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Chemical Energetics: Question 3

Syllabus 5.2

Multiple choice AS 1 mark

Methane reacts with chlorine to form chloromethane:

CH4(g)+Cl2(g)→CH3Cl(g)+HCl(g)\text{CH}_4\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow \text{CH}_3\text{Cl(g)} + \text{HCl(g)}

The table gives some mean bond enthalpies.

Bond Mean bond enthalpy / kJ mol−1\text{kJ mol}^{-1}
C–H 412412
Cl–Cl 244244
C–Cl 338338
H–Cl 428428

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

Comparing reactants and products, only one C–H bond of methane is broken (one H is replaced by Cl), and the Cl–Cl bond in Cl2_2 is fully broken. In the products, a new C–Cl bond and a new H–Cl bond are formed. The other three C–H bonds in methane are unchanged and are not counted.

  • Bonds broken: 1 ×\times C–H, 1 ×\times Cl–Cl
  • Bonds made: 1 ×\times C–Cl, 1 ×\times H–Cl

Step 2: Total the bond enthalpies for each side

Bonds broken=412+244=656 kJ mol−1\text{Bonds broken} = 412 + 244 = 656\ \text{kJ mol}^{-1}

Bonds made=338+428=766 kJ mol−1\text{Bonds made} = 338 + 428 = 766\ \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=656−766=−110 kJ mol−1\Delta H = 656 - 766 = -110\ \text{kJ mol}^{-1}

Check (independent recomputation): 412+244=656412+244=656; 338+428=766338+428=766; 656−766=−110656-766=-110, consistent. The reaction is exothermic overall, since the bonds formed (C–Cl and H–Cl) are collectively stronger than the bonds broken (C–H and Cl–Cl).

Why the other options are wrong

  • B (+110 kJ mol−1+110\ \text{kJ mol}^{-1}): this has the correct magnitude but the wrong sign. It comes from computing (bonds made) −- (bonds broken) instead of the correct order.
  • C (−1422 kJ mol−1-1422\ \text{kJ mol}^{-1}): this comes from adding all four bond enthalpies together (656+766=1422656+766=1422) instead of subtracting one total from the other.
  • D (−354 kJ mol−1-354\ \text{kJ mol}^{-1}): this comes from forgetting to include the Cl–Cl bond among the bonds broken, giving 412−766=−354412-766=-354.

Final answer

  • ΔH=−110 kJ mol−1\Delta H = \boxed{-110\ \text{kJ mol}^{-1}}, option A.