Treating gases as ideal and neglecting the volumes of solids and liquids, for which reaction at 298 K is ΔH equal to ΔU?

2H₂(g) + O₂(g) → 2H₂O(l)

CaCO₃(s) → CaO(s) + CO₂(g)

One mole of an ideal gas at 300 K is expanded isothermally from V to 2V, once reversibly and once against a constant external pressure equal to the final pressure. Which comparison is correct?

ΔS of the gas is the same for both paths, but ΔS of the surroundings is less negative for the irreversible path

ΔS of the gas is larger for the irreversible path, since an irreversible change creates more disorder in the gas

ΔS of the gas is the same for both paths, but ΔS of the surroundings is less negative for the irreversible path

ΔS of the gas is zero for both paths, because the temperature of the gas stays the same in each case

The gas absorbs more heat on the irreversible path, so ΔS of the surroundings is more negative there

A reaction has ΔH° = +40 kJ mol⁻¹ and ΔS° = +125 J K⁻¹ mol⁻¹, both taken as independent of temperature. Under standard conditions, when is the reaction spontaneous?

Only above 320 K, where TΔS° outweighs ΔH°

At every temperature, because its entropy change is positive

Only below 320 K, where the enthalpy term dominates ΔG°

At no temperature, because its enthalpy change is positive

Only above 320 K, where TΔS° outweighs ΔH°

At constant pressure, a reaction has ΔG = −10.0 kJ mol⁻¹ at 300 K and ΔG = −16.0 kJ mol⁻¹ at 330 K. Assuming ΔH and ΔS are constant over this range, which pair of values is correct?

ΔS = +200 J K⁻¹ mol⁻¹ and ΔH = +50 kJ mol⁻¹

ΔS = −200 J K⁻¹ mol⁻¹ and ΔH = −70 kJ mol⁻¹

ΔS = +200 J K⁻¹ mol⁻¹ and ΔH = −70 kJ mol⁻¹

ΔS = +200 J K⁻¹ mol⁻¹ and ΔH = +60 kJ mol⁻¹

A step in a metabolic pathway has ΔG°′ = +7.5 kJ mol⁻¹, yet in the cell it proceeds steadily in the forward direction. Which explanation is consistent with thermodynamics?

The product is removed by the next step, keeping Q far below K′ so that ΔG is negative

The enzyme that catalyses the step lowers its ΔG°′ to a negative value inside the cell

ΔG°′ refers to 1 M H⁺, and at the pH of the cell the sign of the standard value reverses

The product is removed by the next step, keeping Q far below K′ so that ΔG is negative

At 310 K the term TΔS°′ always outweighs ΔH°′, so ΔG°′ becomes negative in the cell

Which of the following statements about reversible and irreversible processes are correct?

In a reversible process the system passes through a continuous series of equilibrium states and can be turned back by an infinitesimal change in a variable

For isothermal expansion of an ideal gas between the same two volumes, the reversible path delivers the most work to the surroundings

In a reversible process the system passes through a continuous series of equilibrium states and can be turned back by an infinitesimal change in a variable

For an ideal gas, w = −nRT ln(V₂/V₁) also gives the work of expansion against a constant external pressure

For isothermal expansion of an ideal gas between the same two volumes, the reversible path delivers the most work to the surroundings

Because q and w are path functions, their sum q + w also depends on the path

Which of the following statements about thermochemistry are correct?

Hess’s law is a consequence of enthalpy being a state function

The standard enthalpy of formation of a compound refers to forming one mole of it from its elements in their reference states

Reversing a thermochemical equation leaves the sign of its ΔH unchanged

Hess’s law is a consequence of enthalpy being a state function

The standard enthalpy of formation of a compound refers to forming one mole of it from its elements in their reference states

Kirchhoff’s equation shows that ΔrH is independent of temperature whenever ΔrCp is large

For which of the following processes is the entropy change of the system positive?

Melting of ice at 273 K and 1 bar

Mixing of two different ideal gases at constant temperature and pressure

Melting of ice at 273 K and 1 bar

Condensation of steam to liquid water at 373 K and 1 bar

The reaction 2H₂(g) + O₂(g) → 2H₂O(l) at 298 K

Mixing of two different ideal gases at constant temperature and pressure

At constant temperature and pressure, which of the following statements about the Gibbs energy are correct?

