AP Chemistry Unit 6: Thermodynamics — Free Review Games.

In an exothermic reaction, the products have lower energy than the reactants, and the difference is released as heat (delta H < 0).

Specific heat capacity is the energy required to raise the temperature of 1 gram of a substance by 1 degree C. Water's value is relatively high.

In calorimetry, the system is the chemical reaction. The solution and calorimeter are the surroundings that absorb or release heat.

Energy must be supplied to overcome the attractive forces holding atoms together. Breaking bonds always requires energy input (endothermic). Bond formation releases energy.

Hess's law states that enthalpy is a state function. The total delta H depends only on initial and final states, not the path.

Bonds broken: \(4(C-H) + 2(O=O) = 4(413) + 2(495) = 1652 + 990 = 2642\) kJ. Bonds formed: \(2(C=O) + 4(O-H) = 2(799) + 4(463) = 1598 + 1852 = 3450\) kJ. \(\Delta H = 2642 - 3450 = -808\) kJ/mol.

q = mc*delta T = (100 g)(4.18 J/g*C)(6.9 C) = 2884 J. Moles = 0.050 mol. delta H = -2884/0.050 = -57,680 J/mol = -57.7 kJ/mol (negative because exothermic).

By definition, the standard enthalpy of formation of any element in its most stable form at standard conditions is zero.

Using Hess's law: reverse the second reaction and add to the first. \(\Delta H = -393.5 + 283.0 = -110.5\) kJ.

For endothermic reactions (delta H > 0), energy is absorbed. Products are at a higher energy level than reactants on the energy diagram.

\(\Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) = 2(-45.9) - [0 + 0] = -91.8\) kJ. Elements in standard states have \(\Delta H_f = 0\).

q = C * delta T = 5.00 kJ/C * 4.60 C = 23.0 kJ total. Per gram = 23.0/1.50 = 15.3 kJ/g.

delta H_solution = -Lattice energy + delta H_hydration. +3.9 = +787 + delta H_hydration. delta H_hydration = 3.9 - 787 = -783.1 kJ/mol.

Activation energy affects kinetics (rate), not thermodynamics (delta H). Two reactions with the same delta H are equally thermodynamically favorable, but the one with lower Ea is faster.

\(\Delta H = [2(-393.5) + 3(-285.8)] - [(-277.7) + 0] = [-787.0 + (-857.4)] - (-277.7) = -1644.4 + 277.7 = -1366.7\) kJ.

Sublimation converts a solid directly to a gas, dramatically increasing molecular disorder and the number of accessible microstates, so ΔS > 0. Freezing, condensation, and crystallization all decrease disorder, making ΔS < 0 for those processes. The key principle is that gases have far more positional and energetic microstates than liquids or solids.

Spontaneity under standard conditions is determined by the sign of ΔG. When ΔG < 0, the reaction is spontaneous as written. Neither ΔH nor ΔS alone determines spontaneity — it is the combination ΔG = ΔH − TΔS that matters. Activation energy is a kinetic concept and has no bearing on thermodynamic spontaneity; a reaction can be spontaneous yet extremely slow.

At equilibrium, there is no net tendency for the reaction to proceed in either direction, so ΔG = 0. This represents the minimum point on the Gibbs free energy curve as a function of reaction progress. It is important to distinguish ΔG (under actual conditions) from ΔG° (under standard conditions) — only the former equals zero at equilibrium.

The third law defines an absolute reference point for entropy: a perfect crystal at 0 K has only one possible microstate, so S = k ln(1) = 0. This allows calculation of absolute standard molar entropies. Choice A is the first law (conservation of energy), choice C describes Hess's law, and choice D describes the second law of thermodynamics.

Enthalpy is a state function, so ΔH depends only on the initial and final states. Reversing a reaction swaps products and reactants, meaning energy that was released is now absorbed — or vice versa — so the sign of ΔH flips while the magnitude is unchanged. This is a foundational rule for applying Hess's law: if you reverse a reaction, negate its ΔH.

Standard molar entropy increases with increasing molecular disorder. Gases have far more accessible translational, rotational, and vibrational microstates than liquids or solids. H2O(g) has the highest entropy among these choices. Solids like Na and NaCl have restricted atomic motion and the lowest entropy. H2O(l) has more disorder than solids but far less than H2O(g). Standard molar entropy values follow the trend: solid < liquid < gas.

