AP Chemistry · Topic 6.7

Bond Enthalpies Practice

Part of Thermodynamics.(ENE-3.B)

Practice questions

8

Want a predicted score for the whole AP CHEM exam? Take the 20-question diagnostic and Lumi will plan the rest.

Sample questions

5 of 8 — sign in to practice the rest with adaptive difficulty and mastery tracking.

  1. Sample 1difficulty 2/5

    Bar heights show bond dissociation energies in kJ/mol.

    C-C 347 C=C 614 C≡C 839 C-O 358

    Which bond requires the most energy to break?

    • A

      C≡C

      check_circle
    • B

      C-C

    • C

      C-O

    • D

      C=C

    Why

    The triple bond C≡C (839 kJ/mol) has the highest bond energy because three shared electron pairs make it the strongest of the bonds shown.

  2. Sample 2difficulty 3/5

    ΔH_rxn ≈

    • A

      Σ(bonds broken) + Σ(bonds formed)

    • B

      Σ(bonds broken) − Σ(bonds formed)

      check_circle
    • C

      0

    • D

      Σ(bonds formed) − Σ(bonds broken)

    Why

    Breaking bonds requires energy (positive); forming bonds releases energy (negative).

  3. Sample 3difficulty 3/5

    For H₂(g) + Cl₂(g) → 2 HCl(g)

    H-H 436 Cl-Cl 243 H-Cl 432

    Using bond energies (kJ/mol), what is ΔH for the reaction?

    • A

      +679 kJ

    • B

      -185 kJ

      check_circle
    • C

      -247 kJ

    • D

      +185 kJ

    Why

    ΔH = (bonds broken) - (bonds formed) = (436 + 243) - 2(432) = 679 - 864 = -185 kJ.

  4. Sample 4difficulty 3/5

    Using bond enthalpies (kJ/mol): H-H 436, Cl-Cl 243, H-Cl 432. For H₂ + Cl₂ → 2 HCl, ΔH ≈

    • A

      +247

    • B

      −185

      check_circle
    • C

      −247

    • D

      +185

    Why

    Bonds broken: 436 + 243 = 679. Bonds formed: 2 × 432 = 864. ΔH ≈ 679 − 864 = −185 kJ/mol (exothermic).

  5. Sample 5difficulty 4/5

    For 2 H₂(g) + O₂(g) → 2 H₂O(g) (each H₂O has 2 O-H bonds)

    H-H 436 O=O 498 O-H 463

    Estimate ΔH using bond energies (kJ/mol).

    • A

      +482 kJ

    • B

      -1852 kJ

    • C

      -482 kJ

      check_circle
    • D

      -241 kJ

    Why

    Bonds broken: 2(436) + 498 = 1370. Bonds formed: 4(463) = 1852. ΔH = 1370 - 1852 = -482 kJ.