6 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 6
The numbered diagrams represent four gas mixtures in identical rigid containers at 298 K. In every diagram, each molecule represents a partial pressure of 0.25 atm. In which diagram are X₂, Y₂ and XY all in their standard states, the conditions to which ΔG° for the reaction X₂(g) + Y₂(g) → 2 XY(g) refers?
Answer and reasoning
ADiagram 1 A student who thinks standard conditions describe the reactants alone, before any product forms, picks this. ΔG° refers to every species, product included, in its standard state, so XY must also be present at 1.0 atm; in Diagram 1 it is absent.
BDiagram 2Correct Standard state for a gas means a partial pressure of 1.0 atm. Diagram 2 contains four molecules each of X₂, Y₂ and XY, so each gas has a partial pressure of 4 × 0.25 atm = 1.0 atm, and every species in the equation is in its standard state.
CDiagram 3 A student who thinks the 1.0 atm of the standard state is the total pressure of the mixture picks this. The four molecules in Diagram 3 give a total of 1.0 atm, but the partial pressures are only 0.50, 0.25 and 0.25 atm; each gas must be at 1.0 atm.
DDiagram 4 A student who thinks standard conditions mean amounts in the ratio of the coefficients picks this: two X₂, two Y₂ and four XY match 1 : 1 : 2. Their partial pressures are 0.50, 0.50 and 1.0 atm, so only XY is in its standard state.
Working Standard state of a gas: partial pressure 1.0 atm. Each molecule is 0.25 atm, so a gas is at 1.0 atm when four of its molecules are present. Only the diagram with four X₂, four Y₂ and four XY meets this for all three gases.
A reference book describes a reaction with ΔG° = −120 kJ/molrxn at 298 K as “spontaneous.” Which statement about the reaction at 298 K is best supported by this value of ΔG°?
Answer and reasoning
AIt is thermodynamically favored under standard conditionsCorrect ΔG° < 0 means the reaction is thermodynamically favored under standard conditions. That is all the sign of ΔG° shows; it says nothing about how fast the reaction is, whether it needs energy to start, or how much heat it releases.
BOnce its reactants are mixed, it occurs very rapidly A student who reads 'spontaneous' as 'quick or sudden' picks this. ΔG° says whether a reaction is favored, not how fast it goes; a favored reaction can be extremely slow.
CIt begins by itself, without needing any input of energy to start A student who reads 'spontaneous' as 'happening without cause' picks this. Many favored reactions need a spark or heating to begin; ΔG° < 0 does not rule that out.
DIt gives off 120 kJ of heat for each mole of reaction A student who treats ΔG° as the heat of reaction picks this. The heat released at constant pressure is given by ΔH°, which differs from ΔG° by the −TΔS° term.
Which equation represents the reaction whose standard Gibbs free energy change is the standard free energy of formation, ΔG°f, of methanol, CH₃OH(l)?
Answer and reasoning
A2 CH₄(g) + O₂(g) → 2 CH₃OH(l) A student who thinks any reaction that produces the compound is its formation reaction picks this. CH₄ is a compound, not an element in its standard state, and the equation forms 2 mol of CH₃OH, so its ΔG° is not ΔG°f of CH₃OH.
BC(s) + 2 H₂(g) + ½ O₂(g) → CH₃OH(l)Correct A formation reaction makes one mole of the compound from its elements, each in its standard state. One CH₃OH needs 1 C, 4 H and 1 O: C(s), 2 H₂(g) and ½ O₂(g).
C2 C(s) + 4 H₂(g) + O₂(g) → 2 CH₃OH(l) A student who clears the fraction to get whole-number coefficients picks this. This equation forms 2 mol of CH₃OH, so its ΔG° is twice ΔG°f; a formation reaction makes exactly one mole.
DC(g) + 4 H(g) + O(g) → CH₃OH(l) A student who thinks a compound is formed from its separate atoms picks this. The elements must be in their standard states, C(s), H₂(g) and O₂(g), not gaseous atoms.
Working No calculation. A formation reaction makes one mole of the compound from its elements in their standard states: carbon as C(s), hydrogen as H₂(g), oxygen as O₂(g). Balancing for 1 mol CH₃OH: C(s) + 2 H₂(g) + ½ O₂(g) → CH₃OH(l).
