Is it correct to say: Writing expansion work as +P_ext ΔV.?

The Maharashtra textbook uses W = −P_ext ΔV with ΔU = Q + W. Work done by an expanding gas is negative.

Is it correct to say: Counting liquids and solids when finding Δn_g.?

Δn_g uses gaseous species only. In combustion reactions that form H₂O(l), water is not counted.

Is it correct to say: Using R = 8.314 J K⁻¹ mol⁻¹ with ΔU in kJ without converting.?

Use R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ when energies are in kJ, or convert everything to J.

Is it correct to say: Using the formation-enthalpy order (products − reactants) for bond enthalpies.?

With bond enthalpies ΔrH° = ΣB.E.(reactants) − ΣB.E.(products).

Is it correct to say: Mixing kJ for ΔH with J K⁻¹ for ΔS in ΔG = ΔH − TΔS.?

Convert ΔS to kJ K⁻¹ (divide by 1000) or ΔH to J before substituting. T must be in kelvin.

Is it correct to say: Judging spontaneity from ΔS of the system alone.?

The criterion is ΔS_total > 0, or ΔG < 0 at constant T and P. A process can be spontaneous even when ΔS_sys is negative.

📄 Formula sheet (PDF)

Chemical Thermodynamics

Pressure–volume work, maximum work, first law, enthalpy, ΔH–ΔU relation, Hess's law, entropy, Gibbs energy and equilibrium constant — Maharashtra HSC Chemistry Ch 4

📐 17 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Write the sign convention at the top of every work or heat numerical. In the Maharashtra textbook W = −P_ext ΔV, so work done by an expanding gas is negative.
→Be able to derive W_max = −2.303 nRT log₁₀(V₂/V₁) for reversible isothermal expansion of an ideal gas, including the integration step.
→In ΔH = ΔU + Δn_g RT count only gaseous species, and use R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ when energies are in kJ.
→For Hess's law problems, rearrange the given equations to match the target equation: reversing an equation changes the sign of ΔH, multiplying it multiplies ΔH.
→For spontaneity questions, learn the four sign combinations of ΔH and ΔS and the crossover temperature T = ΔH/ΔS.
→Keep energy units consistent: ΔS in kJ K⁻¹ when ΔH is in kJ, and ΔG° in J mol⁻¹ when R = 8.314 J K⁻¹ mol⁻¹.

📐 Formulas(17)

Pressure–Volume Work

W = - P_{e x t} Δ V = - P_{e x t} (V_{2} - V_{1})

Symbol	Meaning
$W$	Work (J)
$P_{e x t}$	Constant external pressure (Pa or bar)
$Δ V$	Change in volume V₂ − V₁ (m³ or L)

Maximum Work — Reversible Isothermal Expansion★ Board fav

W_{ma x} = - 2.303 n R T lo g_{10} \frac{V _{2}}{V _{1}}

Symbol	Meaning
$n$	Amount of gas (mol)
$R$	Gas constant = 8.314 J K⁻¹ mol⁻¹
$T$	Constant temperature (K)
$V_{1}, V_{2}$	Initial and final volumes

Maximum Work in Terms of Pressure

W_{ma x} = - 2.303 n R T lo g_{10} \frac{P _{1}}{P _{2}}

First Law of Thermodynamics★ Board fav

Δ U = Q + W

Symbol	Meaning
$Δ U$	Change in internal energy of the system (J)
$Q$	Heat exchanged; + when absorbed by the system (J)
$W$	Work; + when done on the system (J)

First Law: Isothermal and Adiabatic Processes

Isothermal (ideal gas): Δ U = 0, Q = - W Adiabatic: Q = 0, Δ U = W

First Law: Constant Volume and Constant Pressure

Δ U = Q_{V} Δ U = Q_{P} - P_{e x t} Δ V

Enthalpy

H = U + P V, Δ H = Δ U + P Δ V = Q_{P}

Relation between ΔH and ΔU★ Board fav

Δ H = Δ U + Δ n_{g} R T

Symbol	Meaning
$Δ n_{g}$	Change in moles of gas (products − reactants)
$R$	8.314 J K⁻¹ mol⁻¹ = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹

Work Done in a Chemical Reaction

W = - Δ n_{g} R T

Standard Enthalpy of Reaction from Enthalpies of Formation

Δ_{r} H^{\circ} = \sum Δ_{f} H^{\circ} (products) - \sum Δ_{f} H^{\circ} (reactants)

Enthalpy of Reaction from Bond Enthalpies

Δ_{r} H^{\circ} = \sum B.E.(reactants) - \sum B.E.(products)

Hess's Law of Constant Heat Summation★ Board fav

Δ H = Δ H_{1} + Δ H_{2} + Δ H_{3} + \dots

Entropy Change

Δ S = \frac{Q _{r e v}}{T}

Second Law — Total Entropy Change

Δ S_{t o t a l} = Δ S_{sy s} + Δ S_{s u r r} > 0 (spontaneous)

Gibbs Energy Change★ Board fav

G = H - T S, Δ G = Δ H - T Δ S

Criteria of Spontaneity and Crossover Temperature

Δ G < 0 : spontaneous; Δ G = 0 : equilibrium; Δ G > 0 : non-spontaneous; T = \frac{Δ H}{Δ S}

Gibbs Energy and Equilibrium Constant★ Board fav

Δ G^{\circ} = - 2.303 R T lo g_{10} K

✏️ Solved Examples

1Solved Exampleeasy3 steps

A gas expands from 2.0 L to 7.0 L against a constant external pressure of 1.5 bar. Calculate the work done in joules.

