Chap 5 Part 2 Enthalpies of Formation (Delta Hf) - Study Notes

Enthalpies of Formation (Delta Hf) - Study Notes

Definition of enthalpy of formation
- Delta Hf^b0 (denoted as $\Delta Hf^\circ$ ) is the enthalpy change for the reaction in which a compound is formed from its constituent elements in their standard states.
- It is a special type of enthalpy of reaction: the enthalpy of formation (formation reaction).
- For one mole of the compound formed, the reaction is written with a coefficient of 1 in front of the product.
- Example: Formation of ammonia (NH_3)
- Formation reaction: $\tfrac{1}{2} \mathrm{N}2(g) + \tfrac{3}{2} \mathrm{H}2(g) \rightarrow \mathrm{NH}_3(g)$
- The reactant coefficients can be fractional because the reaction is defined to form one mole of the product.
Stable form of elements and why sometimes coefficients are fractional
- If an element exists in more than one form under standard conditions, the most stable form is used in the formation reaction (e.g., for oxygen, the stable form is O2, not O or O3).
- There are seven diatomic elements that are most stably found as diatomic molecules in standard state: H2, N2, O2, F2, Cl2, Br2, I_2.
- Mnemonic: "Have no fear of ice cold beer" to remember the seven diatomic elements.
- For these elements, the standard enthalpy of formation in their standard states is zero: $\Delta H_f^\circ(\text{element in standard state}) = 0.$
Standard state and standard conditions
- The little circle superscript on Delta H_f^\circ denotes standard conditions.
- Standard conditions are defined as 25°C (room temperature) and 1 atm pressure.
- Standard enthalpy of formation values are typically reported in units of $\mathrm{kJ\,mol^{-1}}$ because they are per mole of the formed compound.
Example: Calcium carbonate, CaCO_3
- Composition: Ca, C, O;
- Most stable form for calcium is solid Ca(s).
- Most stable form for carbon is graphite (not diamond).
- Oxygen is written as O_2(g).
- Formation of 1 mole of CaCO_3:
- Equation (balanced with 1 mole of product): $\mathrm{Ca}(s) + \mathrm{C}(\text{graphite}) + \tfrac{3}{2} \mathrm{O}2(g) \rightarrow \mathrm{CaCO}3(s)$
- Delta Hf^b0 for CaCO3:
- $\Delta Hf^\circ(\mathrm{CaCO3}) = -1207\ \text{kJ mol}^{-1}.$
- Interpretation: The enthalpy change associated with forming 1 mole of CaCO_3 from its elements is -1207 kJ.
The standard formation enthalpies table and their use
- The standard enthalpy of formation values can be used to calculate the standard enthalpy change of any reaction (\Delta H^\circ_{rxn}).
- The standard enthalpy change of reaction is computed from formation enthalpies as:
- $\Delta H^\circ{rxn} = \sumi \nui \Delta Hf^\circ(\text{products}) - \sumj \nuj \Delta H_f^\circ(\text{reactants})$
- Here, \nui and \nuj are the stoichiometric coefficients of products and reactants, respectively.
Rules for identifying formation reactions
- Rule 1: The product must be formed with a coefficient of 1 (the chemical equation should form one mole of the compound).
- Rule 2: The reactants must be elements in their most stable, standard states.
- Rule 3: The equation must be balanced.
- Example decision steps (based on the transcript):
- A reaction forming two moles of a compound (e.g., two AgCl) is not a formation reaction (coefficient != 1).
- After ensuring the product coefficient is 1, verify that reactants are in their most stable forms (e.g., Ca(s), F2(g), O2(g)).
- Check that atoms balance on both sides (stoichiometry).
- Applying these rules, some given reactions are valid formation reactions and some are not (as discussed: only some options meet all three rules).
Writing a formation reaction for a compound (example: silver nitrate, AgNO_3)
- Determine the ions: Ag^+, NO3^−, so compound is AgNO3.
- Elements involved: Ag, N, O.
- Most stable elemental forms: Ag(s), N2(g) for nitrogen, O2(g) for oxygen.
- Write one mole of AgNO3 as the product and balance with the simplest possible coefficients to yield one mole of AgNO3:
- Unbalanced starting point: Ag(s) + (1/? ) N2(g) + (1/?) O2(g) -> AgNO_3(s).
- Balance to form 1 mole of AgNO3: use 1/2 N2 and 3/2 O_2 to provide 1 N and 3 O.
- Final formation reaction (balanced): $\mathrm{Ag}(s) + \tfrac{1}{2} \mathrm{N}2(g) + \tfrac{3}{2} \mathrm{O}2(g) \rightarrow \mathrm{AgNO}_3(s).$
- This demonstrates the need for fractional coefficients when forming 1 mole of a compound from its elements.
Another practical example: writing a formation reaction for sodium bicarbonate, NaHCO_3
- Elements present: Na, H, C, O.
- Most stable elemental forms: Na(s), H2(g), C(graphite), O2(g).
- Write a formation reaction to form 1 mole of NaHCO_3:
- Start with Na(s) and other elements in their standard forms to supply 1 Na, 1 H, 1 C, and 3 O in the product.
- Balance with coefficients and allow fractional coefficients if needed:
- One balanced formation reaction: $\mathrm{Na}(s) + \tfrac{1}{2} \mathrm{H}2(g) + \mathrm{C}(s,\text{graphite}) + \tfrac{3}{2} \mathrm{O}2(g) \rightarrow \mathrm{NaHCO}_3(s).$
- Note: The coefficient of the product remains 1 (formation reaction), and fractional coefficients may be required for the elemental reactants.
Practice example: calculation of an enthalpy change from formation enthalpies
- Given a reaction and formation enthalpies for the products and reactants, use the formula
- $\Delta H^\circ{rxn} = \sump \nup \Delta Hf^\circ(\text{products}) - \sumr \nur \Delta H_f^\circ(\text{reactants})$
- Example 1: 2 SO2(g) + O2(g) \rightarrow 2 SO_3(g)
- Given: $\Delta Hf^\circ(\mathrm{SO2}) = -296.9\ \text{kJ mol}^{-1},\quad \Delta Hf^\circ(\mathrm{SO3}) = -395.2\ \text{kJ mol}^{-1},\quad \Delta Hf^\circ(\mathrm{O2}) = 0$
- Compute products: $2 \times (-395.2) = -790.4\ \text{kJ}$
- Compute reactants: $2 \times (-296.9) + 1\times 0 = -593.8\ \text{kJ}$
- $\Delta H^\circ_{rxn} = (-790.4) - (-593.8) = -196.6\ \text{kJ}$
- Interpretation: The reaction is exothermic (negative sign).
- Note: Oxygen is elemental, so its formation enthalpy is zero; this simplifies the calculation.
Worked example from the transcript: a specific concept-check problem
- You may be given products with multiple moles and reactants with their formation enthalpies per mole.
- The general setup is to multiply each formation enthalpy by its coefficient in the balanced equation, add for products, subtract the sum for reactants.
- An example in the transcript yields a final answer of $-1{,}172\ \text{kJ}$ for a particular problem after applying the correct coefficients and summing the per-mole formation enthalpies.
- Practical tip: to avoid mistakes, explicitly write out the calculation for each term, including the multiplication by the coefficients, before summing.
Common pitfalls and tips
- Ensure the product coefficient is 1 when identifying a formation reaction.
- Always use the most stable form of elements in the reactants (e.g., graphite for carbon, O2 for oxygen, H2 for hydrogen, etc.).
- Remember O2, N2, H2, F2, Cl2, Br2, I_2 are the seven diatomic elements in standard state.
- If a formation enthalpy for a pure element is not given, assume it is 0 for its standard state.
- Be careful with balancing when fractional coefficients appear; for formation reactions, fractions are allowed to ensure a single mole of product is formed.
Summary of key concepts to remember
- Delta H_f^b0 represents the molar enthalpy of formation for a compound from its elements in their standard states, at standard conditions, usually in kJ/mol.
- Elements in their standard state have ΔH_f^b0 = 0.
- The standard enthalpy change of a reaction can be calculated from formation enthalpies using the formula above, summing products and subtracting reactants, with their stoichiometric coefficients.
- To determine whether a given equation is a formation reaction, check the product's coefficient (must be 1), the stability of elemental reactants, and that the equation is balanced.
Quick references
- Standard state circle: superscript 0 on ΔH_f indicates standard conditions (25°C, 1 atm).
- Units: $\mathrm{kJ\,mol^{-1}}$ for formation enthalpies.
- Common formation value example: $\Delta Hf^\circ(\mathrm{CaCO3}) = -1207\ \mathrm{kJ\,mol^{-1}}.$
- Example reaction format for formation of a compound from its elements: a) write elements in their standard states, b) form 1 mole of the compound, c) balance using fractional coefficients if needed.

