1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
A student plans to calculate ΔS° at 298 K for the reaction C(s, graphite) + O₂(g) → CO₂(g) and has already looked up the standard molar entropies of O₂(g) and CO₂(g). What other value does the student need?
Answer and reasoning
ANone: S° of graphite is zero A student who thinks an element in its standard state has zero entropy picks this. That rule applies to ΔH°f, not to S°; graphite's absolute entropy is 5.7 J/(mol·K) and must be included.
BThe ΔH°f of carbon dioxide A student who thinks the entropy change can be found from enthalpy data picks this. ΔH°f values give ΔH° for the reaction; ΔS° comes only from the S° values of the species.
CThe mass of graphite used A student who thinks ΔS° depends on the size of the sample picks this. ΔS° is per mole of reaction as written, calculated from S° values and coefficients, so no mass is needed.
DThe S° of graphite, C(s)Correct ΔS° = S°(CO₂) − [S°(C, graphite) + S°(O₂)]. Every species in the equation contributes its absolute entropy, and graphite, although an element, has S° = 5.7 J/(mol·K) at 298 K, so this value is needed.
Working No calculation. ΔS° = S°(CO₂) − [S°(C, graphite) + S°(O₂)]. Graphite, though an element, has a nonzero absolute entropy (5.7 J/(mol·K) at 298 K), so its S° is needed; no ΔH° value or sample mass enters the calculation.
The entropy of one mole of a substance in its standard state at a stated temperature, usually 298 K, in J/(mol·K). S° values are absolute entropies, not changes measured from the elements, so elements as well as compounds have positive S° values at 298 K.
Standard entropy change of a process, ΔS°
The entropy of the species after a process minus the entropy of the species before it: ΔS°reaction = ΣS°products − ΣS°reactants, with each S° multiplied by the coefficient of that species in the balanced equation. It is reported per mole of reaction as written, in J/(molrxn·K).
Sign of ΔS°
A positive ΔS° means the products have greater total entropy than the reactants; a negative ΔS° means they have less. A calculated sign can be checked against particulate reasoning: forming more moles of gas generally gives ΔS° > 0, and forming a liquid or solid from gases generally gives ΔS° < 0.
S° and physical state
For a given substance at the same temperature, S° of the gas is much greater than S° of the liquid, which is greater than S° of the solid, because the particles are more dispersed and freer to move. Example: S° of Br₂(l) is 152.2 J/(mol·K) and S° of Br₂(g) is 245.5 J/(mol·K) at 298 K.
Reading an S° data table
Thermodynamic tables list S° (J/(mol·K)) in a separate column from ΔH°f (kJ/mol). For a ΔS° calculation only the S° column is used; every species in the balanced equation, elements included, needs its S° value, multiplied by its coefficient.
Students often think The standard molar entropy of an element in its standard state is zero, so elements can be left out of ΣS° when ΔS° is calculated. In fact No. S° values are absolute entropies, and every pure substance, element or compound, has a positive S° at 298 K: S° of H₂(g) is 130.7 J/(mol·K) and S° of C(s, graphite) is 5.7 J/(mol·K). It is the standard enthalpy of formation, ΔH°f, that is zero for an element in its standard state.
Students often think ΔS° is found by subtracting the sum of the products' S° values from the sum of the reactants' S° values. In fact The sum for the reactants is subtracted from the sum for the products: ΔS° is the entropy after the process minus the entropy before it. A process whose products are more dispersed than its reactants has a positive ΔS°.
5 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 5
The table gives standard molar entropies at 298 K. What is ΔS° for the reaction N₂(g) + 3 H₂(g) → 2 NH₃(g)?
Answer and reasoning
A+198.1 J/(molrxn·K) A student who subtracts the products' entropies from the reactants' entropies picks this: 583.7 − 385.6. ΔS° is entropy after minus entropy before, products minus reactants, which gives −198.1 J/(molrxn·K).
B−198.1 J/(molrxn·K)Correct ΔS° = 2(192.8) − [191.6 + 3(130.7)] = 385.6 − 583.7 = −198.1 J/(molrxn·K). The value is negative, as expected when 4 mol of gas form 2 mol of gas, so the products are less dispersed than the reactants.
C+385.6 J/(molrxn·K) A student who takes S° of the elements N₂(g) and H₂(g) to be zero picks this: 2(192.8) − 0. S° values are absolute entropies, and N₂ and H₂ have large S° values that must be subtracted.
D−129.5 J/(molrxn·K) A student who uses each S° value once, ignoring the coefficients, picks this: 192.8 − 191.6 − 130.7. S° is per mole, so S°(NH₃) is multiplied by 2 and S°(H₂) by 3.
Working ΔS° = ΣS°products − ΣS°reactants = 2 S°(NH₃) − [S°(N₂) + 3 S°(H₂)] = 2(192.8) − [191.6 + 3(130.7)] = 385.6 − 583.7 = −198.1 J/(molrxn·K). The negative sign fits the particulate picture: 4 mol of gas become 2 mol of gas.
At 298 K the standard molar entropies of H₂(g), O₂(g) and H₂O(l) are 130.7, 205.2 and 70.0 J/(mol·K). A student uses these values to calculate ΔS° for the reaction 2 H₂(g) + O₂(g) → 2 H₂O(l) and reports +326.6 J/(molrxn·K). Which evaluation of the student's result is correct?
Answer and reasoning
AIt is right: the reaction releases much energy, so the products gain entropy A student who thinks an exothermic reaction must have a positive ΔS° picks this. The reaction is strongly exothermic, but ΔS° compares the entropies of the products and reactants, and 2 mol of liquid have far less entropy than 3 mol of gas.
