T like 50°C: $\Delta G = \Delta H - T\Delta S$ becomes positive because TΔS term dominates
- So this is actually Case 3!
Let me use better example:
- Making NH₃ from N₂ and H₂ at very high temperature:
- $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$
- ΔH < 0 (exothermic, wait this would make it case 3)
Actually, finding pure Case 2 examples is hard because:
- If ΔH > 0 (endothermic) usually it means breaking into smaller, more disordered pieces (ΔS > 0) - that’s Case 4
- If ΔS < 0 (disorder decreases) usually it means forming bonds (ΔH < 0) - that’s Case 3
Pure Case 2 example:
- Condensing a simple gas at very low pressure while also absorbing heat (hypothetical)
Better real example:
- Reverse of any spontaneous reaction at T where it’s spontaneous
- Like trying to make diamond from graphite at low T without pressure:
- ΔH > 0 (diamond is higher energy)
- ΔS < 0 (diamond is more ordered)
- Result: Won’t happen spontaneously
Intuition: Both enthalpy AND entropy oppose the reaction - double loss!
Case 3: ΔH < 0, ΔS < 0 (Spontaneous at Low T)
$$\Delta G = \Delta H - T\Delta S = \text{(negative)} - T \times \text{(negative)}$$At low T: $|\Delta H| > T|\Delta S|$ → ΔG < 0 (enthalpy wins) At high T: $|\Delta H| < T|\Delta S|$ → ΔG > 0 (entropy penalty wins)
Equilibrium temperature:
$$T_{eq} = \frac{\Delta H}{\Delta S}$$Examples:
Water freezing: $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(s)$
- ΔH < 0 (exothermic, releases heat)
- ΔS < 0 (liquid → solid, disorder decreases)
- At T < 273 K (0°C): ΔG < 0, freezing is spontaneous
- At T > 273 K (0°C): ΔG > 0, melting is spontaneous
- At T = 273 K: ΔG = 0, equilibrium!
Haber process: $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$
- ΔH < 0 (exothermic, -92 kJ/mol)
- ΔS < 0 (4 gas moles → 2 gas moles)
- Favored at low T: But kinetics too slow!
- Industrial compromise: Moderate T (400-450°C) with catalyst
Intuition: Enthalpy wants it, entropy doesn’t. At low T, enthalpy wins!
Interactive Calculator: Gibbs Energy and Spontaneity
Use this interactive calculator to explore how ΔG changes with enthalpy, entropy, and temperature. See real-time bar charts and temperature dependence graphs!
How to use this calculator:
- Select a preset reaction from the dropdown, or adjust ΔH and ΔS manually
- Change the temperature using the slider to see how spontaneity changes
- Observe the bar chart showing ΔH, -TΔS, and ΔG components
- The temperature graph shows where ΔG crosses zero (equilibrium temperature)
- The shaded green region indicates spontaneous conditions
Case 4: ΔH > 0, ΔS > 0 (Spontaneous at High T)
$$\Delta G = \Delta H - T\Delta S = \text{(positive)} - T \times \text{(positive)}$$At low T: $|\Delta H| > T|\Delta S|$ → ΔG > 0 (enthalpy penalty wins) At high T: $|\Delta H| < T|\Delta S|$ → ΔG < 0 (entropy wins!)
Equilibrium temperature:
$$T_{eq} = \frac{\Delta H}{\Delta S}$$Examples:
Ice melting: $\text{H}_2\text{O}(s) \to \text{H}_2\text{O}(l)$
- ΔH > 0 (endothermic, needs heat)
- ΔS > 0 (solid → liquid, disorder increases)
- At T < 273 K: ΔG > 0, won’t melt
- At T > 273 K: ΔG < 0, melts spontaneously
- At T = 273 K: ΔG = 0, equilibrium!
Decomposition of CaCO₃: $\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$
- ΔH > 0 (endothermic, +178 kJ/mol)
- ΔS > 0 (gas produced, disorder increases)
- High T needed: Above ~1100 K, becomes spontaneous
- Used in cement industry
Evaporation of water: $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(g)$
- ΔH > 0 (endothermic, +44 kJ/mol)
- ΔS > 0 (liquid → gas, huge disorder increase)
- At high T: Boils spontaneously
Intuition: Entropy wants it, enthalpy doesn’t. At high T, entropy term (TΔS) becomes large and wins!