For the dissolution of a salt, MX(s) → M⁺(aq) + X⁻(aq), ΔH° = +20.0 kJ/molrxn and ΔS° = +100.0 J/(molrxn·K). A student adds 2.0 g of MX(s) to 100.0 g of water at 25°C in an insulated cup and stirs. Which prediction about the experiment is correct?
Answer and reasoning
AThe solid does not dissolve, and the temperature stays the same A student who thinks an endothermic process cannot be favored picks this. The entropy term, −TΔS° = −29.8 kJ/molrxn, outweighs ΔH° = +20.0 kJ/molrxn, so ΔG° < 0 and the salt dissolves.
BThe solid dissolves, and the temperature of the water rises A student who reads a positive ΔH° as energy given out picks this. ΔH° > 0 means the dissolving salt absorbs energy from the water, so the water cools.
CThe solid dissolves, and the water temperature stays the same A student who thinks dissolving, being a physical change, involves no energy change picks this. ΔH° = +20.0 kJ/molrxn: energy is absorbed from the water as the salt dissolves, so the temperature falls.
DThe solid dissolves, and the water temperature fallsCorrect ΔG° = 20.0 − 298(0.1000) = −9.8 kJ/molrxn, so dissolution is thermodynamically favored and the solid dissolves. Because ΔH° > 0, the dissolving salt absorbs energy from the water, whose temperature therefore falls.
Working ΔG° = ΔH° − TΔS° = 20.0 − 298(0.1000) = 20.0 − 29.8 = −9.8 kJ/molrxn < 0, so dissolution is thermodynamically favored and the solid dissolves. ΔH° > 0: the process absorbs energy from the water, so the water temperature falls.
A student needs ΔG° at 500 K for a reaction and has a data table compiled at 298 K. Assuming that ΔH° and ΔS° do not change with temperature, which data from the table does the student need?
Answer and reasoning
AEnthalpies of formation and absolute entropies of all the speciesCorrect ΔG° at 500 K = ΔH° − (500 K)ΔS°. ΔH° comes from the ΔH°f values and ΔS° from the S° values of every reactant and product, elements included, since their S° values are not zero.
BFree energies of formation, which apply at any temperature A student who thinks ΔG° does not depend on temperature picks this. ΔG°f values give ΔG° at 298 K only; at 500 K the −TΔS° term is different.
CEnthalpies of formation of the species, since ΔG° equals ΔH° A student who treats ΔG° as the same as ΔH° picks this. ΔG° = ΔH° − TΔS°, so entropy data are needed as well.
DRate constants for the reaction at 298 K and at 500 K A student who links thermodynamic favorability with reaction speed picks this. Rate constants describe how fast a reaction goes; ΔG° at 500 K comes from ΔH° − (500 K)ΔS°, which needs ΔH°f and S° values.
Working No calculation. ΔG° depends on T (ΔG° = ΔH° − TΔS°), so ΔG°f values for 298 K cannot be used at 500 K. Calculate ΔH° from the ΔH°f values and ΔS° from the S° values of all species (elements included), then ΔG°(500 K) = ΔH° − (500 K)ΔS°.
The diagrams represent the contents of a rigid container at the same temperature before and after a reaction of the gas A goes to completion. The reaction is exothermic. Under which temperature conditions is the reaction, as represented, thermodynamically favored?
Answer and reasoning
AFavored at all temperatures A student who thinks every exothermic process is favored picks this. The diagrams show the number of gas particles halving, so ΔS° < 0, and the −TΔS° term makes ΔG° positive at high temperatures.
BNot favored at any temperature A student who thinks a process with a decrease in entropy can never be favored picks this. Because ΔH° < 0, ΔG° is negative at low temperatures, where TΔS° is small.
CFavored only at low temperaturesCorrect The diagrams show 2 A(g) → A₂(g): eight gas particles become four, so the matter becomes less dispersed and ΔS° < 0. With ΔH° < 0 and ΔS° < 0, ΔG° = ΔH° − TΔS° is negative only when T is low enough that the positive −TΔS° term is smaller than |ΔH°|.
DFavored only at high temperatures A student who thinks heating always makes a process more favored picks this. With ΔS° < 0, −TΔS° is positive and grows with T, so heating makes the reaction less favored.
Working No calculation. 8 gas particles (A) become 4 (A₂): 2 A(g) → A₂(g), fewer gas particles, so ΔS° < 0. Exothermic: ΔH° < 0. ΔG° = ΔH° − TΔS° < 0 only when T is low enough that −TΔS° (positive) is smaller than |ΔH°|.
