Thermodynamics
1. Fundamentals and the First Law
Thermodynamics is the study of energy transformations. In JEE/NEET, we focus on macroscopic properties and state functions. A State Function (like $U$, $H$, $S$, $G$) depends only on the current state, not the path taken.
The First Law
Energy cannot be created or destroyed, only transformed:
- Work ($w$): For expansion against constant external pressure, $w = -P_{ext}\Delta V$. For reversible isothermal expansion, $w = -2.303nRT \log(V_2/V_1)$.
- Enthalpy ($H$): Defined as $H = U + PV$. At constant pressure, $\Delta H = q_p$.
💡 Pro Tip: Sign Convention
Remember: Work done on the system is positive ($+w$), and work done by the system is negative ($-w$). This is the IUPAC convention!
2. The Second Law and Spontaneity
The universe tends toward disorder. Entropy ($S$) is a measure of this randomness. For a spontaneous process, $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$.
Gibbs Free Energy ($G$)
The ultimate criterion for spontaneity at constant $T$ and $P$:
- $\Delta G < 0$: Spontaneous (Exergonic).
- $\Delta G > 0$: Non-spontaneous (Endergonic).
- $\Delta G = 0$: Equilibrium.
🎯 JEE/NEET Corner
The relation $\Delta G^\circ = -2.303RT \log K_{eq}$ is frequently tested. Remember that $\Delta G^\circ$ refers to standard states, while $\Delta G$ determines directionality at any state.
3. Relating $\Delta H$ and $\Delta U$
For reactions involving gases, $\Delta H = \Delta U + \Delta n_g RT$. For $N_2 + 3H_2 \to 2NH_3$, $\Delta n_g = -2$, so $\Delta H < \Delta U$. For reactions with only solids and liquids, $\Delta H \approx \Delta U$.
4. Hess's Law and Enthalpies of Reaction
Because $H$ is a state function, the enthalpy change of a reaction is the same whether it happens in one step or many. Thermochemical equations can be added and subtracted like algebra.
- $\Delta_r H^\circ = \sum \Delta_f H^\circ(products) - \sum \Delta_f H^\circ(reactants)$
- $\Delta_r H^\circ = \sum$ bond enthalpies (bonds broken) $- \sum$ bond enthalpies (bonds formed), for gas-phase reactions.
- $\Delta_f H^\circ$ of an element in its standard state is zero (graphite, not diamond, for carbon).
5. Entropy Changes
- $\Delta S = \frac{q_{rev}}{T}$. For a phase change at its transition temperature, $\Delta S = \frac{\Delta H}{T}$.
- Entropy increases on melting, vapourisation, dissolving most solids, and when $\Delta n_g > 0$.
- Third law: the entropy of a perfect crystal at 0 K is zero.
When is a reaction spontaneous?
- $\Delta H < 0$, $\Delta S > 0$: spontaneous at all temperatures.
- $\Delta H > 0$, $\Delta S < 0$: never spontaneous.
- $\Delta H < 0$, $\Delta S < 0$: spontaneous at low temperature.
- $\Delta H > 0$, $\Delta S > 0$: spontaneous at high temperature, above $T = \Delta H / \Delta S$.
Interactive diagram: open this page in a browser to use it.
6. Work in Different Processes
- Isothermal reversible expansion of an ideal gas: $\Delta U = 0$, so $q = -w$.
- Free expansion (into vacuum): $w = 0$.
- Adiabatic process: $q = 0$, so $\Delta U = w$.
- Extensive properties (mass, volume, $U$, $H$) depend on amount; intensive properties (T, P, density) do not.
Practice questions
For an isothermal reversible expansion of an ideal gas:
- Δ U = 0
- q = 0
- w = 0
- Δ H > 0
Answer
A. For an ideal gas U depends only on T; at constant T, Δ U = 0.
A reaction with Δ H > 0 and Δ S > 0 is spontaneous:
- At all temperatures
- At no temperature
- Only at high temperatures
- Only at low temperatures
Answer
C. Δ G = Δ H - TΔ S becomes negative only when T > Δ H/Δ S.
For N₂(g) + 3H₂(g) → 2NH₃(g) at constant temperature:
- Δ H = Δ U
- Δ H > Δ U
- Δ H < Δ U
- Δ H = 0
Answer
C. Δ H = Δ U + Δ n_g RT with Δ n_g = -2, so Δ H < Δ U.
Which of the following is an intensive property?
- Volume
- Enthalpy
- Temperature
- Internal energy
Answer
C. Temperature does not depend on the amount of substance.
The standard enthalpy of formation is zero for:
- Diamond
- Graphite
- O₃(g)
- H₂O(l)
Answer
B. Graphite is the standard (most stable) state of carbon; diamond and ozone are not standard states.