Chemical Thermodynamics and Energetics — MHT-CET Formulas and Traps

Thermodynamic Systems, Properties and Processes

Isothermal, Isochoric, Isobaric, Adiabatic and Reversible

What each process fixes

\text{isochoric: } w = 0;\quad \text{isothermal (ideal): } \Delta U = 0;\quad \text{adiabatic: } q = 0;\quad \text{isobaric: } q_P = \Delta H

Systems, Intensive Versus Extensive, State Versus Path

Property	Intensive or extensive	State or path
Temperature, pressure	Intensive	State
Boiling point, density, surface tension, viscosity, specific heat	Intensive	State SPECIFIC heat (per gram) is intensive; HEAT CAPACITY (of the sample) is extensive.
Mass, volume, number of moles	Extensive	State
Internal energy U, enthalpy H, entropy S	Extensive	State Internal energy is the planted 'not intensive' option.
Heat capacity	Extensive	—
Heat q, work w	—	Path Work is the only path function among U, w, S, H.

Divide the sample in two and ask what changes.

Common traps

Heat capacity as intensive

A bigger sample needs more heat per degree, so heat capacity is EXTENSIVE. Its per-gram version, specific heat, is intensive. The paper pairs them in one option to catch the mix-up.

'Isothermal' read as 'no heat exchange'

That is ADIABATIC. In an isothermal process heat flows freely — it is exactly what keeps the temperature constant while the gas does work.

First Law of Thermodynamics, Internal Energy and Work

Learn this subtopic in the notes

ΔU = q + w and the Sign Convention

First law

\Delta U = q + w,\qquad q_{\text{absorbed}} > 0,\quad w_{\text{on system}} > 0

Work Against a Constant External Pressure: w = −P_ext ΔV

PV work

w = -P_{\text{ext}}\,\Delta V,\qquad 1\ \text{dm}^3\,\text{bar} = 100\ \text{J},\quad 1\ \text{L atm} = 101.3\ \text{J}

ΔU From Heat Absorbed and an Expansion

First law with expansion work

\Delta U = q - P_{\text{ext}}\,\Delta V

Work in a Gas-Phase Reaction: w = −Δn_g RT

Reaction work

w = -\Delta n_g\,RT

Reversible Isothermal Work: −2.303 nRT log(V₂/V₁)

Maximum (reversible) work

w_{\text{rev}} = -2.303\,nRT\log_{10}\frac{V_2}{V_1} = -2.303\,nRT\log_{10}\frac{P_1}{P_2}

Common traps

'Work done BY the system' entered as positive

That is the old convention and it flips every answer's sign. In the current one, work done BY the system LEAVES it: w is negative. 'Work done ON the system' is the positive case.

Leaving the answer in dm³ bar when joules are asked

−12 dm³ bar is −1200 J. Options are printed a factor of 100 apart. In SI, cm³ must become m³ (× 10⁻⁶) before multiplying by pascals.

Adding the work instead of subtracting it

An EXPANSION is work done by the system: it is subtracted from q. 800 + 1000 = 1800 J is not on offer, but 800 − 100 (a slipped conversion) is.

Counting liquids and solids in Δn

Only GASES do PV work. In

\text{C}_2\text{H}_2 + \tfrac{5}{2}\text{O}_2 \to 2\text{CO}_2 + \text{H}_2\text{O}(l)

the water is liquid:

\Delta n_g = 2 - 3.5

, not

3 - 3.5

.

Using the pressure ratio the same way as the volume ratio

V_2/V_1 = P_1/P_2

, inverted. For 10 bar → 1 bar the log argument is 10 (expansion, w negative), not 0.1. Options with the opposite sign are always there.

Enthalpy and the Relation Between ΔH and ΔU

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Enthalpy, Heat at Constant Pressure and Phase Changes

Enthalpy

H = U + PV,\qquad \Delta H = q_P,\qquad \Delta H_{\text{vap}} = \frac{q}{n}

ΔH = ΔU + Δn_g RT

ΔH and ΔU

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

Common traps

Dividing the heat by the grams

5.1 kJ / 13 g is a per-gram figure and matches no option. Convert to moles first: ΔH_vap is per MOLE.

Counting liquid water as a gas

\text{H}_2\text{O}(l)

contributes nothing to

\Delta n_g

. Propane combustion is −3RT because the four waters are liquid; counting them gives +1RT, an offered option.

