NEET Chemistry: Thermodynamics and Equilibrium · Chapter 1

Chemical Thermodynamics

What to remember

Internal energy (U), enthalpy (H), entropy (S) and Gibbs energy (G) are state functions; heat (q) and work (w) are path functions.
First law: ΔU = q + w. Work done on the system is positive. Enthalpy change at constant pressure: ΔH = ΔU + ΔngRT.
Hess's law: the total enthalpy change of a reaction is the same whether it occurs in one step or in several steps.
Spontaneity at constant T and p is decided by ΔG = ΔH − TΔS. ΔG < 0 means spontaneous, ΔG = 0 means equilibrium, and ΔG° = −2.303 RT log K.

System, surroundings and properties

Thermodynamics studies energy changes that accompany physical and chemical processes. The part of the universe chosen for study is the system. Everything else that can exchange energy or matter with it is the surroundings. The system is separated from the surroundings by a boundary (real or imaginary).

Type of system	Exchange of matter	Exchange of energy	Example
Open	Yes	Yes	Hot tea in an open cup
Closed	No	Yes	Gas in a sealed metal cylinder
Isolated	No	No	Tea in an ideal sealed thermos flask

A system is described by its state, which is fixed by measurable properties such as pressure (p), volume (V), temperature (T) and amount (n).

State function: its value depends only on the present state of the system and not on the path used to reach it. Examples: U, H, S, G, p, V, T.
Path function: its value depends on the path. Examples: q and w.
Extensive property: depends on the amount of matter. Examples: mass, volume, internal energy, enthalpy, entropy, heat capacity.
Intensive property: does not depend on the amount of matter. Examples: temperature, pressure, density, molar heat capacity, refractive index.
The ratio of two extensive properties is intensive (density = mass/volume).

Types of processes: isothermal (T constant, ΔT = 0), isobaric (p constant), isochoric (V constant, so w = 0 for pressure-volume work), adiabatic (no heat exchange, q = 0).

Work, heat, internal energy and the first law

The internal energy (U) is the total energy of the system: kinetic energy, potential energy, and energy of electrons and nuclei. Its absolute value cannot be measured, only ΔU. For an ideal gas, U depends only on temperature.

Energy can change in two ways: by heat (q) and by work (w). The sign convention used here is the one where w is positive when work is done on the system and negative when the system does work. Heat q is positive when absorbed by the system.

First law of thermodynamics: the energy of an isolated system is constant. For a closed system:

ΔU = q + w

Adiabatic change (q = 0): ΔU = w.
Isothermal change of an ideal gas: ΔU = 0, so q = −w.

Pressure-volume work. For expansion against a constant external pressure pex:

w = −pex ΔV (ΔV = Vfinal − Vinitial)

If pressure is in bar and volume in litre, then 1 L bar = 100 J. Also 1 L atm = 101.3 J.

Type of expansion	Condition	Work done on the gas (w)
Free expansion (into vacuum)	pex = 0	w = 0
Irreversible (against constant pex)	pex fixed	w = −pex (V₂ − V₁)
Isothermal reversible	pex differs from p by an infinitesimal amount at every stage	w = −2.303 nRT log (V₂/V₁)

In a reversible process the system is in equilibrium with the surroundings at every step, and the work obtained is the maximum work. For the same initial and final states, the magnitude of work in reversible isothermal expansion is greater than in irreversible expansion. For the isothermal free expansion of an ideal gas, q = 0, w = 0 and ΔU = 0.

Worked example 1. A gas absorbs 500 J of heat and does 200 J of work on the surroundings. Find ΔU.

q = +500 J, w = −200 J. ΔU = 500 − 200 = +300 J.

Worked example 2. One mole of an ideal gas expands isothermally and reversibly from 10 L to 100 L at 300 K (log 10 = 1, R = 8.314 J K⁻¹ mol⁻¹).

w = −2.303 × 1 × 8.314 × 300 × log 10 = −5744 J, about −5.74 kJ. Since ΔU = 0, q = +5.74 kJ.

Enthalpy and heat capacity

At constant volume (no pV work), qV = ΔU. In most laboratory and biological processes the pressure is constant, so a new state function is defined:

H = U + pV

At constant pressure, qp = ΔH. ΔH is negative for exothermic and positive for endothermic reactions.

