Thermodynamics Chemistry — Enthalpy, Entropy & Gibbs Free Energy (Class 11)

Thermodynamics: The Science of Energy Flow

Chemistry deals with transformations — but who decides which transformations actually happen? That’s thermodynamics. Every reaction you study, every cell that fires, every fuel that burns — thermodynamics is the rulebook underneath all of it.

The core question thermodynamics answers is deceptively simple: why does anything happen at all? A cube of ice melts. Iron rusts. Glucose burns in our cells. These processes share something — they occur spontaneously, without us pushing them. Thermodynamics tells us exactly when and why a process will run on its own, and when it needs a nudge.

Class 11 thermodynamics covers three big ideas: enthalpy (heat content), entropy (disorder), and Gibbs free energy (the master variable that combines both). Understand these three and you understand why reactions go the direction they do. JEE Main pulls 2-3 questions from this chapter every year. NEET consistently tests Hess’s law and Gibbs energy signs. CBSE board exams love the definitions and the sign conventions.

Let’s work through each concept from the ground up.

Key Terms & Definitions

System — the part of the universe we’re studying. Could be a reaction flask, a gas cylinder, or a living cell.

Surroundings — everything outside the system. The system + surroundings = universe.

Open system: exchanges both matter and energy with surroundings (a boiling pot without a lid).
Closed system: exchanges only energy, not matter (a sealed flask that can heat up or cool down).
Isolated system: exchanges neither (a perfect thermos — idealized, never truly achieved).

Internal energy (U) — total energy stored in a system (kinetic + potential at molecular level). We can’t measure it absolutely, but we can measure ΔU, the change.

\Delta U = q + w

where $q$ = heat absorbed by system, $w$ = work done on the system

Sign convention: heat absorbed by system → $q > 0$ (endothermic); heat released → $q < 0$ (exothermic) Work done on system → $w > 0$ ; work done by system → $w < 0$

Enthalpy (H) — defined as $H = U + pV$ . At constant pressure (which is almost every lab reaction), the heat exchanged equals the enthalpy change: $q_p = \Delta H$ .

Entropy (S) — a measure of randomness or disorder. More precisely, it counts the number of microstates available to a system. Higher disorder = higher entropy.

Gibbs Free Energy (G) — combines enthalpy and entropy into a single spontaneity criterion: $G = H - TS$ .

Core Concepts: Enthalpy

Enthalpy Change (ΔH)

For a reaction at constant pressure:

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

If $\Delta H < 0$ : reaction releases heat → exothermic
If $\Delta H > 0$ : reaction absorbs heat → endothermic

Students flip the sign convention constantly. Remember: ΔH is always from the system’s perspective. “Exothermic” means the system loses heat — energy flows out, so ΔH is negative.

Standard Enthalpy (ΔH°)

The ° symbol means standard conditions: 298 K (25°C) and 1 bar pressure. Standard enthalpies allow us to compare reactions fairly.

Standard enthalpy of formation (ΔH°f) — enthalpy change when 1 mole of a compound forms from its elements in their standard states. By definition, ΔH°f for any element in its standard state = 0.

So ΔH°f(O₂) = 0, ΔH°f(C, graphite) = 0, but ΔH°f(CO₂) = −393.5 kJ/mol.

Standard enthalpy of combustion (ΔH°c) — enthalpy change when 1 mole of substance burns completely in excess oxygen. Always exothermic (negative).

Enthalpy of neutralization — for strong acid + strong base, this is always approximately −57.1 kJ/mol. Why? Because the net reaction is always: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$

NEET 2023 asked why enthalpy of neutralization of weak acid + strong base is less than 57.1 kJ/mol. Answer: some energy is consumed in ionizing the weak acid (endothermic ionization), so the net heat released drops.

Hess’s Law

Hess’s Law: The total enthalpy change of a reaction is the same regardless of whether it occurs in one step or multiple steps — as long as the initial and final states are the same.

This is just the First Law applied to enthalpy. Enthalpy is a state function — path doesn’t matter, only start and end states do.

Practical formula:

\Delta H_{\text{reaction}}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

When using Hess’s law with given equations: if you reverse a reaction, flip the sign of ΔH. If you multiply a reaction by a factor, multiply ΔH by the same factor. These two moves let you combine any set of equations.

Bond Enthalpy

At the molecular level, breaking bonds requires energy (endothermic), forming bonds releases energy (exothermic).

\Delta H_{\text{reaction}} = \sum \text{Bond energies broken} - \sum \text{Bond energies formed}

The subtraction order trips everyone up. It’s “broken minus formed” — NOT “formed minus broken.” Think of it logically: energy IN (breaking) minus energy OUT (forming).

