Thermodynamics
First law, enthalpy & Hess's law, entropy and Gibbs free energy for NEET
First Law and Enthalpy
System, Surroundings and the First LawTopic 1
Thermodynamics studies the energy changes that accompany physical and chemical processes. The part of the universe under study is the system; everything else is the surroundings. A system is open (exchanges matter and energy), closed (energy only) or isolated (neither) — a distinction NEET often opens with.
The total energy stored in a system is its internal energy $U$. We cannot measure $U$ itself, only its change $\Delta U$. The first law of thermodynamics is simply the conservation of energy: $\Delta U = q + W$, where $q$ is heat added to the system and $W$ is work done on it. For a gas expanding or compressing against an external pressure, the work is $W = -P\Delta V$ (work done by the gas is negative for the system).
A vital idea is the difference between state functions and path functions. A state function depends only on the present state, not on how it was reached — $U$, enthalpy $H$, entropy $S$ and Gibbs energy $G$ are all state functions. Heat $q$ and work $W$ are path functions: their values depend on the route taken. Knowing which quantities are state functions is a recurring NEET point.
Sign conventions matter. Heat absorbed by the system is positive, heat released is negative; work done on the system is positive, work done by the system is negative. For special processes the first law simplifies: at constant volume $W = 0$ so $\Delta U = q_V$, and in an isothermal process for an ideal gas $\Delta U = 0$ (internal energy depends only on temperature). Getting these signs right turns most numerical questions into one-line substitutions.
| Idea | Relation |
|---|---|
| First law | $\Delta U = q + W$ |
| Work (gas) | $W = -P\Delta V$ |
| Constant volume | $\Delta U = q_V$ |
| State functions | $U, H, S, G$ |
A system absorbs $150\ \text{J}$ of heat and does $40\ \text{J}$ of work on the surroundings. Find $\Delta U$.
Show solution
Heat in $q = +150\ \text{J}$; work done by the system means $W = -40\ \text{J}$. So $\Delta U = q + W = 150 - 40 = +110\ \text{J}$.
Why is $\Delta U = q$ for a process carried out at constant volume?
Show solution
At constant volume $\Delta V = 0$, so the pressure-volume work $W = -P\Delta V = 0$. The first law then gives $\Delta U = q_V$ — all the heat changes the internal energy.
The first law of thermodynamics is a statement of conservation of:
Which of these is a state function?
A system that exchanges neither matter nor energy is:
For an isothermal process in an ideal gas, $\Delta U$ is:
The pressure-volume work done by a gas is:
NEET tip: Fix signs first — heat in $+$, work done by the gas $-$. $U, H, S, G$ are state functions; $q, W$ are not. At constant volume $\Delta U = q_V$.
Enthalpy, Hess's Law and Reaction EnthalpiesTopic 2
Most chemistry happens at constant (atmospheric) pressure, where it is convenient to define enthalpy $H = U + PV$. At constant pressure the heat exchanged equals the enthalpy change: $q_p = \Delta H$. For reactions involving gases, enthalpy and internal energy are related by $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of gas moles — a formula NEET tests directly.
The sign of $\Delta H$ classifies reactions: exothermic reactions release heat ($\Delta H < 0$, e.g. combustion) and endothermic reactions absorb it ($\Delta H > 0$, e.g. photosynthesis, decomposition). Several standard enthalpies are defined: of formation (forming one mole of a compound from its elements), combustion, neutralisation, and bond enthalpy.
Hess's law of constant heat summation is the most powerful tool here. Because enthalpy is a state function, the total enthalpy change of a reaction is the same whether it occurs in one step or several. This lets you add the enthalpies of individual steps (reversing a step flips the sign; multiplying a step multiplies its $\Delta H$) to find an enthalpy that is hard to measure directly.
