Reactivity 1.2. Energy cycles in reactions

Hydrazine reacts with oxygen.

N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(l), ΔH^o = −623 kJ mol⁻¹

What is the standard enthalpy of formation of N₂H₄(l) in kJ mol⁻¹? The standard enthalpy of formation of H₂O(l) is –286 kJ mol⁻¹.

A. −623 − 286

B. −623 + 572

C. −572 + 623

D. −286 + 623

Answer: C. −572 + 623

Step 1: Write the enthalpy-of-formation expression 

ΔH_reaction = ∑ΔH_f(products) − ∑ΔH_f(reactants) 

So: 

−623 = [ΔH_f(N₂) + 2ΔH_f(H₂O)] − [ΔH_f(N₂H₄) + ΔH_f(O₂)] 

Since: 

ΔH_f(N₂) = 0, ΔH_f(N₂) = 0 

−623 = [0 + 2(−286)] − [ΔH_f(N₂H₄)] 

Step 2: Solve for ΔH_f(N₂H₄)

−623 = −572 − ΔH_f(N₂H₄)

ΔH_f(N₂H₄) = −572 + 623 = +51kJ mol⁻¹ 

Ethanol, C₂H₅OH, has many industrial uses.

a. Define the term average bond enthalpy.

b. Ethanol can be used as a fuel. Determine the enthalpy of combustion of ethanol at 298K, in kJ mol⁻¹, using the values given in section 13 of the IB Chemistry data booklet,assuming all reactants and products are gaseous.

c. Students can also measure the enthalpy of combustion of ethanol in a laboratory using calorimetry. Suggest the major source of systematic error in these procedures.

d. The standard enthalpy change of combustion, `""ΔH_c^o`, of propanoic acid is –1527 kJ mol⁻¹. Determine the standard enthalpy change of formation of propanoic acid, in kJ mol⁻¹, using this information and data from section 13 of the IB Chemistry data booklet.

a. Define average bond enthalpy 

The energy required to break one mole of a given bond in the gas phase, averaged over  that bond in many different molecules. 

b. ΔH° for combustion of ethanol at 298 K (all species gaseous) 

Reaction (gas phase): 

C₂H₅OH(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)

Use standard enthalpies of formation (kJ mol⁻¹, IB Data Booklet):

• `""ΔH_f^o`[CO₂(g)] = −393.5

• `""ΔH_f^o`[H₂O(g)] = −241.8

• `""ΔH_f^o`[C₂H₅OH(g)] ≈ −235

ΔH°_comb = [2(−393.5) + 3(−241.8)] − [(−235) + 0] 

= (−1512.4) − (−235) ≈ −1.28 × 103 kJ mol⁻¹

c. Major source of systematic error in calorimetry

Heat loss to the surroundings (and to the calorimeter) during burning, so the measured ΔH is less exothermic than the true value. (Incomplete combustion is another source, but heat loss is the main systematic error.)

d. ΔHf° of propanoic acid, given `""ΔH_c^o`(C₃H₆O₂(l)) = -1527 kJ mol⁻¹

Combustion:

C₃H₆O₂(l) + O₂(g) → 3CO₂(g) + 3H₂O(l)

Data (kJ mol−1):

• `""ΔH_f^o`[CO₂(g)] = −393.5

• `""ΔH_f^o`[H₂O(l)] = −285.8

−1527 = [3(−393.5) + 3(−285.8)] − `""ΔH_f^o`[C₃H₆O₂(l)]

−1527 = −2037.9 − `""ΔH_f^o`

 `""ΔH_f^o` ≈ −5.11 × 102 kJ mol−1 (≈ −511 kJ mol−1)

Consider the following equations:

2Al(s) + O₂(g) → Al₂O₃(s), ΔH^o = −1670 kJ mol−1

Mn(s) + O₂(g) → MnO₂(s), ΔH^o = −520 kJ mol−1

What is the standard enthalpy change, in kJ mol−1, of the reaction below?

4Al(s) + 3MnO₂(s) → 2Al₂O₃(s) + 3Mn(s)

A. –1670 + 520

B. (`frac{3}{2}`× −1670) + (3 × 520)

C. (2 × −1670) + (3 × −520)

D. (2 × −1670) − (3 × −520)

Step 1: Observe oxidation/reduction

 • In the target equation, MnO₂ → Mn, so the second reaction must be reversed. Reversing changes the sign of ΔH° to +520 kJ mol⁻¹. 

