Exothermic and Endothermic Reactions: Question 9

Syllabus 5.1

Structured Extended 8 marks

Methane burns completely in oxygen according to the equation:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)

The bond energies are:

  • C–H bond: 413 kJ/mol413\,\text{kJ/mol}
  • O=O bond: 498 kJ/mol498\,\text{kJ/mol}
  • C=O bond: 805 kJ/mol805\,\text{kJ/mol}
  • O–H bond: 463 kJ/mol463\,\text{kJ/mol}

(a) One molecule of CH4\text{CH}_4 contains four C–H bonds, and each O2\text{O}_2 molecule contains one O=O bond. Determine the total number of each bond broken when one mole of CH4\text{CH}_4 reacts completely, and calculate the total energy absorbed breaking these bonds. [3]

(b) One molecule of CO2\text{CO}_2 contains two C=O bonds, and each H2O\text{H}_2\text{O} molecule contains two O–H bonds. Determine the total number of each bond formed, and calculate the total energy released forming these bonds. [3]

(c) Calculate the overall energy change, ΔH\Delta H, for the complete combustion of one mole of methane, and state whether the reaction is exothermic or endothermic. [2]

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Worked solution

Part (a): Bonds broken and energy absorbed

One mole of CH4\text{CH}_4 contains four C–H bonds, so all four must be broken. The equation needs two moles of O2\text{O}_2, and each O2\text{O}_2 molecule has one O=O bond, so two O=O bonds must be broken.

4×(C–H)=4×413=1652 kJ/mol4 \times (\text{C–H}) = 4 \times 413 = 1652\,\text{kJ/mol} 2×(O=O)=2×498=996 kJ/mol2 \times (\text{O=O}) = 2 \times 498 = 996\,\text{kJ/mol}

Total energy absorbed breaking bonds: 1652+996=2648 kJ/mol1652 + 996 = 2648\,\text{kJ/mol}

Part (b): Bonds formed and energy released

One mole of CO2\text{CO}_2 is formed, containing two C=O bonds. Two moles of H2O\text{H}_2\text{O} are formed, and each H2O\text{H}_2\text{O} molecule has two O–H bonds, giving four O–H bonds in total.

2×(C=O)=2×805=1610 kJ/mol2 \times (\text{C=O}) = 2 \times 805 = 1610\,\text{kJ/mol} 4×(O–H)=4×463=1852 kJ/mol4 \times (\text{O–H}) = 4 \times 463 = 1852\,\text{kJ/mol}

Total energy released forming bonds: 1610+1852=3462 kJ/mol1610 + 1852 = 3462\,\text{kJ/mol}

Part (c): Overall energy change and classification

ΔH=(energy absorbed breaking bonds)−(energy released forming bonds)\Delta H = (\text{energy absorbed breaking bonds}) - (\text{energy released forming bonds})

ΔH=2648−3462=−814 kJ/mol\Delta H = 2648 - 3462 = -814\,\text{kJ/mol}

Since ΔH\Delta H is negative, considerably more energy is released forming the new bonds in carbon dioxide and water than is absorbed breaking the bonds in methane and oxygen. The reaction is strongly exothermic, consistent with methane being burned as a fuel.

Final answers

  • (a) 4 C–H bonds and 2 O=O bonds broken; energy absorbed == 2648 kJ/mol2648\,\text{kJ/mol}
  • (b) 2 C=O bonds and 4 O–H bonds formed; energy released == 3462 kJ/mol3462\,\text{kJ/mol}
  • (c) ΔH=−814 kJ/mol\Delta H = \boxed{-814}\,\text{kJ/mol}; the reaction is exothermic.