The standard enthalpy of combustion (ΔH°comb) is the enthalpy change for the complete, exothermic reaction of exactly one mole of a substance with excess oxygen under standard conditions (25°C, 1 atm), typically forming CO2(g) and H2O(l). Choice D defines the standard enthalpy of formation (ΔH°f), and choice C describes a decomposition reaction. These three thermodynamic quantities are frequently used together in Hess's law calculations.

Using ΔG = ΔH − TΔS and converting ΔS to kJ: ΔG = −120 kJ/mol − (298 K)(−0.200 kJ/(mol·K)) = −120 + 59.6 = −60.4 kJ/mol. The most common error is forgetting to convert J to kJ before substituting, which would produce −120,000 − (298)(−200) = −60,400 J = −60.4 kJ — the arithmetic still works but only if units are tracked carefully. The reaction is spontaneous because the favorable enthalpy outweighs the unfavorable entropy term at 298 K.

Using ΔG = ΔH − TΔS: when ΔH > 0 and ΔS > 0, at low T the small TΔS term cannot overcome the positive ΔH, so ΔG > 0 (non-spontaneous). At high T, the −TΔS term becomes large and negative, eventually making ΔG < 0 (spontaneous). The reaction becomes spontaneous above the crossover temperature T = ΔH/ΔS. The endothermic dissolution of many salts in water is a real-world example of this behavior.

The dominant factor for ΔS° is the change in moles of gas. Reaction A: liquid → gas, ΔS > 0. Reaction B: 4 mol gas → 2 mol gas, a decrease of 2 mol, giving ΔS << 0. Reaction C: 0 mol gas → 1 mol gas, ΔS > 0. Reaction D: 3 mol gas → 2 mol gas, a decrease of 1 mol, ΔS < 0. Reaction B has the largest decrease in moles of gas (−2 mol) and therefore the most negative ΔS°. This is why the Haber process for ammonia synthesis requires high pressure to favor the product side.

Using Hess's law: reverse reaction (1) to place CH4 as a reactant (ΔH = +74.8 kJ), multiply reaction (2) by 2 to account for 2 mol H2O produced (ΔH = −571.6 kJ), and keep reaction (3) as written (ΔH = −393.5 kJ). Sum: ΔH = +74.8 + (−571.6) + (−393.5) = −890.3 kJ. A common error is multiplying reaction (2) by only 1, giving −743.5 kJ, which accounts for only 1 mol of H2O instead of 2.

Heat gained by water equals heat lost by metal. q_water = (100.0 g)(4.18 J/(g·°C))(24.2 − 20.0°C) = 1755.6 J. Setting q_metal = −q_water: (25.0 g)(c)(24.2 − 95.0°C) = −1755.6 J → (25.0)(c)(−70.8) = −1755.6 → c = 1755.6 / (25.0 × 70.8) ≈ 0.99 J/(g·°C). Choice A (0.45) results from using only the metal mass in the numerator without the correct temperature change. The value 0.99 J/(g·°C) is close to the specific heat of copper (0.385) or possibly zinc (0.388) — real AP problems may ask you to identify the metal from a table.

When ΔH < 0 and ΔS > 0, every term in ΔG = ΔH − TΔS contributes negatively regardless of temperature: ΔH is already negative, and −TΔS is negative because ΔS > 0 and T is always positive. Thus ΔG < 0 at all temperatures. Choice B (ΔH < 0, ΔS < 0) is spontaneous only at low T where enthalpy dominates. Choice D (ΔH > 0, ΔS < 0) is never spontaneous — both terms make ΔG positive.

The crossover temperature is found by setting ΔG = 0: 0 = ΔH − TΔS → T = ΔH/ΔS. Converting ΔH to joules: T = 50,000 J/mol ÷ 100 J/(mol·K) = 500 K. Below 500 K, ΔG > 0 (non-spontaneous). Above 500 K, the −TΔS term dominates and ΔG < 0 (spontaneous). A critical step is ensuring consistent units — mixing kJ for ΔH and J/(mol·K) for ΔS without conversion leads to choice D (5000 K), a common unit error.

The number of moles of gas decreases from 3 (2 mol SO2 + 1 mol O2) to 2 (2 mol SO3). Fewer moles of gas means fewer translational microstates and less disorder, so ΔS° < 0. Choice C incorrectly conflates conservation of atoms with conservation of entropy — these are entirely different concepts. Choice D confuses system entropy with surroundings entropy; ΔS° in thermochemical equations always refers to the system.

