Calculating ΔH from bond energies — answer key

1. JQ8 · Practice

Model answer. ΔH belongs to the chemicals: start with the energy they take in breaking bonds and subtract the energy they give out making bonds. When more is given out, the answer is negative: exothermic.

A full answer includes

• I chose energy taken in breaking bonds − energy given out making bonds.
• I checked that an exothermic reaction then gives a negative ΔH.
• I did not choose energy given out − energy taken in.

Also accept

• Energy in (breaking) − energy out (making)
• Energy absorbed breaking bonds − energy released 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.
• “ΔH = energy in + energy out.” — ΔH is the difference between the two totals. Adding gives a number bigger than either total, and it cannot be negative, so it cannot show an exothermic reaction.
• “ΔH is always written as a positive number.” — The sign is part of the answer: negative for exothermic, positive for endothermic. Dropping it hides which way the energy goes.

2. JQ9 · Practice

Model answer. Bonds broken: one H–H and one Br–Br, 436 + 193 = 629 kJ/mol. Bonds made: two H–Br, 2 × 366 = 732 kJ/mol. ΔH = 629 − 732 = −103 kJ/mol: exothermic.

Checked automatically

• (a) Energy in: 629 kJ/mol (exact value)
• (b) Energy out: 732 kJ/mol (exact value)
  • Common wrong answer 366 : That is one H–Br bond. 2HBr has two: 2 × 366.
• (c) ΔH: -103 kJ/mol (exact value)
  • Common wrong answer 103 : Check the sign: ΔH = energy in − energy out = 629 − 732.
  • Common wrong answer 263 : Energy out is 2 × 366 = 732, so ΔH = 629 − 732.

A full answer includes

• (a) I added one H–H and one Br–Br: 436 + 193.
• (a) I got 629 kJ/mol.
• (b) I multiplied the H–Br bond energy by 2: 2 × 366.
• (b) I got 732 kJ/mol.
• (c) I worked out 629 − 732.
• (c) I got ΔH = −103 kJ/mol.
• I did not use only one H–Br bond for 2HBr.

Also accept

• 629; 732; −103 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.

3. JQ10 · Practice

Model answer. 2H₂ is two hydrogen molecules with one H–H bond each: 2 H–H. O₂ has one O=O bond. Each H₂O has two O–H bonds, and there are two water molecules: 2 × 2 = 4 O–H.

Checked automatically

• (a) H–H bonds broken: 2 (exact value)
  • Common wrong answer 1 : 2H₂ is two molecules, each with one H–H bond.
• (b) O=O bonds broken: 1 (exact value)
  • Common wrong answer 2 : One O₂ molecule has one O=O bond: a double bond counts as one bond.
• (c) O–H bonds made: 4 (exact value)
  • Common wrong answer 2 : Each water molecule has two O–H bonds, and there are two molecules: 2 × 2.

A full answer includes

• (a) I counted 2 H–H bonds.
• (b) I counted 1 O=O bond.
• (c) I counted 2 O–H bonds in each water molecule.
• (c) I got 4 O–H bonds for 2H₂O.
• I did not count the double bond in O₂ as two bonds.

Also accept

• 2 H–H; 1 O=O; 4 O–H

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.

4. JQ11 · Practice

Model answer. Energy in = 2 × 436 + 498 = 1370 kJ/mol. Energy out = 4 × 463 = 1852 kJ/mol. ΔH = 1370 − 1852 = −482 kJ/mol: burning hydrogen is exothermic.

Checked automatically

• ΔH: -482 kJ/mol (exact value)
  • Common wrong answer 482 : Check the sign: ΔH = energy in − energy out = 1370 − 1852.
  • Common wrong answer -918 : 2H₂ has two H–H bonds: energy in = 2 × 436 + 498 = 1370.
  • Common wrong answer 444 : Each water molecule has two O–H bonds: energy out = 4 × 463 = 1852.
  • Common wrong answer 3222 : Subtract, don't add: ΔH = energy in − energy out.

A full answer includes

• I worked out the energy in: 2 × 436 + 498 = 1370 kJ/mol.
• I worked out the energy out: 4 × 463 = 1852 kJ/mol.
• I worked out 1370 − 1852.
• I got ΔH = −482 kJ/mol.
• I did not count only two O–H bonds for 2H₂O.

Also accept

• −482 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.

5. JQ12 · Practice

Model answer. Bonds broken: one N≡N and three H–H, 945 + 1308 = 2253 kJ/mol. Bonds made: 2NH₃ has 2 × 3 = 6 N–H bonds, 6 × 391 = 2346 kJ/mol. ΔH = 2253 − 2346 = −93 kJ/mol for two moles of ammonia. One mole is half of that: −93 ÷ 2 = −46.5 kJ/mol, still negative.

