Lecture 8: Thermochemistry Lecture 8 Topics Brown chapter 5 8.1: Kinetic vs. potential energy 5.1 8.2: Transferring energy as heat & work Thermal energy 8.3: System vs. surroundings Closed systems 8.4: First Law of Thermodynamics 5.2 Internal energy of chemical reactions Energy diagrams E, system & surroundings 8.5: Enthalpy 5.3 Exothermic vs. endothermic Guidelines thermochemical equations 5.4 Hess’s Law 5.6 8.6: Calorimetry Constant pressure calorimetry 8.7: Enthlapy of formation 5.7

Enthalpy Enthalpy is heat flow at constant pressure. Endothermic systems (reactions) require energy input (+ΔH). Exothermic systems (reactions) produce (release) energy (-ΔH). Thermochemical equations include energy as a product or reactant. Hess’s law allows enthalpies of reaction to be summed. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Enthalpy H = Hfinal - Hinitial = Hproducts - Hreactants Enthalpy (H): heat absorbed or released under constant pressure Enthalpy is a state function; only current state matters. As with internal energy, we cannot measure absolute enthalpy at any one point in time. Instead we measure change in enthalpy (H). H = Hfinal - Hinitial = Hproducts - Hreactants = qp Where qp is heat gained or lost by the system under constant pressure. ∆H = ∆(E + PV) = ∆E + P∆V Generally, enthalpy is more useful than internal energy, since most chemical rxns occur at constant pressure. For most reactions, PV is generally a very small factor, so there is little difference between enthalpy and internal energy. If H is positive, the reaction is endothermic; if negative then exothermic. Endothermic when Hproducts > Hreactants Exothermic when Hreactants > Hproducts p. 167-9

Exothermic vs. endothermic surroundings system system heat heat H > 0 (+) endothermic H < 0 (-) exothermic Endothermic: a process in which system absorbs heat from surroundings melting of ice (the container feels cool) Exothermic: a process in which system transfers heat to surroundings combustion of gasoline (the container feels hot) dilution of concentrated acid with water (steam is generated) thermite hand-warmers = Al + Fe2O3 p. 170-4

NH4Cl + H2O + heat  NH4+1 + Cl-1 + H2O Examples: exo- vs. endothermic Fe2O3 + 2Al  2Fe + Al2O3 + heat exothermic NH4Cl + H2O + heat  NH4+1 + Cl-1 + H2O endoothermic p. 167-8

Examples: exo- vs. endothermic For each example, indicate the sign of ΔH and describe the process as wither exo- or endothermic. An ice cube melts. One gram of butane is combusted completely. Energy input is required to melt ice, so ΔH is + & the process is endothermic. Heat is produced; ΔH is negative; process is exothermic. p. 171

Guidelines for thermochemical equations 1. Enthalpy is an extensive property. magnitude is directly proportional to the amount of reactant consumed CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ 2. Enthalpy change (H) for a reaction is equal in magnitude, but opposite in sign, to the H for the reverse reaction. CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) H = +890 kJ 3. Enthalpy change (H) for a reaction depends on the state of reactants and products. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ 2H2O(l) - 2H2O(g) H = 88 kJ Thermochemical equation = a balanced equation showing the H associated associated with the complete chemical reaction p. 173-4

Examples: thermochemical equations How much heat is released when 4.50 g of methane gas is burned at constant pressure? CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l) Hrxn = -890 kJ/mol MW CH4 = 16.042 g/mol 4.50 g 1 mol -890 kJ = -250 kJ 16.042 g 1 mol How many grams of ammonia can be decomposed with 1 MJ of energy? 2NH3 --> 3H2 + N2 Hrxn = +92 kJ 1 MJ = 1000 kJ (1000 kJ)(2 mol NH3/92 kJ)(17 g/mol NH3) = 370 g NH3 p. 174

Hess’s law Hess’s Law allows us to estimate enthalpy changes for complex chemical reactions without doing lab work. If a reaction occurs in a series of steps, the H for the whole reaction is equal to the sum of each step’s H. What property of enthalpy allows Hess’s Law to work? Because enthalpy is a state function (pathway-independent) it doesn’t matter how you get from reactants to products; you can use one step or ten, energy changes will be the same. N2 + 2O2 --> 2NO2 H = 68 kJ N2 + O2 --> 2NO H = 180 kJ 2NO + O2 --> 2NO2 H = -112 kJ N2 + 2O2 --> 2NO2 H = 68 kJ Guidelines for using Hess’s Law 1. If the reaction must be reversed, change the sine of H. (Enthalpy reverses with the direction of the reaction.) 2. If equations must be multiplied or divided by a fudge factor, do the same to H. (Enthlapy is quantitative.) p. 181-3

Examples: Hess’s law Carbon and CO can be combusted in O2 to form CO2. C(s) + O2(g) --> CO2(g) H = -393.5 kJ/mol C CO(g) + 1/2O2(g) --> CO2(g) H = -283.0 kJ/mol CO Use Hess’s Law to calculate H of: C(s) + 1/2O2(g) --> CO(g) C(s) + O2(g) --> CO2(g) H = -393.5 kJ/mol C CO2(g) --> CO(g) + 1/2O2(g) H = +283.0 kJ/mol CO (reversed) C(s) + 1/2O2(g) --> CO(g) H = - 110.5 kJ/mol C Carbon exists in two natural forms: graphite and diamond. Each form can be combusted: C(graphite) + O2(g) --> CO2(g) H = -393.5 kJ/mol C(diamond) + O2(g) --> CO2(g) H = -395.4 kJ/mol Use Hess’s Law to calculate H of the conversion of graphite to diamond: C(graphite) + O2(g) --> CO2(g) H = -393.5 kJ/mol CO2(g) --> C(diamond) + O2(g) H = +395.4 kJ/mol (reversed) C(graphite) --> C(diamond) H = +1.9 kJ/mol p. 181-3

Example: one more Hess’s Calculate H for this reaction: 2C(s) + H2(g) --> C2H2 (gas) Given the following thermochemical reactions: C2H2(g) + 5/2 O2(g) --> 2CO2 + H2O(l) H = -1229.6 kJ/mol C(s) + O2(g) --> CO2(g) H = -393.5 kJ/mol H2(g) + 1/2O2(g) --> H2O(l) H = -285.8 kJ/mol 2CO2 + H2O(l) --> C2H2(g) + 5/2 O2(g) H = +1229.6 kJ/mol (reversed) (2)(C(s) + O2(g) --> CO2(g)) H = (2)(-393.5 kJ/mol) (x2) H2(g) + 1/2O2(g) --> H2O(l) H = -285.8 kJ/mol 2C(s) + H2(g) --> C2H2 (gas) H = +156.8 kJ/mol p. 181-3