In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.3.A.1 Standard state Fix
Standard state
The reference conditions to which ΔG° refers: every reactant and product present as a pure substance (solids and liquids), as a solution of concentration 1.0 M (solutes), or as a gas at a pressure of 1.0 atm (or 1.0 bar). In a gas mixture each gas must have a partial pressure of 1.0 atm.
Standard Gibbs free energy change, ΔG°
The Gibbs free energy change for a process in which all the reactants and products are in their standard states, reported per mole of reaction as written (kJ/molrxn). Its sign is the measure of thermodynamic favorability.
Students often think Standard conditions describe the starting reaction mixture, so the reactants are in their standard states and no product is present. In fact No. ΔG° refers to a mixture in which every reactant and every product is in its standard state at once: for gases, each at a partial pressure of 1.0 atm.
Students often think In a gas mixture at standard state the total pressure is 1.0 atm, shared among the gases. In fact No. Each gas must be at 1.0 atm; in a mixture this means each gas has a partial pressure of 1.0 atm, so a mixture of three gases at standard state has a total pressure of 3.0 atm.
9.3.A.2 Thermodynamically favored Fix
Thermodynamically favored
Describes a process for which ΔG° < 0. The CED prefers this phrase to the historical term 'spontaneous', which is easily misread as 'sudden' or 'happening without cause'. A favored process is not necessarily fast, and it may still need energy to get started.
Students often think A thermodynamically favored (spontaneous) process happens quickly or suddenly, and a process that is fast must be thermodynamically favored. In fact No. ΔG° < 0 shows that a process is thermodynamically favored; it says nothing about how fast the process occurs. A favored reaction may be very slow.
Students often think A thermodynamically favored (spontaneous) process starts by itself, without any cause or input of energy. In fact Not necessarily. ΔG° < 0 means the process is favored, but many favored reactions need an initial input of energy, such as a spark or a flame, before they proceed.
9.3.A.3 Standard free energy of formation, ΔG°fFix
Standard free energy of formation, ΔG°f
ΔG° for the formation of one mole of a compound from its elements, each in its standard state (for example C(s) + 2 H₂(g) + ½ O₂(g) → CH₃OH(l)). ΔG°f of an element in its standard state is zero.
ΔG° from free energies of formation
ΔG°reaction = ΣΔG°f(products) − ΣΔG°f(reactants), with each ΔG°f multiplied by the coefficient of that species in the balanced equation.
Students often think Each species' ΔG°f is used once, whatever its coefficient, because the table already gives the value for that substance. In fact Yes. ΔG°f is per mole of the compound, so each value is multiplied by the coefficient of that species in the balanced equation before the sums are taken.
Students often think ΔG°f of a compound is the free-energy change of any reaction that produces it, so ΔG° for a reaction can be taken from the ΔG°f values of the products alone. In fact No. The formation reaction of a compound makes one mole of it from its elements, each in its standard state (for example C(s), H₂(g), O₂(g)). Reactions that start from other compounds, such as CO(g) or CO₂(g), are not formation reactions.
9.3.A.4 Enthalpy and entropy contributions Fix
Enthalpy and entropy contributions
ΔG° combines an enthalpy term, ΔH°, and an entropy term, −TΔS°. When the two terms have opposite signs, both must be weighed to decide whether a process is favored; the freezing of water (ΔH° < 0, ΔS° < 0) and the dissolution of sodium nitrate (ΔH° > 0, ΔS° > 0) are examples.
Freezing of water
H₂O(l) → H₂O(s) has ΔH° < 0 (hydrogen bonds form as the molecules lock into the ice structure) and ΔS° < 0 (the molecules become less free to move). Below 0°C at 1 atm the enthalpy term outweighs −TΔS°, so freezing is favored; above 0°C it is not.
Dissolution of sodium nitrate
NaNO₃(s) → Na⁺(aq) + NO₃⁻(aq) is endothermic (the solution cools) yet thermodynamically favored at room temperature, because the dispersal of the ions among the water molecules gives a positive ΔS° whose −TΔS° term outweighs ΔH°.
Students often think Releasing energy is what makes a process favored, so an exothermic process is favored at every temperature whatever its entropy change. In fact No. An exothermic process (ΔH° < 0) is favored at every temperature only if ΔS° > 0. If ΔS° < 0, the −TΔS° term is positive and outweighs ΔH° above some temperature, so the process is favored only at low temperature.