Core Concepts: Entropy

Entropy (S) measures the number of ways a system can be arranged — its microstates. More microstates = higher entropy.

S = k_B \ln W

where $k_B$ is Boltzmann’s constant and $W$ is the number of microstates. (Conceptual in Class 11 — not used for calculation.)

Predicting Entropy Changes

For a reaction, we can predict the sign of ΔS without calculating:

Gas produced from solid/liquid → ΔS > 0 (disorder increases dramatically)
Dissolving a solid in water → usually ΔS > 0
Gas absorbed (fewer moles of gas) → ΔS < 0
Crystallization from solution → ΔS < 0

Quick rule: count moles of gas on each side. More gas moles on product side → ΔS > 0.

Second Law of Thermodynamics

The entropy of the universe always increases for a spontaneous process.

\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \geq 0

For a reversible (equilibrium) process, $\Delta S_{\text{universe}} = 0$ . For all real spontaneous processes, it’s strictly positive.

Third Law

At absolute zero (0 K), the entropy of a perfect crystal = 0. This gives us an absolute reference point. Unlike enthalpy (where we can only measure changes), we can actually measure absolute entropy values.

Core Concepts: Gibbs Free Energy

Neither ΔH nor ΔS alone determines spontaneity. A reaction can be exothermic but non-spontaneous. A reaction can increase disorder but still not occur. We need both, combined correctly.

G = H - TS

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

At constant temperature and pressure:

$\Delta G < 0$ : spontaneous (forward reaction favored)
$\Delta G > 0$ : non-spontaneous (reverse reaction favored)
$\Delta G = 0$ : equilibrium

The Four Spontaneity Cases

ΔH	ΔS	Spontaneous?
−	+	Always (ΔG always negative)
+	−	Never (ΔG always positive)
−	−	At low temperature (enthalpy-driven)
+	+	At high temperature (entropy-driven)

The temperature-dependent cases are the ones JEE loves. The crossover temperature where a reaction switches from spontaneous to non-spontaneous is: $T = \frac{\Delta H}{\Delta S}$

JEE Main 2024 Shift 1 had a question: “For a reaction with ΔH = +ve and ΔS = +ve, at what temperature does it become spontaneous?” Answer: above $T = \Delta H / \Delta S$ . This exact case shows up almost every year in some form.

Relation to Equilibrium Constant

At standard conditions:

\Delta G^\circ = -RT \ln K

This is profound. It connects thermodynamics to chemical equilibrium. If K > 1 (products favored), ΔG° < 0 (spontaneous under standard conditions). If K < 1, ΔG° > 0.

Also: $\Delta G = \Delta G^\circ + RT \ln Q$

where Q is the reaction quotient. At equilibrium, Q = K and ΔG = 0 — which gives us the equation above.

Solved Examples

Example 1 — CBSE Level

Calculate ΔH for the reaction:
$\text{C(graphite)} + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$

Given: ΔH°f(CO₂) = −393.5 kJ/mol, ΔH°c(CO) = −283 kJ/mol

Solution:

We need to find ΔH°f(CO). Use Hess’s law by combining:

Reaction 1: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ , ΔH₁ = −393.5 kJ/mol
Reaction 2: $\text{CO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}_2$ , ΔH₂ = −283 kJ/mol

Target: Reaction 1 − Reaction 2:

\text{C} + \text{O}_2 - \text{CO} - \frac{1}{2}\text{O}_2 \rightarrow 0

\text{C} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}

\Delta H = \Delta H_1 - \Delta H_2 = -393.5 - (-283) = -110.5 \text{ kJ/mol}

Example 2 — JEE Main Level

For a reaction, ΔH = −85 kJ/mol and ΔS = −170 J/mol·K. At what temperature will the reaction reach equilibrium (ΔG = 0)?

Solution:

At equilibrium: $\Delta G = 0 \Rightarrow \Delta H = T\Delta S$

T = \frac{\Delta H}{\Delta S} = \frac{-85000 \text{ J/mol}}{-170 \text{ J/mol·K}} = 500 \text{ K}

Watch the units: ΔH in J (not kJ), ΔS in J/K. Below 500 K, enthalpy term dominates (ΔH negative wins) → spontaneous. Above 500 K, entropy term takes over (−TΔS becomes positive and dominates) → non-spontaneous.

Example 3 — JEE Advanced Level

The bond dissociation energies of H₂, Cl₂, and HCl are 436, 242, and 431 kJ/mol respectively. Calculate ΔH for: H₂(g) + Cl₂(g) → 2HCl(g)

Solution:

Bonds broken: 1 mol H−H (436) + 1 mol Cl−Cl (242) = 678 kJ
Bonds formed: 2 mol H−Cl (2 × 431) = 862 kJ

\Delta H = 678 - 862 = -184 \text{ kJ/mol}

Exothermic — makes sense, HCl formation is highly favorable.

