Units of Energy Like we saw with pressure, many different units are used throughout the world for energy. SI unit for energy 1kg m 1J = 2 s 2 Joule (J) calorie (cal) erg (erg) electron volts (ev) British thermal unit (Btu) 4.184 Joules (J) = 1 calorie (cal) Amount of energy it takes to raise one gram of water 1ºC 1 kcal = 1000 calories = 1 Nutritional Calorie (Cal) Energy in Thermochemistry Heat (q): energy transferred from an object with a higher temperature to an object with a lower temperature 100º C heat 50º C Heat cannot be transferred in the opposite direction! Work (w): energy transferred against a force Example: Combustion reactions allow your car to move...performing work Vroom... Thermochemistry In thermochemistry, the universe is divided into two parts: The system: The physical process or chemical reaction in which we are interested. We can define the system anyway we like. The surroundings: Everything else in the universe. System Universe System + Energy flow between system and surroundings Energy (both heat and work) can be transferred between the system and surrounds. The sign + or - tells us the direction of energy flow. _ + System System copper After I add 2.0kJ of heat energy to each sample, will they still be the same temperature? 5.0 g Al 25 C 5.0 g Cu 25 C
Calorimetry The energy released from the reaction (or whatever is in the calorimeter) is gained by the water. So -q rxn = q water Notice that they have opposite signs, which tells us the direction of transfer. However, assuming no heat loss, the magnitudes are the same. The amount of heat lost by the reaction is the same amount gained by the water. Example A 28.2 gram sample of nickel is heated to 99.8 o C and placed in a coffee cup calorimeter containing 150.0 grams of water at 23.5 o C. After the metal cools, the final temperature of the metal and water is 25.0 o C. Which substance absorbed heat? Which substance released heat? Calculate the heat absorbed by the substance you indicated above. Exothermic vs. Endothermic processes Feels warm to the touch q = - Feels cool to the touch q = + Exothermic Endothermic System: chemical reaction System or process Enthalpies of Reaction Determine if the following processes are endothermic or exothermic Combustion of methane Neutralization of HCl Boiling Melting CaCO 3 (s) CaO (s) + CO 2 (g) Heat and Enthalpy So how are q and H related? At constant pressure: q p = H So, using a calorimeter, the H of a chemical reaction can be calculated from the change in heat in the calorimeter. 11
Enthalpies of Physical/Chemical Changes Enthalpy ( H) describes the change in heat from a chemical reaction or process. Enthalpies of Phase Changes Heat of fusion ( H fus ): Amount of heat required to melt (solid liquid) H = H products H reactants Where do these energies come from? Heat of vaporization ( H vap ): Amount of heat required to evaporate (liquid gas) Energy is released if: Energy reactants > Energy products Energy is absorbed if: Energy reactants < Energy products Heat of sublimation ( H sub ): Amount of heat required to sublime (solid gas) Why are there no values for H freezing, H condendsation, or H deposition? Thermodynamic Standard State H The (knot) denotes enthalpy under standard states Standard States A gas at 1 atm An aqueous solution (1 M conc.) at a pressure of 1 atm Pure liquids and solids The most stable form of elements at 1 atm and 25 o C (298 K) Amount of reactant, and H In chemistry, we usually specify energy in terms of a specific amount of reactant or product. For example: 2H 2(g) + O 2(g) 2H 2 O (g) Which means if we react 2 mole of H 2 with 1 mole of O 2, 241.8kJ of heat energy will be released. But what if we react 4 moles of H 2 with 2 moles of O 2? H =-483.6 kj H =-241.8 kj This can be read as: -241.8 kj per 2 moles H 2-241.8 kj per 1 mole O 2-241.8 kj per 2 moles of H 2 O Direction of reaction and H What about the reverse reaction of: Thermochemical Equations and Stoichiometry 2Al (s) + Fe 2 O 3 (s) 2Fe (s) + Al 2 O 3 (s) H o = -852 kj 2H 2(g) + O 2(g) 2H 2 O (g) H =-241.8 kj 2H 2 O (g) 2H 2(g) + O 2(g) H =+241.8 kj All we do is change the sign. Instead of energy being released (as in the first reaction), energy is absorbed when breaking water molecules apart. How much heat is released if 10.0 grams of Fe 2 O 3 reacts with excess Al? 10.0 g Fe 2 O 3 x mol Fe 2 O 3 x mol Fe 2 O 3 852 kj / 1 mol Fe 2 O 3
