Chemical thermodynamics for NEET

Chemical thermodynamics answers two questions about a reaction: how much energy it gives out or takes in, and whether it will go on its own. The chapter overlaps with physics thermodynamics, but the questions are different in style. Chemistry asks about enthalpies of formation and combustion, Hess's law, bond enthalpies and the sign of ΔG, and it uses its own sign convention for work.

NEET questions here are usually short calculations or sign-and-condition questions: ΔH from given data, ΔH − ΔU for a gas reaction, the temperature at which a reaction becomes spontaneous, or which quantity is a state function.

Systems, states and the language of the chapter

The first law and work

The first law says energy is conserved: ΔU = q + w.

NCERT chemistry uses the IUPAC sign convention: q and w are positive when energy goes into the system.

• Heat absorbed by the system: q > 0. Heat released: q < 0.
• Work done on the system (compression): w > 0. Work done by the system (expansion): w < 0.

Work of expansion:

• Against a constant external pressure: w = −p_ext ΔV.
• Free expansion into a vacuum (p_ext = 0): w = 0.
• Reversible isothermal expansion of an ideal gas: w = −2.303 nRT log(V₂/V₁).

For an ideal gas at constant temperature, ΔU = 0, so q = −w.

Enthalpy: heat at constant pressure

Enthalpy is defined as H = U + pV. At constant pressure, the heat exchanged equals ΔH. At constant volume, the heat exchanged equals ΔU.

For reactions involving gases:

ΔH = ΔU + Δn_g RT

where Δn_g = moles of gaseous products − moles of gaseous reactants. Solids and liquids are ignored in Δn_g. When Δn_g = 0, ΔH = ΔU.

For an ideal gas, Cp − Cv = R per mole.

• Exothermic: ΔH < 0, heat released.
• Endothermic: ΔH > 0, heat absorbed.
• A bomb calorimeter works at constant volume and measures ΔU. A coffee-cup calorimeter works at constant pressure and measures ΔH.

Standard enthalpies and Hess's law

Hess's law: the total enthalpy change of a reaction is the same whether it happens in one step or several. So thermochemical equations can be added, subtracted and multiplied like algebraic equations, with their ΔH values.

Two working formulas follow:

• From formation enthalpies: ΔrH° = ∑ΔfH°(products) − ∑ΔfH°(reactants).
• From bond enthalpies (gases): ΔrH° = ∑ bond enthalpies of reactants − ∑ bond enthalpies of products.

The order is reversed between them, and NEET likes to test exactly that. Bonds broken absorb energy; bonds formed release it.

Entropy and the second law

Entropy (S) measures the dispersal of energy and matter, loosely the "disorder" of a system. It increases when:

• a solid melts or a liquid boils (s → l → g)
• a solid dissolves
• the number of gas moles increases in a reaction
• the temperature rises

For a reversible change at temperature T: ΔS = q_rev/T. For a phase change at its transition temperature, ΔS = ΔH/T.

The second law: for a spontaneous process, the total entropy of the system and surroundings increases, ΔS_total > 0. The third law: the entropy of a perfectly crystalline substance approaches zero as the temperature approaches 0 K.

Gibbs energy and spontaneity

Tracking the surroundings' entropy is inconvenient. At constant temperature and pressure, Gibbs energy does the job for the system alone:

ΔG = ΔH − TΔS

A process is spontaneous when ΔG < 0, at equilibrium when ΔG = 0, and non-spontaneous when ΔG > 0.

Link to equilibrium: ΔG° = −RT ln K = −2.303 RT log K. A negative ΔG° means K > 1, so products are favoured at equilibrium. In general, ΔG = ΔG° + RT ln Q.

Link to electrochemistry: ΔG = −nFE_cell, so a positive cell potential means a spontaneous cell reaction. See electrochemistry.

Worked NEET-style questions

Example 1: ΔU from ΔH

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.4 kJ at 298 K. Find ΔU. (R = 8.314 J K⁻¹ mol⁻¹)

• Δn_g = 2 − (1 + 3) = −2.
• Δn_g RT = −2 × 8.314 × 298 ≈ −4955 J ≈ −4.96 kJ.
• ΔU = ΔH − Δn_g RT = −92.4 − (−4.96) ≈ −87.4 kJ.

Example 2: Hess's law

Given C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹, and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹. Find ΔfH° of CO.

• Target: C(graphite) + ½O₂(g) → CO(g).
• Take the first equation and subtract the second: C + ½O₂ → CO.
• ΔfH°(CO) = −393.5 − (−283.0) = −110.5 kJ mol⁻¹.

Example 3: bond enthalpies

Find ΔH for H₂(g) + Cl₂(g) → 2HCl(g). Bond enthalpies: H–H 436, Cl–Cl 242, H–Cl 431 kJ mol⁻¹.

• Bonds broken: 436 + 242 = 678 kJ.
• Bonds formed: 2 × 431 = 862 kJ.
• ΔH = 678 − 862 = −184 kJ.

Example 4: temperature for spontaneity

A reaction has ΔH = +40 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Above what temperature is it spontaneous?

• Both are positive, so the reaction becomes spontaneous at high temperature.
• Convert to the same units: ΔH = 40,000 J mol⁻¹.
• T = ΔH/ΔS = 40,000/100 = 400 K. Above 400 K, ΔG is negative.

Practice MCQs

1. For the melting of ice, ΔS is: (a) positive (b) negative (c) zero (d) positive only below 0 °C
2. A reaction has ΔH < 0 and ΔS > 0. It is: (a) never spontaneous (b) spontaneous at all temperatures (c) spontaneous only at high T (d) spontaneous only at low T
3. Which is a state function? (a) q (b) w (c) q + w (d) q at variable pressure
4. An ideal gas expands freely into a vacuum. The work done is: (a) positive (b) negative (c) zero (d) equal to −pΔV at 1 atm
5. For 2CO(g) + O₂(g) → 2CO₂(g), ΔH − ΔU equals: (a) RT (b) −RT (c) 2RT (d) zero
6. The standard enthalpy of formation is zero for: (a) O₃(g) (b) C(diamond) (c) C(graphite) (d) H₂O(l)
7. Water has ΔvapH = 40.8 kJ mol⁻¹ at 373 K. Its entropy of vaporisation is about: (a) 40.8 J K⁻¹ mol⁻¹ (b) 109 J K⁻¹ mol⁻¹ (c) 373 J K⁻¹ mol⁻¹ (d) 0.109 J K⁻¹ mol⁻¹
8. A gas expands from 1 L to 5 L against a constant external pressure of 2 atm. The work done on the gas is: (a) +8 L atm (b) −8 L atm (c) −10 L atm (d) −2 L atm

Answers

1. (a) Solid to liquid increases disorder.
2. (b) ΔG = ΔH − TΔS is negative at every temperature.
3. (c) q + w = ΔU, which depends only on the states.
4. (c) p_ext = 0, so w = 0.
5. (b) Δn_g = 2 − 3 = −1, so ΔH − ΔU = −RT.
6. (c) Graphite, not diamond, is carbon's standard state; O₃ is not oxygen's standard form.
7. (b) 40,800/373 ≈ 109 J K⁻¹ mol⁻¹.
8. (b) w = −p_ext ΔV = −2 × (5 − 1) = −8 L atm (about −810 J).

What to do next

• Write the chemistry and physics forms of the first law side by side, and solve five questions in each.
• Practise ten Hess's law questions, writing the target equation first every time.
• Learn the four-row spontaneity table and the two ΔrH formulas, noting which order each uses.
• Take ΔG° = −RT ln K straight into chemical equilibrium, and review the NEET chemistry plan for where this chapter fits.