A process with ΔG < 0 is spontaneous, though ΔG says nothing about how fast it proceeds

−ΔG equals the maximum non-expansion work, such as electrical work, that the process can deliver

A reaction with ΔH < 0 and ΔS < 0 is spontaneous at all temperatures

A process with ΔG < 0 is spontaneous, though ΔG says nothing about how fast it proceeds

−ΔG equals the maximum non-expansion work, such as electrical work, that the process can deliver

ΔG° = 0 means that the reaction mixture is at equilibrium whatever its composition

Thermodynamics: First Law, Thermochemistry, Entropy and Free Energy

This chapter answers Section XL0.8 of XL0 Chemistry, the part of the Life Sciences (XL) paper that every XL candidate answers: “Qualitative treatment of state and path functions, First law, reversible and irreversible processes, internal energy, enthalpy, Kirchoff [Kirchhoff] equation, heat of reaction, Hess’s law, heat of formation. Second law, entropy and free energy. Gibbs-Helmholtz equation, free energy change and spontaneity, Free energy changes from equilibrium constant.” It separates state from path functions, applies the first law to reversible and irreversible expansions, builds thermochemistry from heats of formation, Hess’s law and Kirchhoff’s equation, and then uses the second law and entropy to reach the Gibbs energy, its temperature dependence and its link to the equilibrium constant, with biochemical examples. The numericals are expansion work, reaction enthalpies from formation and combustion data, Kirchhoff corrections, entropy changes, crossover temperatures, Gibbs–Helmholtz extrapolation and ΔG° from K.

1. Systems, state functions and path functions

Thermodynamics divides the world into a system and its surroundings. An open system exchanges both matter and energy with its surroundings (a living cell), a closed system exchanges energy only (a sealed flask), and an isolated system exchanges neither. A few variables, such as T, P, V and the amounts of substances, fix the state of a system. A state function depends only on that state, not on how it was reached: T, P, V, internal energy U, enthalpy H, entropy S and Gibbs energy G are all state functions. Its change is final minus initial, and it returns to its starting value round any cycle. A path function describes the journey instead: heat q and work w depend on the route taken, so it is meaningless to speak of the heat 'contained' in a system.

💡 Altitude and the trail

Climbing from a valley to a summit gains the same height whichever trail is taken: height is like a state function. The distance walked and the effort spent depend on the trail, like q and w.

Sign convention. This chapter follows the IUPAC convention: q is positive when the system absorbs heat, and w is positive when work is done on the system, so the first law reads ΔU = q + w. Some older texts write ΔU = q − w, with w as the work done by the system. The physics is the same but the sign of w flips, so read the convention a question uses.

2. The first law, enthalpy, and reversible versus irreversible change

The first law is the conservation of energy: ΔU = q + w. The internal energy U is the total kinetic and potential energy of the particles; for an ideal gas it depends on temperature alone, so ΔU = 0 in any isothermal ideal-gas change. Expansion work against an external pressure is w = −∫P_ext dV, which becomes −P_ext ΔV when P_ext is constant. At constant volume no expansion work is done, so ΔU = q_V, which is what a bomb calorimeter measures. Chemistry and biology mostly run at constant pressure, for which we define enthalpy H = U + PV, with ΔH = q_P. For reactions involving gases, ΔH = ΔU + Δn_g RT, where Δn_g is the change in moles of gas. In the combustion of glucose to CO₂ and liquid water, 6 mol of O₂ is consumed and 6 mol of CO₂ formed, so Δn_g = 0 and ΔH = ΔU. Heat capacities are C_V = (∂U/∂T)_V and C_P = (∂H/∂T)_P, and for one mole of ideal gas C_P − C_V = R.

A reversible process passes through a continuous sequence of equilibrium states and can be turned back at any stage by an infinitesimal change in a variable; it is an idealisation that would take infinite time. A real, irreversible process runs at a finite rate across finite differences of pressure or temperature. In an expansion, the reversible path delivers the maximum work, because the external pressure is kept just below the gas pressure at every stage. For n mol of ideal gas expanding isothermally and reversibly, w = −nRT ln(V₂/V₁), and q = −w because ΔU = 0. Expansion into a vacuum (free expansion) does no work at all.

Worked example. One mole of ideal gas at 300 K doubles its volume isothermally (R = 8.314 J mol⁻¹ K⁻¹). Reversibly, w = −(1)(8.314)(300) ln 2 = −1729 J. Against a constant external pressure equal to the final pressure P₂, w = −P₂(V₂ − V₁) = −P₂V₂/2 = −nRT/2 = −1247 J. Into a vacuum, w = 0. ΔU = 0 in all three, because U is a state function and T is unchanged, but q = −w differs each time: the clearest proof that q and w are path functions.