Hess's law also gives the standard enthalpy of reaction from formation values: $\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$, remembering that the $\Delta H_f$ of an element in its standard state is zero. Reaction enthalpy can likewise be estimated from bond enthalpies as (bonds broken) − (bonds formed). These three routes — steps, formation values, and bond enthalpies — cover essentially every NEET enthalpy calculation.
| Quantity | Relation |
|---|---|
| Enthalpy | $H = U + PV$; $q_p = \Delta H$ |
| $\Delta H$–$\Delta U$ | $\Delta H = \Delta U + \Delta n_g RT$ |
| From formation | $\sum \Delta H_f(\text{prod}) - \sum \Delta H_f(\text{react})$ |
| From bonds | (broken) − (formed) |
For a reaction $\Delta U = -100\ \text{kJ}$ and $\Delta n_g = -1$ at $300\ \text{K}$. Find $\Delta H$. ($R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$)
Show solution
$\Delta H = \Delta U + \Delta n_g RT = -100\,000 + (-1)(8.314)(300) \approx -100\,000 - 2494 = -102.5\ \text{kJ}$.
How does Hess's law let us find an enthalpy that can't be measured directly?
Show solution
Since enthalpy is a state function, we add the $\Delta H$ of a series of measurable steps that sum to the target reaction (reversing a step flips its sign). The total equals the target reaction's $\Delta H$.
At constant pressure, the heat exchanged equals:
$\Delta H$ and $\Delta U$ are related by:
An exothermic reaction has:
Hess's law is a consequence of enthalpy being a:
The standard enthalpy of formation of an element in its standard state is:
NEET tip: Use $\Delta H = \Delta U + \Delta n_g RT$ (count only gas moles). For reaction enthalpy: $\sum\Delta H_f$(products) − $\sum\Delta H_f$(reactants), with elements = 0; or bonds (broken − formed).
Entropy and Spontaneity
Entropy and the Second LawTopic 3
The first law tells us energy is conserved, but it does not say which way a process will go. That direction is governed by entropy $S$, a measure of the disorder or number of ways the energy and particles of a system can be arranged. The more disordered a state, the higher its entropy, and entropy — like enthalpy — is a state function.
Entropy increases with disorder, so for the same substance $S_{gas} > S_{liquid} > S_{solid}$. It rises on melting, boiling and dissolving, when a solid decomposes into gases, or when the number of gas molecules increases in a reaction. NEET frequently asks you to predict the sign of $\Delta S$ from such changes, so reading the disorder of reactants versus products is the key skill.
The second law of thermodynamics states that the total entropy of the universe always increases for a spontaneous process: $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0$. A process can lower the entropy of the system (for instance, freezing) only if the surroundings gain even more entropy, keeping the universe's total rising.
The third law adds a reference point: the entropy of a perfect crystalline solid is zero at absolute zero ($0\ \text{K}$), because there is only one way to arrange a perfectly ordered, motionless crystal. This allows absolute entropies to be tabulated. For NEET the practical takeaways are: predict the sign of $\Delta S$ from changes in state and gas moles, and remember that a spontaneous change always raises the entropy of the universe.
| Change | Effect on entropy |
|---|---|
| Solid to liquid to gas | $S$ increases |
| More gas moles formed | $\Delta S > 0$ |
| Second law | $\Delta S_{universe} > 0$ (spontaneous) |
| Third law | $S = 0$ for perfect crystal at $0\ \text{K}$ |
Predict the sign of $\Delta S$ for: $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$.
Show solution
A gas ($\text{CO}_2$) is produced from a solid, greatly increasing disorder, so $\Delta S > 0$ (positive).
Water freezing into ice lowers the system's entropy. How is this consistent with the second law?
Show solution
The system loses entropy, but the heat released to the surroundings raises their entropy by a larger amount, so $\Delta S_{universe} > 0$ overall — the second law is satisfied.
Entropy is a measure of:
For the same substance, entropy is highest in the:
For a spontaneous process, the entropy of the universe:
The entropy of a perfect crystal at $0\ \text{K}$ is:
$\Delta S$ is positive when:
NEET tip: Predict $\Delta S$ sign from disorder: gas formed / more gas moles / dissolving to positive. Spontaneity is decided by the universe's entropy, not the system's alone.
Gibbs Free Energy and SpontaneityTopic 4
Tracking the entropy of the universe is awkward, so chemists use a property of the system alone that still predicts spontaneity: the Gibbs free energy $G = H - TS$. For a process at constant temperature and pressure, the change is $\Delta G = \Delta H - T\Delta S$. This single equation is the heart of chemical thermodynamics for NEET.