Apply W = −P_ext ΔV.

W = - P_{e x t} (V_{2} - V_{1}) = - 1.5 bar \times (7.0 - 2.0) L = - 7.5 L bar

2Solved Exampleboard3 steps

2 mol of an ideal gas expand isothermally and reversibly at 300 K from 5 L to 20 L. Calculate W, ΔU and Q. (R = 8.314 J K⁻¹ mol⁻¹)

Use the maximum work formula.

W_{ma x} = - 2.303 n R T lo g_{10} \frac{V _{2}}{V _{1}} = - 2.303 \times 2 \times 8.314 \times 300 \times lo g_{10} \frac{20}{5}

3Solved Exampleboard3 steps

For C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), ΔU = −1406.0 kJ mol⁻¹ at 298 K. Calculate ΔH. (R = 8.314 J K⁻¹ mol⁻¹)

Count only gaseous species. H₂O is liquid.

Δ n_{g} = 2 - (1 + 3) = - 2

4Solved ExampleHOTS4 steps

For N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K, ΔH° = −92.4 kJ and ΔS° = −198.3 J K⁻¹. Calculate ΔG° and K at 298 K, and the temperature above which the reaction stops being spontaneous. Assume ΔH° and ΔS° do not change with temperature. (R = 8.314 J K⁻¹ mol⁻¹)

Convert ΔS° to kJ K⁻¹ and apply ΔG° = ΔH° − TΔS°.

Δ G^{\circ} = - 92.4 - 298 \times (- 0.1983) = - 92.4 + 59.09 = - 33.31 kJ

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Writing expansion work as +P_ext ΔV.
✓The Maharashtra textbook uses W = −P_ext ΔV with ΔU = Q + W. Work done by an expanding gas is negative.
2
Counting liquids and solids when finding Δn_g.
✓Δn_g uses gaseous species only. In combustion reactions that form H₂O(l), water is not counted.
3
Using R = 8.314 J K⁻¹ mol⁻¹ with ΔU in kJ without converting.
✓Use R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ when energies are in kJ, or convert everything to J.
4
Using the formation-enthalpy order (products − reactants) for bond enthalpies.
✓With bond enthalpies ΔrH° = ΣB.E.(reactants) − ΣB.E.(products).
5
Mixing kJ for ΔH with J K⁻¹ for ΔS in ΔG = ΔH − TΔS.
✓Convert ΔS to kJ K⁻¹ (divide by 1000) or ΔH to J before substituting. T must be in kelvin.
6
Judging spontaneity from ΔS of the system alone.
✓The criterion is ΔS_total > 0, or ΔG < 0 at constant T and P. A process can be spontaneous even when ΔS_sys is negative.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Given C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ. Use Hess's law to find ΔH for C(graphite) + ½O₂(g) → CO(g).
Q2
Calculate ΔcH° of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), given ΔfH° (kJ mol⁻¹): CH₄ = −74.8, CO₂ = −393.5, H₂O(l) = −285.8.
Q3
Using bond enthalpies H–H = 436, Cl–Cl = 242 and H–Cl = 431 kJ mol⁻¹, calculate ΔH for H₂(g) + Cl₂(g) → 2HCl(g).
Q4
Calculate the work done in the reaction 2H₂(g) + O₂(g) → 2H₂O(l) at 298 K and constant pressure. State whether work is done on or by the system. (R = 8.314 J K⁻¹ mol⁻¹)
Q5
The enthalpy of vaporisation of water at 373 K is 40.7 kJ mol⁻¹. Calculate the entropy change of vaporisation.
Q6
For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178 kJ and ΔS° = +160 J K⁻¹. Above what temperature is the reaction spontaneous? Assume ΔH° and ΔS° are constant.

📝 Notes

Chemical Thermodynamics — Maharashtra HSC Overview

Chapter 4 of the Maharashtra State Board Std XII Chemistry textbook follows one thread: energy transferred as heat and work (first law), heat changes in reactions (enthalpy), and the direction of change (entropy and Gibbs energy).

Fix the sign convention first

The textbook uses the IUPAC convention. Energy that enters the system is positive, so heat absorbed and work done on the system are positive. This is why $W = -P_{ext}\Delta V$ and $\Delta U = Q + W$ . An expanding gas does work on the surroundings, so its $W$ is negative. Keep this convention all the way through. A common sign error is switching to $\Delta U = Q - W$ halfway through a problem.

Work, heat and state functions

$Q$ and $W$ depend on the path, but $\Delta U$ and $\Delta H$ do not. Reversible isothermal expansion gives the largest work, $W_{max} = -2.303\,nRT\log_{10}(V_2/V_1)$ . Because enthalpy is a state function, Hess's law lets you add thermochemical equations like algebra. The relation $\Delta H = \Delta U + \Delta n_g RT$ links constant-pressure and constant-volume heats.

Deciding spontaneity

The second law says a spontaneous process increases the total entropy, $\Delta S_{total} > 0$ . At constant $T$ and $P$ this becomes $\Delta G = \Delta H - T\Delta S < 0$ . When $\Delta H$ and $\Delta S$ have the same sign, the temperature decides, and the crossover is at $T = \Delta H/\Delta S$ . Finally, $\Delta G^\circ = -2.303RT\log_{10}K$ connects thermodynamics with equilibrium: a negative $\Delta G^\circ$ means $K > 1$ .