Students often think A process that absorbs energy cannot be thermodynamically favored; only exothermic processes can occur on their own. In fact Yes. If ΔS° > 0 and the temperature is high enough, −TΔS° outweighs the positive ΔH°, so ΔG° < 0. The dissolution of sodium nitrate and of ammonium nitrate in water are endothermic yet favored at room temperature.
9.3.A.5 ΔG° = ΔH° − TΔS° Fix
ΔG° = ΔH° − TΔS°
Gives ΔG° at a temperature T (in kelvins) from ΔH° and ΔS°. ΔH° is usually in kJ/molrxn and ΔS° in J/(molrxn·K), so one must be converted before they are combined. ΔG° changes with temperature through the −TΔS° term, even when ΔH° and ΔS° are treated as constant.
Students often think Heating always makes a process more thermodynamically favored, because at high temperature the TΔS° term dominates and drives the process forward. In fact No. Raising T makes ΔG° more negative only when ΔS° > 0. When ΔS° < 0, the −TΔS° term is positive and grows with T, so heating makes the process less favored.
Students often think The temperature in degrees Celsius can be substituted directly for T in ΔG° = ΔH° − TΔS°. In fact No. T in ΔG° = ΔH° − TΔS° is the absolute temperature in kelvins, T = °C + 273.15.
9.3.A.6 Signs of ΔH° and ΔS° and temperature Fix
Signs of ΔH° and ΔS° and temperature
ΔH° < 0, ΔS° > 0: favored at all T. ΔH° > 0, ΔS° < 0: favored at no T. ΔH° > 0, ΔS° > 0: favored at high T. ΔH° < 0, ΔS° < 0: favored at low T. Only the last two cases need a calculation to decide favorability at a given temperature.
Crossover temperature
When ΔH° and ΔS° have the same sign, ΔG° = 0 at T = ΔH°/ΔS° (with consistent units); the process is favored on one side of this temperature and unfavored on the other.
Students often think A process in which the entropy of the system decreases can never be thermodynamically favored. In fact Yes. If ΔH° < 0 and the temperature is low enough, ΔH° outweighs the positive −TΔS° term and ΔG° < 0. The freezing of water below 0°C is an example.
Students often think A negative ΔS° means a process is favored at low temperatures, whatever the sign of ΔH°. In fact No. A process with ΔS° < 0 is favored at low temperature only if ΔH° < 0. If ΔH° > 0 and ΔS° < 0, ΔG° is positive at every temperature.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
The table gives standard enthalpies of formation and standard free energies of formation at 298 K. What is ΔG° for the reaction 2 CO(g) + O₂(g) → 2 CO₂(g) at 298 K?
Answer and reasoning
A−257.2 kJ/molrxn A student who uses each ΔG°f once, ignoring the coefficients, picks this: −394.4 − (−137.2). Both CO and CO₂ have coefficient 2, so each value is doubled.
B−566.0 kJ/molrxn A student who treats ΔH° and ΔG° as the same quantity picks this, using the ΔH°f column: 2(−393.5) − 2(−110.5). That gives ΔH°; ΔG° needs the ΔG°f values.
C−788.8 kJ/molrxn A student who thinks ΔG°f of CO₂ is the free-energy change for any reaction that makes CO₂ picks this: 2(−394.4). ΔG°f refers to forming CO₂ from its elements; here CO is the starting material, so ΔG°f of CO must be subtracted.
At a pressure of 1 atm, liquid water freezes when it is placed in surroundings at −10°C, but not when the surroundings are at +10°C. Which particulate-level explanation accounts for these observations?
Answer and reasoning
ABelow 0°C the molecules lack the energy to move at all, so they stop and stay fixed in place A student who thinks particles in a solid do not move picks this. Molecules in ice still vibrate about fixed positions; freezing happens below 0°C because the energy released by forming hydrogen bonds then outweighs the entropy decrease.
BIce occupies more volume than liquid water, so freezing spreads out the molecules, raising entropy A student who judges entropy by volume picks this. Ice is less dense than water, but its molecules are far less free to move, so the entropy decreases on freezing; freezing is driven by the energy released.
