6.8 Enthalpies of Formation: How to Calculate and Apply ΔHf in Chemical Reactions

What is the Enthalpy of Formation?

The standard enthalpy of formation (ΔHf), or standard heat of formation, refers to the change in enthalpy during the formation of 1 mole of a compound from its constituent elements in their most stable states at standard conditions (25°C and 1 atm). Essentially, ΔHf measures the energy involved in forming a compound.

For example, the ΔHf for CO₂ represents the energy change for the reaction:

$$\text{C (s) + O₂ (g) → CO₂ (g)}$$

If we were forming carbon monoxide (CO), the reaction would be:

$$\text{C (s) + 1/2 O₂ (g) → CO (g)}$$

Important Note:

The standard enthalpy of formation for any element in its most stable form (like O₂, N₂, H₂) is zero.

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Using ΔHf to Calculate the Enthalpy of Reaction (ΔHrxn)

You can use the standard enthalpies of formation to determine the overall enthalpy change of a reaction using the following formula:

$$ΔH_{rxn} = ΣnΔHf(\text{products}) – ΣmΔHf(\text{reactants})$$

• n and m are the stoichiometric coefficients for the reactants and products.
• Remember: This formula is products minus reactants. Be careful not to confuse it with bond dissociation energy calculations, which are reactants minus products.

Example Problem

Given the following reaction:

$$8 \text{C (s)} + 10 \text{H₂ (g)} → 2 \text{C₄H₁₀ (g)}$$

Standard Enthalpies of Formation (ΔHf):

• C (s): 0 kJ/mol (element in its standard state)
• H₂ (g): 0 kJ/mol (element in its standard state)
• C₄H₁₀ (g): -147.3 kJ/mol

Step-by-Step Solution:

1. Write down the formula: $$ΔH_{rxn} = ΣnΔHf(\text{products}) – ΣmΔHf(\text{reactants})$$
   
2. Plug in the values: $$ΔH_{rxn} = [2 \times (-147.3)] – [8 \times 0 + 10 \times 0]$$
   
3. Simplify: $$ΔH_{rxn} = -294.6 \text{ kJ}$$
   

Another Example

Reaction:

$$\text{CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2 H₂O (l)}$$

Standard Enthalpies of Formation (kJ/mol):

• CH₄ (g): -74.8
• O₂ (g): 0 (standard state)
• CO₂ (g): -393.5
• H₂O (l): -285.8

Calculation:

1. Write the formula: $$ΔH_{rxn} = ΣnΔHf(\text{products}) – ΣmΔHf(\text{reactants})$$
   
2. Plug in the values: $$ΔH_{rxn} = \left[(1 \times -393.5) + (2 \times -285.8)\right] – \left[(1 \times -74.8) + (2 \times 0)\right]$$
   
3. Calculate: $$ΔH_{rxn} = \left[-393.5 + (-571.6)\right] – \left[-74.8\right]$$
   $$ΔH_{rxn} = -965.1 + 74.8 = -890.3 \text{ kJ/mol}$$
   

Result: The reaction is exothermic, as indicated by the negative ΔH value.

Quick Tips for Using ΔHf

1. Pay Attention to States of Matter: Ensure you use the correct enthalpy values for the specified state (e.g., H₂O (l) vs. H₂O (g)).
2. Use Stoichiometric Coefficients: Multiply the ΔHf values by the number of moles (coefficients) for each compound.
3. Watch Out for Sign Changes: Remember that ΔHf values are given for the formation of compounds, and you need to subtract the sum of reactants from products.

Practice Problem

2014 AP Chemistry Exam Question

In a combustion reaction, the student measures that burning 2.00 mol of vinyl chloride releases 2300 kJ of energy. Compare this with the energy released from burning 2.00 mol of propene using the following balanced equation:

$$2 \text{C₃H₆ (g)} + 9 \text{O₂ (g)} → 6 \text{CO₂ (g)} + 6 \text{H₂O (g)}$$

Standard Enthalpies of Formation:

• CO₂ (g): -394 kJ/mol
• H₂O (g): -242 kJ/mol
• C₃H₆ (g): 21 kJ/mol
• O₂ (g): 0 kJ/mol

Solution:

1. Calculate ΔHrxn for propene combustion: $$ΔH_{rxn} = [6 \times (-242) + 6 \times (-394)] – [9 \times 0 + 2 \times 21]$$
   
2. Simplify: $$ΔH_{rxn} = [-1452 – 2364] – [0 + 42] = -3858 \text{ kJ/mol}$$
   

Result: The combustion of 2.00 mol of propene releases more energy (-3858 kJ) than 2.00 mol of vinyl chloride (-2300 kJ).

Final Thoughts

Enthalpy of formation plays a crucial role in understanding energy changes in chemical reactions. By mastering ΔHf calculations, you can predict whether reactions are exothermic or endothermic, calculate reaction enthalpies, and even assess the feasibility of reactions.