Relation between ΔH and ΔU:

ΔH = ΔU + Δng RT

Δng = (moles of gaseous products) − (moles of gaseous reactants). Solids and liquids are ignored. R = 8.314 J K⁻¹ mol⁻¹.

Worked example 3. For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.4 kJ at 298 K. Find ΔU.

Δng = 2 − 4 = −2. ΔngRT = (−2)(8.314)(298) = −4955 J = −4.96 kJ.

ΔU = ΔH − ΔngRT = −92.4 + 4.96 = −87.4 kJ.

If Δng = 0, then ΔH = ΔU.

Heat capacity. The heat needed to raise the temperature of a system by 1 K is its heat capacity (C), an extensive property. q = C ΔT. The molar heat capacity is an intensive property.

Quantity	Condition	Relation
Cv	constant volume	qV = Cv ΔT = ΔU
Cp	constant pressure	qp = Cp ΔT = ΔH
Cp − Cv	one mole of ideal gas	= R

Cp is larger than Cv because at constant pressure part of the heat is used to do expansion work. For one mole of an ideal gas, ΔU = Cv ΔT and ΔH = Cp ΔT.

Measurement. ΔU is measured in a bomb calorimeter, a sealed steel vessel at constant volume placed in a water bath. The heat released is q = −C ΔT, where C is the heat capacity of the calorimeter and its contents. ΔH is measured at constant pressure, for example in an open calorimeter.

Standard enthalpy changes

The standard state of a substance is its pure form at 1 bar pressure and a specified temperature (usually 298 K). Standard enthalpy changes carry the symbol °.

Enthalpy	Meaning	Example
Reaction (ΔrH°)	Enthalpy change of a reaction when all species are in standard states	N₂ + 3H₂ → 2NH₃
Formation (ΔfH°)	Formation of 1 mol of a compound from its elements in their most stable (reference) forms	C(graphite) + O₂(g) → CO₂(g)
Combustion (ΔcH°)	Complete burning of 1 mol of a substance in oxygen	CH₄ + 2O₂ → CO₂ + 2H₂O(l)
Atomisation (ΔaH°)	Formation of 1 mol of gaseous atoms from the substance	CH₄(g) → C(g) + 4H(g)
Bond dissociation (ΔbondH°)	Breaking of 1 mol of a particular bond in gaseous molecules; always positive	H₂(g) → 2H(g)
Fusion (ΔfusH°)	1 mol of solid becomes liquid at its melting point	H₂O(s) → H₂O(l)
Vaporisation (ΔvapH°)	1 mol of liquid becomes vapour at its boiling point	H₂O(l) → H₂O(g)
Sublimation (ΔsubH°)	1 mol of solid becomes vapour	I₂(s) → I₂(g)
Ionisation (ΔiH°)	Removal of an electron from a gaseous atom	Na(g) → Na⁺(g) + e⁻
Solution (ΔsolH°)	Dissolving 1 mol of substance in a large amount of solvent	NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)
Hydration (ΔhydH°)	Gaseous ions becoming hydrated	Na⁺(g) + aq → Na⁺(aq)
Lattice (ΔlatticeH°)	1 mol of ionic solid dissociating into gaseous ions; always positive	NaCl(s) → Na⁺(g) + Cl⁻(g)

Key points:

ΔfH° of an element in its reference state is zero. Graphite (not diamond) is the reference state of carbon. O₂(g) and H₂(g) have ΔfH° = 0, but O₃(g) does not.
ΔH° of a reaction from formation enthalpies: ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants).
ΔsubH° = ΔfusH° + ΔvapH° (same temperature).
ΔsolH° = ΔlatticeH° + ΔhydH°. A salt dissolves with a small net value, since the two terms are large and opposite in sign.
Enthalpy of reverse reaction has the same magnitude and opposite sign. If a reaction is multiplied by n, ΔH is multiplied by n.
Born-Haber cycle (in brief): the lattice enthalpy of an ionic solid cannot be measured directly. It is found by applying Hess's law to a cycle of steps: sublimation of the metal, ionisation of the metal atom, dissociation of the non-metal into atoms, electron gain by the non-metal atom, and formation of the solid lattice from gaseous ions, which together sum to the enthalpy of formation of the solid.

Hess's law of constant heat summation

Since H is a state function, ΔH depends only on the initial and final states. If a reaction can be written as the sum of other reactions, its ΔH is the sum of their ΔH values.