3. Heat of reaction, heat of formation, Hess’s law and Kirchhoff’s equation

The heat (enthalpy) of reaction ΔrH is the enthalpy change for the reaction as written, in kJ per mole of reaction: negative for an exothermic and positive for an endothermic reaction. Reversing the equation reverses its sign, and doubling the coefficients doubles it. The standard enthalpy (heat) of formation ΔfH° of a compound is ΔH for forming one mole of it from its elements in their reference states, the most stable forms at 1 bar, such as O₂(g), C(graphite) and Br₂(l). For these reference forms ΔfH° = 0 by definition, but not for O₃(g), C(diamond) or Br₂(g).

Hess’s law: because H is a state function, ΔrH is the same whether the reaction occurs in one step or in many, so thermochemical equations can be added and subtracted like algebraic equations. Two working forms follow: ΔrH° = ΣνΔfH°(products) − ΣνΔfH°(reactants), and, from enthalpies of combustion, ΔrH° = ΣνΔcH°(reactants) − ΣνΔcH°(products). Worked example: taking ΔfH° = −393.5 kJ mol⁻¹ for CO₂(g), −285.8 for H₂O(l) and −1273.3 for glucose(s), the combustion C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O(l) has ΔcH° = 6(−393.5) + 6(−285.8) − (−1273.3) = −2802.5 kJ mol⁻¹. As Hess’s law requires, a cell that oxidises glucose to the same products through glycolysis, the citric acid cycle and respiration releases the same total enthalpy, only in many small steps.

Kirchhoff’s equation carries ΔrH to another temperature: d(ΔrH)/dT = ΔrCp, so ΔrH(T₂) = ΔrH(T₁) + ΔrCp(T₂ − T₁) when ΔrCp = ΣνCp(products) − ΣνCp(reactants) is constant over the range and no phase change intervenes. Worked example: for N₂(g) + 3H₂(g) → 2NH₃(g), ΔrH°(298 K) = −92.2 kJ mol⁻¹. Taking Cp = 35.1, 29.1 and 28.8 J K⁻¹ mol⁻¹ for NH₃, N₂ and H₂, ΔrCp = 2(35.1) − 29.1 − 3(28.8) = −45.3 J K⁻¹ mol⁻¹, and ΔrH(398 K) = −92.2 + (−45.3 × 10⁻³)(100) = −96.7 kJ mol⁻¹. Watch the units: Cp is in joules, ΔH in kilojoules.

ℹ️ A biological ΔCp

Protein unfolding has a large positive ΔCp, because nonpolar groups buried in the folded protein become exposed to water. By Kirchhoff’s equation its ΔH therefore rises steeply with temperature, which is why the stability of a protein cannot be described by a single ΔH.

4. The second law and entropy

The first law allows any process that conserves energy; the second law says which direction is natural. In Clausius’s form, heat does not flow spontaneously from a colder to a hotter body; in the Kelvin–Planck form, no cyclic engine can turn heat from a single reservoir entirely into work. In terms of entropy S, a state function defined by dS = dq_rev/T, the total entropy of system plus surroundings rises in every spontaneous (irreversible) process and stays constant in a reversible one: ΔS_univ = ΔS_sys + ΔS_surr ≥ 0. Statistically, S = k_B ln W counts the microscopic arrangements W consistent with the observed state, which is why entropy rises on melting, vaporisation, mixing, dissolving a solid, and in reactions that increase the moles of gas.

Because S is a state function, ΔS_sys for an irreversible change is calculated along any reversible path between the same two states. The standard results are: isothermal ideal-gas change, ΔS = nR ln(V₂/V₁); heating at constant pressure, ΔS = nC_P ln(T₂/T₁); and a phase change at its transition temperature, ΔS = ΔH_trans/T_trans. Ice melting at 273.15 K with ΔfusH = 6.01 kJ mol⁻¹ has ΔS = 6010/273.15 = 22.0 J K⁻¹ mol⁻¹. The surroundings, treated as a large reservoir at temperature T, change by ΔS_surr = −q_sys/T.

Worked example. For the 1 mol, 300 K volume-doubling of Section 2, ΔS_sys = R ln 2 = +5.76 J K⁻¹ whichever path is taken. Reversibly, the surroundings supply 1729 J, so ΔS_surr = −5.76 J K⁻¹ and ΔS_univ = 0. Against constant P₂ they supply only 1247 J, so ΔS_surr = −4.16 J K⁻¹ and ΔS_univ = +1.61 J K⁻¹ > 0, the mark of an irreversible process. In free expansion ΔS_surr = 0 and ΔS_univ = +5.76 J K⁻¹.