• The first reaction stays as written, but must be doubled because the final reaction has 4 Al and 2 Al₂O₃.

Step 2: Multiply and reverse equations

(1) × 2 → 4Al(s) + 3O₂(g) → 2Al₂O₃(s)

ΔH = 2 × (−1670) = −3340 kJ

(2) reversed × 3 → 3MnO₂(s) → 3Mn(s) + 3O₂(g)

ΔH = 3 × (+520) = +1560 kJ

Step 3: Add them together

4Al(s) + 3O₂(g) + 3MnO₂(s) → 2Al₂O₃(s) + 3Mn(s) + 3O₂(g)

→ The 3O₂(g) cancels on both sides

Final reaction:

4Al(s) + 3MnO₂(s) → 2Al₂O₃(s) + 3Mn(s)

Step 4: Add enthalpy changes

ΔH° = (2 × −1670) + (3 × +520)

Which ionic compound has the largest value of lattice enthalpy? 

A. MgS. 

B. MgO. 

C. CaBr₂. 

D. NaF.

Answer: B. MgO. 

Lattice enthalpy depends on:

U `prop``frac{Q_1Q_2}{r}`

where 

• Q₁,Q₂ = ionic charges 

• r = distance between ion centers (sum of ionic radii) 

So lattice enthalpy increases when: 

1. Ion charges are higher, and 

2. Ions are smaller (shorter distance between them). 

A. Incorrect: MgS 

Ions: Mg²⁺ and S²⁻ → both have charge ±2, which gives strong electrostatic attraction. However, sulfide ion (S²⁻) is much larger than oxide (O²⁻), which increases the distance  between ions and weakens the attraction. Therefore, lattice enthalpy is high but less than  MgO. 

B. Correct: MgO 

Ions: Mg²⁺ and O²⁻ → both have ± 2 charges. Oxygen ion is smaller than sulfur ion, so  ions are closer together. Strong charge + small distance = very strong electrostatic  attraction. Among all given options, this combination produces the highest lattice  enthalpy. 

C. Incorrect: CaBr₂ 

Ions: Ca²⁺ and Br⁻. Cation has high charge (+2), but bromide ions are large, increasing  distance. Also, there are two Br⁻ ions, but the overall electrostatic attraction per ion pair  is smaller due to the large radius of Br⁻. Lattice enthalpy is therefore lower than MgO  or MgS. 

D. Incorrect: NaF 

Ions: Na⁺ and F⁻ → both ±1 charges only. Both ions are small, so the distance is short,  which helps, but the charge product (1 × 1 = 1) is much lower than that of ±2 ions (4). Hence, the overall lattice enthalpy is moderate, smaller than MgO’s. 

The Born–Haber cycle for the formation of magnesium oxide is shown below.

What is a correct description of the steps X, Y, and Z in this cycle? 

Answer: C.
Step X:
Goes from Mg(g) + O(g) → Mg²⁺(g) + 2e⁻ + O(g)

→ This step involves removing two electrons from magnesium — that’s the sum of the 1st and 2nd ionization energies of Mg.

`=>` Step X = Sum of the 1st and 2nd ionization energies of Mg

Step Y:

Goes from Mg²⁺(g) + O²⁻(g) → MgO(s)

→ This is the process where gaseous ions form the solid lattice.

`=>` Step Y = Lattice enthalpy of MgO

Step Z:

Goes from Mg(s) + `frac{1}{2}`O2(g) → MgO(s)

→ This is the overall formation reaction, the enthalpy of formation of MgO.

