Questions on mean bond enthalpies

Question 1

Evaluate the use of mean bond enthalpies, rather than standard enthalpies of formation in a Hess cycle, to find the enthalpy change of a reaction.

Accepted Answer

Indicative content

bond enthalpies need only the structures of the molecules, so they give an estimate for compounds whose enthalpies of formation have never been measured
the values are means over many compounds, not the actual bond energies in the molecules reacting, so the answer is only approximate
bond enthalpies apply only to gases, so a reaction with a liquid or solid in its standard state needs a change-of-state correction
where every bond is in a diatomic molecule, as in hydrogen with chlorine, the values are exact and the agreement with the data book is close
a Hess cycle uses measured enthalpies of formation of the actual compounds in their standard states, so it is more accurate where those data exist
bond enthalpies are therefore best used as an estimate, usually within about ten per cent for gases, with a formation cycle preferred whenever its data are available

Levels

Level 3 (5–6 marks) both sides are weighed: the usefulness of mean bond enthalpies where formation data do not exist, their inaccuracy from being averages and applying to gases only, the exact case of diatomic molecules, and the greater accuracy of a formation cycle, ending with a justified view of when each should be used
Level 2 (3–4 marks) the mean-value and gas-phase limits are given and the formation cycle called more accurate; the case for bond enthalpies, or when each method should be used, is not made
Level 1 (1–2 marks) bond enthalpies are called approximate with no reason, or more accurate because each bond is measured separately

Question 2

Calculate ΔH for C₂H₄(g) + Br₂(g) → C₂H₄Br₂(g), counting only the bonds that change, using the mean bond enthalpies in the table.

Accepted Answer

M1 bonds broken: one C=C and one Br–Br, 612 + 193 = 805 M1 bonds made: one C–C and two C–Br, 347 + 2 × 290 = 927; the four C–H bonds are unchanged and may be left out M1 ΔH = bonds broken − bonds made = 805 − 927 A1 ΔH = −122 kJ mol⁻¹ to three significant figures; +122 kJ mol⁻¹, from reversing the subtraction, scores the first two marks only

Question 3

Calculate ΔH for C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g) from the mean bond enthalpies in the table.

Accepted Answer

M1 bonds broken: 2 C–C + 8 C–H + 5 O=O = 694 + 3304 + 2490 = 6488 M1 bonds made: 6 C=O + 8 O–H = 4830 + 3712 = 8542 M1 ΔH = bonds broken − bonds made = 6488 − 8542 A1 ΔH = −2054 kJ mol⁻¹, to four significant figures; allow error carried forward from one miscounted bond

Question 4

Calculate the mean N–H bond enthalpy from ½N₂(g) + 1½H₂(g) → NH₃(g), ΔH = −46 kJ mol⁻¹. Bond enthalpies / kJ mol⁻¹: N≡N 945, H–H 436.

Accepted Answer

M1 bonds broken: ½ × 945 + 1½ × 436 = 472.5 + 654 = 1126.5 M1 bonds made: 3 N–H, so −46 = 1126.5 − 3(N–H) M1 3(N–H) = 1126.5 + 46 = 1172.5 A1 N–H = +391 kJ mol⁻¹ to three significant figures

Question 5

Explain, in terms of bonds, why the combustion of methane is exothermic, stating the sign of the enthalpy change for breaking a bond and for making one.

Accepted Answer

B1 breaking a bond has a positive enthalpy change: energy is taken in to overcome the attraction between the shared pair and the nuclei B1 making a bond has a negative enthalpy change, releasing the same energy B1 energy is taken in to break the C–H and O=O bonds in the reactants and released when the C=O and O–H bonds in the products form B1 more energy is released making the product bonds than is taken in breaking the reactant bonds, so heat is released overall; reject 'breaking bonds releases energy'

Question 6

Compare a bond dissociation enthalpy with a mean bond enthalpy, explaining why the C–H value in a data table is a mean.

Accepted Answer

B1 both are the enthalpy change to break one mole of a bond with everything in the gaseous state, and both are positive B1 a bond dissociation enthalpy refers to one specified bond in one particular molecule B1 a mean bond enthalpy is averaged over that type of bond in a range of compounds, so for a given molecule it is an approximation B1 the energy to break a C–H bond depends on the other atoms attached to the carbon, and even differs between the successive bonds in methane, so the table gives an average

Question 7

Show that the enthalpy change for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), calculated from the mean bond enthalpies in the table, is −818 kJ mol⁻¹, and explain why the data-book value of ΔcH° for methane, −890 kJ mol⁻¹, differs from it.

Accepted Answer

M1 bonds broken: 4 × 413 + 2 × 498 = 1652 + 996 = 2648 kJ mol⁻¹ M1 bonds made: 2 × 805 + 4 × 464 = 1610 + 1856 = 3466 kJ mol⁻¹ A1 ΔH = 2648 − 3466 = −818 kJ mol⁻¹, with the negative sign shown and the working in full B1 mean bond enthalpies are averages over many compounds, not the bonds in methane itself, and ΔcH° refers to liquid water, so it also includes the enthalpy released when the steam condenses

Question 8

Give the meaning of the term mean bond enthalpy.

Accepted Answer

B1 the enthalpy needed to break one mole of a given type of covalent bond B1 with the molecules in the gaseous state B1 averaged over a range of different compounds; ignore 'average' or 'mean' alone without the range

Question 9

Calculate ΔH for H₂(g) + Cl₂(g) → 2HCl(g). Bond enthalpies / kJ mol⁻¹: H–H 436, Cl–Cl 243, H–Cl 432.