Summary Table: The Four Cases
| ΔH | ΔS | ΔG Sign | Spontaneity | Example |
|---|---|---|---|---|
| - | + | Always - | Always spontaneous | Combustion, rusting |
| + | - | Always + | Never spontaneous | Reverse combustion |
| - | - | - at low T, + at high T | Spontaneous at low T | Freezing water, Haber process |
| + | + | + at low T, - at high T | Spontaneous at high T | Melting ice, CaCO₃ decomposition |
Memory trick: “Give Her The Stuff”
- Gibbs = Heat - Temperature × Stuff (entropy)
- $\Delta G = \Delta H - T\Delta S$
Standard Gibbs Energy
Standard Gibbs Energy of Formation
$$\Delta_f G° = \text{Gibbs energy of formation from elements in standard state}$$Standard conditions:
- Temperature: 298 K (25°C)
- Pressure: 1 bar
- Concentration: 1 M (for solutions)
Key fact:
$$\boxed{\Delta_f G°[\text{element in standard state}] = 0}$$Just like enthalpy of formation!
Calculating ΔG° for Reactions
Method 1: From formation data
$$\boxed{\Delta_r G° = \sum \Delta_f G°(\text{products}) - \sum \Delta_f G°(\text{reactants})}$$Method 2: From ΔH° and ΔS°
$$\boxed{\Delta G° = \Delta H° - T\Delta S°}$$Method 3: From equilibrium constant (see below)
$$\boxed{\Delta G° = -RT\ln K = -2.303RT\log K}$$Gibbs Energy and Equilibrium
The Most Important Equation in Chemical Thermodynamics
$$\boxed{\Delta G = \Delta G° + RT\ln Q}$$where:
- Q = Reaction quotient
- R = 8.314 J/mol·K
- T = Temperature (K)
At equilibrium: Q = K (equilibrium constant), and ΔG = 0
$$0 = \Delta G° + RT\ln K$$ $$\boxed{\Delta G° = -RT\ln K = -2.303RT\log K}$$This is HUGE! We can calculate equilibrium constants from thermodynamic data without doing experiments!
Relationship Between ΔG° and K
$$\Delta G° = -RT\ln K$$Rearranging:
$$K = e^{-\Delta G°/RT}$$or
$$\log K = -\frac{\Delta G°}{2.303RT}$$At 298 K:
$$\log K = -\frac{\Delta G°}{2.303 \times 8.314 \times 298} = -\frac{\Delta G°}{5.708}$$where ΔG° is in kJ/mol.
Simplified at 298 K:
$$\boxed{\log K = -\frac{\Delta G° \text{(kJ/mol)}}{5.7}}$$What Different Values Mean
| ΔG° | K | Meaning |
|---|---|---|
| Large negative (< -30 kJ/mol) | K » 1 | Products heavily favored, nearly complete reaction |
| Small negative (-30 to 0 kJ/mol) | K > 1 | Products favored |
| Zero | K = 1 | Equal amounts of products and reactants |
| Small positive (0 to +30 kJ/mol) | K < 1 | Reactants favored |
| Large positive (> +30 kJ/mol) | K « 1 | Reactants heavily favored, almost no reaction |
At 298 K:
- ΔG° = -5.7 kJ/mol → K = 10
- ΔG° = -11.4 kJ/mol → K = 100
- ΔG° = -17.1 kJ/mol → K = 1000
- ΔG° = 0 kJ/mol → K = 1
- ΔG° = +5.7 kJ/mol → K = 0.1
- ΔG° = +11.4 kJ/mol → K = 0.01
Pattern: Every 5.7 kJ/mol change in ΔG° changes K by factor of 10!
ΔG vs ΔG°: What’s the Difference?
ΔG° (Standard Gibbs Energy Change):
- At standard conditions (1 bar, 1 M, 298 K)
- Tells you the equilibrium position
- Related to K
ΔG (Actual Gibbs Energy Change):
- At any conditions (any concentrations/pressures)
- Tells you spontaneity at that moment
- Related to Q
Three cases:
Q < K: System shifts forward (toward products)
- $\Delta G < 0$ (spontaneous forward)
Q = K: System at equilibrium
- $\Delta G = 0$ (no net change)
Q > K: System shifts backward (toward reactants)
- $\Delta G > 0$ (spontaneous backward)
Temperature Dependence of ΔG
Gibbs-Helmholtz Equation
$$\Delta G = \Delta H - T\Delta S$$Assumption: ΔH and ΔS are approximately constant over moderate temperature range.
