Bond Enthalpy Calculations

Energetics Worksheet

Show All Answers

Practice Questions

Mean Bond Dissociation Enthalpy (ΔH_diss^θ) / kJ mol^-1
C-C: 348	C=C: 612	C-H: 412	O=O: 498	O-H: 464
C=O: 740	C-O: 347	H-F: 562	C-H (alkyne): 460	C≡C: 837
S-F: 309	Xe-Xe: 0	F-F: 158	H-H: 436	Br-Br: 193
S=O: 452

1. Define the term mean bond dissociation enthalpy.

The bond dissociation enthalpy is the enthalpy change when breaking one mole of a particular bond to give separated atoms – everything being in the gas state.

2. State which bond is strongest in each pair, and explain why that is the case.

a) C-C; C=C

C=C is stronger due to having a sigma bond and a pi bond.

b) C-Cl; C-Br

C-Cl is stronger than C-Br because Cl has fewer shells than Br. This results in a shorter bond length and therefore stronger electrostatic force of attraction.

3. Calculate the standard enthalpy change of combustion of butanol.

C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O

Bonds Broken (Reactants)

3(C-C) + 9(C-H) + 1(C-O) + 1(O-H) + 6(O=O)
3(348) + 9(412) + 347 + 464 + 6(498) = +8551 kJ

Bonds Formed (Products)

8(C=O) + 10(O-H)
8(740) + 10(464) = -10560 kJ

Answer

+8551 – 10560 = -2009 kJ mol^-1

4. Propose the products of the reaction between hydrogen fluoride and ethyne (C₂H₂). The enthalpy change of this reaction is -179kJmol^-1. Calculate the bond dissociation enthalpy of C-F.

C₂H₂ + 2HF → CH₃CHF₂

Equation

ΔH = Σ(Bonds Broken) – Σ(Bonds Formed)
-179 = [1(C≡C) + 2(H-F) + 2(C-H)] – [1(C-C) + 2(C-F) + 4(C-H)]

Substitution

-179 = [837 + 2(562) + 2(412)] – [348 + 2x + 4(412)]
-179 = [2785] – [1996 + 2x]

Answer

2x = 2785 – 1996 + 179 = 968
x = 484 kJ mol^-1

5. Calculate the bond enthalpy of Xe-F given that when one mole of Xenon releases 52kJmol^-1 when it reacts with excess F_2.

Xe + 2F₂ → XeF₄

Calculation

ΔH = ΣBroken – ΣFormed
-52 = [2(F-F)] – [4(Xe-F)]
-52 = [2(158)] – 4x

Answer

4x = 316 + 52 = 368
x = 92 kJ mol^-1

6. Calculate the bond dissociation energy of C-Br given that the enthalpy change of the reaction between ethene and bromine is -64kJmol^-1.

C₂H₄ + Br₂ → C₂H₄Br₂

Calculation

-64 = [1(C=C) + 1(Br-Br)] – [1(C-C) + 2(C-Br)]
-64 = [612 + 193] – [348 + 2x]

Answer

-64 = 805 – 348 – 2x
2x = 457 + 64 = 521
x = 260.5 kJ mol^-1

7. Calculate the bond strength of the S-O bond in sulfuric acid using the enthalpy change of -195kJmol^-1 for the reaction between sulfur hexafluoride and water to producing sulfuric acid and hydrogen fluoride.

SF₆ + 4H₂O → H₂SO₄ + 6HF

Calculation

-195 = [6(S-F) + 8(O-H)] – [2(S=O) + 2x + 2(O-H) + 6(H-F)]
-195 = [6(309) + 8(464)] – [2(452) + 2x + 2(464) + 6(562)]
-195 = 5566 – [5204 + 2x]

Answer

-195 = 362 – 2x
2x = 557
x = 278.5 kJ mol^-1

8. Calculate the Standard Enthalpy of Combustion (ΔH_c^θ) of Benzene (C₆H₆). You cannot calculate this directly using mean bond enthalpies for benzene because of its delocalized ring structure. Instead, you must use a Square Hess Cycle linking benzene to Cyclohexane, Gaseous Atoms, and Combustion Products.

Data Provided:
Enthalpy of Hydrogenation of Benzene:
C₆H₆(l) + 3H₂(g) → C₆H₁₂(l) ΔH_hydro = -208 kJ mol^-1

Step 1: Calculate ΔH_c(Cyclohexane)

Reaction: C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O
Break: 6(348) + 12(412) + 9(498) = 11514 kJ
Form: 12(740) + 12(464) = 14448 kJ
ΔH_c(Cyclo) = 11514 – 14448 = -2934 kJ mol^-1

Step 2: Calculate ΔH_c(Hydrogen)

Reaction: H₂ + 0.5O₂ → H₂O
Break: 436 + 0.5(498) = 685 kJ
Form: 2(464) = 928 kJ
ΔH_c(H₂) = 685 – 928 = -243 kJ mol^-1

Step 3: Hess’s Law Calculation

Path 1 = Path 2
ΔH_c(Benzene) + 3ΔH_c(H₂) = ΔH_hydro + ΔH_c(Cyclo)
ΔH_c(Benzene) + 3(-243) = -208 + (-2934)
ΔH_c(Benzene) – 729 = -3142
ΔH_c(Benzene) = -3142 + 729 = -2413 kJ mol^-1

Mr Cole Chemistry

Bond Enthalpy Questions

Bond Enthalpy Calculations

Practice Questions