6.9 Hess’s Law

1. AStep I: ΔH > 0; step II: ΔH < 0 Correct
   Step I separates the atoms of the H₂ and Cl₂ molecules; breaking bonds requires energy, so ΔH > 0. Step II brings H and Cl atoms together to form H–Cl bonds, which releases energy, so ΔH < 0.
2. BStep I: ΔH < 0; step II: ΔH > 0
   A student who thinks bonds store energy, releasing it when broken and taking it in when formed, picks this. Breaking bonds requires energy and forming bonds releases energy, so the signs are the other way round.
3. CStep I: ΔH > 0; step II: ΔH > 0
   A student who thinks that forming a bond requires an input of energy picks this. Step I is correctly endothermic, but when H and Cl atoms bond in step II energy is released, so ΔH < 0.
4. DStep I: ΔH < 0; step II: ΔH < 0
   A student who thinks breaking bonds releases energy picks this. Step II is correctly exothermic, but separating the atoms in H₂ and Cl₂ in step I requires energy, so ΔH > 0.

Working No calculation. Step I breaks the H–H and Cl–Cl bonds, which requires energy (ΔH > 0). Step II forms H–Cl bonds, which releases energy (ΔH < 0).

CED 6.9.A.1 · Read this in Fix

1. AThe activation energy of N₂(g) + O₂(g) → 2 NO(g)
   A student who thinks the enthalpy change of a reaction depends on its activation energy picks this. Activation energy affects the rate, not ΔH, and it plays no part in combining the equations.
2. BThe mass of NO(g) that is to be reacted with the O₂(g)
   A student who thinks ΔH depends on the amount that reacts picks this. ΔH° is for the equation as written, one mole of reaction, so no mass is needed.
3. CThe ΔH° of the reaction N₂(g) + 2 O₂(g) → 2 NO₂(g) Correct
   Reversing the given reaction and adding N₂(g) + 2 O₂(g) → 2 NO₂(g) gives the target equation, so ΔH°(target) = ΔH°(N₂ + 2 O₂ → 2 NO₂) − ΔH°(N₂ + O₂ → 2 NO). No other quantity is needed.
4. DThe standard enthalpy of formation, ΔH°_f, of O₂(g)
   A student who thinks every substance, elements included, has a nonzero enthalpy of formation picks this. O₂(g) is an element in its standard state, and in any case the O₂ terms are handled when the equations are combined.

Working No calculation. Reverse the given reaction (2 NO → N₂ + O₂, −ΔH₁) and add N₂ + 2 O₂ → 2 NO₂ (ΔH₂): the N₂ and one O₂ cancel to give 2 NO + O₂ → 2 NO₂, with ΔH° = ΔH₂ − ΔH₁. So ΔH° for N₂(g) + 2 O₂(g) → 2 NO₂(g) is needed.

CED 6.9.B.1 · Read this in Fix

1. AEqual masses of Mg(s) and MgO(s) must be used in the two trials.
   A student who thinks ΔH depends on the amount used, so that equal amounts are needed before results can be combined, picks this. Each ΔH is per mole of solid, so the two masses need not be equal.
2. BEnough HCl(aq) must be added in each trial for all of the solid to react. Correct
   Each ΔH is calculated by dividing the heat absorbed by the solution by the moles of solid weighed. That is valid only if all of the weighed solid reacts, so the HCl(aq) must be in excess in both trials.
3. CThe two reactions must take place at the same rate in the calorimeter.
   A student who thinks the enthalpy change depends on how fast a reaction occurs picks this. The rate does not affect ΔH per mole; the two reactions can occur at different rates.
4. DThe two reactions must produce the same temperature change in the water.
   A student who equates temperature change with energy transferred picks this. q is found from mcΔT for each trial and divided by the moles that reacted, so the two temperature changes need not match.

Working No calculation. ΔH per mole is found as q divided by the moles of solid weighed, so every weighed particle of solid must react: HCl(aq) must be in excess in each trial. Equal masses, equal rates or equal temperature changes are not needed, because each ΔH is calculated per mole.