Checked automatically

• (a) ΔH for N₂ + 3H₂ → 2NH₃: -93 kJ/mol (exact value)
  • Common wrong answer 93 : Check the sign: ΔH = energy in − energy out = 2253 − 2346.
  • Common wrong answer -965 : 3H₂ has three H–H bonds: energy in = 945 + 3 × 436 = 2253.
  • Common wrong answer 1080 : 2NH₃ has six N–H bonds: energy out = 6 × 391 = 2346.
• (b) ΔH for making 1 mol of NH₃: -46.5 kJ/mol (3 significant figures)
  • Common wrong answer 46.5 : Keep the sign: making ammonia is exothermic, so ΔH is negative.
  • Common wrong answer -93 : That is for two moles of ammonia. Halve it for one mole.
  • Common wrong answer -186 : Divide by 2, don't multiply: one mole is half the equation.

A full answer includes

• (a) I worked out the energy in: 945 + 3 × 436 = 2253 kJ/mol.
• (a) I worked out the energy out: 6 × 391 = 2346 kJ/mol.
• (a) I worked out 2253 − 2346.
• (a) I got ΔH = −93 kJ/mol.
• (b) I divided by 2: −93 ÷ 2.
• (b) I got −46.5 kJ/mol, to 3 significant figures.
• I did not count only three N–H bonds for 2NH₃.

Also accept

• −93 kJ/mol; −46.5 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.
• “ΔH stays the same however much is made.” — ΔH is for the amounts in the equation. Half the amounts break and make half the bonds, so ΔH halves too.

6. JQ13 · Practice

Model answer. Bonds broken: 4 C–H and one Cl–Cl, 1648 + 242 = 1890 kJ/mol. Bonds made: 3 C–H, one C–Cl and one H–Cl, 1236 + 338 + 431 = 2005 kJ/mol. ΔH = 1890 − 2005 = −115 kJ/mol. Only one C–H bond really changes; counting only the changed bonds (654 in, 769 out) gives the same ΔH.

Checked automatically

• (a) Energy in, every bond in CH₄ and Cl₂: 1890 kJ/mol (exact value)
  • Common wrong answer 654 : That counts only the bonds that change, which gives the right ΔH. Here count every bond: four C–H and one Cl–Cl.
• (b) Energy out, every bond in CH₃Cl and HCl: 2005 kJ/mol (exact value)
  • Common wrong answer 769 : That counts only the bonds that change. Count every bond: three C–H, one C–Cl and one H–Cl.
• (c) ΔH: -115 kJ/mol (exact value)
  • Common wrong answer 115 : Check the sign: ΔH = energy in − energy out = 1890 − 2005.

A full answer includes

• (a) I worked out 4 × 412 + 242.
• (a) I got 1890 kJ/mol.
• (b) I worked out 3 × 412 + 338 + 431.
• (b) I got 2005 kJ/mol.
• (c) I worked out 1890 − 2005.
• (c) I got ΔH = −115 kJ/mol.
• I did not leave the three unchanged C–H bonds out of (a) and (b).

Also accept

• 1890; 2005; −115 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.

7. JQ14 · Practice

Model answer. Energy in: 4 C–H and 2 O–H, 1648 + 926 = 2574 kJ/mol. Energy out: one C≡O and three H–H, 1077 + 1308 = 2385 kJ/mol. ΔH = 2574 − 2385 = +189 kJ/mol: endothermic, so the reaction takes energy in from its surroundings and must be heated all the time.

Checked automatically

• (a) ΔH: 189 kJ/mol (exact value)
  • Common wrong answer -189 : Check the sign: ΔH = energy in − energy out = 2574 − 2385, which is positive.
  • Common wrong answer -274 : Water has two O–H bonds: energy in = 4 × 412 + 2 × 463.
  • Common wrong answer 1061 : There are three H₂ molecules: energy out = 1077 + 3 × 436.
  • Common wrong answer 461 : Carbon monoxide has a triple bond: use C≡O = 1077, not C=O.

A full answer includes

• (a) I worked out the energy in: 4 × 412 + 2 × 463 = 2574 kJ/mol.
• (a) I worked out the energy out: 1077 + 3 × 436 = 2385 kJ/mol.
• (a) I worked out 2574 − 2385.
• (a) I got ΔH = +189 kJ/mol.
• (b) I wrote that a positive ΔH means the reaction is endothermic.
• (b) I wrote that the reaction takes energy in from its surroundings all the time it runs.
• I did not use the C=O value, 805, for carbon monoxide.

Also accept

• +189 kJ/mol; endothermic, so it must be heated

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.
• “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.