Finding equilibrium temperature:
At equilibrium, ΔG = 0:
$$0 = \Delta H - T_{eq}\Delta S$$ $$\boxed{T_{eq} = \frac{\Delta H}{\Delta S}}$$This is the temperature where:
- Forward and reverse reaction rates are equal
- Reaction switches from spontaneous to non-spontaneous (or vice versa)
Example: Water freezing/melting
$$T_{eq} = \frac{\Delta H_{fusion}}{\Delta S_{fusion}} = \frac{6000 \text{ J/mol}}{22 \text{ J/K·mol}} = 273 \text{ K} = 0°\text{C}$$Perfect!
Van’t Hoff Equation (Temperature and K)
From $\Delta G° = -RT\ln K = \Delta H° - T\Delta S°$:
$$\ln K = -\frac{\Delta H°}{RT} + \frac{\Delta S°}{R}$$Differential form:
$$\frac{d\ln K}{dT} = \frac{\Delta H°}{RT^2}$$Integrated form (between two temperatures):
$$\boxed{\ln\frac{K_2}{K_1} = -\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}$$or in base-10 log:
$$\log\frac{K_2}{K_1} = -\frac{\Delta H°}{2.303R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$Practical use: Calculate how equilibrium constant changes with temperature!
Sign check:
- Exothermic (ΔH° < 0): K decreases with increasing T (Le Chatelier!)
- Endothermic (ΔH° > 0): K increases with increasing T
Memory Tricks & Patterns
Mnemonic for Gibbs Equation
“Give Her The Stuff”
- G = H - T × S
- $\Delta G = \Delta H - T\Delta S$
“Good Hearts Tell Stories”
- G (Gibbs) depends on H (Heat), T (Temperature), S (Stories/entropy)
Spontaneity Decision Tree
Is ΔG known?
├─ Yes → ΔG < 0? Spontaneous!
│ ΔG = 0? Equilibrium!
│ ΔG > 0? Non-spontaneous!
│
└─ No → Know ΔH and ΔS?
├─ ΔH(-), ΔS(+) → Always spontaneous
├─ ΔH(+), ΔS(-) → Never spontaneous
├─ ΔH(-), ΔS(-) → Spontaneous at LOW T
└─ ΔH(+), ΔS(+) → Spontaneous at HIGH T
Quick K Estimation (at 298 K)
$$\Delta G° \text{ (kJ/mol)} = -5.7 \times \log K$$Shortcut:
- Every 6 kJ/mol change in ΔG° ≈ factor of 10 in K
- ΔG° = 0 → K = 1
- ΔG° = -6 → K ≈ 10
- ΔG° = -12 → K ≈ 100
- ΔG° = +6 → K ≈ 0.1
Pattern Recognition for JEE
Given ΔH and ΔS, find temperature range:
| Signs | Formula | When spontaneous |
|---|---|---|
| ΔH(-), ΔS(-) | $T < \frac{\Delta H}{\Delta S}$ | Below equilibrium T |
| ΔH(+), ΔS(+) | $T > \frac{\Delta H}{\Delta S}$ | Above equilibrium T |
Common Mistakes to Avoid
Wrong: “ΔG° < 0, so reaction is spontaneous at all conditions”
Right: “ΔG° < 0 means K > 1 (products favored at equilibrium). Spontaneity depends on ΔG, not ΔG°”
Remember:
- ΔG° tells you equilibrium position (where it ends up)
- ΔG tells you spontaneity (which way it goes now)
Even if ΔG° > 0, reaction can be spontaneous if Q » K!
Wrong:
$$\Delta G = \Delta H - T\Delta S = -100 - 25 \times 0.5 = -112.5 \text{ kJ}$$Right: Convert °C to K first!
$$\Delta G = -100 - 298 \times 0.5 = -100 - 149 = -249 \text{ kJ}$$Always use Kelvin for temperature in Gibbs equation!