Worked example 4. Find ΔH for C(graphite) + ½O₂(g) → CO(g), given:

(i) C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹

(ii) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹

Reverse (ii): CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Add to (i):

ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹.

Worked example 5. Combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), with ΔfH° of CH₄ = −75 kJ mol⁻¹, CO₂ = −394 kJ mol⁻¹, H₂O(l) = −286 kJ mol⁻¹.

ΔcH° = [−394 + 2(−286)] − [−75 + 0] = −966 + 75 = −891 kJ mol⁻¹.

Spontaneity, entropy and Gibbs energy

A spontaneous process proceeds on its own without outside help. Energy lowering (negative ΔH) alone does not decide spontaneity, because some endothermic processes (such as melting of ice above 273 K) are spontaneous.

Entropy (S) is a measure of disorder or randomness. It is a state function and extensive. For a reversible change at temperature T: ΔS = qrev/T.

Second law: the entropy of the universe increases in every spontaneous process: ΔStotal = ΔSsystem + ΔSsurroundings > 0.

Entropy increases with: melting, vaporisation, dissolution of a solid, increase in number of gas molecules in a reaction, and rise in temperature. Order of entropy: solid < liquid < gas.

Entropy of phase changes (at the transition temperature, reversible):

ΔfusS = ΔfusH/Tf
ΔvapS = ΔvapH/Tb
ΔsubS = ΔsubH/Tsub

For example, with ΔfusH = 6.0 kJ mol⁻¹ and Tf = 273 K, ΔfusS = 6000/273 = 22.0 J K⁻¹ mol⁻¹. For a given substance, ΔvapS is larger than ΔfusS.

Third law: the entropy of a perfectly crystalline substance is zero at absolute zero (0 K). Because of this, absolute entropies can be calculated.

Gibbs energy (G) combines both factors: G = H − TS. At constant temperature and pressure:

ΔG = ΔH − TΔS

ΔH	ΔS	ΔG	Spontaneity
−	+	always −	Spontaneous at all temperatures
+	−	always +	Non-spontaneous at all temperatures
−	−	− at low T	Spontaneous at low temperature only
+	+	− at high T	Spontaneous at high temperature only

When ΔG = 0, ΔH = TΔS, so the temperature at which a reaction changes from non-spontaneous to spontaneous (or the reverse) is T = ΔH/ΔS.

Worked example 6. ΔH = +30 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. The reaction becomes spontaneous above T = 30000/100 = 300 K.

Worked example 7. ΔH = −92.2 kJ, ΔS = −198.6 J K⁻¹ at 298 K.

TΔS = 298 × (−198.6) = −59183 J = −59.2 kJ.

ΔG = −92.2 − (−59.2) = −33.0 kJ. The reaction is spontaneous.

Gibbs energy and equilibrium. ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 and Q = K, so

ΔG° = −RT ln K = −2.303 RT log K

If ΔG° < 0, then K > 1 and products are favoured.
If ΔG° = 0, then K = 1.
If ΔG° > 0, then K < 1 and reactants are favoured.

Worked example 8. K = 10⁵ at 298 K. ΔG° = −2.303 × 8.314 × 298 × 5 = −28530 J, about −28.5 kJ mol⁻¹.

Constants: R = 8.314 J K⁻¹ mol⁻¹ = 0.0831 L bar K⁻¹ mol⁻¹; 1 L bar = 100 J; N_A = 6.022 × 10²³ mol⁻¹.

NEET traps

1. Heat and work are not state functions, although q + w (that is, ΔU) is.
2. Work done on the system is positive here. In expansion, w is negative.
3. Free expansion of an ideal gas: w = 0 because pex = 0, and q = 0 too. Do not confuse it with reversible isothermal expansion, where w ≠ 0.
4. ΔH = ΔU only when Δng = 0 (or for reactions of solids and liquids with negligible volume change).
5. Cp − Cv = R applies per mole of an ideal gas, not for liquids and solids.
6. Heat capacity is extensive, molar heat capacity is intensive.
7. Bond dissociation, atomisation, sublimation, fusion and vaporisation enthalpies are positive. Combustion enthalpies are negative.
8. ΔfH° of O₃, diamond and Br₂(g) is not zero. The reference form of carbon is graphite.
9. ΔS of the system can be negative for a spontaneous process, but ΔS of the universe must be positive.
10. ΔG° and ΔG are not the same. ΔG = 0 at equilibrium, but ΔG° = 0 only when K = 1.