A negative ΔG°f means that forming 1 mol of liquid water from its elements (H2(g) and 1/2 O2(g)) in their standard states is thermodynamically spontaneous under standard conditions (25°C, 1 atm). Choice D is a common confusion: ΔG°f and ΔH°f are different quantities; ΔH°f for H2O(l) is −285.8 kJ/mol, not −237 kJ/mol. The difference arises from the TΔS term.

If K >> 1, then ln K >> 0. Substituting into ΔG° = −RT ln K gives ΔG° << 0 (a large negative value), indicating the reaction strongly favors products. Conversely, K << 1 would give ΔG° >> 0 (favoring reactants). ΔG° = 0 corresponds to K = 1, not K >> 1. This relationship is key for connecting thermodynamics to equilibrium: a reaction with a very negative ΔG° has a very large K and lies almost entirely toward products.

Using ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants): ΔH° = [ΔH°f(Al2O3) + 2×ΔH°f(Fe)] − [ΔH°f(Fe2O3) + 2×ΔH°f(Al)] = [(−1675.7) + 2(0)] − [(−824.2) + 2(0)] = −1675.7 + 824.2 = −851.5 kJ/mol. The enthalpies of formation of Fe(s) and Al(s) are both zero because they are elements in their standard states. The large negative value explains the intensely exothermic nature of thermite reactions.

Use Hess's law with three manipulations: keep reaction (1) as written (ΔH = +180.5 kJ), multiply reaction (2) by 3/2 so that 3NO is produced (ΔH = 3/2 × −113.0 = −169.5 kJ), and keep reaction (3) as written (ΔH = −137.0 kJ). Adding the three equations: N2 + O2 + 3NO + 3/2 O2 + 3NO2 + H2O → 2NO + 3NO2 + 2HNO3 + NO. The 3NO and 3NO2 cancel on both sides, and O2 terms combine to give 5/2 O2, yielding the target equation. ΔH° = 180.5 + (−169.5) + (−137.0) = −126.0 kJ.

ΔS° = ΣS°(products) − ΣS°(reactants) = 2(69.9) − [2(130.7) + 205.2] = 139.8 − [261.4 + 205.2] = 139.8 − 466.6 = −326.8 J/(mol·K). The large negative value is physically sensible: three moles of gas are converted to two moles of liquid, which has vastly fewer microstates. A common error is using choice A (+326.8) by subtracting in the wrong order. Choice C results from ignoring stoichiometric coefficients.

ΔG = ΔH − TΔS = −40,000 J/mol − T(−150 J/(mol·K)) = −40,000 + 150T. At low T, the favorable enthalpy term dominates and ΔG < 0 (spontaneous). Setting ΔG = 0 to find the crossover: 150T = 40,000 → T = 267 K. Above 267 K, the +150T term exceeds 40,000 and ΔG > 0 (non-spontaneous). This is an enthalpy-favored, entropy-disfavored reaction that loses spontaneity at high temperatures — the opposite of an endothermic reaction driven by entropy.

Using ΔG = ΔG° + RT ln Q: RT = (8.314 × 10⁻³ kJ/(mol·K))(298 K) = 2.478 kJ/mol. RT ln Q = (2.478)(ln 10) = (2.478)(2.303) = 5.71 kJ/mol. ΔG = −15.0 + 5.71 = −9.3 kJ/mol. Since Q > 1, there are more products than at standard state, which shifts ΔG above ΔG° toward less negative values. The reaction remains spontaneous but is less thermodynamically favorable than under standard conditions. Choice A results from incorrectly subtracting RT ln Q instead of adding it.

q_solution = (100.0 + 10.0 g)(4.18 J/(g·°C))(21.4 − 25.0°C) = (110.0)(4.18)(−3.6) = −1655 J. Because the solution loses heat (cools down), the dissolution process absorbed that heat from the solution, making the reaction endothermic: q_rxn = +1655 J. Moles of NH4NO3 = 10.0/80.0 = 0.125 mol. ΔH_dissolution = +1655 J / 0.125 mol ≈ +13,240 J/mol = +13.2 kJ/mol. A negative sign (choice B) would incorrectly describe the process as exothermic, contradicting the observed temperature decrease.