⚠️ q = 0 does not mean ΔS = 0

Entropy is dq_rev/T, not dq_actual/T. Free expansion absorbs no heat, yet the gas gains R ln(V₂/V₁) of entropy. Only a reversible adiabatic change is isentropic.

5. Free energy, spontaneity and the Gibbs–Helmholtz equation

Tracking the surroundings is awkward, so at constant T and P we use the Gibbs energy G = H − TS. Since ΔS_surr = −ΔH/T, the condition ΔS_univ > 0 becomes ΔG = ΔH − TΔS < 0. A process with ΔG < 0 is spontaneous, one with ΔG = 0 is at equilibrium, and one with ΔG > 0 is non-spontaneous as written (its reverse is spontaneous). −ΔG is also the maximum non-expansion work, electrical, chemical or mechanical, that the process can deliver. At constant T and V the matching function is the Helmholtz energy A = U − TS, with ΔA < 0 for a spontaneous change. Spontaneity says nothing about speed: glucose in air is spontaneous to burn yet keeps for years.

Signs of ΔH and ΔS and the sign of ΔG = ΔH − TΔS (ΔH and ΔS taken as constant)
ΔH	ΔS	Spontaneous when	Example
Negative	Positive	At all temperatures	Combustion of glucose
Positive	Negative	At no temperature; the reverse is spontaneous	Forming glucose and O₂ from CO₂ and water without an energy input
Negative	Negative	Below T = ΔH/ΔS	Freezing of water below 273 K
Positive	Positive	Above T = ΔH/ΔS	Melting of ice above 273 K

How ΔG changes with temperature follows from (∂G/∂T)_P = −S, so (∂ΔG/∂T)_P = −ΔS: a reaction with positive ΔS becomes more favourable as T rises. Combining this with G = H − TS gives the Gibbs–Helmholtz equation, [∂(ΔG/T)/∂T]_P = −ΔH/T². With ΔH constant over the range it integrates to ΔG₂/T₂ = ΔG₁/T₁ + ΔH(1/T₂ − 1/T₁), so a free energy measured at one temperature can be extended to another using ΔH alone. Worked example: a reaction has ΔG = +20 kJ mol⁻¹ at 298 K and ΔH = +200 kJ mol⁻¹. At 310 K, ΔG₂ = 310[20/298 + 200(1/310 − 1/298)] = 310(0.06711 − 0.02598) = +12.8 kJ mol⁻¹. The same value follows from ΔS = (200 − 20)/298 = 0.604 kJ K⁻¹ mol⁻¹ and ΔG = 200 − 310(0.604): with ΔH constant, the two routes are equivalent.

6. Free energy and the equilibrium constant

For a reaction mixture of any composition, ΔG = ΔG° + RT ln Q, where Q is the reaction quotient (activities, approximated by concentrations in mol L⁻¹ for solutes and partial pressures in bar for gases) and ΔG° refers to every species in its standard state. At equilibrium ΔG = 0 and Q = K, so ΔG° = −RT ln K, or K = e^(−ΔG°/RT). Thus ΔG° < 0 means K > 1, ΔG° = 0 means K = 1 and ΔG° > 0 means K < 1. At 298 K, RT ln 10 = 5.71 kJ mol⁻¹, so every 5.7 kJ mol⁻¹ of ΔG° changes K tenfold. Combined with the Gibbs–Helmholtz equation this gives the van ’t Hoff equation, d ln K/dT = ΔH°/RT², by which K of an endothermic reaction rises with temperature.

ΔG° fixes K; ΔG, which depends on the actual concentrations, decides the direction. Since ΔG = RT ln(Q/K), a reaction runs forward whenever Q < K. Biochemists use ΔG°′, the standard free-energy change at pH 7 (with [H⁺] = 10⁻⁷ M treated as standard) and water at unit activity. Worked example: taking ΔG°′ = −30.5 kJ mol⁻¹ for ATP hydrolysis at 310 K, with [ATP] = 5.0 mM, [ADP] = 0.50 mM and [Pᵢ] = 5.0 mM, ΔG = −30.5 + (8.314 × 10⁻³)(310) ln[(5.0 × 10⁻⁴)(5.0 × 10⁻³)/(5.0 × 10⁻³)] = −30.5 − 19.6 = −50.1 kJ mol⁻¹. Cells keep ATP far from equilibrium, and because the ΔG values of coupled reactions add, its hydrolysis can drive an unfavourable step such as glucose + Pᵢ → glucose 6-phosphate (ΔG°′ = +13.8 kJ mol⁻¹), giving a net ΔG°′ of −16.7 kJ mol⁻¹ for the coupled reaction.