One-liners

1. An open system exchanges both matter and energy; a closed system exchanges only energy.
2. Internal energy of an ideal gas depends only on temperature.
3. First law: ΔU = q + w.
4. In an adiabatic process ΔU = w.
5. Isothermal reversible work: w = −2.303 nRT log (V₂/V₁).
6. 1 L bar = 100 J.
7. H = U + pV and ΔH = ΔU + ΔngRT.
8. qV = ΔU and qp = ΔH.
9. A bomb calorimeter measures ΔU.
10. ΔfH° of graphite is zero.
11. ΔsubH° = ΔfusH° + ΔvapH°.
12. Lattice enthalpy is always positive in the dissociation direction.
13. Hess's law follows from H being a state function.
14. Entropy of a perfect crystal at 0 K is zero.
15. ΔG° = −2.303 RT log K, and ΔG = 0 at equilibrium.

Practice questions

Which of the following is a state function?
1. Enthalpy
2. Work done by the gas
3. Heat absorbed
4. Work done in an expansion
Answer
A. Enthalpy
Enthalpy depends only on the state of the system; heat and work depend on the path.
Which of the following is an intensive property?
1. Entropy
2. Internal energy
3. Density
4. Volume
Answer
C. Density
Density is a ratio of two extensive properties (mass/volume), so it does not depend on the amount.
An isolated system exchanges
1. energy only with its surroundings
2. matter only with its surroundings
3. neither energy nor matter with its surroundings
4. both energy and matter with its surroundings
Answer
C. neither energy nor matter with its surroundings
An isolated system has a boundary that allows no exchange of matter or energy.
A gas expands against a constant external pressure of 2 bar from 1 L to 6 L. The work done on the gas is (1 L bar = 100 J)
1. −10 J
2. −1000 J
3. −500 J
4. +1000 J
Answer
B. −1000 J
w = −pex ΔV = −2 × 5 = −10 L bar = −1000 J.
A system absorbs 500 J of heat and does 200 J of work on the surroundings. The change in internal energy is
1. +700 J
2. +500 J
3. −300 J
4. +300 J
Answer
D. +300 J
ΔU = q + w = 500 + (−200) = +300 J.
For 2CO(g) + O₂(g) → 2CO₂(g) at 300 K, ΔH − ΔU is closest to (R = 8.314 J K⁻¹ mol⁻¹)
1. −2.49 kJ
2. 0 kJ
3. +2.49 kJ
4. −4.99 kJ
Answer
A. −2.49 kJ
Δng = 2 − 3 = −1, so ΔH − ΔU = ΔngRT = −8.314 × 300 = −2494 J.
For an ideal gas Cv = 12.5 J K⁻¹ mol⁻¹. Taking R = 8.314 J K⁻¹ mol⁻¹, the value of Cp is nearest to
1. 4.2 J K⁻¹ mol⁻¹
2. 20.8 J K⁻¹ mol⁻¹
3. 12.5 J K⁻¹ mol⁻¹
4. 29.1 J K⁻¹ mol⁻¹
Answer
B. 20.8 J K⁻¹ mol⁻¹
Cp = Cv + R = 12.5 + 8.314 = 20.8 J K⁻¹ mol⁻¹.
Given ΔfH° (kJ mol⁻¹): CH₄(g) = −75, CO₂(g) = −394, H₂O(l) = −286. The enthalpy of combustion of CH₄(g) to CO₂(g) and H₂O(l) is
1. −966 kJ mol⁻¹
2. −1041 kJ mol⁻¹
3. +891 kJ mol⁻¹
4. −891 kJ mol⁻¹
Answer
D. −891 kJ mol⁻¹
ΔcH° = [−394 + 2(−286)] − (−75) = −966 + 75 = −891 kJ mol⁻¹.
A reaction has ΔH = +30 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. It becomes spontaneous above
1. 30 K
2. 3000 K
3. 0.3 K
4. 300 K
Answer
D. 300 K
At ΔG = 0, T = ΔH/ΔS = 30000/100 = 300 K; above this TΔS exceeds ΔH.
For a system at equilibrium at constant temperature and pressure, the change in Gibbs energy ΔG is
1. positive
2. equal to ΔG° always
3. zero
4. negative
Answer
C. zero
At equilibrium there is no net driving force, so ΔG = 0.
For a reaction at 298 K the equilibrium constant is 10⁵. Taking R = 8.314 J K⁻¹ mol⁻¹, ΔG° is nearest to
1. −28.5 kJ mol⁻¹
2. +28.5 kJ mol⁻¹
3. −57.1 kJ mol⁻¹
4. −5.7 kJ mol⁻¹
Answer
A. −28.5 kJ mol⁻¹
ΔG° = −2.303 RT log K = −2.303 × 8.314 × 298 × 5 = −28530 J.
The enthalpy of fusion of ice is 6.0 kJ mol⁻¹ at 273 K. The entropy of fusion is nearest to
