AP Chemistry: Endothermic vs Exothermic, Enthalpy

What This Is

Endothermic and exothermic reactions describe whether a process absorbs or releases heat energy. Enthalpy (?H) is the heat content of a system at constant pressure—it tells us how much energy is gained or lost in a reaction. This is critical for the AP exam because it appears in thermochemistry FRQs, bond energy calculations, and Hess’s Law problems. Real-world example: Instant cold packs (endothermic) use ammonium nitrate dissolving in water to absorb heat, while hand warmers (exothermic) release heat when iron oxidizes.

Key Terms & Concepts

• Enthalpy (H): Total heat content of a system at constant pressure. ?H = H_products – H_reactants (change in enthalpy).
• Exothermic reaction: Releases heat to surroundings (?H < 0). Example: Combustion of methane (CH? + 2O?-CO? + 2H?O + heat).
• Endothermic reaction: Absorbs heat from surroundings (?H > 0). Example: Photosynthesis (6CO? + 6H?O + light-C?HO? + 6O?).
• Standard enthalpy of formation (?H°f): Enthalpy change when 1 mole of a compound forms from its elements in standard states (e.g., ?H°f for H?O(l) = –285.8 kJ/mol).
• Bond enthalpy (bond energy): Energy required to break 1 mole of bonds in gaseous molecules. ?H = ?(bonds broken) – ?(bonds formed).
• Hess’s Law: The total enthalpy change for a reaction is the same whether it occurs in one step or multiple steps. Used to calculate ?H for reactions indirectly.
• Calorimetry: Experimental method to measure heat flow using q = mc?T (q = heat, m = mass, c = specific heat, ?T = temperature change).
• State function: Property that depends only on initial and final states (e.g., ?H, ?G, ?S), not the path taken.
• Surroundings vs. system: System = the reaction; surroundings = everything else (e.g., water in a calorimeter).
• First Law of Thermodynamics: Energy is conserved (?E = q + w, where q = heat, w = work). For reactions at constant pressure, ?H-?E (since work is minimal).

Step-by-Step: Solving Enthalpy Problems

1. Identify the type of problem

• Bond enthalpy?-Use ?H = ?(bonds broken) – ?(bonds formed).
• Formation enthalpies?-Use ?H°rxn = H°f(products) – H°f(reactants).
• Calorimetry?-Use q = mc?T, then relate to moles of reactant.
• Hess’s Law?-Manipulate given reactions to match the target reaction.

2. Write the balanced equation

• Always start with a balanced chemical equation (AP loves testing this!).
• Example: For the combustion of propane (C?H? + 5O?-3CO? + 4H?O), ensure coefficients are correct.

3. Calculate ?H using the appropriate method

• Bond enthalpy example:
• Break all bonds in reactants (C?H?: 2 C–C + 8 C–H; 5O?: 5 O=O).
• Form all bonds in products (3CO?: 6 C=O; 4H?O: 8 O–H).
• ?H = (energy to break bonds) – (energy to form bonds).
• Formation enthalpy example:
• ?H°rxn = [3?H°f(CO?) + 4?H°f(H?O)] – [?H°f(C?H?) + 5?H°f(O?)].
• Note: ?H°f for elements (e.g., O?) = 0 kJ/mol.

4. Check units and signs

• ?H < 0-Exothermic (heat released).
• ?H > 0-Endothermic (heat absorbed).
• Units: kJ/mol (per mole of reaction as written).

5. Apply Hess’s Law (if needed)

• Reverse a reaction?-Flip the sign of ?H.
• Multiply coefficients?-Multiply ?H by the same factor.
• Add reactions?-Add their ?H values.

Common Mistakes

Mistake 1: Forgetting to balance the equation before calculating ?H.

• Correction: Always balance first! Coefficients affect ?H (e.g., 2H? + O?-2H?O has twice the ?H of H? + ½O?-H?O).

Mistake 2: Mixing up bond breaking (endothermic) and bond forming (exothermic).

• Correction: Breaking bonds requires energy (?H > 0), forming bonds releases energy (?H < 0). Think: "Break in, form out."

Mistake 3: Using ?H°f for elements (e.g., O?, N?) in calculations.

• Correction: ?H°f for elements in their standard states = 0 kJ/mol. Don’t include them in H°f(reactants)!

Mistake 4: Ignoring the sign of ?H in calorimetry problems.

• Correction: If the system loses heat, q_system is negative (exothermic). If the surroundings gain heat, q_surroundings is positive.

Mistake 5: Confusing ?H with ?E (internal energy).

• Correction: For reactions at constant pressure, ?H-?E (since work is negligible). At constant volume, ?E = q_v.

AP Exam Insights

1. FRQs often test Hess’s Law and bond enthalpy.

• Example: Given 3 reactions and their ?H values, calculate ?H for a target reaction.
• Trick: Watch for reversed reactions (flip ?H sign) or scaled coefficients (multiply ?H).

2. Multiple-choice traps:

• ?H vs. ?S vs. ?G: ?H is heat, ?S is entropy (disorder), ?G is spontaneity. Don’t mix them up!
• Units: kJ vs. kJ/mol. AP often asks for per mole of reaction (e.g., "per mole of CO? produced").
• State of matter: ?H°f for H?O(l)-H?O(g). Always check the phase (solid, liquid, gas).

3. Calorimetry problems require careful attention to signs.

• Example: If a reaction in a calorimeter warms the water, the reaction is exothermic (?H < 0), but q_water is positive (since it gains heat).

4. Bond enthalpy problems are less precise than ?H°f.

• Why? Bond enthalpies are averages (e.g., C–H bond energy varies slightly in different molecules). ?H°f is more accurate.

Quick Check Questions

1. Multiple Choice:

For the reaction: 2H?(g) + O?(g)-2H?O(l) ?H = –571.6 kJ What is ?H for the formation of 1 mole of H?O(l)? (A) –285.8 kJ (B) –571.6 kJ (C) +285.8 kJ (D) +571.6 kJ

Answer: (A) –285.8 kJ Explanation: The given ?H is for 2 moles of H?O. Divide by 2 to get ?H per mole.

2. Short FRQ:

Given the following data: - ?H°f for CO?(g) = –393.5 kJ/mol - ?H°f for H?O(l) = –285.8 kJ/mol - ?H°f for C?H?(g) = –103.8 kJ/mol

Calculate ?H°rxn for the combustion of propane: C?H?(g) + 5O?(g)-3CO?(g) + 4H?O(l)

Answer: ?H°rxn = [3(–393.5) + 4(–285.8)] – [–103.8 + 5(0)] = –2219.9 kJ Explanation: Use ?H°rxn = H°f(products) – H°f(reactants). ?H°f for O? = 0.

Last-Minute Cram Sheet

1. Exothermic: ?H < 0 (heat released, feels hot).
2. Endothermic: ?H > 0 (heat absorbed, feels cold).
3. ?H = ?(bonds broken) – ?(bonds formed) (bond enthalpy).
4. ?H°rxn = H°f(products) – H°f(reactants) (formation enthalpy).
5. ?H°f for elements (e.g., O?, N?) = 0 kJ/mol .
6. Hess’s Law: Flip reaction-flip ?H sign; multiply coefficients-multiply ?H.
7. q = mc?T (calorimetry: q = heat, m = mass, c = specific heat, ?T = temp change).
8. State function: ?H depends only on initial/final states, not the path.
9. Combustion reactions are always exothermic (?H < 0).
10. Units matter! kJ vs. kJ/mol (AP often asks for per mole).