AP Chemistry Quiz: Bond Enthalpies

14 questions · exam conditions

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Bond EnthalpiesQuestion 1 of 14

Use average bond enthalpies to estimate $\Delta H$ for the reaction $\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)}$ . The average bond enthalpies are: $\mathrm{H-H}=436\ \text{kJ/mol}$ , $\mathrm{Cl-Cl}=243\ \text{kJ/mol}$ , and $\mathrm{H-Cl}=431\ \text{kJ/mol}$ . What is the estimated enthalpy change, $\Delta H$ , for the reaction?

\Delta H = -62\ \text{kJ/mol}

\Delta H = -183\ \text{kJ/mol}

\Delta H = -305\ \text{kJ/mol}

\Delta H = +183\ \text{kJ/mol}

\Delta H = +62\ \text{kJ/mol}

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AP Chemistry Quiz

AP Chemistry Quiz: Bond Enthalpies

Practice Bond Enthalpies in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Bond Enthalpies, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\Delta H = -62\ \text{kJ/mol}$
$\Delta H = -183\ \text{kJ/mol}$ (correct answer)
$\Delta H = -305\ \text{kJ/mol}$
$\Delta H = +183\ \text{kJ/mol}$
$\Delta H = +62\ \text{kJ/mol}$

Explanation: This question tests the skill of using average bond enthalpies to estimate the enthalpy change for a chemical reaction. To estimate ΔH, calculate the total energy required to break the bonds in the reactants and subtract the total energy released when forming the bonds in the products. For the reaction H2 + Cl2 → 2HCl, the bonds broken are one H-H bond (436 kJ/mol) and one Cl-Cl bond (243 kJ/mol), totaling 679 kJ/mol. The bonds formed are two H-Cl bonds (2 × 431 kJ/mol = 862 kJ/mol), so ΔH = 679 - 862 = -183 kJ/mol, indicating an exothermic reaction. A tempting distractor is +183 kJ/mol, which results from the misconception of subtracting the energy of bonds broken from the energy of bonds formed instead of the reverse. A transferable strategy is to always remember that ΔH ≈ Σ(bond enthalpies broken) - Σ(bond enthalpies formed), where a negative value means heat is released.

Question 2

Use the average bond enthalpies given to estimate $\Delta H$ for the reaction: $\mathrm{CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g)}$ . Bond enthalpies: $\mathrm{C-H}=413\ \text{kJ/mol}$ , $\mathrm{Cl-Cl}=243\ \text{kJ/mol}$ , $\mathrm{C-Cl}=328\ \text{kJ/mol}$ , $\mathrm{H-Cl}=431\ \text{kJ/mol}$ . What is the estimated enthalpy change, $\Delta H$ , for the reaction?

$+103\ \text{kJ/mol}$
$+87\ \text{kJ/mol}$
$-103\ \text{kJ/mol}$ (correct answer)
$-87\ \text{kJ/mol}$
$-431\ \text{kJ/mol}$

Explanation: This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g), so bonds broken are one C-H bond at 413 kJ/mol and one Cl-Cl bond at 243 kJ/mol, totaling 656 kJ/mol of energy absorbed. Bonds formed are one C-Cl bond at 328 kJ/mol and one H-Cl bond at 431 kJ/mol, totaling 759 kJ/mol of energy released. Therefore, the estimated ΔH is 656 - 759 = -103 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +103 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

Question 3

Use average bond enthalpies (in kJ/mol) to estimate $\Delta H$ for combustion: $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}$ . Bonds broken: 4 C–H (413), 2 O=O (498). Bonds formed: 2 C=O in CO $_2$ (799), 4 O–H (463). What is the estimated enthalpy change?