Rearranging ΔG° = ΔH° − TΔS° to solve for ΔS°: ΔS° = (ΔH° − ΔG°)/T = (120 − 50) kJ/mol ÷ 500 K = 70 kJ/mol ÷ 500 K = 0.140 kJ/(mol·K) = +140 J/(mol·K). The positive ΔS° is consistent with the stoichiometry — 1 mol of gas producing 2 mol of gas increases disorder. Choice D (+70) results from forgetting to divide by T. Choice A (+340) results from adding ΔH° and ΔG° instead of subtracting.

By definition, ideal gases have no intermolecular forces, so mixing produces no enthalpy change: ΔH_mix ≈ 0. However, the number of spatial and energetic arrangements increases enormously when gases mix — each molecule now has access to the entire volume — giving ΔS_mix > 0. With ΔH ≈ 0 and ΔS > 0, ΔG = ΔH − TΔS = −TΔS < 0 at all positive temperatures. This makes the mixing of ideal gases a purely entropy-driven spontaneous process, illustrating that enthalpy is not required for spontaneity.

By Hess's law, ΔH°f = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 = 108 + 496 + 121 + (−349) + (−788) = 725 − 1137 = −412 kJ/mol. This is a Born-Haber cycle: despite the large energy costs of sublimation, ionization, and bond dissociation, the large negative lattice enthalpy (−788 kJ/mol) drives the overall formation to be exothermic. Choice C (−788) uses only the lattice enthalpy and ignores all other steps. Choice D (−214) results from incorrectly summing only the electron affinity and lattice enthalpy steps.

Total mass = 100.0 mL × 1.00 g/mL = 100.0 g. ΔT = 28.9 − 22.0 = 6.9°C. Heat absorbed by solution: q = (100.0)(4.18)(6.9) = 2884 J = 2.884 kJ. Heat released by reaction = −2.884 kJ. Moles of H2O formed = 0.0500 L × 1.00 mol/L = 0.0500 mol. ΔH = −2.884 kJ ÷ 0.0500 mol = −57.7 kJ/mol. The negative sign reflects heat released by the neutralization. Choice A (−28.8) uses 0.100 mol instead of 0.0500 mol for moles of water. Choice B has the correct magnitude but wrong sign.

At 298 K: ΔG° = −200,000 − (298)(−400) = −200,000 + 119,200 = −80,800 J/mol ≈ −80.8 kJ/mol. Since ΔG° << 0, K = exp(80800 / (8.314 × 298)) = exp(32.6) ≈ 10^14 (K >> 1). At 1000 K: ΔG° = −200,000 − (1000)(−400) = −200,000 + 400,000 = +200,000 J/mol = +200 kJ/mol. Now ΔG° > 0, so K = exp(−200000 / (8.314 × 1000)) = exp(−24.1) ≈ 10^−11 (K << 1). K drops from ~10^14 to ~10^−11 — an enormous decrease. The negative ΔS° penalty, magnified by the factor T, eventually overwhelms the favorable ΔH°. Choice A incorrectly predicts K increases with temperature for this exothermic reaction.

Enthalpy is a state function equal to \(H = U + PV\), and at constant pressure the change in enthalpy equals the heat exchanged with the surroundings, \(q_p = \Delta H\). The choice "A path function that depends on how a reaction occurs" is wrong because state functions like \(H\) depend only on initial and final states, not the pathway taken. Students should remember that state functions (H, S, G, U) are path-independent, which is what makes Hess's law valid.

A positive \(\Delta S\) means the final state has more accessible microstates than the initial state, corresponding to increased disorder, such as when a solid becomes a gas or moles of gas increase. The distractor "The reaction is always spontaneous at all temperatures" is wrong because spontaneity depends on both \(\Delta H\) and \(\Delta S\) through \(\Delta G = \Delta H - T\Delta S\), not entropy alone. Students should associate entropy strictly with the statistical distribution of energy and matter, not directly with heat release or mass.

By convention, a process that releases heat to the surroundings loses enthalpy from the system, giving \(\Delta H < 0\), which defines it as exothermic. The option "Endothermic, with \(\Delta H < 0\)" is inconsistent because endothermic processes absorb heat and have \(\Delta H > 0\) by definition. Students should memorize that negative \(\Delta H\) always signals heat flowing out of the system into the surroundings.

Hess's law states that because enthalpy is a state function, the total \(\Delta H\) for a reaction equals the sum of the \(\Delta H\) values of any sequence of steps that add up to the overall reaction. The option "The enthalpy change of only the slowest step" confuses thermodynamics with kinetics, where the slowest step controls reaction rate, not overall energy change. Students should use this additive property to combine given reactions, reversing signs or scaling coefficients as needed to match the target equation.