⚠️ Three different questions

ΔG° tells where equilibrium lies, ΔG tells which way the reaction goes now, and neither tells how fast. A step with positive ΔG°′ still runs forward in a cell if its product is removed quickly enough to keep Q below K′.

Key takeaways

U, H, S and G are state functions: their changes depend only on the initial and final states. Heat and work are path functions. With the IUPAC convention, ΔU = q + w, where w is work done on the system.
ΔU = q_V and ΔH = q_P, with ΔH = ΔU + Δn_g RT. Reversible isothermal expansion of an ideal gas gives the maximum work, −nRT ln(V₂/V₁); free expansion does no work.
Hess’s law: ΔrH° = ΣνΔfH°(products) − ΣνΔfH°(reactants), with ΔfH° = 0 for elements in their reference states. Kirchhoff: ΔrH(T₂) = ΔrH(T₁) + ΔrCp(T₂ − T₁).
Second law: ΔS_univ ≥ 0, with equality only for reversible change. Compute ΔS_sys along a reversible path, using nR ln(V₂/V₁) or ΔH_trans/T_trans, even when the actual change is irreversible.
At constant T and P, ΔG = ΔH − TΔS < 0 means spontaneous; the crossover temperature is ΔH/ΔS. Gibbs–Helmholtz: [∂(ΔG/T)/∂T]_P = −ΔH/T², and (∂ΔG/∂T)_P = −ΔS.
ΔG = ΔG° + RT ln Q and ΔG° = −RT ln K; at 298 K each 5.7 kJ mol⁻¹ is a factor of 10 in K. Biochemistry uses ΔG°′ at pH 7, and the ΔG values of coupled reactions add.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which of the following quantities depends on the path by which a system is taken from one state to another?
1. Change in internal energy, ΔU
2. Heat absorbed, q
3. Change in enthalpy, ΔH
4. Change in Gibbs energy, ΔG
Show answer
Answer: B — Heat absorbed, q
Heat is a path function: the same change of state can be reached with different amounts of heat and work. In the isothermal doubling of the volume of 1 mol of ideal gas at 300 K, q is 1729 J reversibly, 1247 J against the final pressure and 0 into a vacuum. ΔU is the tempting choice because ΔU = q + w, but the sum is path-independent even though its two parts are not; ΔH and ΔG are likewise changes of state functions.
One mole of an ideal gas expands isothermally into an evacuated vessel (free expansion). Which set of values is correct for the gas?
1. q = 0, w = 0, ΔU = 0, ΔS = 0
2. q > 0, w < 0, ΔU = 0, ΔS > 0
3. q = 0, w < 0, ΔU < 0, ΔS = 0
4. q = 0, w = 0, ΔU = 0, ΔS > 0
Show answer
Answer: D — q = 0, w = 0, ΔU = 0, ΔS > 0
Against zero external pressure w = 0; T is unchanged, so ΔU = 0 for an ideal gas and q = ΔU − w = 0. Entropy is evaluated along a reversible isothermal path between the same states, ΔS = R ln(V₂/V₁) > 0. The set with ΔS = 0 is the trap of using the actual q (zero) instead of q_rev; q > 0 with w < 0 describes reversible isothermal expansion, and w < 0 with ΔU < 0 and ΔS = 0 describes reversible adiabatic expansion.
Treating gases as ideal and neglecting the volumes of solids and liquids, for which reaction at 298 K is ΔH equal to ΔU?
1. H₂(g) + Cl₂(g) → 2HCl(g)
2. 2H₂(g) + O₂(g) → 2H₂O(l)
3. CaCO₃(s) → CaO(s) + CO₂(g)
4. N₂(g) + 3H₂(g) → 2NH₃(g)
Show answer
Answer: A — H₂(g) + Cl₂(g) → 2HCl(g)
ΔH = ΔU + Δn_g RT, so ΔH = ΔU only when the moles of gas do not change. For H₂ + Cl₂ → 2HCl, Δn_g = 2 − 2 = 0. The others have Δn_g = −3 (the water is liquid), +1 and −2, so ΔH and ΔU differ by −3RT, +RT and −2RT respectively.
At 298 K and 1 bar, the standard enthalpy of formation is zero by definition for which substance?
1. O₃(g)
2. C(diamond)
3. O₂(g)
4. Br₂(g)
Show answer
Answer: C — O₂(g)
ΔfH° is zero for an element in its reference state, its most stable form at 1 bar and the stated temperature: O₂(g) for oxygen, graphite for carbon and the liquid for bromine at 298 K. Ozone, diamond and gaseous bromine are not reference forms, so each has a non-zero ΔfH° (diamond about +1.9 kJ mol⁻¹, the tempting distractor because it is pure carbon).