+802 kJ/mol
−1204 kJ/mol
−401 kJ/mol
−802 kJ/mol (correct answer)
+401 kJ/mol

Explanation: This question tests the skill of estimating the enthalpy change of a reaction using average bond enthalpies. For the combustion of methane, the bonds broken are four C–H bonds (413 kJ/mol each) and two O=O bonds (498 kJ/mol each), totaling 2648 kJ/mol. The bonds formed are two C=O bonds (799 kJ/mol each) and four O–H bonds (463 kJ/mol each), totaling 3450 kJ/mol. Thus, ΔH ≈ 2648 - 3450 = -802 kJ/mol. A tempting distractor is +802 kJ/mol, arising from the misconception of subtracting in the wrong order. A transferable strategy is to list all bonds in reactants and products systematically to ensure none are missed in complex molecules.

Question 4

Use average bond enthalpies to estimate $\Delta H$ for the reaction $\mathrm{H_2(g) + F_2(g) \rightarrow 2HF(g)}$ . The average bond enthalpies are: $\mathrm{H-H}=436\ \text{kJ/mol}$ , $\mathrm{F-F}=159\ \text{kJ/mol}$ , and $\mathrm{H-F}=565\ \text{kJ/mol}$ . What is the estimated enthalpy change, $\Delta H$ , for the reaction?

$\Delta H = -535\ \text{kJ/mol}$ (correct answer)
$\Delta H = +535\ \text{kJ/mol}$
$\Delta H = -29\ \text{kJ/mol}$
$\Delta H = +29\ \text{kJ/mol}$
$\Delta H = -406\ \text{kJ/mol}$

Explanation: This question tests the skill of using average bond enthalpies to estimate the enthalpy change for a chemical reaction. To estimate ΔH, calculate the total energy required to break the bonds in the reactants and subtract the total energy released when forming the bonds in the products. For the reaction H2 + F2 → 2HF, the bonds broken are one H-H bond (436 kJ/mol) and one F-F bond (159 kJ/mol), totaling 595 kJ/mol. The bonds formed are two H-F bonds (2 × 565 kJ/mol = 1130 kJ/mol), so ΔH = 595 - 1130 = -535 kJ/mol, indicating a strongly exothermic reaction. A tempting distractor is +535 kJ/mol, which results from the misconception of reversing the subtraction, calculating bonds formed minus bonds broken. A transferable strategy is to double-check the arithmetic and ensure all bonds are accounted for according to the balanced equation.

Question 5

Use the average bond enthalpies given to estimate $\Delta H$ for the reaction: $\mathrm{H_2(g) + Br_2(g) \rightarrow 2HBr(g)}$ . Bond enthalpies: $\mathrm{H-H}=436\ \text{kJ/mol}$ , $\mathrm{Br-Br}=193\ \text{kJ/mol}$ , $\mathrm{H-Br}=366\ \text{kJ/mol}$ . What is the estimated enthalpy change, $\Delta H$ , for the reaction?

$-103\ \text{kJ/mol}$ (correct answer)
$+103\ \text{kJ/mol}$
$-69\ \text{kJ/mol}$
$+69\ \text{kJ/mol}$
$-732\ \text{kJ/mol}$

Explanation: This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is H₂(g) + Br₂(g) → 2HBr(g), so bonds broken are one H-H bond at 436 kJ/mol and one Br-Br bond at 193 kJ/mol, totaling 629 kJ/mol of energy absorbed. Bonds formed are two H-Br bonds at 366 kJ/mol each, totaling 732 kJ/mol of energy released. Therefore, the estimated ΔH is 629 - 732 = -103 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +103 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

Question 6

Use the average bond enthalpies given to estimate $\Delta H$ for the reaction: $\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)}$ . Bond enthalpies: $\mathrm{H-H}=436\ \text{kJ/mol}$ , $\mathrm{Cl-Cl}=243\ \text{kJ/mol}$ , $\mathrm{H-Cl}=431\ \text{kJ/mol}$ . What is the estimated enthalpy change, $\Delta H$ , for the reaction?