Standard state conditions are defined as 1 atm (or 1 bar) pressure with the substance in its most stable physical form at a specified reference temperature, typically 298 K, which allows tabulated thermodynamic values to be compared consistently. The option "0 atm pressure and 0 K" is incorrect because thermodynamic standard states require a defined nonzero pressure and a specific reference temperature for meaningful comparisons. Students should recognize that standard state values like \(\Delta H^\circ_f\) and \(S^\circ\) are only valid under these agreed-upon reference conditions.

Breaking a chemical bond requires input of energy to overcome the attractive forces between bonded atoms, making bond breaking an endothermic process by definition. The distractor "Energy is released when bonds are broken because atoms become more stable when separated" is wrong because separated atoms are higher in potential energy, not more stable, than atoms held together in a bond. Students should remember the rule: breaking bonds absorbs energy, forming bonds releases energy, and the net \(\Delta H\) of a reaction depends on the balance between these two processes.

In this reaction, 4 moles of gas react to form 2 moles of gas, decreasing the number of gas particles and therefore decreasing the system's disorder, giving a negative \(\Delta S\). The option "$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$" is incorrect because a solid decomposing to produce a gas increases the number of gas particles and thus increases entropy. Students should focus on changes in the number of moles of gas as the dominant factor when predicting the sign of \(\Delta S\) for a reaction.

Using \(\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds formed})\), one \(C{-}H\) (413) and one \(Cl{-}Cl\) (242) bond are broken (655 kJ total) while one \(C{-}Cl\) (328) and one \(H{-}Cl\) (431) bond are formed (759 kJ total), giving \(\Delta H = 655 - 759 = -104\) kJ/mol. The option "\(+655\) kJ/mol" only accounts for the energy needed to break bonds and ignores the energy released upon forming new bonds, which is a common calculation error. Students should always subtract the energy of bonds formed from the energy of bonds broken, remembering that bond breaking costs energy and bond forming releases energy.

When \(\Delta H^\circ\) is negative and \(\Delta S^\circ\) is negative, \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) becomes more positive as \(T\) increases, so the reaction is only spontaneous when \(T\) is small enough that the favorable \(\Delta H\) term dominates. The option "At all temperatures" is incorrect because at sufficiently high temperature the \(-T\Delta S^\circ\) term (positive, since \(\Delta S^\circ\) is negative) will overcome the negative \(\Delta H^\circ\) and make \(\Delta G^\circ\) positive. Students should memorize the four sign combinations of \(\Delta H\) and \(\Delta S\) and how each interacts with temperature in the Gibbs equation.

Using \(q = mc\Delta T\), solving for \(\Delta T\) gives \(\Delta T = \frac{450}{25.0 \times 0.900} = 20.0\,^\circ C\). The option "\(12.5\,^\circ C\)" likely results from dividing by mass alone or misapplying the specific heat value rather than using both mass and specific heat in the denominator. Students should remember that calorimetry problems require correctly rearranging \(q = mc\Delta T\) and using consistent units for mass, heat, and specific heat capacity.

Reversing equation (2) to give \(CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g)\), \(\Delta H = +283.0\) kJ, and adding it to equation (1) cancels \(CO_2\) and \(\frac{1}{2}O_2\), yielding \(\Delta H = -393.5 + 283.0 = -110.5\) kJ. The option "\(-676.5\) kJ" comes from simply adding the two original enthalpies without reversing the second equation, which is a common Hess's law error. Students should always check that reversing or scaling a given equation requires reversing the sign or scaling the corresponding \(\Delta H\) value accordingly.

When a system releases heat, that energy flows into the surroundings and disperses among a huge number of particles, increasing the surroundings' entropy, so $\Delta S_{surroundings} = -\Delta H_{system}/T$ is positive for exothermic reactions. The option "Negative, because heat released by the system decreases the thermal disorder of the surroundings" reverses the correct relationship, since adding thermal energy to the surroundings increases, not decreases, molecular motion and disorder there. Students should remember that $\Delta S_{surroundings}$ is inversely related to $\Delta H_{system}$ and temperature, forming the basis for why exothermic reactions tend to be entropically favored in the surroundings.