1. 2.2 J K⁻¹ mol⁻¹
2. 22.0 J K⁻¹ mol⁻¹
3. 0.22 J K⁻¹ mol⁻¹
4. 44.0 J K⁻¹ mol⁻¹
Answer
B. 22.0 J K⁻¹ mol⁻¹
ΔS = ΔH/T = 6000/273 = 22.0 J K⁻¹ mol⁻¹.
A bomb calorimeter with heat capacity 12 kJ K⁻¹ shows a temperature rise of 0.50 K when 0.5 mol of a compound is burnt. ΔU of combustion per mole is
1. −6 kJ mol⁻¹
2. +12 kJ mol⁻¹
3. −12 kJ mol⁻¹
4. −24 kJ mol⁻¹
Answer
C. −12 kJ mol⁻¹
Heat released = 12 × 0.50 = 6 kJ for 0.5 mol, so ΔU = −6/0.5 = −12 kJ mol⁻¹ (constant volume).
The standard enthalpy of formation is zero for
1. O₂(g)
2. O₃(g)
3. C (diamond)
4. H(g)
Answer
A. O₂(g)
Only an element in its most stable reference form has ΔfH° = 0; O₃, diamond and atomic H are not reference forms.
The change NaCl(s) → Na⁺(g) + Cl⁻(g) represents the
1. enthalpy of formation
2. enthalpy of solution
3. ionisation enthalpy
4. lattice enthalpy
Answer
D. lattice enthalpy
Dissociation of one mole of an ionic solid into gaseous ions is the lattice enthalpy.
Given below are two statements: Statement I: Internal energy is a state function. Statement II: Work done in a process is a state function.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Internal energy depends only on the state, but work depends on the path.
Given below are two statements: Statement I: The work obtained in a reversible isothermal expansion of an ideal gas is greater than in an irreversible expansion between the same states. Statement II: In a reversible process the system is in equilibrium with the surroundings at every stage.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Reversible expansion gives maximum work, and it proceeds through a series of equilibrium states.
Given below are two statements: Statement I: In an isothermal expansion of an ideal gas the internal energy decreases. Statement II: The enthalpy of an ideal gas depends on its volume.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
U and H of an ideal gas depend only on temperature, so both stay constant when T is constant.
Given below are two statements: Statement I: The enthalpy of combustion of a substance is always negative. Statement II: The enthalpy of formation of every compound is negative.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Combustion is exothermic, but some compounds, such as NO(g), have positive enthalpy of formation.
Given below are two statements: Statement I: Cp is greater than Cv for an ideal gas. Statement II: At constant pressure part of the heat supplied is used for expansion work.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Cp − Cv = R because heat at constant pressure also supplies pV work.
Which of the following statements are correct? 1. ΔH = ΔU when Δng = 0. 2. ΔH = ΔU for every gaseous reaction. 3. Enthalpy is a state function.
1. 1, 2 and 3
2. 1 and 3 only
3. 2 and 3 only
4. 1 and 2 only
Answer
B. 1 and 3 only
ΔH = ΔU + ΔngRT, so they are equal only when Δng = 0; H is a state function.
Given below are two statements: Statement I: For a spontaneous process the entropy of the universe increases. Statement II: For a spontaneous process the entropy of the system must always increase.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Only ΔS of the universe must be positive; the system entropy can decrease if the surroundings gain more.