Exam-Specific Tips

CBSE Board (3-mark and 5-mark questions)

Always define the system type before solving. Examiners specifically check whether you’ve stated “open/closed/isolated” in definitional questions. Write the First Law equation and define all terms — 1 mark is often for the formula itself.

For Hess’s law numericals: show the intermediate steps explicitly. Don’t just write the answer. CBSE marking schemes award marks for each step, so skipping steps costs marks even if your answer is correct.

JEE Main Weightage

Thermodynamics gets 1-2 questions per attempt. Highest probability topics: ΔG spontaneity cases, Hess’s law calculation, bond enthalpy calculation, and the ΔG° = −RT ln K relation. The question on temperature-dependent spontaneity (ΔH and ΔS same sign) has appeared in 7 of the last 10 JEE Main sessions.

JEE Main never asks you to derive. It gives numbers and asks you to calculate or identify the correct statement. Practice recognizing the four spontaneity cases in table form so you can answer in under 30 seconds.

NEET Strategy

NEET focuses more on conceptual understanding than calculation. Common question types: “Which of the following has highest entropy?”, “For which process is ΔS negative?”, “Enthalpy of neutralization of weak acid is less than 57.1 kJ/mol because…”

For enthalpy of bond dissociation questions, NEET usually provides bond energies in the question — you’re not expected to memorize them.

Common Mistakes to Avoid

Mistake 1: Confusing q and ΔH. ΔH = q only at constant pressure. At constant volume, ΔU = q. Most reactions occur in open containers (constant pressure), so ΔH = q_p. But bomb calorimetry measures ΔU, not ΔH directly.

Mistake 2: Forgetting to convert units in ΔG = ΔH − TΔS. ΔH is typically given in kJ/mol, ΔS in J/mol·K. You must convert one to match the other before subtracting. Multiplying 298 × (−170 J/K) and subtracting from −85 kJ without unit conversion gives a completely wrong answer.

Mistake 3: Assuming exothermic = spontaneous. Many students learn “heat is released = spontaneous” and never unlearn it. The combustion of diamond to graphite is exothermic but takes millions of years under normal conditions (kinetics, not thermodynamics — but the ΔG argument holds: it is spontaneous, just extremely slow). More critically: dissolving many salts in water is endothermic yet spontaneous because ΔS is large and positive.

Mistake 4: Wrong sign when reversing a Hess’s law equation. When you flip a reaction to use it in Hess’s law, multiply ΔH by −1. Students often forget this when the algebra is multi-step.

Mistake 5: Applying standard values at non-standard conditions. ΔG° applies at 298 K and 1 bar. If a question asks about spontaneity at 500 K, you cannot directly use ΔG° = −RT ln K unless you recalculate at the new temperature. Use ΔG = ΔH − TΔS with the new T instead.

Practice Questions

Q1. The enthalpy of combustion of carbon and carbon monoxide are −393.5 and −283 kJ/mol respectively. The enthalpy of formation of carbon monoxide is:

Using Hess’s law: C + O₂ → CO₂ (ΔH = −393.5), CO + ½O₂ → CO₂ (ΔH = −283) Subtract: C + ½O₂ → CO, ΔH = −393.5 − (−283) = −110.5 kJ/mol

Q2. For the reaction N₂(g) + O₂(g) → 2NO(g), ΔH = +180 kJ and ΔS = +25 J/K. Is this reaction spontaneous at 298 K? At what temperature does it become spontaneous?

At 298 K: ΔG = 180000 − 298(25) = 180000 − 7450 = +172550 J/mol = +172.6 kJ/mol ΔG > 0 → non-spontaneous at 298 K.

Crossover temperature: T = ΔH/ΔS = 180000/25 = 7200 K Above 7200 K, the reaction becomes spontaneous. This explains why NO forms in lightning (extremely high temperature) but not at room temperature.

Q3. Which of the following processes has ΔS < 0?
(a) Vaporization of water
(b) Dissolution of NaCl in water
(c) 2NO₂(g) → N₂O₄(g)
(d) Melting of ice

(c) — Two moles of gas combining into one mole of gas. Fewer gas molecules = fewer microstates = lower entropy. All other options increase disorder.

Q4. Calculate the bond enthalpy of C−H bond given: ΔH°f(CH₄) = −74.8 kJ/mol, bond enthalpy of H−H = 436 kJ/mol, enthalpy of sublimation of C = 718 kJ/mol.

