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InkFlaskCambridge IGCSE
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Answers · Bond energies

Bond energies — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. JK1 · Final check

Model answer. ΔH = energy in − energy out = 6186 − 7446 = −1260 kJ/mol, so the products' line is 1260 kJ/mol below the reactants' line. The top of the hump is 1600 kJ/mol above the products, so it is 1600 − 1260 = 340 kJ/mol above the reactants. Ea is measured from the reactants: Ea = 340 kJ/mol.

Checked automatically

  • (a) ΔH: -1260 kJ/mol (exact value)
    • Common wrong answer 1260 : Check the sign: ΔH = energy in − energy out = 6186 − 7446.
    • Common wrong answer 13632 : Subtract, don't add: ΔH = 6186 − 7446.
  • (b) Ea: 340 kJ/mol (exact value)
    • Common wrong answer 1600 : That is measured from the products. Ea is measured from the reactants, which are 1260 kJ/mol higher.
    • Common wrong answer 2860 : The products are below the reactants (ΔH is negative), so subtract: 1600 − 1260.
    • Common wrong answer 6186 : That is the energy to break every bond, a total for the sum. Ea is much smaller: use the diagram.

A full answer includes

  • (a) I wrote ΔH = 6186 − 7446.
  • (a) I got ΔH = −1260 kJ/mol.
  • (b) I placed the products' line 1260 kJ/mol below the reactants' line.
  • (b) I worked out Ea as 1600 − 1260.
  • (b) I got Ea = 340 kJ/mol, measured up from the reactants.
  • I did not give 6186 kJ/mol, the energy taken in, as Ea.

Also accept

  • −1260 kJ/mol; Ea = 340 kJ/mol

Watch for

  • “ΔH = energy out − energy in.” — That flips the sign. ΔH = energy taken in breaking bonds − energy given out making bonds, so an exothermic reaction comes out negative.
  • “The energy needed to break every bond is the activation energy.” — Breaking every bond is only a total for the sum. In a real reaction old bonds loosen while new ones form, so the activation energy is much smaller.
  • “Ea is measured from the products' line.” — Ea is the energy the reactants need to start reacting, so it is measured from the reactants' line up to the top of the hump.

2. JK2 · Final check

Model answer. ΔH = 2057 − 1949 = +108 kJ/mol. More energy is taken in breaking the bonds in methanol than is given out making the bonds in carbon monoxide and hydrogen, so the difference, 108 kJ/mol, comes from the surroundings: the reaction is endothermic, which is why the catalyst has to be kept hot.

Checked automatically

  • (a) ΔH: 108 kJ/mol (exact value)
    • Common wrong answer -108 : Check the sign: ΔH = energy in − energy out = 2057 − 1949, which is positive.
    • Common wrong answer 4006 : Subtract, don't add: ΔH = 2057 − 1949.

A full answer includes

  • (a) I wrote ΔH = 2057 − 1949.
  • (a) I got ΔH = +108 kJ/mol.
  • (b) I deduced that the reaction is endothermic.
  • (b) I explained that breaking the bonds in methanol takes in more energy than making the bonds in CO and H2 gives out.
  • (b) I wrote that the difference is taken in from the surroundings, so the catalyst must be kept hot.
  • I did not explain it only by saying that bonds break.

Also accept

  • +108 kJ/mol; endothermic, more taken in breaking bonds than given out making bonds

Watch for

  • “ΔH = energy out − energy in.” — That flips the sign. ΔH = energy taken in breaking bonds − energy given out making bonds, so an exothermic reaction comes out negative.
  • “A positive ΔH means the reaction gives out energy.” — A positive ΔH means the chemicals gain energy: more is taken in breaking bonds than is given out making bonds. The reaction is endothermic.

3. JK3 · Final check

Model answer. Energy in: N–N 158, 4 N–H 1564 and O=O 498, total 2220 kJ/mol. Energy out: N≡N 945 and 2H2O with 4 O–H, 1852, total 2797 kJ/mol. ΔH = 2220 − 2797 = −577 kJ/mol: exothermic, as a fuel should be.

Checked automatically

  • (a) Energy in: 2220 kJ/mol (exact value)
    • Common wrong answer 2062 : Hydrazine has an N–N single bond (158) as well as four N–H bonds.
  • (b) Energy out: 2797 kJ/mol (exact value)
    • Common wrong answer 1871 : 2H2O has four O–H bonds: 945 + 4 × 463.
  • (c) ΔH: -577 kJ/mol (exact value)
    • Common wrong answer 577 : Check the sign: ΔH = energy in − energy out = 2220 − 2797.