Accepted Answer

M1 bonds broken: 436 + 243 = 679 M1 bonds made: 2 × 432 = 864 A1 ΔH = 679 − 864 = −185 kJ mol⁻¹, to three significant figures

Question 10

Calculate ΔH for C₂H₄(g) + H₂O(g) → C₂H₅OH(g), counting only the bonds that change, from the mean bond enthalpies in the table.

Accepted Answer

M1 bonds broken: C=C + one O–H = 612 + 464 = 1076 M1 bonds made: C–C + C–H + C–O = 347 + 413 + 358 = 1118 A1 ΔH = 1076 − 1118 = −42 kJ mol⁻¹, to two significant figures

Question 11

Calculate the C=C bond enthalpy in propene from C₃H₆(g) + H₂(g) → C₃H₈(g), ΔH = −124 kJ mol⁻¹, using the mean bond enthalpies in the table.

Accepted Answer

M1 changes: C=C and H–H broken; one C–C and two C–H made M1 −124 = (C=C + 436) − (347 + 2 × 413) = C=C − 737 A1 C=C = +613 kJ mol⁻¹, to three significant figures

Question 12

Explain why ΔH for the combustion of ethane calculated from mean bond enthalpies, −1436 kJ mol⁻¹, differs from the standard enthalpy of combustion of ethane, −1560 kJ mol⁻¹.

Accepted Answer

B1 mean bond enthalpies are averages over many compounds, not the actual values for the bonds in ethane B1 bond enthalpies apply to gases, so the calculation gives the three moles of water as steam B1 the standard value forms liquid water, and condensing it releases extra energy, so the standard value is more exothermic; reject heat loss, since nothing was measured

Question 13

Suggest why breaking the first O–H bond in a water molecule takes 498 kJ mol⁻¹ but breaking the second takes 428 kJ mol⁻¹, and calculate the value that would be used as the O–H bond enthalpy for water.

Accepted Answer

B1 the second bond is broken in an OH fragment, not in H₂O, so its environment is different B1 the arrangement of electrons around the oxygen has changed once one hydrogen has gone, so the bond has a different strength B1 the mean for water is (498 + 428) ÷ 2 = 463 kJ mol⁻¹

Question 14

Describe what happens to energy, step by step, in an enthalpy level diagram for a reaction calculated from bond enthalpies.

Accepted Answer

B1 the reactants rise to separate gaseous atoms as all their bonds are broken, taking in energy B1 the atoms fall to the products as the new bonds form, releasing energy B1 ΔH is the difference between the energy taken in and the energy released, from the reactant level to the product level

Question 15

Calculate the C–C bond enthalpy in ethane from the data in the table.

Accepted Answer

M1 C₂H₆(g) → 2C(g) + 6H(g): ΔH = 2(717) + 6(218) − (−84) = 1434 + 1308 + 84 = +2826 kJ mol⁻¹ M1 this breaks one C–C and six C–H bonds, and 6 × 413 = 2478 kJ mol⁻¹ A1 C–C = 2826 − 2478 = 348 kJ mol⁻¹ to three significant figures

Question 16

Explain, in terms of bond length and bond enthalpy, why nitrogen gas reacts with very few substances at room temperature.

Accepted Answer

B1 the two nitrogen atoms are joined by a triple bond, which is short, holding three shared pairs close to both nuclei B1 the N≡N bond enthalpy is very high, 945 kJ mol⁻¹, the strongest bond in the table B1 any reaction must begin to break this bond, so the activation energy is high and few collisions at room temperature have enough energy

Question 17

Suggest why ΔH for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g) calculated from mean bond enthalpies, −122 kJ mol⁻¹, is more exothermic than the measured value of −99 kJ mol⁻¹, given that the first C–H bond broken in methane takes 439 kJ mol⁻¹.

Accepted Answer

B1 only one C–H bond is broken, and it is the first one in methane, CH₃–H, not an average C–H bond B1 breaking that bond takes 439 kJ mol⁻¹, more than the mean of 413 kJ mol⁻¹, so more energy is taken in than the calculation allows B1 using 439 makes the answer 26 kJ mol⁻¹ less exothermic, −96 kJ mol⁻¹, close to the measured value

Question 18

Explain why ΔH for H₂(g) + Cl₂(g) → 2HCl(g) calculated from bond enthalpies agrees with the data-book value.

Accepted Answer

B1 every bond involved is in a diatomic molecule, so each bond enthalpy is an exact value, not a mean B1 all the species are gases in their standard states, so no change of state is left out of the calculation

Question 19

Suggest why ΔH for the combustion of liquid hexane calculated from mean bond enthalpies is less exothermic than its standard enthalpy of combustion, apart from the use of mean values.

Accepted Answer

B1 the bond enthalpy calculation treats hexane and the water as gases B1 condensing the seven moles of water formed releases more energy than vaporising the hexane takes in, so the standard value, with both liquid, is more exothermic

Quantity	Value / kJ mol⁻¹
ΔfH°(C₂H₆, g)	−84
Enthalpy of atomisation of carbon, per mole of C(g)	+717
Enthalpy of atomisation of hydrogen, per mole of H(g)	+218
Mean C–H bond enthalpy	413