CED 6.9.B.2.iii · Read this in Fix

1. AStep 1 releases energy, and step 2 absorbs more energy than step 1 releases.
   A student who thinks attractions store energy, releasing it when particles separate and taking it in when they come together, picks this. Separating ions requires energy and forming ion–water attractions releases energy.
2. BStep 1 absorbs energy, and step 2 also absorbs energy from the water.
   A student who thinks that forming an attraction between particles requires energy picks this. When ions become surrounded by water molecules, the ion–water attractions release energy, so step 2 is exothermic.
3. CStep 1 absorbs energy, and step 2 releases a greater amount than step 1 absorbs.
   A student who thinks a fall in temperature means the process released energy picks this. That would make the overall process exothermic and the solution would warm up; the solution cools because the dissolving absorbs energy from the water.
4. DStep 1 absorbs energy, and step 2 releases less energy than step 1 absorbs. Correct
   Separating oppositely charged ions requires energy, and the attraction of the ions to water molecules releases energy. The solution cools, so the overall process is endothermic: the energy released in step 2 must be smaller than the energy absorbed in step 1.

Working No calculation. Step 1 overcomes the attractions between ions: energy absorbed (ΔH₁ > 0). Step 2 forms ion–water attractions: energy released (ΔH₂ < 0). The solution cools, so the overall process absorbs energy: ΔH₁ + ΔH₂ > 0, so |ΔH₂| < ΔH₁.

CED 6.9.A.1 · Read this in Fix

1. A+150 kJ/mol_rxn
   A student who gives a positive sign to energy released picks this. The overall process releases 150 kJ per mole of reaction, so the enthalpy of the system falls and ΔH is negative.
2. B−150 kJ/mol_rxn Correct
   Step A → B has ΔH = 400 − 250 = +150 kJ/mol_rxn and step B → C has ΔH = 100 − 400 = −300 kJ/mol_rxn. The overall ΔH is the sum of the steps, +150 + (−300) = −150 kJ/mol_rxn.
3. C+100 kJ/mol_rxn
   A student who reads ΔH as the enthalpy level of the products picks this. C sits at 100 kJ/mol, but A starts at 250 kJ/mol, so ΔH = 100 − 250 = −150 kJ/mol_rxn.
4. D−300 kJ/mol_rxn
   A student who thinks the energy absorbed in step A → B is used up and does not count picks this, using only step B → C. Energy is conserved: the +150 kJ/mol_rxn of the first step must be added to the −300 kJ/mol_rxn of the second.

Working ΔH(A → B) = 400 − 250 = +150 kJ/mol_rxn; ΔH(B → C) = 100 − 400 = −300 kJ/mol_rxn. Overall ΔH = +150 + (−300) = −150 kJ/mol_rxn (equivalently H(C) − H(A) = 100 − 250).

CED 6.9.B.1 · Read this in Fix

1. A+30 kJ/mol_rxn, because the slow step controls the overall energy change
   A student who thinks the slowest step determines the overall enthalpy change picks this. The slow step limits the rate, but ΔH for the overall process is the sum of both steps, −50 kJ/mol_rxn.
2. B+50 kJ/mol_rxn, because energy released by a process is counted as positive
   A student who gives a positive sign to energy released picks this. The process releases a net 50 kJ per mole of reaction to the surroundings, so the system's enthalpy falls and ΔH = −50 kJ/mol_rxn.
3. C−50 kJ/mol_rxn, because energy is conserved, so the ΔH values of the steps add Correct
   The net thermal energy transferred in a sequence of steps equals the sum of the transfers in the steps, because total energy is conserved. ΔH = +30 + (−80) = −50 kJ/mol_rxn. The speed of each step does not affect ΔH.
4. D−110 kJ/mol_rxn, because energy taken in by step 1 is given out again in step 2
   A student who thinks energy absorbed in one step is stored and given out again later, on top of the next step's change, picks this. ΔH₂ already measures the change from the higher enthalpy of B, so the 30 kJ is counted once: +30 + (−80) = −50 kJ/mol_rxn.