`=>` Step Z = Enthalpy of formation of MgO

a. Two chemistry students wished to determine the enthalpy of hydration of anhydrous magnesium sulfate. They measured the initial and the highest temperature reached when anhydrous magnesium sulfate, MgSO4 (s), was dissolved in water. They presented their results in this table:

i. Calculate the amount, in mol, of anhydrous magnesium sulfate.
ii. Calculate the enthalpy change, ΔH₁, for anhydrous magnesium sulfate dissolving in water, in kJ mol⁻¹. State your answer to the correct number of significant figures.

b. The students repeated the experiment using 6.16 g of solid hydrated magnesium sulfate, MgSO4⋅7H₂O(s), and 50 cm³ of water. They found the enthalpy change, ΔH₂, to be +18 kJ mol⁻¹. The enthalpy of hydration of solid magnesium sulfate is difficult to determine experimentally, but can be determined using the diagram below:

i. Determine the enthalpy change, ΔH, in kJ mol⁻¹, for the hydration of solid anhydrous magnesium sulfate, MgSO4.

ii. The literature value for the enthalpy of hydration of anhydrous magnesium sulfate is –103 kJ mol⁻¹. Calculate the percentage difference between the literature value and the value determined from the experimental result, giving your answer to one decimal place.

a.i. Moles of anhydrous MgSO₄ 

M_r(MgSO₄) = 24.3 + 32.1 + 4 × 16.0 = 120.4 g mol⁻¹ 

n = `frac{3.01}{120.4}`3.01 = 2.50 × 10⁻² mol 

Answer: 2.50 × 10⁻² mol (3 s.f.) 

a.ii. Enthalpy change for dissolving anhydrous MgSO₄, ΔH₁

ΔT = 26.7 − 17.0 = 9.7 K

Take the mass of water = 50.0 g(density ≈ 1 g cm⁻³) and c = 4.18 J g⁻¹K^−1.

Heat released to the water:

q = mcΔT = 50.0 × 4.18 × 9.7 = 2.03 × 103

J = 2.03 kJ

Molar enthalpy (exothermic):

ΔH₁ = `-frac{q}{n}`= `-frac{2.03}{2.50*10^(-2)}`= −81.2 kJ mol⁻¹ 

Answer: ΔH1 ≈ −81 kJ mol⁻¹ (2 s.f. from ΔT)

b.i. Enthalpy change for hydration of solid anhydrous MgSO4, ΔH

From the Hess diagram: 

MgSO₄(s) + 7H₂O(l)MgSO₄`*`7H₂O(s) 

and both solids dissolve to the same ions:

ΔH₁ = ΔH + ΔH₂

with ΔH₂ = +18 kJ mol⁻¹.

→ ΔH = ΔH₁ − ΔH₂ ≈ (−81) − (+18) = −99 kJ mol⁻¹

b.ii. % difference vs. literature value −103 kJ mol⁻¹

%difference = `frac{∣ −99 − (−103) ∣}{103}*100 = frac{4}{103}*100`= 3.9%

Which value represents the lattice enthalpy, in kJ mol^-1, of strontium chloride, SrCl₂?

A. −(−829) + 164 + 243 + 550 + 1064 − (−698) 

B. −829 + 164 + 243 + 550 + 1064 − 698 

C. −(−829) + 164 + 243 + 550 + 1064 − 698 

D. −829 + 164 + 243 + 550 + 1064 − (−698) 

Answer: C. −(−829) + 164 + 243 + 550 + 1064 − 698

−829 = 164 + 243 + 550 + 1064 − 698 + U

U = −(−829) + 164 + 243 + 550 + 1064 − 698

The Born–Haber cycle for potassium oxide is shown below:

Which expression represents the lattice enthalpy, in kJ mol⁻¹?

A. −361 + 428 + 838 + 612

B. −(−361) + 428 + 838 + 612

C. −361 + 428 + 838 − 612

D. −(−361) + 428 + 838 − 612

Answer: B. −(−361) + 428 + 838 + 612

Using Hess’s Law

The formation enthalpy (−361 kJ mol⁻¹) = sum of all steps leading to the gaseous ions plus the lattice enthalpy.

−361 = 428 + 838 + 612 + U

Rearranging for U:

U = −(−361) + 428 + 838 + 612

What is the enthalpy change of the reaction? 

C₆H₁₄(l) → C₂H₄(g) + C₄H₁₀(g)

A. +1411 + 2878 + 4163 

B. +1411 − 2878 − 4163 

C. +1411 + 2878 − 4163 

D. −1411 − 2878 + 4163 

Answer: C. +1411 + 2878 − 4163 

Reasoning (Hess via combustions): 

ΔH_rxn = ΔH_c(reactant) − ∑ΔH_c(products) = (−4163) − [(−1411) + (−2878)]

What is the enthalpy of combustion of urea, (NH₂)₂CO, in kJ mol⁻¹?

