NEET Chemistry: Thermodynamics & Thermochemistry – Complete Revision Guide, Formulas & Solved PYQs

Introduction to Thermodynamics & Thermochemistry for NEET UG

Thermodynamics is one of the most fundamental and high-yielding chapters in Physical Chemistry for the NEET UG examination. Derived from the Greek words therme (heat) and dynamis (power), thermodynamics is the branch of science that deals with the quantitative relationships between heat, work, and other forms of energy. Thermochemistry, a specialized sub-discipline, focuses specifically on the energy transformations and heat changes occurring during chemical reactions.

For a prospective medical student, understanding thermodynamic principles is not merely an academic requirement for cracking NEET. Thermodynamic laws govern biological bioenergetics—such as how adenosine triphosphate (ATP) hydrolysis drives metabolic pathways, how cell membranes maintain electrochemical gradients, and how enzyme-catalyzed reactions achieve thermodynamic stability. A thorough grasp of state functions, internal energy, enthalpy, entropy, and Gibbs free energy equips NEET aspirants with the analytical tools needed to solve complex physical chemistry numericals and conceptual questions effortlessly.

NEET Weightage & Expected Questions

In the NEET UG exam pattern set by the NTA (National Testing Agency), Physical Chemistry holds a major share of marks, and Thermodynamics along with Thermochemistry consistently accounts for 2 to 3 direct questions (8 to 12 marks) every year. The distribution of questions across sub-topics typically follows a distinct trend:

• First Law of Thermodynamics & Work Done Calculations: ~30% weightage (Focus on isothermal, adiabatic, isobaric, and isochoric expansions/compressions).
• Thermochemistry & Hess's Law: ~30% weightage (Enthalpy of formation, combustion, neutralization, and bond dissociation energies).
• Second Law, Entropy & Gibbs Free Energy: ~40% weightage (Criteria for spontaneity, equilibrium constant connection $\Delta G^\circ = -RT \ln K$, temperature dependence of spontaneity).

Mastering this single chapter provides a significant boost to your All India Rank (AIR), as numerical problems in thermodynamics are highly standardized and predictable once key formulas and sign conventions are understood.

Core Concepts & Key Mechanisms Explained

1. Thermodynamic Systems, Surroundings, and State Functions

A system is defined as the specific portion of the universe under thermodynamic study, while everything outside the system constitutes the surroundings. Universe = System + Surroundings.

• Open System: Can exchange both energy and matter with surroundings (e.g., liquid in an open beaker).
• Closed System: Can exchange energy but NOT matter (e.g., liquid in a sealed metallic flask).
• Isolated System: Can exchange neither energy nor matter (e.g., liquid in a perfectly insulated thermos flask).

Properties of a system are categorized into:

• State Functions: Properties that depend strictly on the initial and final states of the system, independent of the path taken. Examples: Internal energy ($U$), Enthalpy ($H$), Entropy ($S$), Gibbs Free Energy ($G$), Pressure ($P$), Temperature ($T$), Volume ($V$).
• Path Functions: Properties whose values depend on the specific path or mechanism chosen during the transformation. Examples: Work done ($w$) and Heat ($q$).

2. The First Law of Thermodynamics (FLOT) & Sign Conventions

The First Law of Thermodynamics is the principle of conservation of energy. It states that energy can neither be created nor destroyed, only transformed from one form to another. Mathematically, it is expressed as:

$$\Delta U = q + w$$

Where $\Delta U$ is the change in internal energy, $q$ is the heat added to the system, and $w$ is the work done on the system. It is critical to adhere to the IUPAC sign convention strictly in Chemistry:

• Heat absorbed by the system: $q > 0$ (positive)
• Heat released by the system: $q < 0$ (negative)
• Work done ON the system (compression): $w > 0$ (positive)
• Work done BY the system (expansion): $w < 0$ (negative)