BIt is right: H₂O molecules contain more atoms than H₂ or O₂ molecules A student who thinks larger molecules always have more entropy, whatever their state, picks this. The change from gas to liquid outweighs molecular size: S° of H₂O(l), 70.0 J/(mol·K), is smaller than S° of H₂(g) or O₂(g).
CIt is wrong: here 3 mol of gas become 2 mol of liquid, so ΔS° is negativeCorrect Gas molecules are far more dispersed than molecules in a liquid, and here 3 mol of gas are replaced by 2 mol of liquid, so the entropy of the products is less than that of the reactants. The correct value is 140.0 − 466.6 = −326.6 J/(molrxn·K); the student subtracted in the wrong order.
DIt is wrong: each S° value is used only once, so ΔS° = −265.9 J/(molrxn·K) A student who uses each species' S° value once, whatever its coefficient, picks this: 70.0 − (130.7 + 205.2) = −265.9. Each S° value is multiplied by its coefficient in the balanced equation: 2(70.0) − [2(130.7) + 205.2] = −326.6 J/(molrxn·K).
Working ΔS° = 2(70.0) − [2(130.7) + 205.2] = 140.0 − 466.6 = −326.6 J/(molrxn·K). The student subtracted in the wrong order. A negative value fits the particulate picture: 3 mol of gas form 2 mol of liquid.
At 298 K the standard molar entropy of Br₂(l) is 152.2 J/(mol·K) and that of Br₂(g) is 245.5 J/(mol·K), so ΔS° for Br₂(l) → Br₂(g) is +93.3 J/(molrxn·K). Which particulate-level description best accounts for the positive value of ΔS°?
Answer and reasoning
AIn the gas, Br₂ molecules are far apart and move freely through a larger volumeCorrect Vaporization leaves each Br₂ molecule intact but separates the molecules: in the gas they are much farther apart and free to move throughout a much larger volume. This greater dispersal of matter is why S° of Br₂(g) exceeds S° of Br₂(l).
BBr–Br bonds break when the liquid boils, so the vapor contains Br atoms A student who thinks boiling breaks the covalent bonds within molecules picks this. Boiling overcomes the attractions between Br₂ molecules; the Br–Br bond is unchanged, and the vapor consists of Br₂ molecules.
CBr₂ molecules expand as they enter the gas phase, so each takes up more space A student who pictures molecules growing when a liquid vaporizes picks this. The molecules keep their size; the gas occupies a larger volume because the molecules are farther apart.
DA mole of the gas contains more Br₂ molecules than does a mole of the liquid A student who thinks a larger volume means more particles picks this. A mole of Br₂ contains 6.022 × 10²³ molecules in any state; the entropy increases because the same molecules become more dispersed.
Working ΔS° = 245.5 − 152.2 = +93.3 J/(molrxn·K). The same number of Br₂ molecules, still bonded as Br₂, becomes far more dispersed: the molecules are much farther apart and move freely through a much larger volume.
For the reaction 2 SO₂(g) + O₂(g) → 2 SO₃(g), ΔS° = −188.0 J/(molrxn·K) at 298 K. The table gives the standard molar entropies of SO₂(g) and O₂(g) at 298 K. What is the standard molar entropy of SO₃(g)?
Answer and reasoning
A265.4 J/(mol·K) A student who uses each S° value once, ignoring the coefficients, picks this: −188.0 + 248.2 + 205.2. Both SO₂ and SO₃ have coefficient 2, so S°(SO₂) is doubled and the result is divided by 2.
C154.2 J/(mol·K) A student who takes S° of the element O₂(g) to be zero picks this, solving 2 S°(SO₃) = −188.0 + 2(248.2). O₂(g) has S° = 205.2 J/(mol·K), which must be included with the reactants.
D444.8 J/(mol·K) A student who writes ΔS° as reactants minus products picks this, solving −188.0 = 701.6 − 2 S°(SO₃). ΔS° is products minus reactants, which gives 256.8 J/(mol·K).
The table gives thermodynamic data at 298 K for the species in the reaction 2 C(s, graphite) + O₂(g) → 2 CO(g). Based on the data in the table, what is ΔS° for the reaction?
Answer and reasoning
A+395.4 J/(molrxn·K) A student who takes the entropy of an element in its standard state to be zero, as its ΔH°f is, picks this: 2(197.7) − 0. S° values are absolute entropies, so 2(5.7) + 205.2 = 216.6 J/K for the graphite and O₂ must be subtracted.
B−221.0 J/(molrxn·K) A student who thinks the entropy change is found from the enthalpy-of-formation data picks this: 2(−110.5) − 0 = −221.0. That number is ΔH° for the reaction in kJ/molrxn; ΔS° comes from the S° column.
C+178.8 J/(molrxn·K)Correct ΔS° is calculated from the S° column alone: 2(197.7) − [2(5.7) + 205.2] = 395.4 − 216.6 = +178.8 J/(molrxn·K). The elements have ΔH°f = 0 but nonzero absolute entropies, and each S° is multiplied by its coefficient.
D−178.8 J/(molrxn·K) A student who subtracts the products' entropies from the reactants' entropies picks this: 216.6 − 395.4. ΔS° is the entropy after the process minus the entropy before it, 395.4 − 216.6 = +178.8 J/(molrxn·K).
Working Only the S° column is used. ΔS° = ΣS°products − ΣS°reactants = 2 S°(CO) − [2 S°(C, graphite) + S°(O₂)] = 2(197.7) − [2(5.7) + 205.2] = 395.4 − 216.6 = +178.8 J/(molrxn·K). The positive sign fits the particulate picture: 1 mol of gas and a solid form 2 mol of gas.
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