CO(NH₂)₂(s) +`frac{3}{2}`O₂(g) → CO₂(g) + N₂(g) + 2H₂O(l)

A. (−333) − (−394) − 2 × (−286) 

B. [−394 + 2 × (−286) − `frac{3}{2}`× (−333)] 

C. (−394) + 2 × (−286) + (−333) 

D. (−394) + 2 × (−286) − (−333) 

Answer: D. (−394) + 2 × (−286) − (−333) 

Step 1. Formula 

ΔH_reaction = ∑ΔH_f(products) − ∑ΔH_f(reactants) 

Step 2. Apply data 

ΔH = [(−394) + 2(−286) + 0] − [(−333) + 0] 

What is the enthalpy change, in kJ mol⁻¹, of the following reaction? 3H2(g) + N2(g) ⇌ 2NH3(g) 

A.(6 × 391) − [(3 × 436) + 945] 

B. (3 × 391) − (436 + 945) 

C. −[(3 × 436) + 945] + (3 × 391) 

D. −(6 × 391) + [(3 × 436) + 945] 

Answer: A. (6 × 391) − [(3 × 436) + 945] 

Step 1. Bonds broken (energy absorbed) 

   • 3 × H–H = 3 × 436 = 1308 kJ 

   • 1 × N≡N = 945 kJ 

Total energy absorbed = 1308 + 945 = 2253 kJ 

Step 2. Bonds formed (energy released) 

Each NH₃ molecule has 3 N–H bonds, and 2 NH₃ are formed: 

6 × 391 = 2346 kJ released 

Step 3. Enthalpy change 

 ΔH = energy absorbed − energy released 

 ΔH = (3 × 436 + 945) − (6 × 391) 

 ΔH = 2253 − 2346 = −93 kJ mol⁻¹ 

⇒ Correct expression: 

(6 × 391) − [(3 × 436) + 945] 

a. Define the term average bond enthalpy. 

b. Use the information from Table 10 in the Data Booklet to calculate the enthalpy change for the complete combustion of but-1-ene according to the following equation.

C₄H₈(g) + 6O₂(g) → 4CO₂(g) + 4H₂O(g)

c. Predict, giving a reason, how the enthalpy change for the complete combustion of but 2-ene would compare with that of but-1-ene based on average bond enthalpies. 

a. Define average bond enthalpy 

The average bond enthalpy is the energy required to break one mole of a specified bond in gaseous molecules, averaged over that bond in many different compounds (so it’s an  average value, not compound-specific). 

b. ΔH for the complete combustion of but-1-ene 

C₄H₈(g) + 6O₂(g) → 4CO₂(g) + 4H₂O(g)

Count bonds: 

   • Reactants: 

     o But-1-ene (CH₂=CH–CH₂–CH₃): 1 × C=C, 2 × C–C, 8 × C–H 

     o Oxygen: 6 × O=O 

   • Products: 

     o 4 CO₂: 8 × C=O (in CO₂) 

     o 4 H₂O(g): 8 × O–H 

Use average bond enthalpies (kJ mol^-1): 

C=C: 614, C–C: 348, C–H: 413, O=O: 498, C=O (in CO₂): 805, O–H: 463.

Energy broken = 1(614) + 2(348) + 8(413) + 6(498) 

                        = 614 + 696 + 3304 + 2988 = 7602 kJ 

Energy formed = 8(805) + 8(463) = 6440 + 3704 = 10144 kJ

ΔH ≈ ΣH_broken − ΣH_formed = 7602 − 10144 = −2.54 × 103 kJ mol⁻¹

c. Compare with but-2-ene 

On a bond-enthalpy basis, but-2-ene has the same numbers of each bond type as but-1-ene (1 C=C, 2 C–C, 8 C–H), so the calculated combustion enthalpy would be essentially  the same.

Reason: Average bond enthalpy depends on bond type and count, not on isomer  arrangement. (In reality there may be a small difference, but the bond-enthalpy method  will not capture it.) 