Working Total energy is conserved, so ΔH(overall) = ΔH₁ + ΔH₂ = +30 + (−80) = −50 kJ/mol_rxn. The rates of the steps do not affect ΔH.

CED 6.9.B.1 · Read this in Fix

1. A+200 kJ/mol_rxn Correct
   A → B has ΔH = −150 − (−50) = −100 kJ/mol_rxn. Reversing the reaction changes the sign, +100 kJ/mol_rxn, and doubling the equation doubles ΔH, giving +200 kJ/mol_rxn for 2 B → 2 A.
2. B−200 kJ/mol_rxn
   A student who thinks reversing a reaction leaves ΔH unchanged picks this, doubling −100 kJ/mol_rxn. B is lower in enthalpy than A, so converting B to A absorbs energy and ΔH is positive.
3. C+500 kJ/mol_rxn
   A student who takes the height from B to the top of the curve as the ΔH of the reverse reaction picks this: 2 × (100 − (−150)) = +500. That height is the activation energy of B → A; ΔH depends only on the levels of A and B.
4. D−100 kJ/mol_rxn
   A student who reads ΔH as the enthalpy level of the products picks this, doubling the level of A, −50 kJ/mol. ΔH is the difference between the levels of A and B: 2 × [−50 − (−150)] = +200 kJ/mol_rxn.

Working From the profile, A → B: ΔH = −150 − (−50) = −100 kJ/mol_rxn. Reversed (B → A): +100 kJ/mol_rxn. Multiplied by 2 (2 B → 2 A): +200 kJ/mol_rxn.

CED 6.9.B.2.ii · Read this in Fix

1. AYes. The same bonds change in both directions, so ΔH is −60 kJ/mol_rxn.
   A student who thinks reversing a reaction leaves ΔH unchanged picks this. Bonds formed in the forward reaction are broken in the reverse, so energy released one way is absorbed the other way.
2. BYes. Breaking the bonds in Y releases energy, so the reverse ΔH is negative.
   A student who thinks breaking bonds releases energy picks this. Breaking bonds requires energy; Y is the lower-enthalpy species, so converting it to X absorbs energy.
3. CNo. The reverse ΔH is positive but smaller in size, since energy was lost.
   A student who thinks some energy is lost each time a reaction occurs picks this. Energy is conserved: the reverse reaction absorbs exactly as much energy as the forward reaction releases, 60 kJ per mole of reaction.
4. DNo. Y is lower in enthalpy than X, so Y → X absorbs 60 kJ/mol_rxn. Correct
   The forward diagram shows Y 60 kJ/mol_rxn below X. Reversing a reaction keeps the magnitude of ΔH and changes its sign, so Y → X absorbs energy: ΔH = +60 kJ/mol_rxn. The student's diagram wrongly puts Y above X.

Working No calculation. Forward: Y is 60 kJ/mol_rxn below X. Reverse Y → X goes from the lower level to the higher one: ΔH = +60 kJ/mol_rxn. The student has placed Y above X and kept the negative sign, which is inconsistent.

CED 6.9.B.2.i · Read this in Fix

1. A−110.5 kJ/mol_rxn Correct
   The second equation is reversed and halved, CO₂(g) → CO(g) + ½ O₂(g), so its ΔH° becomes −½ × (−566.0) = +283.0 kJ/mol_rxn. Adding it to the first equation cancels CO₂(g) and ½ O₂(g): ΔH° = −393.5 + 283.0 = −110.5 kJ/mol_rxn.
2. B−676.5 kJ/mol_rxn
   A student who halves the second equation but does not change the sign of its ΔH° on reversing it picks this: −393.5 + (−283.0) = −676.5. Reversing an equation changes the sign of ΔH°.
3. C+172.5 kJ/mol_rxn
   A student who reverses the second equation but does not multiply its ΔH° by ½ picks this: −393.5 + 566.0 = +172.5. The equation must be halved to give one mole of CO, so its ΔH° is halved too.
4. D−959.5 kJ/mol_rxn
   A student who uses each ΔH° unchanged, however its equation has been rearranged, picks this: −393.5 + (−566.0) = −959.5. Both the reversal and the factor of ½ must be applied to the second ΔH°.

