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Chemical Energetics: Question 8

Syllabus 5.2

Multiple choice AS 1 mark

Methane burns completely in oxygen, with all species treated as gases:

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

The table gives some mean bond enthalpies.

Bond Mean bond enthalpy / kJ mol−1\text{kJ mol}^{-1}
C–H 412412
O=O 496496
C=O (in CO2\text{CO}_2) 805805
O–H 463463

Using these mean bond enthalpies, what is the enthalpy change, ΔH\Delta H, for this reaction?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Identify which bonds break and which form

In one mole of methane, all four C–H bonds break, and both O=O bonds in the two moles of O2\text{O}_2 break. In the products, two C=O bonds form within the single CO2\text{CO}_2 molecule, and four O–H bonds form across the two H2O\text{H}_2\text{O} molecules (22 bonds each).

  • Bonds broken: 4×4 \times C–H, 2×2 \times O=O
  • Bonds made: 2×2 \times C=O, 4×4 \times O–H

Step 2: Total the bond enthalpies for each side

Bonds broken=(4×412)+(2×496)=1648+992=2640 kJ mol−1\text{Bonds broken} = (4\times412) + (2\times496) = 1648 + 992 = 2640\ \text{kJ mol}^{-1}

Bonds made=(2×805)+(4×463)=1610+1852=3462 kJ mol−1\text{Bonds made} = (2\times805) + (4\times463) = 1610 + 1852 = 3462\ \text{kJ mol}^{-1}

Step 3: Apply ΔH=\Delta H = (bonds broken) −- (bonds made)

Breaking bonds requires energy (endothermic, positive), and forming bonds releases energy (exothermic, negative), so overall:

ΔH=2640−3462=−822 kJ mol−1\Delta H = 2640 - 3462 = -822\ \text{kJ mol}^{-1}

Check (independent recomputation): 4×412=16484\times412=1648, 2×496=9922\times496=992, 1648+992=26401648+992=2640; 2×805=16102\times805=1610, 4×463=18524\times463=1852, 1610+1852=34621610+1852=3462; 2640−3462=−8222640-3462=-822, consistent. The reaction is exothermic overall, since the bonds formed (C=O and O–H) are collectively stronger than the bonds broken (C–H and O=O).

Why the other options are wrong

  • B (+822 kJ mol−1+822\ \text{kJ mol}^{-1}): correct magnitude but wrong sign, comes from computing (bonds made) −- (bonds broken) instead of the correct order.
  • C (−1318 kJ mol−1-1318\ \text{kJ mol}^{-1}): comes from breaking only one O=O bond instead of two, giving bonds broken =1648+496=2144= 1648+496=2144, then 2144−3462=−13182144-3462=-1318.
  • D (−17 kJ mol−1-17\ \text{kJ mol}^{-1}): comes from counting only one C=O bond in CO2\text{CO}_2 instead of two, giving bonds made =805+1852=2657=805+1852=2657, then 2640−2657=−172640-2657=-17.

Final answer

  • ΔH=−822 kJ mol−1\Delta H = \boxed{-822\ \text{kJ mol}^{-1}}, option A.