MC test: Bond enthalpies

Multiple choice test on 5.3 Bond enthalpies

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

Which statements concerning bond enthalpies are correct?

I. Bond enthalpies refer to both bond breaking and bond formation.

II. The formation of covalent bonds is an endothermic process.

III. Average covalent bond enthalpies refer only to the gaseous state.

I and II only

II and III only

I and III only

I, II and III

Breaking bonds is an endothermic process. Energy is always evolved when new bonds are formed. All covalent bond enthalpies, whether absolute values or average values, refer to the gaseous state.

Which process is endothermic?

NaOH(aq) + HCl(aq) → NaOH(aq) + H₂O(l)

Cl(g) + e− → Cl⁻(g)

2Na(s) + Cl₂(g) → 2NaCl(s)

Cl₂(g) → 2Cl(g)

Breaking the bond in chlorine molecules to form chlorine atoms requires an input of energy. The other three processes are all exothermic.

Which bond enthalpy has an absolute value rather than an average value?

N−N

N=N

N≡N

N−H

There is only one substance that contains the N≡N bond and that is nitrogen, N₂, itself. The other bonds occur in different compounds and the value is the average value found in similar compounds containing the same bond.

Which is the correct order for the carbon to carbon bond enthalpies (weakest first)?

C≡C < C−C < C=C

C−C < C=C < C≡C

C≡C < C=C < C−C

C=C < C−C < C≡C

Because there are more pairs of electrons being shared, triple bonds between the same two atoms are always stronger than double bonds which in turn are always stronger than single bonds.

Which enthalpy change can be calculated using bond enthalpies and/or average bond enthalpies only?

2C₃H₆(g) + 9O₂(g) → 6CO₂(g) + 6H₂O(l)

C₅H₁₀(l) + Cl₂(g) → C₅H₁₀Cl₂(l)

C₅H₁₀(l) + H₂O(l) → C₅H₁₁OH(l)

C₃H₆(g) + H₂(g) → C₃H₈(g)

Other than the hydrogenation of propene, all the other changes do not have all the reactants and products in the gaseous state so enthalpy values for changes of state will also be required for the calculation.

What is the enthalpy change for the hydrogenation of propene to form propane?

C₃H₆(g) + H₂(g) → C₃H₈(g)

Bond enthalpies / kJ mol⁻¹: C−C 346; C=C 614; H−H 436; C−H 414

+ 124 kJ mol⁻¹

− 290 kJ mol⁻¹

+ 290 kJ mol⁻¹

− 124 kJ mol⁻¹

Energy required to break C=C and H−H bonds = 614 + 436 = 1050 kJ mol⁻¹
Energy given out by formation of C−C and 2C−H bonds = 346 + 828 = 1174 kJ mol⁻¹
ΔH = − (1174 − 1050) = − 124 kJ mol⁻¹

Hydrazine gas, N₂H₄(g), reacts exothermically with chlorine

N₂H₄(g) + 2Cl₂(g) → N₂(g) + 4HCl(g) ΔH^⦵ = − 463 kJ

Some relevant bond energies in kJ mol⁻¹ are: N−N 158; N≡N 945; N−H 391 and Cl−Cl 242.

What is the bond enthalpy of the H−Cl bond?

371 kJ mol⁻¹

235 kJ mol⁻¹

255 kJ mol⁻¹

431 kJ mol⁻¹

Energy in to break N−N, 4N−H and Cl−Cl bonds = 158 + (4 x 391) + (2 x 242) = 2206 kJ
Energy given out to form N≡N and 4H−Cl bonds = 945 + 4H−Cl kJ
From the value for ΔH, 463 kJ more energy is given out so 4H−Cl = 2206 − 945 + 463 = 1724 kJ
H−Cl bond enthalpy = 1724 ÷ 4 = 431 kJ mol−1.

Benzene can be hydrogenated to form cyclohexane

The enthalpy of vaporization of benzene is + 34 kJ mol⁻¹ and the enthalpy of vaporization of cyclohexane is 32 kJ mol⁻¹.
The C−C, C−H and H−H bond enthalpies are 346, 414 and 436 kJ mol⁻¹ respectively.

Which is the correct value for the average carbon to carbon bond enthalpy in benzene?

542 kJ mol⁻¹

507 kJ mol⁻¹

731 kJ mol⁻¹

659 kJ mol⁻¹

Energy in (in kJ mol⁻¹) = 34 to vaporize the benzene + 3 x 436 to break the H−H bonds + x (where x is the energy required to break 6 C to C bonds) = 1342 + x
Energy out = 6 x 346 (to form 6 C−C bonds) + 6 x 414 (to account for the extra 6 C−H bonds formed) + 32 (gaseous cyclohexane condenses to liquid cyclohexane) = 4592 kJ mol⁻¹.
Energy out = 208 kJ mol⁻¹ more than energy in hence x = 4592 − (1342 + 208) = 3042 kJ mol⁻¹.
C to C bond energy in benzene = 3042 ÷ 6 = 507 kJ mol⁻¹.

Why does the steady state equilibrium between oxygen and ozone in the ozone layer prevent much of the sun's uv radiation reaching the Earth's surface?

Short wavelength UV light is required to break the bonds homolytically in ozone and longer wavelength UV light is required to break the bonds heterolytically in oxygen

Short wavelength UV light is required to break the bonds homolytically in ozone and longer wavelength UV light is required to break the bonds homolytically in oxygen.

Short wavelength UV light is required to break the bonds homolytically in oxygen and longer wavelength UV light is required to break the bonds homolytically in ozone.

Short wavelength UV light is required to break the bonds heterolytically in oxygen and longer wavelength UV light is required to break the bonds homolytically in ozone.

Both reactions are homolytic. O=O is the stronger bond so requires shorter wavelength (higher energy) UV light to break compared to the O to O bond in ozone which has a bond order of 1.5.

Which statements explain why the combustion reactions of hydrocarbons are always exothermic?

I. The O−H and C=O bonds in the products have relatively high bond enthalpies.

II. The products are more thermodynamically stable than the reactants.

III. The C-H bonds in the hydrocarbons are very weak.

I and III only

II and III only

I, II and III

I and II only

The O-H and C=O are particularly strong with values of 463 and 804 kJ mol⁻¹ respectively. The C−H bond is 414 kJ mol⁻¹ which is also quite strong as is the O=O bond (498 kJ mol⁻¹) but the C−C bond is weaker (346 kJ mol⁻¹). Overall the bonds in the products are considerably stronger than the bonds in the reactants making the products more thermodynamically stable.

Total Score:

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MC test: Bond enthalpies

Multiple choice test on 5.3 Bond enthalpies

MC test: Bond enthalpies