One mole of an ideal gas at 300 K is expanded isothermally from V to 2V, once reversibly and once against a constant external pressure equal to the final pressure. Which comparison is correct?
1. ΔS of the gas is larger for the irreversible path, since an irreversible change creates more disorder in the gas
2. ΔS of the gas is the same for both paths, but ΔS of the surroundings is less negative for the irreversible path
3. ΔS of the gas is zero for both paths, because the temperature of the gas stays the same in each case
4. The gas absorbs more heat on the irreversible path, so ΔS of the surroundings is more negative there
Show answer
Answer: B — ΔS of the gas is the same for both paths, but ΔS of the surroundings is less negative for the irreversible path
S is a state function, so ΔS_gas = R ln 2 = +5.76 J K⁻¹ for both paths. The surroundings lose q = −w: 1729 J reversibly (ΔS_surr = −5.76 J K⁻¹, total 0) but only RT/2 = 1247 J irreversibly (ΔS_surr = −4.16 J K⁻¹, total +1.61 J K⁻¹). The gas absorbs less heat, not more, on the irreversible path, and ΔS of an isothermal expansion is not zero because the volume changes.
A reaction has ΔH° = +40 kJ mol⁻¹ and ΔS° = +125 J K⁻¹ mol⁻¹, both taken as independent of temperature. Under standard conditions, when is the reaction spontaneous?
1. At every temperature, because its entropy change is positive
2. Only below 320 K, where the enthalpy term dominates ΔG°
3. At no temperature, because its enthalpy change is positive
4. Only above 320 K, where TΔS° outweighs ΔH°
Show answer
Answer: D — Only above 320 K, where TΔS° outweighs ΔH°
ΔG° = ΔH° − TΔS° is negative when T > ΔH°/ΔS° = 40,000/125 = 320 K. With both terms positive, the entropy term wins only at high temperature. A positive ΔS alone does not guarantee spontaneity (that needs ΔH < 0 as well), and a positive ΔH does not forbid it.
At constant pressure, a reaction has ΔG = −10.0 kJ mol⁻¹ at 300 K and ΔG = −16.0 kJ mol⁻¹ at 330 K. Assuming ΔH and ΔS are constant over this range, which pair of values is correct?
1. ΔS = +200 J K⁻¹ mol⁻¹ and ΔH = +50 kJ mol⁻¹
2. ΔS = −200 J K⁻¹ mol⁻¹ and ΔH = −70 kJ mol⁻¹
3. ΔS = +200 J K⁻¹ mol⁻¹ and ΔH = −70 kJ mol⁻¹
4. ΔS = +200 J K⁻¹ mol⁻¹ and ΔH = +60 kJ mol⁻¹
Show answer
Answer: A — ΔS = +200 J K⁻¹ mol⁻¹ and ΔH = +50 kJ mol⁻¹
(∂ΔG/∂T)_P = −ΔS, so ΔS = −(−16.0 − (−10.0))/(330 − 300) = +0.200 kJ K⁻¹ mol⁻¹ = +200 J K⁻¹ mol⁻¹. Then ΔH = ΔG + TΔS = −10.0 + 300(0.200) = +50.0 kJ mol⁻¹; check at 330 K: −16.0 + 330(0.200) = +50.0. A ΔS of −200 comes from dropping the minus sign, ΔH = −70 from writing ΔH = ΔG − TΔS, and +60 from forgetting the ΔG term.
A step in a metabolic pathway has ΔG°′ = +7.5 kJ mol⁻¹, yet in the cell it proceeds steadily in the forward direction. Which explanation is consistent with thermodynamics?
1. The enzyme that catalyses the step lowers its ΔG°′ to a negative value inside the cell
2. ΔG°′ refers to 1 M H⁺, and at the pH of the cell the sign of the standard value reverses
3. The product is removed by the next step, keeping Q far below K′ so that ΔG is negative
4. At 310 K the term TΔS°′ always outweighs ΔH°′, so ΔG°′ becomes negative in the cell
Show answer
Answer: C — The product is removed by the next step, keeping Q far below K′ so that ΔG is negative
Direction is set by ΔG = ΔG°′ + RT ln Q = RT ln(Q/K′), not by ΔG°′. If the next enzyme removes product fast enough, Q stays below K′ and ΔG < 0 even though ΔG°′ > 0. An enzyme cannot change ΔG°′ or K′; ΔG°′ is already defined at pH 7 (it is ΔG° that refers to 1 M H⁺); and there is no rule that TΔS outweighs ΔH at 310 K.
At 298 K the reaction quotient of a reaction mixture is one-hundredth of the equilibrium constant (Q = K/100). Taking R = 8.314 J mol⁻¹ K⁻¹, what is ΔG for the reaction in this mixture?