The standard enthalpy changes for the following reactions can be found in Table 13 of the Data Booklet: 

C(s) + O₂(g) → CO₂(g)

H₂(g) +O₂(g) → H₂O(l)

C₈H₁₈(l) + `12*frac{1}{2}`O₂(g) → 8CO₂(g) + 9H₂O(l)

a. Use this information to determine the standard enthalpy change for the formation of octane from its elements. 

8C(s) + 9H₂(g) → C₈H₁₈(l)

b. Predict which of the following reactions has the most negative enthalpy change, and explain your choice. 

I. H₂(g) + `frac{1}{2}`O₂(g) → H₂O(g)

II. H₂(g) + `frac{1}{2}`O₂(g) → H₂O(l)

a. Step 1: Write an energy cycle using Hess’s law

Formation = [combustion of elements] – [combustion of compound]

`"ΔH_f^o` = [8ΔH(C) + 9ΔH(H₂)] − [ΔH(C₈H₁₈)]

Step 2: Substitute values

`"ΔH_f^o` = [8(−394) + 9(−286)] − (−5470) = [−3152 − 2574] + 5470

= −5726 + 5470 = −256 kJ mol⁻¹

b. Predict which of the following has the most negative enthalpy change

Water (l) formation releases more energy because of extra hydrogen bonding/condensation energy when gas condenses to liquid. 

→ Reaction II has the most negative (most exothermic) enthalpy change. (Formation of  liquid water releases additional energy due to intermolecular forces.) 

Two students were asked to use information from the Data Booklet to calculate a value  for the enthalpy of hydrogenation of ethene to form ethane. 

C₂H₄(g) + H₂(g) → C₂H₆(g)

John used the average bond enthalpies from Table 10. Marit used the values of  enthalpies of combustion from Table 12.

a. Calculate the value for the enthalpy of hydrogenation of ethene obtained using the  average bond enthalpies given in Table 10. 

b. Marit arranged the values she found in Table 12 into an energy cycle.

Calculate the value for the enthalpy of hydrogenation of ethene from the energy cycle.

c. Suggest one reason why John’s answer is slightly less accurate than Marit’s answer.

d. John then decided to determine the enthalpy of hydrogenation of cyclohexene to  produce cyclohexane. 

C₆H₁₀(l) + H₂(g) → C₆H₁₂(l)

  i. Use the average bond enthalpies to deduce a value for the enthalpy of hydrogenation  of cyclohexene. 

  ii. The percentage difference between these two methods (average bond enthalpies and  enthalpies of combustion) is greater for cyclohexene than it was for ethene. John’s  hypothesis was that it would be the same. Determine why the use of average bond  enthalpies is less accurate for the cyclohexene equation shown above than it was for  ethene. Deduce what extra information is needed to provide a more accurate answer. 

a. Using average bond enthalpies (Table 10) for 

C₂H₄(g) + H₂(g) → C₂H₆(g)

Bonds broken: 

1 × C=C (614) + 1 × H–H (436) = 1050 kJ mol^-1 

Bonds formed: 

1 × C–C (348) + 2 × C–H (2×413 = 826) = 1174 kJ mol^-1

ΔH = ΣH_broken − ΣH_formed = 1050 − 1174 = −1.24 × 102 kJ mol⁻¹ So the hydrogenation is exothermic (~–124 kJ mol^-1). 

b. Using the enthalpies of combustion (Table 12) from Marit’s cycle From the diagram (standard values): 

• ΔH_c (C₂H₄) = −1411 kJ mol⁻¹ 

• ΔH_c(H₂) = −286 kJ mol⁻¹ 

• ΔH_c (C₂H₆) = −1560 kJ mol⁻¹ 

By Hess’ law: 

ΔH_hydrog = [(−1411) + (−286)] − [(−1560)] = −1697 + 1560 = −1.37 × 102 kJ mol⁻¹ 

c. Why John’s answer is slightly less accurate than Marit’s 

Average bond enthalpies are mean gas-phase values taken over many molecules, so they  ignore the actual molecular environment and states. Enthalpies of combustion are compound-specific experimental data, so they give a more accurate ΔH.

d. Hydrogenation of cyclohexene → cyclohexane

C₆H₁₀(l) + H₂(g) → C₆H₁₂(l)

i. Estimate with average bond enthalpies 

The reaction changes one C=C to one C–C and adds two C–H (same local change as for  ethene): 

ΔH ≈ −124 kJ mol⁻¹(exothermic) 

ii. Why the % difference between methods is larger here, and what extra data are needed

   • Average bond enthalpies apply to gas-phase, average bonds and do not account  for ring strain/conformational effects in cyclohexene/cyclohexane.