$-183\ \text{kJ/mol}$ (correct answer)
$+183\ \text{kJ/mol}$
$-110\ \text{kJ/mol}$
$+110\ \text{kJ/mol}$
$-862\ \text{kJ/mol}$

Explanation: This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is H₂(g) + Cl₂(g) → 2HCl(g), so bonds broken are one H-H bond at 436 kJ/mol and one Cl-Cl bond at 243 kJ/mol, totaling 679 kJ/mol of energy absorbed. Bonds formed are two H-Cl bonds at 431 kJ/mol each, totaling 862 kJ/mol of energy released. Therefore, the estimated ΔH is 679 - 862 = -183 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +183 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

Question 7

Use average bond enthalpies (in kJ/mol) to estimate $\Delta H$ for: $\mathrm{C_2H_4(g) + Cl_2(g) \rightarrow C_2H_4Cl_2(g)}$ . Bonds broken: 1 C=C (614), 1 Cl–Cl (243). Bonds formed: 1 C–C (347), 2 C–Cl (338). What is the estimated enthalpy change?

−166 kJ/mol (correct answer)
+166 kJ/mol
−83 kJ/mol
+83 kJ/mol
−249 kJ/mol

Explanation: This question tests the skill of estimating the enthalpy change of a reaction using average bond enthalpies. For this addition reaction, the bonds broken are one C=C bond (614 kJ/mol) and one Cl–Cl bond (243 kJ/mol), summing to 857 kJ/mol. The bonds formed are one C–C bond (347 kJ/mol) and two C–Cl bonds (338 kJ/mol each), summing to 1023 kJ/mol. Thus, ΔH ≈ 857 - 1023 = -166 kJ/mol. A tempting distractor is +166 kJ/mol, stemming from the misconception of reversing the formula's subtraction. A transferable strategy is to confirm that the bond counts match the molecular changes in addition reactions to multiple bonds.

Question 8

Use the average bond enthalpies in the table to estimate $\Delta H$ for the reaction: $\mathrm{H_2(g) + Br_2(g) \rightarrow 2HBr(g)}$ . Assume $\Delta H \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$ .

Average bond enthalpies (kJ/mol): H–H = 436, Br–Br = 193, H–Br = 366.

−629 kJ/mol
−173 kJ/mol
−103 kJ/mol (correct answer)
+629 kJ/mol
+103 kJ/mol

Explanation: This question tests the ability to estimate the enthalpy change (ΔH) of a reaction using average bond enthalpies. For the reaction H₂(g) + Br₂(g) → 2HBr(g), the bonds broken are one H–H bond (436 kJ/mol) and one Br–Br bond (193 kJ/mol), totaling 629 kJ/mol. The bonds formed are two H–Br bonds (366 kJ/mol each), totaling 732 kJ/mol. Therefore, ΔH ≈ 629 - 732 = -103 kJ/mol, showing the reaction is exothermic due to stronger bonds formed compared to those broken. A tempting distractor is +103 kJ/mol, stemming from the misconception of subtracting broken from formed enthalpies, leading to an incorrect positive sign. A transferable strategy is to account for the stoichiometry by multiplying bond enthalpies by the number of bonds, then compute ΔH as sum(broken) - sum(formed) for reliable estimates in gaseous reactions.

Question 9

Use average bond enthalpies (in kJ/mol) to estimate $\Delta H$ for the reaction: $\mathrm{H_2(g) + Cl_2(g) \rightarrow 2\,HCl(g)}$ . Bonds broken: 1 H–H (436), 1 Cl–Cl (243). Bonds formed: 2 H–Cl (431). What is the estimated enthalpy change, using $\Delta H \approx \sum E_{\text{broken}}-\sum E_{\text{formed}}$ ?

