Bond enthalpiesEdexcel A-Level Chemistry: Subtopic test

1. 1
   A student is estimating the enthalpy change for the reaction H₂(g) + Cl₂(g) → 2HCl(g) using mean bond enthalpies (kJ mol⁻¹): H–H 436, Cl–Cl 243, H–Cl 432.

   (a)

   Which statement defines the mean bond enthalpy of the H–Cl bond?

   [1 mark]

   • AThe enthalpy change when one mole of H–Cl bonds is formed from gaseous atoms in HCl only
   • BThe energy released when one mole of H–Cl bonds is broken
   • CThe average enthalpy change to break one mole of H–Cl bonds in gaseous molecules, averaged over a range of compounds
   • DThe enthalpy change to break one mole of H–Cl bonds in liquid hydrogen chloride

   (b)

   What is the enthalpy change for H₂(g) + Cl₂(g) → 2HCl(g)?

   [1 mark]

   • A−185 kJ mol⁻¹
   • B+185 kJ mol⁻¹
   • C+247 kJ mol⁻¹
   • D+679 kJ mol⁻¹

   (c)

   Use the bond enthalpy values to explain why this reaction is exothermic.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Ethene reacts with hydrogen in the presence of a nickel catalyst: C₂H₄(g) + H₂(g) → C₂H₆(g). Mean bond enthalpies (kJ mol⁻¹): C=C 612, C–C 347, C–H 413, H–H 436. The experimental enthalpy change for the reaction is −137 kJ mol⁻¹.

   (a)

   Which row correctly shows the bonds broken and the bonds made in this reaction, using the mean bond enthalpies given?

   [1 mark]

   • ABroken: C–C and H–H. Made: C=C and 2 × C–H
   • BBroken: C=C and H–H. Made: C–C and 2 × C–H
   • CBroken: C=C and 4 × C–H. Made: C–C and 6 × C–H
   • DBroken: H–H only. Made: C–C and 2 × C–H

   (b)

   The calculated value from mean bond enthalpies is −125 kJ mol⁻¹, but the experimental value is −137 kJ mol⁻¹. What is the best explanation?

   [1 mark]

   • AThe reaction is endothermic, which makes the calculated value inaccurate
   • BHydrogen is used in excess, so the enthalpy change is larger
   • CThe enthalpy change includes the activation energy of the catalysed reaction
   • DMean bond enthalpies are averages over many compounds, so they differ from the actual bond enthalpies in ethene and ethane

   (c)

   Calculate the enthalpy change for the reaction using the mean bond enthalpies.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Ammonia has a standard enthalpy change of formation of −46 kJ mol⁻¹. Mean bond enthalpies (kJ mol⁻¹): N≡N 945, H–H 436. Hydrazine, H₂N–NH₂, which contains one N–N bond and four N–H bonds, decomposes: N₂H₄(g) → N₂(g) + 2H₂(g), ΔH = −95 kJ mol⁻¹.

   (a)

   Calculate the mean bond enthalpy of the N–H bond from the standard enthalpy change of formation of ammonia, for ½N₂(g) + 3/2 H₂(g) → NH₃(g).

   [3 marks]

   (b)

   Use a mean N–H bond enthalpy of 391 kJ mol⁻¹ to calculate the mean bond enthalpy of the N–N bond in hydrazine.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   Ethanol, CH₃CH₂OH, is burned as a fuel: C₂H₅OH(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(g). Mean bond enthalpies (kJ mol⁻¹): C–H 413, C–C 347, C–O 358, O–H 464, O=O 498, C=O 805. The experimental enthalpy change of combustion of ethanol is −1367 kJ mol⁻¹, measured with liquid ethanol and liquid water.

   (a)

   Use the mean bond enthalpies to calculate the enthalpy change for the combustion of gaseous ethanol to carbon dioxide and water vapour.

   [6 marks]

   (b)

   The calculation using mean bond enthalpies gives −1276 kJ mol⁻¹, which differs from the experimental value. The enthalpy change of vaporisation of ethanol is +42 kJ mol⁻¹ and the enthalpy change for condensing one mole of water vapour is −44 kJ mol⁻¹. Evaluate whether the difference can be explained by the limitations of the method.

   [6 marks]

   Total for question 4: 12 marks

End of questions