   • The equation involves liquids, so using gas-phase bond enthalpies neglects  intermolecular forces; you would need to add enthalpies of vaporization of cyclohexene and cyclohexane (and of H₂ if applicable) to compare properly.

   • Also, the C=C bond enthalpy in a ring can differ from the average value.

Therefore: To obtain a more accurate answer, you need compound-specific thermochemical data – e.g. standard enthalpies of combustion (as in part b), or, if using  bond enthalpies, the enthalpies of vaporization to convert liquids to gases and bond  enthalpies specific to the cyclic double bond. 

In December 2010, researchers in Sweden announced the synthesis of N,N dinitronitramide, N(NO₂)₃. They speculated that this compound, more commonly called trinitramide, may have significant potential as an environmentally friendly rocket fuel  oxidant. 

a. Methanol reacts with trinitramide to form nitrogen, carbon dioxide, and water. Deduce  the coefficients required to balance the equation for this reaction. 

N(NO₂)₃(g) + CH₃OH(l) → N₂(g) + CO₂(g) + H₂O(l) 

b. Suggest one reason why trinitramide might be more environmentally friendly than  other rocket fuel oxidants such as ammonium perchlorate (NH₄ClO₄).

c. Calculate the enthalpy change, in kJ mol^-1, when one mole of trinitramide decomposes  to its elements, using bond enthalpies from section 12 of the data booklet. Assume that  all the N–O bonds in this molecule have a bond enthalpy of 305 kJ mol^-1. 

a. Balance the reaction with methanol 

Let 

aN(NO₂)₃(g) + bCH₃OH(l) → cN₂(g) + dCO₂(g) + eH₂O(l) 

Atom balances give:

N: 4a = 2c ⇒ c = 2a
C: b = d
H: 4b = 2e ⇒ e = 2b
O: 6a + b = 2d + e = 2b + 2b = 4b ⇒ b = 2a

Choose a = 1:

N(NO₂)₃(g) + 2CH₃OH(l) → 2N₂(g) + 2CO₂(g) + 4H₂O(l) 

b. Why trinitramide may be more environmentally friendly than ammonium perchlorate

• Combustion of trinitramide produces only N₂, CO₂, and H₂O (no chlorine).

• Ammonium perchlorate releases HCl/Cl-containing species, which are corrosive  

and environmentally harmful (e.g., ozone depletion, acid deposition). → Trinitramide yields no chlorine-containing toxic gases. 

c. Enthalpy change for decomposition of 1 mol of trinitramide to its elements Formula of trinitramide: N₄O₆ 

Assume all N–O bonds = 305 kJ mol⁻¹ (given). 

Use IB bond enthalpies for the rest (typical values): 

• N–N (single) ≈ 163 kJ mol⁻¹ 

• N≡N = 945 kJ mol⁻¹ 

• O=O = 498 kJ mol⁻¹ 

Bonds broken in N(NO₂)₃: 

• 6 × N–O: 6 × 305 = 1830 kJ

• 3 × N–N: 3 × 163 = 489 kJ

Total broken = 2319 kJ

Methanol can be produced according to the following equation:

CO(g) + 2H₂(g) → CH₃OH

Calculate the standard enthalpy change of this reaction using the following data: 

Use Hess’ law with the given combustion enthalpies. 

The idea: enthalpy of reaction = (sum of combustion enthalpies of reactants) – (combustion enthalpy of product). 