+183 kJ/mol
−183 kJ/mol (correct answer)
−62 kJ/mol
+62 kJ/mol
−305 kJ/mol

Explanation: This question tests the skill of estimating the enthalpy change of a reaction using average bond enthalpies. To estimate ΔH, calculate the total energy required to break the bonds in the reactants and subtract the total energy released when forming the bonds in the products. For this reaction, the bonds broken are one H–H bond (436 kJ/mol) and one Cl–Cl bond (243 kJ/mol), summing to 679 kJ/mol. The bonds formed are two H–Cl bonds (431 kJ/mol each), summing to 862 kJ/mol, so ΔH ≈ 679 - 862 = -183 kJ/mol. A tempting distractor is +183 kJ/mol, which results from the misconception of reversing the subtraction, treating bond formation as energy input instead of release. A transferable strategy is to consistently remember that bond breaking is endothermic and bond forming is exothermic, ensuring the correct sign for ΔH.

Question 10

Use the average bond enthalpies given to estimate $\Delta H$ for the reaction: $\mathrm{CH_2=CH_2(g) + HCl(g) \rightarrow CH_3CH_2Cl(g)}$ . Bond enthalpies: $\mathrm{C=C}=614\ \text{kJ/mol}$ , $\mathrm{H-Cl}=431\ \text{kJ/mol}$ , $\mathrm{C-C}=348\ \text{kJ/mol}$ , $\mathrm{C-H}=413\ \text{kJ/mol}$ , $\mathrm{C-Cl}=328\ \text{kJ/mol}$ . What is the estimated enthalpy change, $\Delta H$ , for the reaction?

$-741\ \text{kJ/mol}$
$-44\ \text{kJ/mol}$ (correct answer)
$+170\ \text{kJ/mol}$
$-170\ \text{kJ/mol}$
$+44\ \text{kJ/mol}$

Explanation: This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is CH₂=CH₂(g) + HCl(g) → CH₃CH₂Cl(g), so bonds broken are one C=C bond at 614 kJ/mol and one H-Cl bond at 431 kJ/mol, totaling 1045 kJ/mol of energy absorbed. Bonds formed are one C-C bond at 348 kJ/mol, one C-H bond at 413 kJ/mol, and one C-Cl bond at 328 kJ/mol, totaling 1089 kJ/mol of energy released. Therefore, the estimated ΔH is 1045 - 1089 = -44 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +44 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

Question 11

Use the average bond enthalpies given to estimate $\Delta H$ for the reaction: $\mathrm{C_2H_4(g) + Cl_2(g) \rightarrow C_2H_4Cl_2(g)}$ (addition of $\mathrm{Cl_2}$ across the double bond). Bond enthalpies: $\mathrm{C=C}=614\ \text{kJ/mol}$ , $\mathrm{Cl-Cl}=243\ \text{kJ/mol}$ , $\mathrm{C-C}=348\ \text{kJ/mol}$ , $\mathrm{C-Cl}=328\ \text{kJ/mol}$ . What is the estimated enthalpy change, $\Delta H$ , for the reaction?

$-656\ \text{kJ/mol}$
$-147\ \text{kJ/mol}$ (correct answer)
$-25\ \text{kJ/mol}$
$+147\ \text{kJ/mol}$
$+25\ \text{kJ/mol}$

Explanation: This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is C₂H₄(g) + Cl₂(g) → C₂H₄Cl₂(g), so bonds broken are one C=C bond at 614 kJ/mol and one Cl-Cl bond at 243 kJ/mol, totaling 857 kJ/mol of energy absorbed. Bonds formed are one C-C bond at 348 kJ/mol and two C-Cl bonds at 328 kJ/mol each, totaling 1004 kJ/mol of energy released. Therefore, the estimated ΔH is 857 - 1004 = -147 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +147 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

Question 12

Use average bond enthalpies to estimate $\Delta H$ for the reaction: $\mathrm{C_2H_4(g) + Cl_2(g) \rightarrow C_2H_4Cl_2(g)}$ (addition of $\mathrm{Cl_2}$ across the double bond). Given average bond enthalpies: $\mathrm{C=C}=614\ \text{kJ/mol}$ , $\mathrm{Cl-Cl}=243\ \text{kJ/mol}$ , $\mathrm{C-C}=347\ \text{kJ/mol}$ , $\mathrm{C-Cl}=328\ \text{kJ/mol}$ . What is the estimated $\Delta H$ for the reaction?

