Hess’s Law and Standard Enthalpies of Formation Thermochemistry Pages 242 - 252

Hess’s Law Enthalpy changes are state functions. It does not matter if ΔH for a reaction is calculated in one step or a series of steps = Hess’s Law By using values of ΔH of known reactions, we can use Hess’s law to solve for enthalpies of reactions whose values we do not know.

Hess’s Laws Problem-Solving Manipulate equations so that they add up to the desired equation. There are 3 ways we can manipulate equations: 1. We can reverse the entire equation (switch reactants and products). 2. We can multiply the entire equation by a factor (such as 3, 2, 1/2 ) 3. We can do both #1 and #2. **When you manipulate an equation you must manipulate the ΔH value in the EXACT same way!!**

Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal. Reverse any reactions as needed to give the required reactants and products Multiply reactions to give the correct numbers of reactants and products (Trial and error- allow the final reaction to guide you) **Start by finding a substance that only appears once in the reactants.

Hess’s Law Example Calculate ΔH for N2 (g) + O2 (g) 2NO(g) Given: N2(g) + 2O2 (g) 2NO2(g) ΔH = 66.4 kJ/mol 2NO(g) + O2(g) 2NO2(g) ΔH = -114.1kJ/mol A: ΔH = 180.5 kJ/mol

Hess’s Law Practice Together Given 2H2 (g) + C(s) CH4(g) ΔH = -74.81 kJ/mol 2H2(g) + O2(g) 2H2O (l) ΔH = -571.66 kJ/mol C(s) + O2(g) CO2(g) ΔH = -393.52 kJ/mol Calculate ΔH for CH4(g) + 2O2(g) CO2(g) + 2H2O(l) A: ΔH = -890.37 kJ/mol

Hess’s Law Practice Problem Given the following thermochemical data, calculate the ΔH° for: Ca(s) + 2H2O(l) Ca(OH)2(s) + H2(g) H2(g) + ½ O2(g) H2O(l) ΔH° = -285kJ CaO(s) + H2O(l) Ca(OH)2(s) ΔH° = -64 kJ Ca(s) + ½ O2(g) CaO(s) ΔH° = -635 kJ A: -414 kJ

Hess’s Law Practice Problem Calculate ΔH for 4NH3 (g) + 3O2(g) 2N2(g) + 6H2O(l) Given the following reactions and ΔH values, a. 2N2O(g) O2(g) + 2N2(g) ΔH= -164 kJ b. 2NH2(g) + 3N2O(g) 4N2(g) + 3H2O(l) ΔH= -1012 kJ A: ΔH= -1532 kJ

Hess’s Law Practice Problem For the reaction: H2O (l) H2O(g) ΔH = +44kJ How much heat is evolved when 9.0 grams of water vapor is condensed to liquid water? A: 22kJ evolved or -22kJ

Standard Enthalpy of Formation ΔH°f The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. ΔH°f The degree symbol (°) on a thermodynamic function indicates the process has been carried out under standard conditions. Standard state is not the same as standard temperature and pressure (STP)

ΔH°f Key Ideas ΔH°f is always given per mole of compound formed. ΔH°f involves formation of a compound from its elements with the substances in their standard states. For an element: It is the form which the element exists in at 25°C and 1 atm. For a compound: For a gas it is a pressure of exactly 1 atm (IUPAC – 1bar ) For a substance in solution, it is a concentration of exactly 1M. For a pure solid or liquid, is the pure solid or liquid. ΔH°f for an element in its standard state, such as Ba(s) or N2(g), equals 0.

ΔH°reaction = Σnp ΔH°f (products) - Σnr ΔH°f (reactants) The enthalpy change for a given reaction can be calculated by subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products. When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes. When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer. Elements in their standard states are not included in the ΔHreaction calculations. That is, ΔH°f for an element in its standard state is zero. ΔH°reaction = Σnp ΔH°f (products) - Σnr ΔH°f (reactants)

Working with Standard Enthalpy of Formation… Consulting your textbook Appendix (4) and knowledge on standard states, list the standard enthalpy of formation for each of the following substances. Al2O3(s) Ti(s) P4(g) SO42-(aq) F2 (g)

Using the standard heats of formation, calculate ΔH for the following reactions. HCl (g) H+(aq) + Cl-(aq) 2NO2(g) N2O4 (g) C2H2(g) + H2(g) C2H4(g) 2NaOH(s) + CO2(g) Na2CO3(s) + H2O(g)

Standard Enthalpy of Formation Practice The heat released when HNO3 reacts with NaOH is 56 kJ/mole of water produced. How much energy is released when 400.0mL of 0.200M HNO3 is mixed with 500.0mL of 0.150 M NaOH? The enthalpy of neutralization for the reaction of a strong acid with a strong base is -56kJ/mol of water produced. How much energy will be released when 200.0mL of 0.400M HCl is mixed with 150.0mL of 0.500M NaOH? How does this compare with the answer in the first part? Why?

Bond Energies – Calculating Heat of Reaction All substances must be in the gaseous form. Breaking bonds requires energy (Endothermic) Forming bonds releases energy (Exothermic) ΔH = ΣEbonds broken - ΣEbonds formed

Bond Energies and Heat of Reaction Example Calculate the value of ΔH for the reaction below using the average bond energies. H2(g) + F2(g) 2HF(g) Bond Average Bond Energy (kJ/mol) H-H 432 H-F 565 F-F 160

Bond Energies and Heat of Reaction Practice Calculate the value of ΔH for the reaction below using the average bond energies. H2(g) + ½ O2(g) H2O(g) Bond Average Bond Energy (kJ/mol) H-H 432 O=O 496 H-O 463

Individual Practice 57, 58, 59, 60, 61, 63, 65, 67, 71

Reminders Hand-warmer Lab Report Due ___________! Practice and Review Ch. 6 Concepts and Problems Begin reading Ch. 7