CMore hydrogen bonds form as ice forms, releasing energy that outweighs the entropy decrease only at low TCorrect As molecules lock into the ice structure, hydrogen bonds form and energy is released (ΔH° < 0), while the molecules lose freedom of motion (ΔS° < 0). In ΔG° = ΔH° − TΔS°, the unfavorable −TΔS° term grows with temperature, so the energy released outweighs it only below 0°C.
DFreezing absorbs energy from the surroundings, which happens only when the surroundings are cold A student who thinks freezing takes in energy, or 'cold', picks this. Freezing releases energy as hydrogen bonds form (ΔH° < 0); the surroundings must be colder than the water for that energy to flow out.
Working No calculation. H₂O(l) → H₂O(s): hydrogen bonds form as molecules lock into ice, so ΔH° < 0; the molecules lose freedom of motion, so ΔS° < 0. ΔG° = ΔH° − TΔS° is negative only when T is low enough that the positive −TΔS° term is smaller than |ΔH°|: below 0°C at 1 atm.
The table gives thermodynamic data at 298 K. Assuming that ΔH° and ΔS° do not change with temperature, what is ΔG°, in kJ/molrxn, for the reaction CaCO₃(s) → CaO(s) + CO₂(g) at 900°C?
Answer and reasoning
A−8.7Correct ΔH° = −634.9 − 393.5 + 1207.6 = +179.2 kJ/molrxn; ΔS° = 38.1 + 213.8 − 91.7 = +160.2 J/(molrxn·K). At 1173 K, ΔG° = 179.2 − 1173(0.1602) = −8.7 kJ/molrxn: at 900°C the entropy term just outweighs the enthalpy term.
B+35.0 A student who substitutes the Celsius temperature for T picks this: 179.2 − 900(0.1602). T in ΔG° = ΔH° − TΔS° must be in kelvins, 1173 K.
C+179.2 A student who treats ΔG° as equal to ΔH° picks this, leaving out the −TΔS° term. At 1173 K that term is −187.9 kJ/molrxn, which makes ΔG° negative.
D−1.88 × 10⁵ A student who combines ΔH° in kJ with ΔS° in J without converting picks this: 179.2 − 1173(160.2). ΔS° must be converted to 0.1602 kJ/(molrxn·K) first.
The table gives ΔH° and ΔS° at 298 K for four processes. Which process has the most negative ΔG° at 298 K?
Answer and reasoning
AProcess W A student who judges favorability by ΔH° alone picks this, because Process W has the most negative ΔH°, −45.0 kJ/molrxn. Its ΔS° is −100.0 J/(molrxn·K), so −TΔS° = +29.8 kJ/molrxn and ΔG° is only −15.2 kJ/molrxn.
BProcess XCorrect ΔG° = ΔH° − TΔS°, with ΔS° converted to kJ. Process X: −35.0 − 298(0.0400) = −46.9 kJ/molrxn. The others: W −15.2, Y −38.8, Z +9.7. Process X has the most negative ΔG°: it is exothermic and its entropy increases, so both terms favor it.
CProcess Y A student who judges favorability by ΔS° alone picks this, because Process Y has the largest ΔS°, +80.0 J/(molrxn·K). Its ΔH° is only −15.0 kJ/molrxn, so ΔG° = −15.0 − 23.8 = −38.8 kJ/molrxn, less negative than for Process X.
DProcess Z A student who treats a negative ΔS° as favorable, like a negative ΔH°, picks this, calculating ΔH° + TΔS° = −35.0 − 44.7 = −79.7 kJ/molrxn. With the correct sign, ΔG° = −35.0 + 44.7 = +9.7 kJ/molrxn: Process Z is not even favored.
Working ΔG° = ΔH° − TΔS° at T = 298 K with ΔS° in kJ/(molrxn·K). (−35.0, +40.0): −35.0 − 11.9 = −46.9. (−45.0, −100.0): −45.0 + 29.8 = −15.2. (−15.0, +80.0): −15.0 − 23.8 = −38.8. (−35.0, −150.0): −35.0 + 44.7 = +9.7. Most negative: the process with ΔH° = −35.0 kJ/molrxn and ΔS° = +40.0 J/(molrxn·K), Process X.
For a hypothetical reaction, ΔH° = +40.0 kJ/molrxn and ΔS° = −50.0 J/(molrxn·K) at 298 K, and these values change little with temperature. A student predicts that heating the reaction mixture to 1000 K will make the reaction thermodynamically favored. Which evaluation of the prediction is correct?