$-146\ \text{kJ/mol}$ (correct answer)
$-66\ \text{kJ/mol}$
$+146\ \text{kJ/mol}$
$-590\ \text{kJ/mol}$
$+66\ \text{kJ/mol}$

Explanation: This question tests your ability to calculate enthalpy change for an addition reaction across a double bond. Breaking bonds requires: 1 C=C double bond (614 kJ/mol) + 1 Cl-Cl bond (243 kJ/mol) = 857 kJ/mol total. Forming bonds releases: 1 C-C single bond (347 kJ/mol) + 2 C-Cl bonds (2 × 328 = 656 kJ/mol) = 1003 kJ/mol total. Therefore, ΔH = 857 - 1003 = -146 kJ/mol. A common error is forgetting that when Cl₂ adds across the double bond, two new C-Cl bonds form (one Cl atom bonds to each carbon). Always visualize the molecular structure change to count bonds correctly.

Question 13

Use average bond enthalpies to estimate $\Delta H$ for the reaction: $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ Given average bond enthalpies: $\mathrm{N\equiv N}=945\ \text{kJ/mol}$ , $\mathrm{H-H}=436\ \text{kJ/mol}$ , $\mathrm{N-H}=391\ \text{kJ/mol}$ . What is the estimated $\Delta H$ for the reaction?

$-93\ \text{kJ/mol}$ (correct answer)
$+93\ \text{kJ/mol}$
$-111\ \text{kJ/mol}$
$+111\ \text{kJ/mol}$
$-204\ \text{kJ/mol}$

Explanation: This question tests your ability to calculate enthalpy change for the Haber process using bond enthalpies. Breaking bonds requires: 1 N≡N triple bond (945 kJ/mol) + 3 H-H bonds (3 × 436 = 1308 kJ/mol) = 2253 kJ/mol total. Forming bonds releases: 6 N-H bonds in 2 NH₃ molecules (6 × 391 = 2346 kJ/mol). Therefore, ΔH = 2253 - 2346 = -93 kJ/mol. A common mistake is forgetting that each NH₃ molecule contains 3 N-H bonds, so 2 NH₃ molecules contain 6 N-H bonds total. Always count bonds carefully by drawing out the molecular structures if needed.

Question 14

Use average bond enthalpies to estimate $\Delta H$ for the reaction: $\mathrm{H_2(g) + Br_2(g) \rightarrow 2HBr(g)}$ Given average bond enthalpies: $\mathrm{H-H}=436\ \text{kJ/mol}$ , $\mathrm{Br-Br}=193\ \text{kJ/mol}$ , $\mathrm{H-Br}=366\ \text{kJ/mol}$ . What is the estimated $\Delta H$ for the reaction?

$-103\ \text{kJ/mol}$ (correct answer)
$+103\ \text{kJ/mol}$
$-176\ \text{kJ/mol}$
$+176\ \text{kJ/mol}$
$-629\ \text{kJ/mol}$

Explanation: This question tests your ability to calculate enthalpy change using bond enthalpies for halogen reactions. Breaking bonds requires: 1 H-H bond (436 kJ/mol) + 1 Br-Br bond (193 kJ/mol) = 629 kJ/mol total. Forming bonds releases: 2 H-Br bonds (2 × 366 = 732 kJ/mol). Therefore, ΔH = 629 - 732 = -103 kJ/mol. A common mistake is using the wrong bond energy values or forgetting to multiply by 2 when forming two HBr molecules. Remember that the reaction stoichiometry determines how many bonds of each type are broken and formed.