Common error: ΔH in kJ/mol, ΔS in J/K·mol
Example:
- ΔH = -100 kJ/mol
- ΔS = -200 J/K·mol
- T = 298 K
Wrong:
$$\Delta G = -100 - 298 \times (-200) = -100 + 59600 = +59500 \text{ kJ/mol}$$Right: Convert to same units!
$$\Delta G = -100 - 298 \times (-0.200) = -100 + 59.6 = -40.4 \text{ kJ/mol}$$Always check units: Either both in kJ or both in J!
Wrong: $\Delta G° = RT\ln K$ (forgetting negative sign)
Right: $\Delta G° = -RT\ln K$
Check:
- If K > 1 (products favored), then ln K > 0, so ΔG° < 0 ✓
- If K < 1 (reactants favored), then ln K < 0, so ΔG° > 0 ✓
The negative sign makes sense!
Wrong: “ΔG = 0, so nothing happens”
Right: “ΔG = 0 means system is at equilibrium - forward and reverse rates are equal, but both reactions still occur!”
At equilibrium:
- ΔG = 0 (no net change)
- Both forward and reverse reactions continue
- Concentrations/pressures remain constant
- Dynamic equilibrium, not static!
Practice Problems
Level 1: Foundation (NCERT)
Question: Calculate ΔG at 298 K for a reaction with:
- ΔH = -100 kJ/mol
- ΔS = -200 J/K·mol
Is the reaction spontaneous?
Solution:
$$\Delta G = \Delta H - T\Delta S$$Convert ΔS to kJ/K·mol: $\Delta S = -0.200$ kJ/K·mol
$$\Delta G = -100 - 298 \times (-0.200)$$ $$= -100 + 59.6$$ $$= -40.4 \text{ kJ/mol}$$Since ΔG < 0, reaction is spontaneous at 298 K.
Answer: ΔG = -40.4 kJ/mol, spontaneous
Note: This is Case 3 (ΔH < 0, ΔS < 0), spontaneous at low T.
Question: Find the temperature at which the reaction becomes non-spontaneous if:
- ΔH = -50 kJ/mol
- ΔS = -100 J/K·mol
Solution:
At equilibrium: $\Delta G = 0$
$$0 = \Delta H - T_{eq}\Delta S$$ $$T_{eq} = \frac{\Delta H}{\Delta S}$$Convert to same units: ΔH = -50,000 J/mol, ΔS = -100 J/K·mol
$$T_{eq} = \frac{-50000}{-100} = 500 \text{ K}$$Interpretation:
- Case 3 (ΔH < 0, ΔS < 0): Spontaneous below 500 K
- At T < 500 K: ΔG < 0 (spontaneous)
- At T > 500 K: ΔG > 0 (non-spontaneous)
- At T = 500 K: ΔG = 0 (equilibrium)
Answer: 500 K (227°C)
Question: Calculate ΔG° at 298 K for a reaction with equilibrium constant K = 100.
Solution:
$$\Delta G° = -RT\ln K = -2.303RT\log K$$ $$\Delta G° = -2.303 \times 8.314 \times 298 \times \log 100$$ $$= -2.303 \times 8.314 \times 298 \times 2$$ $$= -11,414 \text{ J/mol}$$ $$= -11.4 \text{ kJ/mol}$$Quick check: $\Delta G° = -5.7 \times \log K = -5.7 \times 2 = -11.4$ kJ/mol ✓
Answer: -11.4 kJ/mol
Interpretation: K > 1, so products are favored (ΔG° < 0).
Level 2: JEE Main
Question: For the reaction: $\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$
Given:
- $\Delta H° = +178$ kJ/mol
- $\Delta S° = +160$ J/K·mol
(a) Is the reaction spontaneous at 298 K? (b) At what temperature does it become spontaneous?
Solution:
(a) At 298 K:
$$\Delta G° = \Delta H° - T\Delta S°$$ $$= 178 - 298 \times 0.160$$ $$= 178 - 47.7$$ $$= +130.3 \text{ kJ/mol}$$ΔG° > 0, so non-spontaneous at 298 K.
(b) Equilibrium temperature:
$$T_{eq} = \frac{\Delta H°}{\Delta S°} = \frac{178000}{160} = 1112.5 \text{ K}$$Above this temperature, ΔG < 0 (spontaneous).