Pressure-Volume ($P$-$V$) work done during an irreversible expansion against constant \texternal pressure ($P_{\text{ext}}$) is calculated as:

$$w = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1)$$

For an isothermal reversible expansion of an ideal gas from volume $V_1$ to $V_2$:

$$w_{\text{rev}} = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right) = -2.303 nRT \log_{10}\left(\frac{P_1}{P_2}\right)$$

3. Enthalpy Changes ($ΔH$) and Thermal Capacity

Enthalpy ($H$) is defined as the total heat content of a system at constant pressure: $H = U + PV$. The change in enthalpy during a process at constant pressure is related to internal energy change by:

$$\Delta H = \Delta U + \Delta(PV)$$

For reactions involving ideal gases, this simplifies to the high-yield NEET formula:

$$\Delta H = \Delta U + \Delta n_g RT$$

Where $\Delta n_g = \text{(Moles of gaseous products)} - \text{(Moles of gaseous reactants)}$.

• If $\Delta n_g = 0$, then $\Delta H = \Delta U$.
• If $\Delta n_g > 0$, then $\Delta H > \Delta U$.
• If $\Delta n_g < 0$, then $\Delta H < \Delta U$.

4. Second Law of Thermodynamics, Entropy, and Spontaneity

The First Law predicts energy conservation but fails to predict the direction of spontaneous change. The Second Law of Thermodynamics states that for any spontaneous process, the total entropy of the universe ($ΔS_{\text{total}}$ or $\Delta S_{\text{univ}}$) must always increase:

$$\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$$

Entropy ($S$) is a quantitative measure of randomness or molecular disorder. Entropy change is defined as:

$$\Delta S = \frac{q_{\text{rev}}}{T}$$

5. Gibbs Free Energy ($ΔG$) and Chemical Equilibrium

Gibbs Free Energy ($G$) combines enthalpy and entropy into a single state function to assess reaction spontaneity at constant temperature and pressure:

$$G = H - TS \implies \Delta G = \Delta H - T\Delta S$$

Criteria for Spontaneity at constant $T$ and $P$:

• $\Delta G < 0$ (Negative): The process is spontaneous (exergonic).
• $\Delta G > 0$ (Positive): The process is non-spontaneous in the forward direction (endergonic).
• $\Delta G = 0$: The system has reached dynamic equilibrium.

The standard free energy change $\Delta G^\circ$ is related to the chemical equilibrium constant ($K_{\text{eq}}$) by the relation:

$$\Delta G^\circ = -RT \ln K_{\text{eq}} = -2.303 RT \log_{10} K_{\text{eq}}$$

6. Key Thermochemical Laws

Hess's Law of Constant Heat Summation: The overall enthalpy change for a chemical reaction is identical whether the reaction takes place in one single step or in a series of multiple steps. Enthalpy is a state function!

$$\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$$

Bond Enthalpy Formula:

$$\Delta_r H = \sum \text{Bond Energy of Reactants} - \sum \text{Bond Energy of Products}$$

Important Formulas & Key Terms Table

Solved Step-by-Step Previous Years Questions (PYQs)

PYQ 1 (NEET 2020): Work Done in Isothermal Free Expansion

Question: An ideal gas expands isothermally and reversibly from $2\text{ L}$ to $20\text{ L}$ at $300\text{ K}$, doing work when expanding against vacuum. What is the work done?

• (A) $-2.303 \times 300 \text{ J}$
• (B) Zero
• (C) $-8.314 \text{ J}$
• (D) $+18 \text{ J}$

Detailed Solution:

1. Identify key thermodynamic terms given: 'expanding against vacuum'.

2. Expansion against a vacuum means \texternal opposing pressure $P_{\text{ext}} = 0$.

3. The general expression for work done is $w = -P_{\text{ext}} \Delta V$.

4. Since $P_{\text{ext}} = 0$, $w = -(0) \times (20 - 2) = 0$.

5. Expansion against zero \texternal pressure is called free expansion, where work done is always zero whether the process is isothermal or adiabatic.