The sign of $\Delta G$ decides the direction: a process is spontaneous (feasible) when $\Delta G < 0$, non-spontaneous when $\Delta G > 0$, and at equilibrium when $\Delta G = 0$. The magnitude of $\Delta G$ also represents the maximum useful (non-expansion) work a reaction can deliver.
Because $\Delta G$ combines enthalpy and entropy with temperature, four cases arise. If $\Delta H < 0$ and $\Delta S > 0$, $\Delta G$ is negative at all temperatures (always spontaneous). If $\Delta H > 0$ and $\Delta S < 0$, it is positive at all temperatures (never spontaneous). The two mixed cases are temperature-dependent: an exothermic reaction with $\Delta S < 0$ is spontaneous only at low temperature, while an endothermic reaction with $\Delta S > 0$ becomes spontaneous only at high temperature. This explains why an endothermic reaction can still proceed.
Gibbs energy also links to equilibrium. The standard free-energy change relates to the equilibrium constant by $\Delta G^{\circ} = -RT\ln K$, so a large negative $\Delta G^{\circ}$ means a large $K$ (products favoured). Mastering $\Delta G = \Delta H - T\Delta S$, the four spontaneity cases, and $\Delta G^{\circ} = -RT\ln K$ covers nearly every spontaneity question NEET asks.
| $\Delta H$ / $\Delta S$ | Spontaneity |
|---|---|
| $\Delta H<0,\ \Delta S>0$ | spontaneous at all T |
| $\Delta H>0,\ \Delta S<0$ | never spontaneous |
| $\Delta H<0,\ \Delta S<0$ | spontaneous at low T |
| $\Delta H>0,\ \Delta S>0$ | spontaneous at high T |
A reaction has $\Delta H = +50\ \text{kJ}$ and $\Delta S = +0.2\ \text{kJ K}^{-1}$. Is it spontaneous at $300\ \text{K}$?
Show solution
$\Delta G = \Delta H - T\Delta S = 50 - (300)(0.2) = 50 - 60 = -10\ \text{kJ}$. Since $\Delta G < 0$, it is spontaneous at $300\ \text{K}$.
What is the value of $\Delta G$ for a reaction at equilibrium?
Show solution
At equilibrium $\Delta G = 0$ (the system has no tendency to move in either direction). Note this is $\Delta G$, not the standard $\Delta G^{\circ}$.
The Gibbs free energy is defined as:
A reaction is spontaneous when:
At equilibrium, $\Delta G$ equals:
A reaction with $\Delta H < 0$ and $\Delta S > 0$ is spontaneous:
The standard free energy relates to $K$ by:
NEET tip: $\Delta G = \Delta H - T\Delta S$; spontaneous if $< 0$, equilibrium at $= 0$. Memorise the four sign cases and $\Delta G^{\circ} = -RT\ln K$ (negative $\Delta G^{\circ}$ to large $K$).
Quick Revision — Thermodynamics
- First law: $\Delta U = q + W$ (energy conserved). For a gas $W = -P\Delta V$. $U$ and $H$ are state functions; $q$ and $W$ are not.
- Enthalpy: $H = U + PV$; at constant pressure $q_p = \Delta H$. $\Delta H = \Delta U + \Delta n_g RT$.
- Sign: exothermic $\Delta H < 0$; endothermic $\Delta H > 0$.
- Hess's law: total enthalpy change is the same whatever the path — add steps' $\Delta H$.
- Entropy $S$ measures disorder; gas $>$ liquid $>$ solid. $\Delta S_{universe} > 0$ for a spontaneous change (second law).
- Gibbs energy: $\Delta G = \Delta H - T\Delta S$. Spontaneous if $\Delta G < 0$; equilibrium at $\Delta G = 0$.
- $\Delta H<0,\ \Delta S>0$ to spontaneous at all T; $\Delta H>0,\ \Delta S<0$ to never spontaneous; the mixed cases are temperature-dependent.
- $\Delta G^\circ = -RT\ln K$ — links free energy to the equilibrium constant.
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