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 (?H°f).

Key Terms & Concepts

• Enthalpy (H): The total heat content of a system at constant pressure. Measured in kJ/mol.
• Enthalpy change (?H): The heat absorbed or released in a reaction (?H = H_products – H_reactants).
• Standard enthalpy of formation (?H°f): The enthalpy change when 1 mole of a compound forms from its elements in their standard states (e.g., O?(g), C(s, graphite)). ?H°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., H?O(l), not H?O(g)).
• Hess’s Law: The total ?H for a reaction is the sum of ?H for each step, regardless of the pathway.
• State function: A property (like ?H) that depends only on the initial and final states, not the path taken (e.g., altitude change on a hike, not the route).
• ?H°rxn =-?H°f(products) –-?H°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 ?H.
• Multiplying a reaction by a coefficient-Multiply ?H by the same coefficient.
• Bond enthalpy: The energy required to break 1 mole of bonds in a gaseous molecule (not the same as ?H°f; used in a different Hess’s Law approach).

Step-by-Step / Process Flow

Using Hess’s Law to Find ?H for a Target Reaction

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

Using Standard Enthalpies of Formation (?H°f)

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

Common Mistakes

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

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

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

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

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

AP Exam Insights

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

Quick Check Questions

1. Multiple Choice: Given the following reactions:
2. C(s) + O?(g)-CO?(g) ?H = –393.5 kJ
3. H?(g) + ½ O?(g)-H?O(l) ?H = –285.8 kJ What is ?H for the reaction C(s) + 2H?(g) + 2O?(g)-CO?(g) + 2H?O(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 ?H of the first reaction to twice the ?H of the second reaction (since 2H?O is needed).

1. Short FRQ: The standard enthalpy of formation for NH?(g) is –45.9 kJ/mol. Write the balanced equation for the formation of NH?(g) from its elements, and calculate ?H°rxn for the reaction: 2NH?(g)-N?(g) + 3H?(g).

Answer: Formation equation: ½ N?(g) + ³ H?(g)-NH?(g) ?H°f = –45.9 kJ/mol. For 2NH?-N? + 3H?, reverse the formation reaction and multiply by 2: ?H°rxn = 2 × (+45.9 kJ) = +91.8 kJ.

Last-Minute Cram Sheet

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