NeetCrackers Chemistry Notes
Complete Notes

Chemical Thermodynamics

Every topic explained simply, with practice questions and answers after each one.

📑 Topics Covered

  1. Basic Thermodynamic Terms
  2. Internal Energy & the First Law
  3. Enthalpy & Hess's Law
  4. Types of Enthalpy Changes
  5. Heat Capacity: Cv and Cp
  6. Spontaneity & Entropy
  7. Gibbs Energy & Criteria for Spontaneity
  8. Gibbs Energy & Equilibrium; the Third Law

Basic Thermodynamic Terms

System & Surroundings

The system is the specific part of the universe under study; everything else is the surroundings. A boundary (real or imaginary) separates the two.

Type of SystemExchanges with SurroundingsExample
OpenBoth matter and energyHot tea in an open cup
ClosedOnly energy, not matterSealed flask being heated
IsolatedNeither matter nor energyAn ideal thermos flask

State Functions vs Path Functions

Key Distinction

A state function depends only on the initial and final state of the system, not on the path taken between them — e.g., internal energy, enthalpy, entropy, Gibbs energy. A path function depends on the route taken — e.g., heat (q) and work (w).

Q. Classify the following as state or path functions: internal energy, heat, enthalpy, work.
Ans: Internal energy — state function. Heat — path function. Enthalpy — state function. Work — path function.

Internal Energy & the First Law

Internal Energy (U)

The sum of all forms of kinetic and potential energy of the particles within a system. Only the change in internal energy ($\Delta U$) between two states can be measured, not its absolute value.

First Law of Thermodynamics $$\Delta U = q + w$$

Energy can neither be created nor destroyed, only transformed from one form to another. Here $q$ = heat absorbed by the system (positive if absorbed, negative if released) and $w$ = work done on the system (positive if done on the system, negative if done by the system).

Work in Expansion/Compression

Pressure-Volume Work $$w = -p_{ext}\,\Delta V$$

For work done against a constant external pressure. The negative sign reflects that work done by the system (during expansion, $\Delta V$ positive) reduces its internal energy.

Solved Example

A gas absorbs 200 J of heat and does 80 J of work on the surroundings during expansion. Calculate ΔU.

$q = +200\ \text{J}$ (absorbed). $w = -80\ \text{J}$ (work done by system, so work done on system is negative).

$\Delta U = 200 + (-80) = 120\ \text{J}$

Q. A system releases 150 J of heat while 60 J of work is done on it. Calculate the change in internal energy.
Ans: q = −150 J (released), w = +60 J (done on system). ΔU = −150 + 60 = −90 J.

Enthalpy & Hess's Law

Enthalpy (H)

Defined as $H = U + pV$. At constant pressure (the most common lab condition), the heat exchanged equals the change in enthalpy: $q_p = \Delta H$. This makes enthalpy more practically useful than internal energy for most reactions carried out in open vessels.

Relating ΔH and ΔU $$\Delta H = \Delta U + \Delta n_g RT$$

where $\Delta n_g$ = (moles of gaseous products) − (moles of gaseous reactants). If $\Delta n_g = 0$ (no change in gas moles), $\Delta H = \Delta U$.

Hess's Law of Constant Heat Summation

Statement

The total enthalpy change for a reaction is the same whether it occurs in one step or through several intermediate steps — since enthalpy is a state function, only the initial and final states matter, not the path. This allows enthalpies of reactions that are hard to measure directly to be calculated by combining known enthalpies of other, related reactions.

Solved Example

Given $\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$, $\Delta H = -393.5$ kJ, and $\text{CO(g)} + \tfrac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$, $\Delta H = -283.0$ kJ, find $\Delta H$ for $\text{C(s)} + \tfrac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}$.

Subtracting the second equation from the first: $\Delta H = -393.5 - (-283.0) = -110.5\ \text{kJ}$

Q. For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), calculate Δn_g and state when ΔH would equal ΔU for this type of reaction.
Ans: Δn_g = 2 − (1+3) = −2. ΔH equals ΔU only when Δn_g = 0, i.e., when the moles of gaseous products equal the moles of gaseous reactants — which is not the case here, so ΔH ≠ ΔU for this reaction.

Types of Enthalpy Changes

TypeDefinition
Enthalpy of formation (ΔfH°)Enthalpy change when 1 mole of a compound forms from its elements in their standard states
Enthalpy of combustion (ΔcH°)Enthalpy change when 1 mole of a substance is completely burnt in oxygen
Enthalpy of atomizationEnthalpy change to break 1 mole of bonds to obtain atoms in the gas phase
Bond enthalpyAverage energy needed to break one mole of a specific bond in the gaseous state
Enthalpy of sublimationEnthalpy change when 1 mole of a solid converts directly to gas
Enthalpy of solutionEnthalpy change when 1 mole of a solute dissolves completely in a specified amount of solvent
Lattice enthalpyEnthalpy change when 1 mole of an ionic solid separates into its gaseous ions
Enthalpy of Reaction from Bond Enthalpies $$\Delta H_{rxn} = \sum(\text{Bond enthalpies of reactants}) - \sum(\text{Bond enthalpies of products})$$

Energy is absorbed to break bonds (reactants) and released when new bonds form (products).