Small Summary and Quick Points:

Enthalpy of Formation (ΔHf∘ΔHf∘): The enthalpy change when one mole of a compound is formed from its constituent elements in their most stable standard states.
- Denoted with a small circle superscript (∘∘) for standard conditions (25°C and 1 atm pressure).
- Units are typically kJ mol−1kJmol−1.
Standard States of Elements:
- Elements in their most stable forms under standard conditions have a standard enthalpy of formation of zero (ΔHf∘=0ΔHf∘=0).
- Examples: O22(g), C(graphite), H22(g), Ag(s).
- Remember diatomic elements: H22, N22, O22, F22, Cl22, Br22, I22.
Rules for Identifying Formation Reactions:
1. The product must be formed with a coefficient of 1 mole.
2. Reactants must be elements in their most stable standard states.
3. The equation must be balanced.
  - Fractional coefficients for reactants are allowed to ensure 1 mole of product is formed.
Calculating Standard Enthalpy of Reaction (ΔHrxn∘ΔHrxn∘):
- Can be calculated using standard enthalpies of formation of products and reactants: ΔHrxn∘=∑νproductsΔHf∘(products)−∑νreactantsΔHf∘(reactants)ΔHrxn∘=∑νproductsΔHf∘(products)−∑νreactantsΔHf∘(reactants)
  - νν represents the stoichiometric coefficients from the balanced chemical equation.
Key Tips:
- Always ensure the correct states (g, l, s) for elements and compounds.
- Be careful with reaction stoichiometry when applying the calculation formula, multiplying each ΔHf∘ΔHf∘ by its coefficient.