Working Keep equation 1 (−393.5). Reverse equation 2 and multiply it by ½: CO₂(g) → CO(g) + ½ O₂(g), ΔH° = +283.0. Add: C(s) + ½ O₂(g) → CO(g), ΔH° = −393.5 + 283.0 = −110.5 kJ/mol_rxn.

CED 6.9.B.2.iii · Read this in Fix

1. A−18.4 kJ
   A student who keeps the negative sign when the reaction is reversed picks this. Forming NH₃ releases energy, so decomposing it absorbs energy and ΔH is positive.
2. B+92.2 kJ
   A student who takes ΔH° as the enthalpy change whatever amount reacts picks this. +92.2 kJ is for 2 mol of NH₃; only 0.400 mol decomposes here.
3. C+18.4 kJ Correct
   Decomposition is the reverse of the given reaction, so ΔH° = +92.2 kJ per mole of reaction, which decomposes 2 mol of NH₃. 0.400 mol of NH₃ is 0.200 mol of reaction, so ΔH = 0.200 × 92.2 = +18.4 kJ.
4. D−92.2 kJ
   A student who uses the ΔH° value unchanged, whatever the direction and amount, picks this. Both the reversal (sign) and the amount (0.200 mol of reaction) must be taken into account.

Working Reverse: 2 NH₃(g) → N₂(g) + 3 H₂(g), ΔH° = +92.2 kJ/mol_rxn. One mole of reaction decomposes 2 mol NH₃, so 0.400 mol NH₃ = 0.200 mol_rxn. ΔH = 0.200 × (+92.2) = +18.4 kJ.

CED 6.9.B.2.i · Read this in Fix

1. A−420 kJ/mol_rxn
   A student who doubles reaction 2 but keeps the sign of its ΔH when reversing it picks this: −300 = ΔH₁ + 490 − 370 gives ΔH₁ = −420. Reversing reaction 2 makes its ΔH negative.
2. B+560 kJ/mol_rxn Correct
   To obtain the overall equation, reaction 2 is reversed and doubled (2 C + 2 D → 2 E, ΔH = −490) and added to reactions 1 and 3. So −300 = ΔH₁ − 490 − 370, and ΔH₁ = +560 kJ/mol_rxn.
3. C+315 kJ/mol_rxn
   A student who reverses reaction 2 but does not multiply its ΔH by 2 picks this: −300 = ΔH₁ − 245 − 370 gives +315. Reaction 2 must be doubled to cancel 2 C and 2 D, so its ΔH is doubled too.
4. D−175 kJ/mol_rxn
   A student who uses each ΔH unchanged, however its equation is rearranged, picks this: −300 = ΔH₁ + 245 − 370 gives −175. Reaction 2 must be both reversed and doubled.

Working Overall = (1) + 2 × reverse(2) + (3): A + B → 2 C; 2 C + 2 D → 2 E; 2 E → F. So ΔH(overall) = ΔH₁ − 2(+245) + (−370) = −300, giving ΔH₁ = −300 + 490 + 370 = +560 kJ/mol_rxn.

CED 6.9.B.2 · Read this in Fix

1. AHeat escaped from the calorimeter while MgO(s) reacted with HCl(aq).
   A student who adds ΔH₂ without changing its sign, although reaction 2 must be reversed, picks this. With the reversal, a less negative ΔH₂ makes −ΔH₂ smaller, so this error would make the result more negative, not less.
2. BThe temperature rise of the solution was smaller with Mg(s) than with MgO(s).
   A student who equates temperature change with energy transferred picks this. Each ΔH is calculated from q = mcΔT divided by the moles of solid that reacted, so a smaller temperature rise in one trial (for example, from a smaller sample) is not an error and does not change ΔH per mole.
3. CA larger volume of excess HCl(aq) was used with MgO(s) than with Mg(s).
   A student who thinks using more of an excess reagent gives a larger ΔH picks this. The heat is calculated from the mass of solution actually used and divided by the moles of solid, so ΔH₂ per mole is unchanged.
4. DHeat escaped from the calorimeter while Mg(s) reacted with HCl(aq). Correct
   ΔH°_f = ΔH₁ − ΔH₂ + ΔH₃. If heat escaped during reaction 1, less heat was measured and ΔH₁ came out less negative than its true value, which makes the calculated ΔH°_f less negative.