Answer:
- (a) Non-spontaneous at 298 K (ΔG° = +130.3 kJ/mol)
- (b) Spontaneous above 1113 K (840°C)
Industrial relevance: This is why lime kilns operate at ~900°C!
Question: Calculate ΔG° for combustion of methane at 298 K:
$$\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$Given:
- $\Delta_f G°[\text{CH}_4(g)] = -50.8$ kJ/mol
- $\Delta_f G°[\text{CO}_2(g)] = -394.4$ kJ/mol
- $\Delta_f G°[\text{H}_2\text{O}(l)] = -237.2$ kJ/mol
- $\Delta_f G°[\text{O}_2(g)] = 0$ (element)
Solution:
$$\Delta_r G° = \sum \Delta_f G°(\text{products}) - \sum \Delta_f G°(\text{reactants})$$Products:
$$= \Delta_f G°[\text{CO}_2] + 2\Delta_f G°[\text{H}_2\text{O}]$$ $$= -394.4 + 2(-237.2) = -868.8 \text{ kJ}$$Reactants:
$$= \Delta_f G°[\text{CH}_4] + 2\Delta_f G°[\text{O}_2]$$ $$= -50.8 + 0 = -50.8 \text{ kJ}$$ $$\Delta_r G° = -868.8 - (-50.8) = -818.0 \text{ kJ/mol}$$Calculate K:
$$\log K = -\frac{\Delta G°}{5.7} = -\frac{-818}{5.7} = 143.5$$ $$K = 10^{143.5} \approx \infty$$Answer: ΔG° = -818 kJ/mol, K ≈ 10¹⁴³ (essentially irreversible!)
Insight: Huge negative ΔG° means combustion is extremely favorable!
Question: A reaction has ΔG° = +10 kJ/mol at 298 K.
(a) Is the reaction spontaneous under standard conditions? (b) Can it ever be spontaneous?
Solution:
(a) Under standard conditions:
ΔG° > 0, so reaction is non-spontaneous when all species at 1 M/1 bar.
(b) Can it be spontaneous?
Yes! Using $\Delta G = \Delta G° + RT\ln Q$:
For ΔG < 0:
$$\Delta G° + RT\ln Q < 0$$ $$RT\ln Q < -\Delta G°$$ $$\ln Q < -\frac{\Delta G°}{RT}$$At 298 K:
$$\ln Q < -\frac{10000}{8.314 \times 298} = -4.03$$ $$Q < e^{-4.03} = 0.018$$Interpretation: If Q < 0.018 (reactants in large excess), reaction is spontaneous!
Real-life example: Even though ΔG° > 0, we can drive reaction forward by:
- Increasing reactant concentrations
- Removing products
- Le Chatelier’s principle!
Answer:
- (a) Non-spontaneous at standard conditions
- (b) Spontaneous when Q < 0.018 (reactants » products)
Level 3: JEE Advanced
Question: The equilibrium constant for a reaction is K₁ = 100 at 300 K. If ΔH° = -50 kJ/mol, calculate K₂ at 400 K.
Solution:
Using Van’t Hoff equation:
$$\ln\frac{K_2}{K_1} = -\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ $$\ln\frac{K_2}{100} = -\frac{-50000}{8.314}\left(\frac{1}{400} - \frac{1}{300}\right)$$ $$= +\frac{50000}{8.314}\left(\frac{1}{400} - \frac{1}{300}\right)$$ $$= 6014 \times (0.0025 - 0.00333)$$ $$= 6014 \times (-0.00083)$$ $$= -4.99$$ $$\frac{K_2}{100} = e^{-4.99} = 0.00674$$ $$K_2 = 100 \times 0.00674 = 0.674$$Answer: K₂ = 0.67 at 400 K
Check: ΔH° < 0 (exothermic), so K decreases with increasing T ✓ (Le Chatelier!)
Insight: Increasing T from 300 K to 400 K decreased K from 100 to 0.67 - huge change!
