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90780 Describe properties of
particles and thermochemical
        principles 2010

  Questions 3 and 4 thermochemistry
90780 Describe properties of
       particles and thermochemical
             principles Q3 2010
QUESTION THREE
(a) (i) Write the equation for which the enthalpy change is
    the enthalpy of formation, Δf H°, for zinc oxide.

(a) (ii) Zinc oxide is formed when zinc sulfide is heated in air.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

Calculate the enthalpy change, ΔrH°, for this reaction, using
the following data.
Δf H° (ZnS(s)) = –200 kJ mol–1
Δf H° (ZnO(s)) = –348 kJ mol–1
Δf H° (SO2(g)) = –297 kJ mol–1
90780 Describe properties of particles and
       thermochemical principles Q3 2010
(b) The equation for the combustion of ethanol
is:
CH3CH2OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔrH° = –
1379 kJ mol–1
Calculate the bond enthalpy for the O–H bond
using the enthalpy of the reaction above and the
bond enthalpy data in the table.
     Bond               Bond enthalpy / kJ mol–1
     C-H                +412
     C-O                +360
     O=O                +496
     C-C                +348
     C=O                +743
90780 Describe properties of particles
 and thermochemical principles Q4
                2010
(a) (i) Write the equation for which the enthalpy
    change is the enthalpy of fusion, ΔfusH°, for
    zinc sulfide.

(ii) Give a reason why ΔfusH° is always greater
than zero.
90780 Describe properties of particles
 and thermochemical principles Q4
                2010
(b) Ethanol (CH3CH2OH) and propane
(CH3CH2CH3) have similar molar masses but
ethanol is a liquid at room temperature, while
propane is a gas.
Identify the types of intermolecular forces for
each of these substances and explain why
ethanol has a higher boiling point than propane.
90780 Describe properties of particles
   and thermochemical principles Q4
                  2010
  © Account for the difference in the boiling
  points of the two substances in the table below
  by comparing all the intermolecular forces.
Name                  Structure      Boiling Point / °C


butan-1-ol            CH3CH2CH2CH2OH 117.7


2-methylpropan-2-ol                  82.6

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90780 2010 particles and thermochemistry

  • 1. 90780 Describe properties of particles and thermochemical principles 2010 Questions 3 and 4 thermochemistry
  • 2. 90780 Describe properties of particles and thermochemical principles Q3 2010 QUESTION THREE (a) (i) Write the equation for which the enthalpy change is the enthalpy of formation, Δf H°, for zinc oxide. (a) (ii) Zinc oxide is formed when zinc sulfide is heated in air. 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) Calculate the enthalpy change, ΔrH°, for this reaction, using the following data. Δf H° (ZnS(s)) = –200 kJ mol–1 Δf H° (ZnO(s)) = –348 kJ mol–1 Δf H° (SO2(g)) = –297 kJ mol–1
  • 3. 90780 Describe properties of particles and thermochemical principles Q3 2010 (b) The equation for the combustion of ethanol is: CH3CH2OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔrH° = – 1379 kJ mol–1 Calculate the bond enthalpy for the O–H bond using the enthalpy of the reaction above and the bond enthalpy data in the table. Bond Bond enthalpy / kJ mol–1 C-H +412 C-O +360 O=O +496 C-C +348 C=O +743
  • 4. 90780 Describe properties of particles and thermochemical principles Q4 2010 (a) (i) Write the equation for which the enthalpy change is the enthalpy of fusion, ΔfusH°, for zinc sulfide. (ii) Give a reason why ΔfusH° is always greater than zero.
  • 5. 90780 Describe properties of particles and thermochemical principles Q4 2010 (b) Ethanol (CH3CH2OH) and propane (CH3CH2CH3) have similar molar masses but ethanol is a liquid at room temperature, while propane is a gas. Identify the types of intermolecular forces for each of these substances and explain why ethanol has a higher boiling point than propane.
  • 6. 90780 Describe properties of particles and thermochemical principles Q4 2010 © Account for the difference in the boiling points of the two substances in the table below by comparing all the intermolecular forces. Name Structure Boiling Point / °C butan-1-ol CH3CH2CH2CH2OH 117.7 2-methylpropan-2-ol 82.6