Heat of Fusion (Hf) Q = mHf Fusion means melting/freezing amount of energy needed to melt/freeze 1g of a substance Different for every substance – look on reference tables Q = mHf

Heat of Vaporization(Hv) Vaporization means boiling/condensing amount of energy needed to boil/condense 1g of a substance Different for every substance – look on reference tables Q = mHv

Examples: Calculate the mass of water that can be frozen by releasing 49370 J. Calculate the heat required to boil 8.65 g of alcohol (Hv = 855 J/g). Calculate the heat needed to raise the temperature of 100. g of water from 25 C to 63 C .

Phase Change Diagram The flat points represent a phase change – temperature does not change while a phase change is occurring even though heat is being added. Diagonal points represent the 3 phases

Enthalpy Thermodynamics

Enthalpy (H) Measures 2 things in a chemical reaction: Energy change Amount of work Most chemical reactions happen at constant pressure (atmospheric pressure)—open container

Enthalpy (ΔH) 2 types of chemical reactions: Exothermic—heat released to the surroundings, getting rid of heat, -ΔΗ Endothermic—heat absorbed from surroundings, bringing heat in, +ΔΗ **Enthalpy of reaction— amount of heat from a chemical reaction which is given off or absorbed, units = kJ/mol

Enthalpy of Reaction and the process is ENDOTHERMIC DH = Hfinal - Hinitial Hinitial = reactants Hfinal = products If Hfinal > Hinitial then DH is positive and the process is ENDOTHERMIC If Hfinal < Hinitial then DH is negative and the process is EXOTHERMIC

Enthalpy of Reaction Hfinal < Hinitial and DH is negative

Enthalpy of Reaction Hfinal > Hinitial and DH is positive

Water Formation 2H2 + O2 2H20 ΔΗ = -967.28 kJ/mol

More Enthalpy The reverse of a chemical reaction will have an EQUAL but OPPOSITE enthalpy change HgO  Hg + ½ O2 ΔH = + 90.83 kJ Hg + ½ O2  HgO ΔH = - 90.83 kJ SOOO-----total ΔH = 0

Stoichiometry Returns

Example 1: Calculate the ΔH for the following reaction when 12.8 grams of hydrogen gas combine with excess chlorine gas to produce hydrochloric acid. H2 + Cl2  2HCl ΔH = -184.6 kJ/mol

Methods for determining ΔH Calorimetry Application of Hess’ Law Enthalpies of Formation

Hess’ Law Enthalpy change for a chemical reaction is the same whether it occurs in multiple steps or one step ΔHrxn = ΣΔHA+B+C (sum of ΔH for each step) Break a chemical reaction down into multiple steps Add the enthalpies (ΔH) of the steps for the enthalpy for the overall chemical reaction

Guidelines for using Hess’ Law Use data and combine each step to give total reaction Chemical compounds not in the final reaction should cancel Reactions CAN be reversed but remember to reverse the SIGN on ΔH

USING ENTHALPY Calculate DH for S(s) + 3/2O2(g)  SO3(g) knowing that S(s) + O2(g)  SO2(g) DH1 = -296.8 kJ SO2(g) + 1/2O2(g)  SO3(g) DH2 = -98.9 kJ The two equations add up to give the desired equation, so DHnet = DH1 + DH2 = -395.7 kJ

Example 3: H2O(l)  H2O (g) ΔH° = ? Based on the following: H2 + ½ O2  H2O(l) ΔH° = -285.83 kJ/mol H2 + ½ O2  H2O(g) ΔH° = -241.82 kJ/mol

Example 4: NO(g) + ½ O2  NO2 (g) ΔH° = ? Based on the following: ½ N2(g) + ½ O2  NO (g) ΔH° = + 90.29 kJ ½ N2(g) + O2  NO2 (g) ΔH° = +33.2 kJ -57.1 kJ

Methods for determining ΔH Calorimetry Application of Hess’ Law Enthalpies of Formation

Enthalpy of Formation (ΔHf°) Enthalpy for the reaction forming 1 mole of a chemical compound from its elements in a thermodynamically stable state. A chemical compound is formed from its basic elements present at a standard state (25°C, 1 atm) Enthalpy change for this reaction = ΔHf° ΔHf°= 0 for ALL elements in their standard/stable state.

Enthalpy of Formation cont. DHrxn = Hfinal – Hinitial Really, ΔHf (products) - ΔHf (reactants) Calculate ΔHrxn based on enthalpy of formation (ΔHf) aA + bB  cC + dD ΔH° =[c (ΔHf°)C + d(ΔHf°)D] - [a (ΔHf°)A + b (ΔHf°)B ]

Calculate the ΔH° for the reaction 2Mg(s) + O2(g)  2MgO(s) DHof for MgO = -601.6 kJ/mol Recall that DHof for elements in their standard state = 0 kJ/mol DHrxn = ∑(DHof products)(moles of products) – ∑(DHof reactants)(moles of reactants) = (-601.6kJ/mol)(2) – [(0kJ/mol)(2) + (1)(0kJ/ mol)] = - 1203.2kJ/mol

Enthalpy of Formation Values for #6 and 7 Chemical Compound Enthalpy of Formation (ΔHf°) (kJ/mol) CO -110.5 CO2 -393.5 Fe2O3 -824.2 Al2O3 -1676