Question: Calculate ΔG° for the synthesis:
$$\text{C(graphite)} + 2\text{H}_2(g) \to \text{CH}_4(g)$$Using coupled reactions:
- $\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$, $\Delta G°_1 = -818$ kJ
- $\text{C(graphite)} + \text{O}_2(g) \to \text{CO}_2(g)$, $\Delta G°_2 = -394$ kJ
- $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{H}_2\text{O}(l)$, $\Delta G°_3 = -237$ kJ
Solution:
Target: $\text{C} + 2\text{H}_2 \to \text{CH}_4$
Manipulate:
Reverse reaction (1):
$$\text{CO}_2 + 2\text{H}_2\text{O} \to \text{CH}_4 + 2\text{O}_2, \quad \Delta G° = +818 \text{ kJ}$$Use reaction (2) as is:
$$\text{C} + \text{O}_2 \to \text{CO}_2, \quad \Delta G° = -394 \text{ kJ}$$Multiply reaction (3) by 2:
$$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}, \quad \Delta G° = 2(-237) = -474 \text{ kJ}$$Add all three:
$$\text{CO}_2 + 2\text{H}_2\text{O} + \text{C} + \text{O}_2 + 2\text{H}_2 + \text{O}_2 \to \text{CH}_4 + 2\text{O}_2 + \text{CO}_2 + 2\text{H}_2\text{O}$$Cancel common terms:
$$\text{C} + 2\text{H}_2 \to \text{CH}_4$$ $$\Delta_f G°[\text{CH}_4] = +818 - 394 - 474 = -50 \text{ kJ/mol}$$Answer: ΔG° = -50 kJ/mol
Verification: This matches standard formation data! ✓
Question: For the reaction:
$$2\text{NO}(g) + \text{O}_2(g) \to 2\text{NO}_2(g)$$Given at 298 K:
- $\Delta H° = -114$ kJ/mol
- $\Delta S° = -146$ J/K·mol
(a) Determine the temperature range for spontaneity (b) Calculate K at 298 K (c) Explain why this reaction is favored at low T despite being exothermic
Solution:
(a) Temperature range:
Case 3: ΔH < 0, ΔS < 0 → Spontaneous at low T
Equilibrium temperature:
$$T_{eq} = \frac{\Delta H}{\Delta S} = \frac{-114000}{-146} = 781 \text{ K}$$Spontaneous when: T < 781 K (508°C) Non-spontaneous when: T > 781 K
(b) K at 298 K:
$$\Delta G° = \Delta H° - T\Delta S°$$ $$= -114 - 298 \times (-0.146)$$ $$= -114 + 43.5$$ $$= -70.5 \text{ kJ/mol}$$ $$\log K = -\frac{\Delta G°}{5.7} = -\frac{-70.5}{5.7} = 12.4$$ $$K = 10^{12.4} \approx 2.5 \times 10^{12}$$(c) Explanation:
This reaction:
- Exothermic (ΔH < 0): Releases heat, favored by enthalpy
- Decreases disorder (ΔS < 0): 3 gas moles → 2 gas moles, opposed by entropy
At low T (like 298 K):
- TΔS term is small: $298 \times 0.146 = 43.5$ kJ
- ΔH term dominates: $|-114| > 43.5$
- Net ΔG < 0, spontaneous!
At high T (above 781 K):
- TΔS term becomes large: $800 \times 0.146 = 117$ kJ
- Now $117 > |-114|$
- Net ΔG > 0, non-spontaneous!
Environmental relevance: This is why NO₂ formation from NO is favored in cool parts of atmosphere and car exhausts, contributing to air pollution!