Given below are two statements: Statement I: The entropy of a perfect crystal is zero at 0 K. Statement II: For a given substance, the entropy of vaporisation is greater than the entropy of fusion.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Third law and ΔvapS > ΔfusS (gas is much more disordered than liquid) are both true.
Which of the following statements are correct? 1. A reaction with ΔH < 0 and ΔS < 0 is spontaneous at low temperature. 2. A reaction with ΔH > 0 and ΔS > 0 is spontaneous at high temperature. 3. A reaction with ΔH > 0 and ΔS < 0 is spontaneous at all temperatures.
1. 1, 2 and 3
2. 2 and 3 only
3. 1 and 3 only
4. 1 and 2 only
Answer
D. 1 and 2 only
ΔG = ΔH − TΔS. Statements 1 and 2 are true; with ΔH > 0 and ΔS < 0, ΔG is always positive.
Given below are two statements: Statement I: The enthalpy of vaporisation of water is negative. Statement II: Evaporation of a liquid requires absorption of heat.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Vaporisation is endothermic, so ΔvapH is positive; Statement II is true.
Assertion (A): ΔU = 0 for the isothermal expansion of an ideal gas. Reason (R): The internal energy of an ideal gas depends only on its temperature.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Temperature does not change, so U does not change.
Assertion (A): The enthalpy of vaporisation of water is positive. Reason (R): The entropy of a gas is greater than that of the liquid.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but the sign of ΔvapH is due to the energy needed to overcome attractions, not to entropy.
Assertion (A): The work done by an ideal gas in free expansion is zero. Reason (R): Free expansion takes place against a constant external pressure of 1 atm.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Free expansion is against zero external pressure; so R is false.
Assertion (A): Heat is a state function. Reason (R): The heat absorbed at constant pressure is equal to the enthalpy change.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Heat is a path function, so A is false; qp = ΔH is true.
Assertion (A): The standard enthalpy of formation of CO₂(g) equals the standard enthalpy of combustion of graphite. Reason (R): Both refer to the reaction C(graphite) + O₂(g) → CO₂(g).
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The two definitions describe the same reaction.
Match List I with List II. List I: a. Extensive property b. Intensive property c. State function d. Path function List II: i. Pressure ii. Mass iii. Work iv. Internal energy
1. a-ii, b-i, c-iii, d-iv
2. a-i, b-ii, c-iv, d-iii
3. a-iv, b-i, c-ii, d-iii
4. a-ii, b-i, c-iv, d-iii
Answer
D. a-ii, b-i, c-iv, d-iii
Mass is extensive, pressure intensive, internal energy a state function and work a path function.
Match List I with List II. List I: a. Enthalpy of combustion b. Enthalpy of fusion c. Enthalpy of atomisation d. Enthalpy of sublimation List II: i. H₂O(s) → H₂O(l) ii. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) iii. CH₄(g) → C(g) + 4H(g) iv. I₂(s) → I₂(g)
1. a-i, b-ii, c-iii, d-iv
2. a-ii, b-i, c-iii, d-iv
3. a-ii, b-i, c-iv, d-iii
4. a-iii, b-i, c-ii, d-iv
Answer
B. a-ii, b-i, c-iii, d-iv
Each term is matched with its defining change.
Match List I with List II. List I: a. ΔH < 0, ΔS > 0 b. ΔH > 0, ΔS < 0 c. ΔH < 0, ΔS < 0 d. ΔH > 0, ΔS > 0 List II: i. Spontaneous at low temperature only ii. Spontaneous at all temperatures iii. Non-spontaneous at all temperatures iv. Spontaneous at high temperature only
1. a-iii, b-ii, c-i, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-ii, b-iii, c-iv, d-i