A full answer includes

  • (a) I worked out 158 + 4 × 391 + 498.
  • (a) I got 2220 kJ/mol.
  • (b) I worked out 945 + 4 × 463.
  • (b) I got 2797 kJ/mol.
  • (c) I worked out 2220 − 2797.
  • (c) I got ΔH = −577 kJ/mol.
  • I did not leave out the N–N single bond in hydrazine.

Also accept

  • 2220; 2797; −577 kJ/mol

Watch for

  • “The number in front of a formula can be ignored when you count bonds.” — Count the bonds in one molecule, then multiply by the number in front of the formula: every bond in the molecule is counted that many times.
  • “ΔH = energy out − energy in.” — That flips the sign. ΔH = energy taken in breaking bonds − energy given out making bonds, so an exothermic reaction comes out negative.

4. JK4 · Final check

Model answer. Counting every bond: energy in = C=C 612 + 4 C–H 1648 + Cl–Cl 242 = 2502 kJ/mol; energy out = C–C 348 + 4 C–H 1648 + 2 C–Cl 676 = 2672 kJ/mol. ΔH = 2502 − 2672 = −170 kJ/mol. The four C–H bonds don't change, so counting only C=C and Cl–Cl in (854) and C–C and 2 C–Cl out (1024) gives the same −170 kJ/mol. Negative, so exothermic.

Checked automatically

  • (a) ΔH: -170 kJ/mol (exact value)
    • Common wrong answer 170 : Check the sign: ΔH = energy in − energy out.
    • Common wrong answer -434 : Ethene has a C=C double bond: use 612 for the bond broken, not 348.
    • Common wrong answer 168 : The product has two C–Cl bonds, one on each carbon atom.

A full answer includes

  • (a) I treated both sides the same way: every bond, or only the bonds that change.
  • (a) I worked out the energy in: 2502 kJ/mol for every bond, or 854 kJ/mol for the changed bonds only (either is right).
  • (a) I worked out the energy out: 2672 kJ/mol for every bond, or 1024 kJ/mol for the changed bonds only (either is right).
  • (a) I subtracted the energy out from the energy in.
  • (a) I got ΔH = −170 kJ/mol.
  • (b) I deduced exothermic, because ΔH is negative.
  • I did not use the C–C value, 348, for the C=C bond broken.

Also accept

  • −170 kJ/mol; exothermic

Watch for

  • “A double bond has the same bond energy as a single bond.” — A double bond has its own, larger bond energy. Use the value for C=C or C=O from the table, not the single bond.
  • “ΔH = energy out − energy in.” — That flips the sign. ΔH = energy taken in breaking bonds − energy given out making bonds, so an exothermic reaction comes out negative.

5. JK5 · Final check

Model answer. 2H2O2 has four O–H bonds and two O–O bonds; 2H2O has four O–H bonds and O2 has one O=O. The four O–H bonds are on both sides. 1852 + 2 × (O–O) − 2350 = −206, so 2 × (O–O) = 2350 − 206 − 1852 = 292 and O–O = 146 kJ/mol. Check: 1852 + 292 − 2350 = −206.

Checked automatically

  • O–O bond energy: 146 kJ/mol (exact value)
    • Common wrong answer 292 : That is two O–O bonds: 2H2O2 has two. Divide by 2.
    • Common wrong answer 352 : ΔH = energy in − energy out: 1852 + 2 × (O–O) − 2350 = −206.

A full answer includes

  • I wrote the energy in as 4 × 463 + 2 × (O–O) = 1852 + 2 × (O–O).
  • I worked out the energy out: 4 × 463 + 498 = 2350 kJ/mol.
  • I wrote 1852 + 2 × (O–O) − 2350 = −206.
  • I found 2 × (O–O) = 292.
  • I got O–O = 146 kJ/mol.
  • I did not stop at 292, the energy of two O–O bonds.

Also accept

  • O–O = 146 kJ/mol

Watch for

  • “The number in front of a formula can be ignored when you count bonds.” — Count the bonds in one molecule, then multiply by the number in front of the formula: every bond in the molecule is counted that many times.
  • “ΔH = energy out − energy in.” — That flips the sign. ΔH = energy taken in breaking bonds − energy given out making bonds, so an exothermic reaction comes out negative.