Working Target = (1) − (2) + (3): ΔH°_f = ΔH₁ − ΔH₂ + ΔH₃. Heat lost in reaction 1 makes the measured ΔH₁ less negative, so the result is less negative. Heat lost in reaction 2 makes ΔH₂ less negative, so −ΔH₂ is smaller and the result is more negative. The size of the temperature rise and the volume of excess acid do not change ΔH per mole.

CED 6.9.B.2.iii · Read this in Fix

1. A+2
   A student who thinks reversing a reaction leaves its ΔH unchanged picks this, doubling ΔH₁ but keeping its sign. SO₂(g) is a reactant in the target equation, so the first equation is reversed and its enthalpy change becomes −2ΔH₁.
2. B−2 Correct
   The first equation must be reversed, which changes the sign of its enthalpy change, and multiplied by 2, which multiplies its enthalpy change by 2. Adding the result to the second equation gives the target equation, so ΔH = ΔH₂ − 2ΔH₁ and n = −2.
3. C−1
   A student who thinks ΔH stays the same when an equation is multiplied by a factor picks this, reversing the sign of ΔH₁ but not doubling it. Two moles of SO₂(g) are needed, so the reversed equation and its enthalpy change are multiplied by 2.
4. D+1
   A student who uses each ΔH value unchanged, whatever the direction and coefficients of the equation, picks this, simply adding the two given values. The first equation must be reversed and doubled, which changes its enthalpy change to −2ΔH₁.

Working Reverse the first equation and multiply it by 2: 2 SO₂(g) → 2 S(s) + 2 O₂(g), ΔH = −2ΔH₁. Add the second equation: 2 S(s) + 3 O₂(g) → 2 SO₃(g), ΔH₂. The 2 S(s) cancel and 2 O₂(g) cancel, leaving 2 SO₂(g) + O₂(g) → 2 SO₃(g), so ΔH = ΔH₂ − 2ΔH₁ and n = −2.

CED 6.9.B.2.iii · Read this in Fix

1. AΔH° is twice as negative, and the heat released is twice as large
   A student who takes the ΔH value of an equation as the heat transferred whatever amount reacts picks this, reading 571.6 kJ as the heat released by the 1.00 mol of H₂(g). The 571.6 kJ is released when 2 mol of H₂(g) reacts, so 1.00 mol releases half of it.
2. BΔH° is unchanged, and the heat released is the same
   A student who thinks ΔH is a fixed per-mole value that stays the same when an equation is multiplied picks this. The heat released by 1.00 mol of H₂(g) is the same, but ΔH° for the doubled equation is doubled, because it describes twice as much reaction.
3. CΔH° is twice as negative, and the heat released is the same Correct
   Multiplying an equation by 2 multiplies its enthalpy change by 2, so ΔH° for equation II is −571.6 kJ/mol_rxn. That value is for 2 mol of H₂(g), so 1.00 mol of H₂(g) still releases 285.8 kJ.
4. DΔH° is unchanged, and the heat released is half as large
   A student who uses one ΔH value for a reaction whatever its coefficients picks this, assigning −285.8 kJ to equation II and so 142.9 kJ to each mole of its 2 mol of H₂(g). ΔH° for equation II is −571.6 kJ/mol_rxn, so 1.00 mol of H₂(g) releases 285.8 kJ.

Working Multiplying the equation by 2 multiplies ΔH° by 2: ΔH° = 2(−285.8) = −571.6 kJ/mol_rxn. One mole of reaction II consumes 2 mol H₂, so 1.00 mol H₂ releases 571.6/2 = 285.8 kJ, the same as from equation I.

CED 6.9.B.2.ii · Read this in Fix