Correct Answer: (B) Zero

PYQ 2 (NEET 2019): Relation between $ΔH$ and $ΔU$

Question: For the reaction $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$ at $298\text{ K}$, the value of $\Delta H - \Delta U$ is equal to:

• (A) $-3 RT$
• (B) $+3 RT$
• (C) $-1 RT$
• (D) $+1 RT$

Detailed Solution:

1. Recall the fundamental relationship: $\Delta H = \Delta U + \Delta n_g RT$. Therefore, $\Delta H - \Delta U = \Delta n_g RT$.

2. Calculate $\Delta n_g = n_g(\text{products}) - n_g(\text{reactants})$.

3. Note the physical states carefully: $\text{CO}_2$ is gas, $\text{H}_2\text{O}$ is liquid, $\text{C}_3\text{H}_8$ is gas, $\text{O}_2$ is gas.

4. Moles of gaseous products = $3$ (from $3\text{CO}_2(g)$). Water is liquid, so its moles are excluded!

5. Moles of gaseous reactants = $1$ (from $\text{C}_3\text{H}_8$) $+ 5$ (from $\text{O}_2$) = $6$.

6. $\Delta n_g = 3 - 6 = -3$.

7. Substituting $\Delta n_g = -3$ into the equation: $\Delta H - \Delta U = -3 RT$.

Correct Answer: (A) $-3 RT$

PYQ 3 (NEET 2021): Spontaneity Conditions

Question: For a reaction to be spontaneous at all temperatures, the required conditions are:

• (A) $\Delta H > 0$ and $\Delta S < 0$
• (B) $\Delta H < 0$ and $\Delta S > 0$
• (C) $\Delta H < 0$ and $\Delta S < 0$
• (D) $\Delta H > 0$ and $\Delta S > 0$

Detailed Solution:

1. Use the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$.

2. For a reaction to be spontaneous, $\Delta G$ must be strictly negative ($ΔG < 0$).

3. If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in randomness), then $-T\Delta S$ will also be negative since temperature $T$ in Kelvin is always positive.

4. Resulting in $\Delta G = (-\text{ve}) + (-\text{ve}) = -\text{ve}$ at all values of $T$. Thus, the reaction is universally spontaneous at all temperatures.

Correct Answer: (B) $\Delta H < 0$ and $\Delta S > 0$

Common NEET Traps & Mistakes to Avoid

• Trap 1: Confusion between Physics & Chemistry Sign Conventions for Work. In Chemistry, IUPAC defines work done ON the system as positive ($w = -P \Delta V$). In Physics, work done BY the system is taken as positive ($w = +P \Delta V$). Always use Chemistry sign convention in the Chemistry section of NEET!
• Trap 2: Counting Liquid/Solid Moles in $Δn_g$. Candidates frequently add moles of liquid or solid products/reactants while calculating $\Delta n_g$. Only count gaseous species! For example, in $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$, $\Delta n_g = 1 - 0 = 1$.
• Trap 3: Standard Enthalpy of Formation ($Δ_f H^\circ$) of Pure Elements. By convention, standard enthalpy of formation of pure elements in their most stable natural reference states is assigned as zero ($0\text{ kJ/mol}$). Examples: $\text{O}_2(g) = 0$, $\text{C}(\text{graphite}) = 0$, $\text{S}_8(\text{rhombic}) = 0$, $\text{Br}_2(l) = 0$. Note: $\text{C}(\text{diamond}) \neq 0$.
• Trap 4: Mismatch of Units in $ΔG = ΔH - TΔS$. $\Delta H$ is usually given in $\text{kJ/mol}$, while $\Delta S$ is given in $\text{J K}^{-1} \text{mol}^{-1}$. Before performing calculations, convert $\Delta S$ to $\text{kJ K}^{-1} \text{mol}^{-1}$ by dividing by $1000$ (or multiply $\Delta H$ by $1000$ to convert to Joules). Failing to match units leads to incorrect choices carefully placed in NEET answer options.
• Trap 5: Assuming Reversible and Irreversible Work are Equal. $w_{\text{rev}}$ is always greater in magnitude than $w_{\text{irrev}}$ for expansion processes. Make sure to use $-2.303 nRT \log(V_2/V_1)$ for reversible processes and $-P_{\text{ext}}\Delta V$ for irreversible processes.

