R1.2 part 1 Energy cycles in reactions Exercise

1. Which of the following processes are endothermic? I. H₂O(s) → H₂O(g) II. CO₂(g) → CO₂(s) III. O₂(g) → 2O(g) [1 mark]

• I and II only
• I and III only
• II and III only
• I, II and III

Answer explanation

I and III only (1 mark). Process I involves melting/vaporization which requires energy input (endothermic). Process III involves bond breaking which requires energy (endothermic). Process II involves condensation/sublimation which releases energy (exothermic).

2. Identify the equation that represents the bond enthalpy for the H−Cl bond. [1 mark]

• HCl(g) → H(g) + Cl(g)
• HCl(g) → ½H₂(g) + ½Cl₂(g)
• HCl(g) → H⁺(g) + Cl⁻(g)
• HCl(aq) → H⁺(aq) + Cl⁻(aq)

Answer explanation

HCl(g) → H(g) + Cl(g) (1 mark). Bond enthalpy is defined as the energy required to break one mole of bonds in gaseous molecules to form gaseous atoms.

3. Identify the bonds that are broken in the following process: C₂H₆(g) → 2C(g) + 6H(g) [1 mark]

Answer explanation

1 × C−C + 6 × C−H (1 mark). The process breaks all bonds in ethane: one carbon-carbon bond and six carbon-hydrogen bonds.

4. Which of the following enthalpy changes corresponds to the bond enthalpy of the carbon−oxygen bond in carbon monoxide? [1 mark]

• CO(g) → C(s) + O(g)
• CO(g) → C(g) + O(g)
• CO(g) → C(s) + ½O₂(g)
• CO(g) → C(g) + ½O₂(g)

Answer explanation

CO(g) → C(g) + O(g) (1 mark). Bond enthalpy requires breaking bonds to form gaseous atoms, not elements in their standard states.

5. The hydrogenation of the alkene double bond in unsaturated oils is an important reaction in margarine production. Calculate the enthalpy change when one mole of C=C bonds is hydrogenated from the bond energy data given earlier. [1 mark]

• −290 kJ mol⁻¹
• −124 kJ mol⁻¹
• +124 kJ mol⁻¹
• +290 kJ mol⁻¹

Answer explanation

−124 kJ mol⁻¹ (1 mark). The hydrogenation reaction forms stronger bonds (C−C and C−H) than those broken (C=C and H−H), resulting in a net energy release.

6. Use bond enthalpy data to calculate the enthalpy change of the reaction between methane and fluorine: C₂H₄(g) + F₂(g) → CH₂FCH₂F(g) [1 mark]

• +557 kJ mol⁻¹
• +65 kJ mol⁻¹
• −65 kJ mol⁻¹
• −557 kJ mol⁻¹

Answer explanation

+65 kJ mol⁻¹ (1 mark). Calculate by breaking bonds (C=C and F−F) and forming bonds (C−C, C−H, and C−F). The energy required to break bonds is less than energy released in forming new bonds.

7. Use bond enthalpy data to calculate ΔH⊖ for the reaction: 2H₂(g) + O₂(g) → 2H₂O(g) [1 mark]

Answer explanation

ΔH⊖ = −482 kJ mol⁻¹ (1 mark). Bonds broken: 2×H−H (+436 each) + 1×O=O (+498) = +1370 kJ. Bonds formed: 4×H−O (−463 each) = −1852 kJ. Net: +1370 − 1852 = −482 kJ mol⁻¹

8. Use the bond enthalpies given earlier to estimate the enthalpy of combustion of ethanol and comment on the reliability of your result. [3 marks]

Answer explanation

−1263 kJ mol⁻¹ (2 marks). C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g). Bonds broken: C−C (+346), O−H (+463), C−O (+358), 5×C−H (+414 each), 3×O=O (+498 each) = +4731 kJ. Bonds formed: 4×C=O (−804 each), 6×H−O (−463 each) = −5994 kJ. Net = −1263 kJ mol⁻¹. Less exothermic than literature value (−1367 kJ mol⁻¹) because calculation assumes gaseous states while ethanol and water are actually liquids (1 mark).

9. [REFER DIAGRAM] The diagram illustrates the enthalpy changes for a set of reactions. Which of the following statements are correct? I. P → S ΔH = −10 kJ II. R → Q ΔH = +90 kJ III. P → R ΔH = +20 kJ [1 mark]

• I and II only
• I and III only
• II and III only
• I, II and III

Answer explanation

I and III only (1 mark). From the diagram: I. P → S = −60 + 50 = −10 kJ (correct). II. R → Q = +30 + 60 = +90 kJ (incorrect, should be R → Q via P). III. P → R = +50 − 30 = +20 kJ (correct).

10. The thermochemical equations for three related reactions are shown: CO(g) + ½O₂(g) → CO₂(g) ΔH₁ = −283 kJ mol⁻¹; 2H₂(g) + O₂(g) → 2H₂O(l) ΔH₂ = −572 kJ mol⁻¹; CO₂(g) + H₂(g) → CO(g) + H₂O(l) ΔH₃ = ? What is the value of ΔH₃? [1 mark]

• +289 kJ mol⁻¹
• −3 kJ mol⁻¹
• −289 kJ mol⁻¹
• −855 kJ mol⁻¹

Answer explanation

+289 kJ mol⁻¹ (1 mark). Using Hess's law: ΔH₃ = −ΔH₁ + ½ΔH₂ = −(−283) + ½(−572) = +283 − 286 = +289 kJ mol⁻¹