8. JQ15 · Practice

Model answer. Energy in = 436 + 151 = 587 kJ/mol; energy out = 2 × (H–I). So 587 − 2 × (H–I) = −9, which gives 2 × (H–I) = 587 + 9 = 596 and H–I = 298 kJ/mol. Check: 587 − 596 = −9.

Checked automatically

• H–I bond energy: 298 kJ/mol (exact value)
  • Common wrong answer 596 : That is the energy of two H–I bonds. Divide by 2.
  • Common wrong answer 289 : ΔH is −9, so 2 × (H–I) = 587 + 9 = 596, not 587 − 9.
  • Common wrong answer 587 : That is the energy in. Use 587 − 2 × (H–I) = −9.

A full answer includes

• I worked out the energy in: 436 + 151 = 587 kJ/mol.
• I wrote 587 − 2 × (H–I) = −9.
• I found 2 × (H–I) = 596.
• I got H–I = 298 kJ/mol.
• I did not stop at 596, the energy of two H–I bonds.

Also accept

• H–I = 298 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.

9. JQ16 · Practice

Model answer. Ethene's double bond needs the C=C value, 612, not C–C. 3O₂ has three O=O bonds, not one. And ΔH = energy in − energy out. Energy in = 612 + 1648 + 1494 = 3754 kJ/mol; energy out = 3220 + 1852 = 5072 kJ/mol, which the pupil had right. ΔH = 3754 − 5072 = −1318 kJ/mol: a burning fuel, so negative.

Checked automatically

• (b) The correct ΔH: -1318 kJ/mol (exact value)
  • Common wrong answer 1318 : Check the sign: ΔH = energy in − energy out = 3754 − 5072.
  • Common wrong answer -2578 : Fix the energy in as well: C=C is 612, and 3O₂ has three O=O bonds.
  • Common wrong answer -1582 : Fix the bond energy too: ethene has a C=C double bond, 612, not C–C, 348.
  • Common wrong answer -2314 : Fix the number of O=O bonds too: 3O₂ has three.

A full answer includes

• (a) I identified that C–C (348) was used instead of C=C (612).
• (a) I identified that only one O=O bond was broken, not three for 3O₂.
• (a) I identified that energy in was subtracted from energy out, instead of energy out from energy in.
• (b) I worked out the energy in: 612 + 4 × 412 + 3 × 498 = 3754 kJ/mol.
• (b) I worked out 3754 − 5072.
• (b) I got ΔH = −1318 kJ/mol.
• I did not change the energy out: 5072 kJ/mol was right.

Also accept

• C=C not C–C; three O=O; in − out; ΔH = −1318 kJ/mol

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.
• “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.

10. JT2 · Apply

Model answer. Energy in: C–C 348, 5 C–H 2060, C–O 358, O–H 463 and 3 O=O 1494, total 4723 kJ/mol. Energy out: 4 C=O 3220 and 6 O–H 2778, total 5998 kJ/mol. ΔH = 4723 − 5998 = −1275 kJ/mol. It differs from −1367 because bond energies are averages taken over many molecules, and because they are for gases while the data book measures with liquid ethanol and liquid water.

Checked automatically

• (a) ΔH: -1275 kJ/mol (exact value)
  • Common wrong answer 1275 : Check the sign: ΔH = energy in − energy out = 4723 − 5998.
  • Common wrong answer -1738 : Ethanol has an O–H bond too (463): energy in = 4723.
  • Common wrong answer -2271 : 3O₂ has three O=O bonds: 3 × 498.
  • Common wrong answer 114 : Each of the three water molecules has two O–H bonds: 6 × 463.
  • Common wrong answer 335 : Each CO₂ has two C=O bonds, and there are two CO₂: 4 × 805.

A full answer includes

• (a) I counted one C–C, five C–H, one C–O and one O–H bond in ethanol.
• (a) I worked out the energy in: 348 + 5 × 412 + 358 + 463 + 3 × 498 = 4723 kJ/mol.
• (a) I worked out the energy out: 4 × 805 + 6 × 463 = 5998 kJ/mol.
• (a) I worked out 4723 − 5998.
• (a) I got ΔH = −1275 kJ/mol.
• (b) I suggested that bond energies are average values, not the exact values in these molecules (tick this line or the next).
• (b) I suggested that bond energies are for gases, but the data-book value is for liquid ethanol and liquid water (tick this line or the one before).
• (b) I did not blame heat lost to the air: ΔH from bond energies is calculated, not measured.

Also accept

• −1275 kJ/mol; bond energies are averages (or: they are for gases)

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.
• “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.
• “The calculated value is different because heat is lost to the air.” — No heat is lost in a calculation. The difference comes from the data: bond energies are averages, and they are for gases.