Answer and reasoning
AIt is right: at a high temperature the TΔS° term dominates, so ΔG° becomes negative A student who thinks heating always makes a process more favored picks this. TΔS° does grow with T, but ΔS° is negative, so −TΔS° adds a positive amount: ΔG° = +90.0 kJ/molrxn at 1000 K.
BIt is right: heating makes the reaction go faster, so it becomes more favored A student who equates a fast reaction with a favored one picks this. Heating does increase the rate, but favorability depends on ΔG°, which is positive at every temperature here.
CIt is wrong: ΔG° stays positive at every temperature, as ΔH° > 0 and ΔS° < 0Correct Both terms of ΔG° = ΔH° − TΔS° are positive: ΔH° = +40.0 kJ/molrxn and −TΔS° = +T(0.0500 kJ/(molrxn·K)). At 1000 K, ΔG° = +90.0 kJ/molrxn; heating only makes ΔG° more positive.
DIt is wrong: with ΔS° < 0 the reaction is favored only at low temperatures, so cool it A student who remembers 'negative ΔS°, favored at low T' without the condition ΔH° < 0 picks this. Here ΔH° > 0, so ΔG° = 40.0 + T(0.0500) is positive at low temperatures as well.
Working ΔG° = ΔH° − TΔS° = 40.0 + T(0.0500). Both terms are positive at every T: at 298 K, +54.9; at 1000 K, +90.0 kJ/molrxn. ΔH° > 0 and ΔS° < 0: unfavored at all temperatures.
For a reaction, ΔH° < 0 and ΔS° < 0. Which change would make ΔG° for the reaction more negative?
Answer and reasoning
ARaising the temperature A student who thinks heating always makes a process more favored picks this. Because ΔS° < 0, raising T makes −TΔS° larger and positive, so ΔG° becomes less negative.
BAdding a suitable catalyst A student who thinks a catalyst makes a reaction more favored picks this. A catalyst lowers the activation energy and speeds the reaction but leaves ΔH°, ΔS° and ΔG° unchanged.
CUsing more of each reactant A student who thinks ΔG° grows with the amount of reactant picks this. ΔG° is per mole of reaction with every species in its standard state, so it does not depend on the amounts used.
DLowering the temperatureCorrect With ΔS° < 0, the term −TΔS° is positive and proportional to T. Lowering T shrinks this unfavorable term, so ΔG° = ΔH° − TΔS° becomes more negative.
Working No calculation. ΔG° = ΔH° − TΔS°; with ΔS° < 0, −TΔS° is positive and proportional to T. Lowering T makes this unfavorable term smaller, so ΔG° becomes more negative. A catalyst and the amount of reactant do not change ΔG°.
Solid sodium nitrate, NaNO₃, dissolves readily when it is stirred into water at 25°C. Which particulate-level account explains why the dissolution is thermodynamically favored?
Answer and reasoning
AWhole NaNO₃ units, which stay intact, spread through the water and so raise the entropy of the system A student who thinks an ionic compound dissolves as whole formula units picks this. NaNO₃ is ionic; its lattice separates into Na⁺ and NO₃⁻ ions, which are what spread through the water.
BSeparated Na⁺ and NO₃⁻ ions disperse through the water; the entropy gain outweighs the energy absorbedCorrect NaNO₃ separates into Na⁺ and NO₃⁻ ions that become dispersed among the water molecules, so ΔS° > 0. The dissolution absorbs energy overall (ΔH° > 0; the solution becomes colder), but at 25°C the −TΔS° term outweighs ΔH°, so ΔG° < 0.
CThe ions form covalent bonds with water molecules, which releases the energy that drives dissolving A student who pictures hydration as a chemical reaction picks this. The ions are attracted to polar water molecules (ion–dipole attractions), not covalently bonded, and dissolving NaNO₃ absorbs energy overall (the solution becomes colder).
DBreaking up the crystal lattice of the solid releases energy, and this energy release drives dissolving A student who thinks breaking attractions releases energy picks this. Separating oppositely charged ions requires energy; dissolving NaNO₃ absorbs energy overall (the solution becomes colder) and is driven by the entropy increase.
Working No calculation. NaNO₃(s) → Na⁺(aq) + NO₃⁻(aq). ΔH° > 0 (the solution cools). The ions become dispersed among the water molecules, so ΔS° > 0. ΔG° = ΔH° − TΔS° < 0 at 25°C because −TΔS° outweighs the positive ΔH°.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