Q. Which enthalpy term describes the energy released when 1 mole of methane is completely burnt in oxygen?
Ans: Enthalpy of combustion (ΔcH°) of methane.

Heat Capacity: Cv and Cp

Definitions

Heat capacity is the heat required to raise a system's temperature by 1°C (or 1 K). $C_v$ = heat capacity at constant volume; $C_p$ = heat capacity at constant pressure. $C_p$ is always greater than $C_v$ because at constant pressure, some of the added heat also does expansion work rather than only raising the temperature.

Relation Between Cp and Cv (Ideal Gas) $$C_p - C_v = R$$
Q. Why is Cp always greater than Cv for a gas?
Ans: At constant volume, all the heat supplied goes entirely into raising internal energy (temperature). At constant pressure, the gas also expands and does work against the surroundings, so extra heat is needed for the same temperature rise — making Cp > Cv.

Spontaneity & Entropy

Spontaneous Process

A process that occurs on its own, without continuous external intervention, once started (though it may need initial activation energy). Spontaneity is about the natural direction of a process, not its speed — a spontaneous reaction can still be very slow.

Entropy (S)

Definition

Entropy is a measure of the randomness or disorder of a system — a state function. Gases have higher entropy than liquids, which have higher entropy than solids, for the same substance.

Second Law of Thermodynamics $$\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0 \ \ (\text{for a spontaneous process})$$

The total entropy of the universe always increases for any spontaneous process.

Q. Predict whether entropy increases or decreases when ice melts into water, and explain why.
Ans: Entropy increases. In the solid (ice), molecules are held in a fixed, ordered lattice; in the liquid state, molecules move more freely and randomly, so disorder — and hence entropy — increases.

Gibbs Energy & Criteria for Spontaneity

Why Gibbs Energy Is Needed

Entropy change alone isn't a convenient spontaneity test because it requires tracking both system and surroundings. Gibbs energy (G) combines both effects into one system-only quantity: $G = H - TS$.

Gibbs Energy Equation $$\Delta G = \Delta H - T\Delta S$$
ΔHΔSΔGSpontaneity
−+Always −Spontaneous at all temperatures
+−Always +Non-spontaneous at all temperatures
−−− at low T, + at high TSpontaneous only at low temperature
+++ at low T, − at high TSpontaneous only at high temperature
Solved Example

For a reaction, ΔH = −40 kJ/mol and ΔS = −100 J/K/mol at 300 K. Determine if the reaction is spontaneous.

$\Delta G = \Delta H - T \Delta S = -40{,}000 - (300)(-100) = -40{,}000 + 30{,}000 = -10{,}000\ \text{J} = -10\ \text{kJ}$

Since ΔG is negative, the reaction is spontaneous at 300 K.

Q. A reaction has ΔH = +50 kJ/mol and ΔS = +150 J/K/mol. Is it spontaneous at 400 K?
Ans: ΔG = 50,000 − (400)(150) = 50,000 − 60,000 = −10,000 J = −10 kJ. Since ΔG is negative, the reaction is spontaneous at 400 K.

Gibbs Energy & Equilibrium; the Third Law

Gibbs Energy and the Equilibrium Constant $$\Delta G^\circ = -RT \ln K$$

Links the standard Gibbs energy change of a reaction to its equilibrium constant K — a large K (products favoured) corresponds to a large negative ΔG°.

Third Law of Thermodynamics

Statement

The entropy of a perfectly crystalline substance is zero at absolute zero temperature (0 K) — at this point, there is no thermal disorder, and the particles are in a single, perfectly ordered arrangement. This law provides a reference point for calculating absolute entropy values at other temperatures.

Q. According to the third law of thermodynamics, what is the entropy of a perfect crystal at 0 K?
Ans: Zero — a perfectly ordered crystalline arrangement at absolute zero has no positional or thermal disorder.

⚡ Formula Cheat-Sheet

First lawΔU = q + w
PV workw = −p_ext ΔV
EnthalpyH = U + pV
ΔH vs ΔUΔH = ΔU + Δn_g RT
Cp − Cv= R
Gibbs energyΔG = ΔH − TΔS
Gibbs & equilibriumΔG° = −RT ln K
SpontaneityΔG < 0