Thermochemistry, Hess's Law and Bond Enthalpy

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Enthalpy of Formation and the Reaction Enthalpy From It

Reaction enthalpy from formation enthalpies

\Delta_r H^\circ = \sum \nu_p\,\Delta_f H^\circ_{\text{products}} - \sum \nu_r\,\Delta_f H^\circ_{\text{reactants}}

Hess's Law and the Enthalpy of Solution

Hess's law; enthalpy of solution

\Delta H_{\text{overall}} = \sum \Delta H_{\text{steps}},\qquad \Delta_{\text{soln}}H = \Delta_L H + \Delta_{\text{hyd}}H

Bond Enthalpy: Bonds Broken Minus Bonds Formed

Reaction enthalpy from bond enthalpies

\Delta_r H = \sum \Delta H_{\text{broken}} - \sum \Delta H_{\text{formed}}

Common traps

Reporting the equation's ΔH as the formation enthalpy

\Delta_f H

is per MOLE of compound. An equation that makes 2 mol of NH₃ at −92 kJ gives −46 kJ mol⁻¹; −92 is the first option every time.

Subtracting hydration from lattice enthalpy

Hydration enthalpy is already NEGATIVE; the enthalpy of solution is the SUM. 700 − (−680) = 1380 is the planted option.

Using a whole N₂ for the formation enthalpy of NH₃

Formation enthalpy is per mole of product, so the equation is

\tfrac{1}{2}\text{N}_2 + \tfrac{3}{2}\text{H}_2 \to \text{NH}_3

. The whole-equation answer, −85 kJ, is offered beside the right −42.5.

Entropy and the Second Law

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Predicting the Sign of ΔS

Rule of thumb

\Delta n_g > 0 \Rightarrow \Delta S > 0;\qquad \Delta n_g < 0 \text{ or gas} \to \text{liquid/solid} \Rightarrow \Delta S < 0

ΔS = q_rev/T, ΔS_surr = −ΔH/T and ΔS_total

Entropy of surroundings and total

\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T},\qquad \Delta S_{\text{total}} = \Delta S_{\text{sys}} - \frac{\Delta H_{\text{sys}}}{T}

Common traps

Counting moles without looking at phases

2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)

has 2 → 3 moles overall, but what matters is that a GAS appears from liquids: ΔS > 0. Conversely 3 gas moles to 2 moles of LIQUID water is a large decrease.

Dividing kilojoules by kelvin

28.1/300 = 0.094 is in kJ K⁻¹ and cannot be added to 108.7 J K⁻¹. Convert ΔH to joules first: 28100/300 = 93.7 J K⁻¹.

Gibbs Free Energy and Spontaneity

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ΔG = ΔH − TΔS and the Sign Table

Gibbs energy

\Delta G = \Delta H - T\Delta S

ΔG = 0: Boiling Points, Transition Temperatures and K

Equilibrium conditions

T_{\text{eq}} = \frac{\Delta H}{\Delta S},\qquad \Delta G^\circ = -2.303\,RT\log_{10} K

Common traps

Subtracting joules from kilojoules

7 − 300 × 24.8 = −7433 'kJ' is nonsense; 24.8 J K⁻¹ is 0.0248 kJ K⁻¹, giving −0.44 kJ. The mismatched-unit answer is on the option list.

Inverting to T = ΔS/ΔH

75/30000 = 0.0025 is not a temperature. ΔH (J) over ΔS (J K⁻¹) gives kelvin; the inverted ratio gives K⁻¹, and the wrong option is built from it.

Chemical Thermodynamics and Energetics formulas

Thermodynamic Systems, Properties and Processes

Isothermal, Isochoric, Isobaric, Adiabatic and Reversible

Systems, Intensive Versus Extensive, State Versus Path

Common traps

First Law of Thermodynamics, Internal Energy and Work

ΔU = q + w and the Sign Convention

Work Against a Constant External Pressure: w = −P_ext ΔV

ΔU From Heat Absorbed and an Expansion

Work in a Gas-Phase Reaction: w = −Δn_g RT

Reversible Isothermal Work: −2.303 nRT log(V₂/V₁)

Common traps

Enthalpy and the Relation Between ΔH and ΔU

Enthalpy, Heat at Constant Pressure and Phase Changes

ΔH = ΔU + Δn_g RT

Common traps

Thermochemistry, Hess's Law and Bond Enthalpy

Enthalpy of Formation and the Reaction Enthalpy From It

Hess's Law and the Enthalpy of Solution

Bond Enthalpy: Bonds Broken Minus Bonds Formed

Common traps

Entropy and the Second Law

Predicting the Sign of ΔS

ΔS = q_rev/T, ΔS_surr = −ΔH/T and ΔS_total

Common traps

Gibbs Free Energy and Spontaneity

ΔG = ΔH − TΔS and the Sign Table

ΔG = 0: Boiling Points, Transition Temperatures and K

Common traps

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