Answer:
- (a) Spontaneous below 781 K (508°C)
- (b) K = 2.5 × 10¹² at 298 K
- (c) Enthalpy term dominates entropy term at low T
Quick Revision Box
Essential Formulas
| Concept | Formula | Use |
|---|---|---|
| Gibbs definition | $G = H - TS$ | Fundamental |
| Gibbs change | $\Delta G = \Delta H - T\Delta S$ | Spontaneity |
| Spontaneity | $\Delta G < 0$ | Forward spontaneous |
| Formation method | $\Delta G° = \sum \Delta_f G°_{\text{prod}} - \sum \Delta_f G°_{\text{react}}$ | From tables |
| Equilibrium | $\Delta G° = -RT\ln K$ | K from thermodynamics |
| Reaction quotient | $\Delta G = \Delta G° + RT\ln Q$ | Any conditions |
| Equilibrium temp | $T_{eq} = \frac{\Delta H}{\Delta S}$ | Where ΔG = 0 |
| Van’t Hoff | $\ln\frac{K_2}{K_1} = -\frac{\Delta H°}{R}(\frac{1}{T_2}-\frac{1}{T_1})$ | K vs temperature |
Quick Reference at 298 K
$$\Delta G° = -5.7 \times \log K \quad \text{(kJ/mol)}$$| ΔG° (kJ/mol) | K | Products vs Reactants |
|---|---|---|
| -17 | 1000 | Strongly favor products |
| -11 | 100 | Favor products |
| -6 | 10 | Favor products |
| 0 | 1 | Equal |
| +6 | 0.1 | Favor reactants |
| +11 | 0.01 | Favor reactants |
| +17 | 0.001 | Strongly favor reactants |
The Four Cases Summary
| ΔH | ΔS | Low T | High T | Example |
|---|---|---|---|---|
| - | + | Spont | Spont | Combustion |
| + | - | Non-spont | Non-spont | Reverse combustion |
| - | - | Spont | Non-spont | Freezing H₂O |
| + | + | Non-spont | Spont | Melting H₂O |
Teacher’s Summary
1. Gibbs Energy: The Ultimate Spontaneity Criterion
$$\Delta G = \Delta H - T\Delta S$$Combines enthalpy and entropy into one number:
- ΔG < 0 → Spontaneous
- ΔG = 0 → Equilibrium
- ΔG > 0 → Non-spontaneous
2. ΔG is Related to Entropy of Universe
$$\Delta G = -T\Delta S_{universe}$$So ΔG < 0 automatically means Second Law is satisfied!
3. The Four Cases: Enthalpy vs Entropy
Temperature determines which factor wins:
- Both favor: Always spontaneous
- Both oppose: Never spontaneous
- Compete: Temperature decides winner
4. Equilibrium Constant from Thermodynamics
$$\Delta G° = -RT\ln K$$We can predict K without experiments! At 298 K:
$$\log K = -\frac{\Delta G°}{5.7}$$5. ΔG° vs ΔG
- ΔG°: Equilibrium position (standard conditions)
- ΔG: Spontaneity now (actual conditions)
6. Temperature Changes Everything
Equilibrium temperature: $T_{eq} = \frac{\Delta H}{\Delta S}$
This is where reaction switches from spontaneous to non-spontaneous.
“Gibbs energy is the final judge - it combines heat, disorder, and temperature to predict the future of every reaction.”
JEE Weightage: 4-5 questions per paper. Highest-yield thermodynamics topic!
Time-saving tip: For MCQs, memorize $\log K = -\Delta G°/5.7$ at 298 K for instant K calculations!
Cross-Links to Related Concepts
Prerequisites
- First Law - Internal energy, work, heat
- Enthalpy - Heat content, ΔH calculations
- Entropy - Disorder, Second Law
- Thermochemistry - Bond energies, Hess’s Law
Equilibrium Applications
- Chemical Equilibrium - K, Q, reaction quotient
- Le Chatelier’s Principle - Temperature effects on equilibrium
- Ionic Equilibrium - Acid-base equilibria
Electrochemistry Connections
- Electrochemical Cells - ΔG° = -nFE°
- Nernst Equation - Cell potential at non-standard conditions
- Electrode Potentials - Standard potentials
Kinetics Links
- Arrhenius Equation - Activation energy
- Rate of Reaction - Kinetics vs thermodynamics
Solution Chemistry
- Colligative Properties - Dissolving spontaneity
- Raoult’s Law - Vapor pressure equilibrium
Advanced Applications
- Ionic Bonding - Born-Haber Cycle - Lattice energy
- p-Block Elements - Ellingham diagrams for metallurgy
What’s Next?
Thermochemistry - Apply everything you’ve learned! Calculate bond energies, heat of combustion, heat of formation, and solve complex energy problems using Born-Haber cycles and lattice energies!
What you’ll master:
- Bond dissociation energies and their applications
- Heat of combustion and heat of formation calculations
- Born-Haber cycle for ionic compounds
- Lattice energy calculations
- Flame temperatures and calorimetry
Connection: Now that you know WHEN reactions happen (ΔG), you’ll learn HOW MUCH energy they involve in different contexts!