We need: C(graphite) + 2H₂(g) → CH₄(g), ΔH = −74.8 kJ/mol

Atomization: C(graphite) → C(g), ΔH = +718 kJ/mol
2H₂ → 4H(g), ΔH = 2 × 436 = +872 kJ/mol
Formation from atoms: C(g) + 4H(g) → CH₄(g)

This step’s ΔH = −74.8 − 718 − 872 = −1664.8 kJ/mol

This equals −4 × (C−H bond enthalpy), so:
C−H bond enthalpy = 1664.8/4 = 416.2 kJ/mol

Q5. For a reaction at 500 K, ΔG° = −9.6 kJ/mol. Calculate the equilibrium constant K. (R = 8.314 J/mol·K)

ΔG° = −RT ln K
−9600 = −8.314 × 500 × ln K
ln K = 9600/4157 = 2.309
K = e^2.309 ≈ 10.06 ≈ 10

(Note: ΔG° converted to J/mol before plugging in.)

Q6. The enthalpy of neutralization of HCN (weak acid) with NaOH is −12.1 kJ/mol. Given that neutralization of strong acid with strong base releases 57.1 kJ/mol, calculate the enthalpy of ionization of HCN.

Net neutralization: H⁺ + OH⁻ → H₂O releases 57.1 kJ/mol

For HCN: HCN → H⁺ + CN⁻ (ionization, ΔH_ionization) then H⁺ + OH⁻ → H₂O (−57.1 kJ/mol)

Total = ΔH_ionization + (−57.1) = −12.1
ΔH_ionization = −12.1 + 57.1 = +45 kJ/mol

Positive — ionization of HCN is endothermic, which makes physical sense for a weak acid.

Q7. Predict spontaneity for each case:
(a) ΔH = −ve, ΔS = +ve
(b) ΔH = +ve, ΔS = +ve, high T
(c) ΔH = −ve, ΔS = −ve, high T

(a) ΔG = (−) − T(+) = always negative → always spontaneous
(b) ΔG = (+) − T(+): at high T, the TΔS term dominates and becomes large positive, making ΔG negative → spontaneous at high T
(c) ΔG = (−) − T(−) = (−) + T(|ΔS|): at high T, the positive TΔS term overwhelms the negative ΔH → ΔG becomes positive, non-spontaneous at high T

Q8. Arrange in order of increasing entropy: NaCl(s), NaCl(aq), Na(g), H₂O(l)

Entropy order: NaCl(s) < H₂O(l) < NaCl(aq) < Na(g)

Reasoning: Solids have lowest entropy. Liquids are higher. Dissolved salt (aqueous) gives more disorder than pure liquid water. Gas has by far the highest entropy (Na gas means freely moving atoms).

FAQs

Why is enthalpy a state function but heat is not?

Heat (q) depends on the path — you can release the same amount of energy slowly or all at once, and q differs. But enthalpy is defined as H = U + pV, where U, p, V are all state properties. So H is a state function, and ΔH only depends on initial and final states. This is why Hess’s law works.

What’s the difference between ΔG and ΔG°?

ΔG° (standard) is calculated at 298 K and 1 bar with all species at unit activity. ΔG is the actual free energy change under whatever conditions exist. They’re related by ΔG = ΔG° + RT ln Q. At equilibrium (Q = K), ΔG = 0, which gives ΔG° = −RT ln K.

Can a reaction with positive ΔG ever proceed forward?

Under standard conditions, no — it’s thermodynamically unfavorable. But if Q < K (the system is far from equilibrium, with less product than equilibrium requires), the reaction will proceed forward even if ΔG° > 0. The ΔG (not ΔG°) at those specific concentrations is what determines direction.

Why is entropy of a gas much larger than a solid?

Gas molecules occupy far more volume, can move in all three dimensions with high kinetic energy, and have a huge number of accessible translational, rotational, and vibrational states. A crystal is locked in fixed positions — the atoms vibrate but can’t move freely. Fewer possible arrangements = fewer microstates = lower entropy.

Does ΔH change with temperature?

Technically yes — through Kirchhoff’s law involving heat capacities. But for Class 11 and most JEE/NEET problems, ΔH is treated as temperature-independent unless stated otherwise. Only account for temperature dependence if the question explicitly gives Cp data.

What does it mean for a system to be “at equilibrium” thermodynamically?

ΔG = 0. The Gibbs energy is at its minimum. The system has no driving force to move in either direction. Note this is a thermodynamic equilibrium — the reaction is still happening at the molecular level, but the forward and reverse rates are equal (dynamic equilibrium).

Is the universe always moving toward higher entropy?

Yes — this is the Second Law. Local entropy can decrease (living organisms build ordered structures, crystals form), but always at the cost of greater entropy increase in the surroundings. The entropy of the universe strictly increases for any real process. At thermodynamic equilibrium, total entropy is maximized.