AP Chemistry: Hess’s Law and Standard Enthalpies of Formation

AP Chemistry – Hess’s Law and Standard Enthalpies of Formation

AP Chemistry: Hess’s Law and Standard Enthalpies of Formation

What This Is

Hess’s Law states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction takes. This is crucial for calculating enthalpy changes for reactions that are hard to measure directly (e.g., combustion of glucose in the body). A real-world example: Engineers use Hess’s Law to design energy-efficient industrial processes, like calculating the heat released when converting coal to synthetic fuels. On the AP exam, you’ll use Hess’s Law to manipulate and combine reactions to find unknown enthalpy changes, often involving standard enthalpies of formation (ΔH°f).

Key Terms & Concepts

• Enthalpy (H): The total heat content of a system at constant pressure. Measured in kJ/mol.
• Enthalpy change (ΔH): The heat absorbed or released in a reaction (ΔH = H_products – H_reactants).
• Standard enthalpy of formation (ΔH°f): The enthalpy change when 1 mole of a compound forms from its elements in their standard states (e.g., O₂(g), C(s, graphite)). ΔH°f for any element in its standard state = 0 kJ/mol.
• Standard state: The most stable form of a substance at 25°C (298 K) and 1 atm pressure (e.g., H₂O(l), not H₂O(g)).
• Hess’s Law: The total ΔH for a reaction is the sum of ΔH for each step, regardless of the pathway.
• State function: A property (like ΔH) that depends only on the initial and final states, not the path taken (e.g., altitude change on a hike, not the route).
• ΔH°rxn = Σ ΔH°f(products) – Σ ΔH°f(reactants): The formula to calculate the standard enthalpy change of a reaction using standard enthalpies of formation.
• Manipulating reactions for Hess’s Law:
• Reversing a reaction → Change the sign of ΔH.
• Multiplying a reaction by a coefficient → Multiply ΔH by the same coefficient.
• Bond enthalpy: The energy required to break 1 mole of bonds in a gaseous molecule (not the same as ΔH°f; used in a different Hess’s Law approach).

Step-by-Step / Process Flow

Using Hess’s Law to Find ΔH for a Target Reaction

1. Write the target reaction (the one you’re solving for).
2. List given reactions and their ΔH values. Identify how they relate to the target (e.g., reversed, scaled).
3. Manipulate reactions to match the target:
4. Reverse reactions → Flip ΔH sign.
5. Multiply/divide reactions → Scale ΔH by the same factor.
6. Cancel intermediates (substances that appear on both sides of the combined reactions).
7. Add the manipulated ΔH values to get the total ΔH for the target reaction.

Using Standard Enthalpies of Formation (ΔH°f)

1. Write the balanced target reaction.
2. Look up ΔH°f values for all products and reactants (elements in standard states = 0).
3. Apply the formula:
   ΔH°rxn = Σ ΔH°f(products) – Σ ΔH°f(reactants).
4. Multiply each ΔH°f by its stoichiometric coefficient in the balanced equation.
5. Calculate the total ΔH°rxn.

Common Mistakes

• Mistake: Forgetting to reverse the sign of ΔH when reversing a reaction.
  Correction: If you flip a reaction (e.g., A → B becomes B → A), change the sign of ΔH (e.g., +100 kJ → –100 kJ).

• Mistake: Not scaling ΔH when multiplying a reaction by a coefficient.
  Correction: If you multiply a reaction by 2 (e.g., A → B becomes 2A → 2B), multiply ΔH by 2.

• Mistake: Using ΔH°f for elements in non-standard states (e.g., O(g) instead of O₂(g)).
  Correction: ΔH°f = 0 only for elements in their standard states (e.g., O₂(g), C(s, graphite)).

• Mistake: Confusing ΔH°f with bond enthalpies.
  Correction: ΔH°f is for forming 1 mole of a compound from elements; bond enthalpies are for breaking bonds in gaseous molecules.

• Mistake: Forgetting to multiply ΔH°f by stoichiometric coefficients in the formula ΔH°rxn = Σ ΔH°f(products) – Σ ΔH°f(reactants).
  Correction: Always multiply each ΔH°f by its coefficient in the balanced equation (e.g., for 2H₂O, use 2 × ΔH°f(H₂O)).

AP Exam Insights

• FRQs often ask you to:
• Combine 2–3 reactions to find ΔH for a target reaction (Hess’s Law).
• Calculate ΔH°rxn using ΔH°f values (memorize the formula!).
• Explain why ΔH°f for an element is zero (standard state).
• Multiple-choice traps:
• Giving ΔH°f for an element in a non-standard state (e.g., O(g) instead of O₂(g)).
• Asking for ΔH°rxn but providing bond enthalpies instead of ΔH°f (trick question!).
• Reversing a reaction without changing the sign of ΔH.
• Tricky distinction: ΔH°f vs. ΔH°rxn
• ΔH°f is for forming 1 mole of a compound from elements.
• ΔH°rxn is for any reaction (calculated using ΔH°f or Hess’s Law).

Quick Check Questions

1. Multiple Choice:
   Given the following reactions:
2. C(s) + O₂(g) → CO₂(g) ΔH = –393.5 kJ
3. H₂(g) + ½ O₂(g) → H₂O(l) ΔH = –285.8 kJ
   What is ΔH for the reaction C(s) + 2H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)?
   (A) –679.3 kJ (B) –965.1 kJ (C) –393.5 kJ (D) –571.6 kJ

Answer: (B) –965.1 kJ
Explanation: Add the ΔH of the first reaction to twice the ΔH of the second reaction (since 2H₂O is needed).

1. Short FRQ:
   The standard enthalpy of formation for NH₃(g) is –45.9 kJ/mol. Write the balanced equation for the formation of NH₃(g) from its elements, and calculate ΔH°rxn for the reaction:
   2NH₃(g) → N₂(g) + 3H₂(g).

Answer:
Formation equation: ½ N₂(g) + ³⁄₂ H₂(g) → NH₃(g) ΔH°f = –45.9 kJ/mol.
For 2NH₃ → N₂ + 3H₂, reverse the formation reaction and multiply by 2:
ΔH°rxn = 2 × (+45.9 kJ) = +91.8 kJ.

Last-Minute Cram Sheet

1. Hess’s Law: ΔH is path-independent; add ΔH for steps to get total ΔH.
2. ΔH°f for elements in standard states = 0 kJ/mol (e.g., O₂(g), C(s, graphite)).
3. ΔH°rxn = Σ ΔH°f(products) – Σ ΔH°f(reactants).
4. Reverse a reaction → Flip ΔH sign.
5. Multiply a reaction → Multiply ΔH by the same factor.
6. Standard state = 25°C (298 K) and 1 atm pressure.
7. ⚠️ Don’t forget stoichiometric coefficients when using ΔH°f!
8. ΔH°f is for forming 1 mole of a compound from elements.
9. Bond enthalpies ≠ ΔH°f (used in a different Hess’s Law approach).
10. ⚠️ Elements in non-standard states (e.g., O(g)) have ΔH°f ≠ 0!