1. +11.4 kJ mol⁻¹
2. −11.4 kJ mol⁻¹
3. −5.7 kJ mol⁻¹
4. −5.0 kJ mol⁻¹
Show answer
Answer: B — −11.4 kJ mol⁻¹
ΔG = ΔG° + RT ln Q = −RT ln K + RT ln Q = RT ln(Q/K) = 8.314 × 298 × ln(0.01) = 2477.6 × (−4.605) = −11,410 J mol⁻¹ ≈ −11.4 kJ mol⁻¹. Since Q < K the reaction runs forward, so ΔG is negative and +11.4 has the sign reversed; −5.7 uses ln 10 instead of ln 100, and −5.0 uses log₁₀ 100 without the factor 2.303.
Which of the following statements about reversible and irreversible processes are correct?
1. In a reversible process the system passes through a continuous series of equilibrium states and can be turned back by an infinitesimal change in a variable
2. For an ideal gas, w = −nRT ln(V₂/V₁) also gives the work of expansion against a constant external pressure
3. For isothermal expansion of an ideal gas between the same two volumes, the reversible path delivers the most work to the surroundings
4. Because q and w are path functions, their sum q + w also depends on the path
Show answer
Answer: A — In a reversible process the system passes through a continuous series of equilibrium states and can be turned back by an infinitesimal change in a variable; C — For isothermal expansion of an ideal gas between the same two volumes, the reversible path delivers the most work to the surroundings
A reversible process is a sequence of equilibrium states, reversible by an infinitesimal change, and in expansion it extracts the maximum work because P_ext is always just below the gas pressure. Against a constant external pressure the work is −P_ext(V₂ − V₁), smaller in magnitude than nRT ln(V₂/V₁). And q + w = ΔU, a state function, even though q and w separately depend on the path.
Which of the following statements about thermochemistry are correct?
1. Reversing a thermochemical equation leaves the sign of its ΔH unchanged
2. Hess’s law is a consequence of enthalpy being a state function
3. The standard enthalpy of formation of a compound refers to forming one mole of it from its elements in their reference states
4. Kirchhoff’s equation shows that ΔrH is independent of temperature whenever ΔrCp is large
Show answer
Answer: B — Hess’s law is a consequence of enthalpy being a state function; C — The standard enthalpy of formation of a compound refers to forming one mole of it from its elements in their reference states
Hess’s law works because ΔH depends only on the initial and final states, and ΔfH° is defined per mole of compound from elements in their reference states. Reversing an equation reverses the sign of ΔH. By Kirchhoff’s equation, d(ΔrH)/dT = ΔrCp, so ΔrH is independent of temperature only when ΔrCp = 0; a large ΔrCp makes it change strongly.
For which of the following processes is the entropy change of the system positive?
1. Melting of ice at 273 K and 1 bar
2. Condensation of steam to liquid water at 373 K and 1 bar
3. The reaction 2H₂(g) + O₂(g) → 2H₂O(l) at 298 K
4. Mixing of two different ideal gases at constant temperature and pressure
Show answer
Answer: A — Melting of ice at 273 K and 1 bar; D — Mixing of two different ideal gases at constant temperature and pressure
Melting increases the number of accessible arrangements (ΔS = ΔfusH/T = +22.0 J K⁻¹ mol⁻¹), and mixing two ideal gases gives each more volume to occupy, ΔS = −nR Σ x ln x > 0. Condensation is the reverse of vaporisation and lowers S, and the formation of liquid water from three moles of gas has a large negative ΔS.
At constant temperature and pressure, which of the following statements about the Gibbs energy are correct?
1. A reaction with ΔH < 0 and ΔS < 0 is spontaneous at all temperatures
2. A process with ΔG < 0 is spontaneous, though ΔG says nothing about how fast it proceeds