4. a-i, b-iii, c-ii, d-iv
Answer
B. a-ii, b-iii, c-i, d-iv
From ΔG = ΔH − TΔS: a is always negative, b always positive, c negative at low T, d negative at high T.
Match List I with List II. List I: a. Cp − Cv (one mole of ideal gas) b. Work of isothermal reversible expansion c. ΔG° d. Entropy of fusion List II: i. −2.303 nRT log (V₂/V₁) ii. ΔfusH/Tf iii. −2.303 RT log K iv. R
1. a-iv, b-iii, c-i, d-ii
2. a-iv, b-i, c-ii, d-iii
3. a-iv, b-i, c-iii, d-ii
4. a-i, b-iv, c-iii, d-ii
Answer
C. a-iv, b-i, c-iii, d-ii
Standard relations: Cp − Cv = R, w = −2.303 nRT log (V₂/V₁), ΔG° = −2.303 RT log K, ΔS = ΔH/T.
Match List I with List II. List I: a. Isothermal process b. Adiabatic process c. Isochoric process d. Isobaric process List II: i. q = 0 ii. Pressure constant iii. ΔT = 0 iv. Volume constant
1. a-i, b-iii, c-iv, d-ii
2. a-ii, b-i, c-iv, d-iii
3. a-iii, b-i, c-ii, d-iv
4. a-iii, b-i, c-iv, d-ii
Answer
D. a-iii, b-i, c-iv, d-ii
Isothermal means ΔT = 0, adiabatic q = 0, isochoric constant volume, isobaric constant pressure.
Match List I with List II. List I: a. Ionisation enthalpy b. Lattice enthalpy c. Hydration enthalpy d. Enthalpy of solution List II: i. NaCl(s) → Na⁺(g) + Cl⁻(g) ii. Na(g) → Na⁺(g) + e⁻ iii. NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq) iv. Na⁺(g) + aq → Na⁺(aq)
1. a-i, b-ii, c-iv, d-iii
2. a-ii, b-iv, c-i, d-iii
3. a-ii, b-i, c-iii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
D. a-ii, b-i, c-iv, d-iii
Each term corresponds to the change shown for sodium chloride.
One mole of an ideal gas expands isothermally and reversibly from 10 L to 100 L at 300 K. Using R = 8.314 J K⁻¹ mol⁻¹ and log 10 = 1, the work done on the gas is
1. −5.74 kJ
2. 0 kJ
3. −57.4 kJ
4. +5.74 kJ
Answer
A. −5.74 kJ
w = −2.303 nRT log (V₂/V₁) = −2.303 × 8.314 × 300 × 1 = −5744 J.
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. The enthalpy change for C(graphite) + ½O₂(g) → CO(g) is
1. +110.5 kJ mol⁻¹
2. −110.5 kJ mol⁻¹
3. −676.5 kJ mol⁻¹
4. −283.0 kJ mol⁻¹
Answer
B. −110.5 kJ mol⁻¹
Subtract the second equation from the first: −393.5 − (−283.0) = −110.5 kJ mol⁻¹.
For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.4 kJ at 298 K. Taking R = 8.314 J K⁻¹ mol⁻¹, ΔU is nearest to
1. −97.4 kJ
2. −89.9 kJ
3. −87.4 kJ
4. −92.4 kJ
Answer
C. −87.4 kJ
Δng = −2; ΔU = ΔH − ΔngRT = −92.4 + 2 × 8.314 × 298/1000 = −92.4 + 4.96 = −87.4 kJ.
The enthalpy of atomisation of CH₄(g) is 1665 kJ mol⁻¹. The mean C–H bond enthalpy in methane is nearest to
1. 416 kJ mol⁻¹
2. 6660 kJ mol⁻¹
3. 208 kJ mol⁻¹
4. 1665 kJ mol⁻¹
Answer
A. 416 kJ mol⁻¹
CH₄ has four C–H bonds; 1665/4 = 416 kJ mol⁻¹.
A reaction has ΔH = −40 kJ mol⁻¹ and ΔS = −100 J K⁻¹ mol⁻¹. It is spontaneous
1. above 400 K only
2. at all temperatures
3. below 400 K only
4. at no temperature
Answer
C. below 400 K only
ΔG = −40000 + 100T; ΔG < 0 when T < 400 K.
An ideal gas expands into an evacuated container, inside a thermally insulated rigid vessel. For this process
1. q = 0, w > 0 and ΔU > 0
2. q = 0, w = 0 and ΔU = 0
3. q > 0, w = 0 and ΔU > 0
4. q = 0, w < 0 and ΔU < 0
Answer
B. q = 0, w = 0 and ΔU = 0
Free expansion: pex = 0 so w = 0; insulated so q = 0; hence ΔU = 0.
For a reaction ΔH = −92.2 kJ and ΔS = −198.6 J K⁻¹ at 298 K. The value of ΔG is nearest to
1. +33.0 kJ
2. −92.2 kJ
3. −151.4 kJ
4. −33.0 kJ
Answer
D. −33.0 kJ
TΔS = 298 × (−198.6) = −59183 J; ΔG = −92.2 − (−59.2) = −33.0 kJ.

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