High-Yield NEET Practice MCQs with Answer Keys

Practice MCQ 1

Calculate the work done when $2\text{ moles}$ of an ideal gas expand reversibly and isothermally from a volume of $10\text{ L}$ to $100\text{ L}$ at $300\text{ K}$. ($R = 8.314\text{ J K}^{-1} \text{mol}^{-1}$)

• (A) $-11.48 \text{ kJ}$
• (B) $+11.48 \text{ kJ}$
• (C) $-5.74 \text{ kJ}$
• (D) $-114.8 \text{ J}$

Solution: $w = -2.303 nRT \log_{10}(V_2/V_1) = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10}(100/10) = -2.303 \times 2 \times 8.314 \times 300 \times 1 = -11488.2\text{ J} = -11.488\text{ kJ}$. Correct Option: (A).

Practice MCQ 2

For a reaction $\text{A}(g) + 2\text{B}(g) \rightarrow 2\text{C}(g) + \text{D}(s)$, $\Delta U^\circ = -10.5\text{ kJ}$ at $298\text{ K}$. Calculate $\Delta H^\circ$ for the reaction. ($R = 8.314\text{ J K}^{-1} \text{mol}^{-1}$)

• (A) $-12.98 \text{ kJ}$
• (B) $-8.02 \text{ kJ}$
• (C) $-10.50 \text{ kJ}$
• (D) $+12.98 \text{ kJ}$

Solution: $\Delta n_g = n_g(\text{products}) - n_g(\text{reactants}) = 2 - (1 + 2) = -1$. Using $\Delta H^\circ = \Delta U^\circ + \Delta n_g RT$: $\Delta H^\circ = -10.5\text{ kJ} + (-1) \times (8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1}) \times 298\text{ K} = -10.5 - 2.48 = -12.98\text{ kJ}$. Correct Option: (A).

Practice MCQ 3

A system absorbs $500\text{ J}$ of heat and performs $200\text{ J}$ of work on the surroundings. The net change in internal energy ($ΔU$) of the system is:

• (A) $+700 \text{ J}$
• (B) $+300 \text{ J}$
• (C) $-300 \text{ J}$
• (D) $-700 \text{ J}$

Solution: According to FLOT: $\Delta U = q + w$. Heat absorbed by system $q = +500\text{ J}$. Work done BY system on surroundings $w = -200\text{ J}$. Therefore, $\Delta U = (+500) + (-200) = +300\text{ J}$. Correct Option: (B).

Practice MCQ 4

Under what conditions of $ΔH$ and $ΔS$ will a non-spontaneous reaction become spontaneous upon INCREASING the temperature?

• (A) Both $\Delta H$ and $\Delta S$ are negative.
• (B) Both $\Delta H$ and $\Delta S$ are positive.
• (C) $\Delta H$ is positive and $\Delta S$ is negative.
• (D) $\Delta H$ is negative and $\Delta S$ is positive.

Solution: $\Delta G = \Delta H - T\Delta S$. If both $\Delta H > 0$ and $\Delta S > 0$, at low temperatures $\Delta H > T\Delta S$, making $\Delta G > 0$ (non-spontaneous). As $T$ increases, the magnitude of $T\Delta S$ increases until $T\Delta S > \Delta H$, turning $\Delta G < 0$ (spontaneous). Correct Option: (B).

Practice MCQ 5

The standard enthalpy of combustion of carbon (graphite), hydrogen gas, and methane at $298\text{ K}$ are $-393.5\text{ kJ/mol}$, $-285.8\text{ kJ/mol}$, and $-890.3\text{ kJ/mol}$ respectively. The standard enthalpy of formation of $\text{CH}_4(g)$ is:

• (A) $-74.8 \text{ kJ/mol}$
• (B) $+74.8 \text{ kJ/mol}$
• (C) $-52.6 \text{ kJ/mol}$
• (D) $+52.6 \text{ kJ/mol}$