3. −ΔG equals the maximum non-expansion work, such as electrical work, that the process can deliver
4. ΔG° = 0 means that the reaction mixture is at equilibrium whatever its composition
Show answer
Answer: B — A process with ΔG < 0 is spontaneous, though ΔG says nothing about how fast it proceeds; C — −ΔG equals the maximum non-expansion work, such as electrical work, that the process can deliver
ΔG < 0 is the spontaneity criterion at constant T and P, and kinetics is separate. −ΔG is the maximum non-PV work, the basis of cell EMF (ΔG = −nFE). With ΔH < 0 and ΔS < 0, ΔG is negative only below ΔH/ΔS. ΔG° = 0 means K = 1; a particular mixture is at equilibrium only if its Q equals 1.
The standard enthalpy of combustion of liquid ethanol, C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), is −1367.0 kJ mol⁻¹. The standard enthalpies of formation of CO₂(g) and H₂O(l) are −393.5 and −285.8 kJ mol⁻¹. The standard enthalpy of formation of liquid ethanol is ____ kJ mol⁻¹ (to one decimal place; include the sign).
Numerical answer — type the value.
Show answer
Answer: -277.4
Hess’s law on the combustion equation: ΔcH° = 2ΔfH°(CO₂) + 3ΔfH°(H₂O) − ΔfH°(ethanol) − 3ΔfH°(O₂), with ΔfH°(O₂) = 0. So ΔfH°(ethanol) = 2(−393.5) + 3(−285.8) − (−1367.0) = −787.0 − 857.4 + 1367.0 = −277.4 kJ mol⁻¹. Forgetting to reverse the sign of the combustion term gives −3011.4.
The unfolding of a small protein has ΔH = +250 kJ mol⁻¹ at 330 K. Its heat-capacity change of unfolding is ΔCp = +6.0 kJ K⁻¹ mol⁻¹, taken as constant between 310 K and 330 K. Using Kirchhoff’s equation, ΔH of unfolding at 310 K is ____ kJ mol⁻¹ (to the nearest integer).
Numerical answer — type the value.
Show answer
Answer: 130
ΔH(T₂) = ΔH(T₁) + ΔCp(T₂ − T₁) = 250 + 6.0 × (310 − 330) = 250 − 120 = 130 kJ mol⁻¹. Going down in temperature with a positive ΔCp lowers ΔH; adding 120 instead (370) is the sign slip. The large positive ΔCp reflects nonpolar groups becoming exposed to water on unfolding.
An isomerisation A ⇌ B reaches equilibrium at 310 K with [B]/[A] = 0.050. Taking R = 8.314 J mol⁻¹ K⁻¹, the standard free-energy change ΔG° of the reaction is ____ kJ mol⁻¹ (to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 7.7
K = [B]/[A] = 0.050. ΔG° = −RT ln K = −8.314 × 310 × ln(0.050) = −2577.3 × (−2.9957) = +7721 J mol⁻¹ = +7.7 kJ mol⁻¹. It is positive because K < 1: at equilibrium A is favoured. Using log₁₀ without 2.303 gives 3.4.
For a reaction, ΔG° = −30.0 kJ mol⁻¹ at 300 K and ΔH° = −60.0 kJ mol⁻¹, with ΔH° taken as independent of temperature. Using the integrated Gibbs–Helmholtz equation, ΔG° at 360 K is ____ kJ mol⁻¹ (to the nearest integer; include the sign).
Numerical answer — type the value.
Show answer
Answer: -24
ΔG₂/T₂ = ΔG₁/T₁ + ΔH(1/T₂ − 1/T₁) = −30.0/300 + (−60.0)(1/360 − 1/300) = −0.10000 + 0.03333 = −0.06667 kJ K⁻¹ mol⁻¹. ΔG₂ = 360 × (−0.06667) = −24.0 kJ mol⁻¹. Check: ΔS° = (ΔH° − ΔG°)/T = (−60.0 + 30.0)/300 = −0.100 kJ K⁻¹ mol⁻¹, and −60.0 − 360(−0.100) = −24.0. An exothermic reaction with negative ΔS becomes less favourable on heating.
2.00 mol of an ideal gas expands isothermally at 300 K from 10.0 L to 40.0 L against a constant external pressure equal to its final pressure. Taking R = 8.314 J mol⁻¹ K⁻¹, the entropy change of the gas is ____ J K⁻¹ (to one decimal place).
Numerical answer — type the value.
Show answer
Answer: 23.1
S is a state function, so use a reversible isothermal path between the same states: ΔS = nR ln(V₂/V₁) = 2.00 × 8.314 × ln 4 = 16.628 × 1.3863 = 23.05 J K⁻¹ ≈ 23.1 J K⁻¹. Dividing the actual heat by T is the trap: q = P₂ΔV = nRT(1 − V₁/V₂) = 3741 J, and 3741/300 = 12.5 J K⁻¹ is the entropy lost by the surroundings, not the entropy gained by the gas.