Freezing converts a relatively disordered liquid into a highly ordered crystalline solid, reducing the number of accessible microstates and giving \(\Delta S < 0\) for the system. The option "\(\Delta S > 0\), because heat is released during freezing" incorrectly conflates the exothermic nature of freezing (a \(\Delta H\) effect) with the entropy change of the system, which is actually governed by the change in molecular order. Students should treat \(\Delta H\) and \(\Delta S\) as related but distinct quantities, since heat release does not automatically dictate the sign of $\Delta S_{system}$.

For an endothermic reaction, \(\Delta H^\circ\) is positive, so as \(T\) increases the term \(-T\Delta S^\circ\) becomes more negative (assuming \(\Delta S^\circ > 0\)), making \(\Delta G^\circ\) more negative and, through \(\Delta G^\circ = -RT\ln K\), causing \(K\) to increase. The option "\(K\) remains unchanged with temperature because \(\Delta G^\circ\) is temperature-independent" is false because \(\Delta G^\circ\) explicitly depends on \(T\) through the \(-T\Delta S^\circ\) term. Students should connect Le Chatelier's qualitative prediction, that raising temperature shifts endothermic reactions toward products, with the quantitative Gibbs free energy explanation.

Adding equation (1) as written to the reverse of equation (2), \(Fe_2O_3(s) \rightarrow 2Fe(s) + \frac{3}{2}O_2(g)\) with \(\Delta H = +824\) kJ, cancels the \(\frac{3}{2}O_2(g)\) terms and gives \(\Delta H = -1676 + 824 = -852\) kJ for the target reaction. The option "\(-2500\) kJ" results from adding both enthalpies without reversing equation (2), which fails to cancel the oxygen and iron oxide terms correctly. Students should carefully track which species must cancel between the given equations and the target reaction, reversing signs whenever an equation is flipped.

Rearranging \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) gives \(\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} = \frac{75.0 - 32.5}{298} \text{ kJ/(mol·K)} = 0.143\) kJ/(mol·K) \(= +143\) J/(mol·K), a positive value indicating increased disorder. The option "\(+254\) J/(mol·K)" comes from dividing \(\Delta H^\circ\) alone by \(T\) without subtracting \(\Delta G^\circ\) first, ignoring the correct algebraic rearrangement of the Gibbs equation. Students should practice rearranging the Gibbs free energy equation to solve for any one of the three variables when the other two are given at a specific temperature.

Reversing equation (1) to \(C_2H_2(g) \rightarrow 2C(s) + H_2(g)\), \(\Delta H = -227\) kJ, then adding \(2\times\) equation (2) (\(-787\) kJ) and equation (3) (\(-285.8\) kJ) gives a total of \(-227 - 787 - 285.8 = -1299.8\) kJ, matching the closest listed value of \(-1299.6\) kJ within rounding. The option "\(-1526.6\) kJ" would result from forgetting to reverse the sign on equation (1) when flipping it to place \(C_2H_2\) on the reactant side. Students should carefully identify which equations need reversal or scaling by comparing the position and coefficients of each species in the target reaction versus the given equations.

Since \(\Delta G^\circ\) is a state function, the overall free energy change for two coupled reactions is simply the sum, \(45 + (-60) = -15\) kJ/mol, and a negative overall \(\Delta G^\circ\) means the coupled process is spontaneous. The option "\(+105\) kJ/mol; nonspontaneous, because the free energies of the two reactions must be added without regard to sign" incorrectly adds magnitudes instead of algebraically summing signed values, which violates how Gibbs energies combine for sequential or coupled reactions. Students should recognize that biological and industrial systems often drive unfavorable reactions by coupling them to strongly favorable ones, as long as the net \(\Delta G^\circ\) is negative.

Since the reaction becomes spontaneous only as temperature increases, \(\Delta S^\circ\) must be positive, and the crossover temperature where \(\Delta G^\circ = 0\) satisfies \(T = \Delta H^\circ / \Delta S^\circ\), so \(\Delta S^\circ\) must be at least \(25000/280 \approx 89.3\) J/(mol·K) to make the reaction just spontaneous by 280 K (setting the tighter bound); this matches option one's reasoning about the crossover lying between the two temperatures. The option "\(\Delta S^\circ\) is negative, with a minimum magnitude of about \(89.3\) J/(mol·K)" is incorrect because a negative \(\Delta S^\circ\) combined with positive \(\Delta H^\circ\) would make the reaction nonspontaneous at all temperatures, contradicting the observation that it becomes spontaneous at higher \(T\). Students should use the temperature dependence of spontaneity, via \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\), to deduce the sign and estimate the magnitude of \(\Delta S^\circ\) from crossover behavior.