From the data: 

• (II) 2CO + O₂ → 2CO₂, ΔH^o = −566 kJ

→ per 1 CO: –283 kJ

• (III) 2H₂ + O₂ → 2H₂O(l), ΔH^o = −572 kJ

→ for 2 H₂: –572 kJ

• (I) 2CH₃OH(l) + 3O₂ → 2CO₂ + 4H₂O(l), ΔH^o = −1452 kJ

→ per 1 CH₃OH: –726 kJ

Now apply Hess:

ΔH^o = [(−283) + (−572)] − [(−726)] = −855 + 726 = −129 kJ mol⁻¹ 

In the gas phase, A reacts with hydrogen to form D. 

 a. Use bond enthalpies given in section 12 of the IB Chemistry data booklet to determine  the enthalpy change, in kJ mol^-1, of the reaction. State whether the reaction is exothermic  or endothermic. 

b. The standard enthalpy change of combustion of A is –4000 kJ mol^-1. Calculate the amount of A, in mol, that would have to be burned to raise the temperature of 1 dm³ of  water from 20°C to 100°C.

a. Enthalpy change for the hydrogenation A → D 

Treat it with average bond enthalpies (IB data booklet): 

• Bonds broken: 1 × C=C (614), 1 × H–H (436) 

ΣE_broken = 614 + 436 = 1050 kJ mol⁻¹ 

• Bonds formed: 1 × C–C (348), 2 × C–H (2 × 413 = 826) 

ΣE_formed = 348 + 826 = 1174 kJ mol⁻¹ 

ΔH = ΣE_broken − ΣE_formed = 1050 − 1174 = −1.2 × 102 kJ mol⁻¹ So the reaction is exothermic (~ –120 kJ mol⁻¹). 

b. Moles of A needed to heat 1 dm³ of water from 20°C to 100°C 

Heat needed for water: 

q = mcΔT = (1000 g)(4.18 g⁻¹K⁻¹)(80 K) = 3.344 × 105 = 334.4 kJ ∘

Given ΔH_comb = (of A) = −4000 kJ mol⁻¹

n(A) = `frac{q}{∣ ΔH_"comb" ∣}` = `frac{334.4}{4000}`= 8.4 × 10⁻² mol 

In which order does the oxygen–oxygen bond enthalpy increase? 

A. H₂O₂ < O₂ < O₃ 

B. H₂O₂ < O₃ < O₂ 

C. O₂ < O₃ < H₂O₂ 

D. O₃ < H₂O₂ < O₂ 

Answer: B. H₂O₂ < O₃ < O₂ 

Bond enthalpy trend 

• Higher bond order → stronger bond → higher bond enthalpy. 

⇒ Single < Resonance (1.5) < Double 

Order of increasing bond enthalpy

H₂O₂ < O₃  < O₂  

Which combination will give you the enthalpy change for the hydrogenation of ethene  to ethane, ΔH_1?

A. ΔH₂ + ΔH₁ - ΔH₄

B. ΔH₂ - ΔH₄ + ΔH₁

C. -ΔH₂ + ΔH₁ - ΔH₄

D. -ΔH₂ - ΔH₁ + ΔH₄

Answer: C. -ΔH₂ + ΔH₁ - ΔH₄

Use Hess’ law around the cycle. 

Top path (left → right): 

Use Hess’ law around the cycle. 

Top path (left → right): 

ΔH₂ + ΔH₃ 

Bottom path (left → bottom → right): 

ΔH₁ + (−ΔH₄) 

(We use –ΔH₄ because going from bottom to right is the reverse of the arrow labeled  ΔH₄.) 

Equate paths: 

ΔH₂ + ΔH₃ = ΔH₁ − ΔH₄ 

⇒ΔH₃ = −ΔH₂ + ΔH₁ − ΔH₄ 

The C=C bond has a bond length of 134 pm and an average bond enthalpy of 614 kJ  mol⁻¹. Which values would be most likely for the C–C bond? 

A. 154 pm, 346 kJ mol^-1

B. 154 pm, 780 kJ mol^-1

C. 116 pm, 346 kJ mol^-1 

D. 116 pm, 780 kJ mol^-1

Answer: A. 154 pm, 346 kJ mol^-1

Trends in bond length and strength: 

• More bonds (multiple bonds) → shorter and stronger. 

• Fewer bonds (single bonds) → longer and weaker. 

So for a C–C single bond, compared to C=C: 

• Bond length → longer than 134 pm 

• Bond enthalpy → smaller than 614 kJ mol⁻¹