Hess s Law If a compound cannot be directly synthesized from its elements, we must use multiple reactions to calculate the enthalpy of reaction Hess s Law H is a state function it doesn t matter what route we take to get there. Same change in energy, H. Hess s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps Look at direction of reaction and amount of reactants/products These two steps (solid liquid gas) should have the same change in energy as this one step (solid gas) 20 Hess s Law Values of enthalpy change For a reaction in the reverse direction, enthalpy is numerically equal but opposite in sign Reverse direction, heat flow changes; endothermic becomes exothermic (and vice versa) Proportional to the amount of reactant consumed Twice as many moles = twice as much heat Depends on the physical state of the reactants and products (i.e., elemental state) Hess s Law Does the energy change in a reaction depend on the number of steps in the reaction? H 2 (g) + I 2 (s) 2HI(g) H = 53.00 kj I 2 (s) I 2 (g) H = 62.44 kj H 2 (g) + I 2 (g) 2HI(g) H =? How can we add the first two reactions to get the third? Hess s Law: H sum of steps = H 1 + H 2 21 22 Application of Hess s Law We can use known values of H o to calculate unknown values for other reactions P 4 (s) + 3 O 2 (g) P 4 O 6 (s) H = -1640.1 kj P 4 (s) + 5 O 2 (g) P 4 O 10 (s) H = -2940.1 kj What is H o for the following reaction? P 4 O 6 (s) + 2 O 2 (g) P 4 O 10 (s) H =? Hess Law Problem Given: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) H = -802 kj H 2 O (l) H 2 O (g) H = 88 kj Find the enthalpy for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) H =? 23
Standard Enthalpies of Formation Standard enthalpy of formation ( H o f): heat needed to make 1 mole of a substance from its stable elements in their standard states H o f = 0 for a stable (naturally occurring) element Which of these have H o f = 0? CO(g), Cu(s), Br 2 (l), Cl(g), O 2 (g), O 3 (g), O 2 (s) Standard Enthalpies of Formation Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book) H o rxn = Σn H o f (products) Σn H o f (reactants) Σ = sum n = number of moles (coefficients) Do the following equations represent standard enthalpies of formation? Why or why not? 2Ag (l) + Cl 2 (g) 2AgCl (s) Ca (s) + F 2 (g) CaF 2 (s) 25 Heats of Formation H o rxn = Σ n H o f,products - Σ n H o f,reactants Calculate values of H o for the following rxns: 1) CaCO 3 (s) CaO (s) + CO 2 (g) 2) 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H of (in kj/mol): CaCO3 (-1207.1) CaO (-635.5) CO2 (-393.5) C6H6 (49.0) CO2 (-393.5) H2O (-285.8) Group Work Use Standard Heat of Formation values to calculate the enthalpy of reaction for: C 6 H 12 O 6 (s) 2 C 2 H 5 OH(l) + 2 CO 2 (g) H o f (C 6 H 12 O 6 (s)) = -1260.0 kj/mol H o f (C 2 H 5 OH(l)) = -277.7 kj/mol H o f (CO 2 (g)) = -393.5 kj/mol 27 Bond Dissociation Energies Bond dissociation energy (BDE). Energy needed to break one 1 mol of a specific bond (eg. H-H, C-H, C=C, C-C, etc.) Bond dissociation energy varies somewhat from one molecule to another, or even within one molecule, so in most cases, we have use an average bond energy (D). Bond Energy For example: H-OH H-O H-OOH Average = 502 kj/mol 427 kj/mol 431 kj/mol 453.3 kj/mol for O-H Note: there are certain bonds where the BDE is not and average. These are bonds that only exist within diatomic molecules (eg. H-H, H-Br, Cl-Cl, etc.) 30
Bond Dissociation Energies H o rxn = ΣBDE (reactants) + - ΣBDE (products) endothermic energy input exothermic energy released Or: H o rxn = ΣBDE (reactants) ΣBDE (products) ΣBDE(react) > ΣBDE(prod) endothermic ΣBDE(react) < ΣBDE(prod) exothermic We use only when heats of formation are not available, since bond energies are average values for gaseous molecules Enthalpy of Reaction Use bond energies to calculate the enthalpy change for the following reaction: NΞN = 945 kj/mol N 2 (g) + 3H 2 (g) 2NH 3 (g) H rxn = [BE N N + 3BE H-H ] [6BE N-H ] H rxn = [945 + 3(436)] [6(390)] = -87 kj measured value = -92.2 kj H-H = 436 kj/mol N-H = 390 kj/mol Why are the calculated and measured values different? 31 32 Enthalpy of Reaction Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl 3 (g) N 2 (g) + Cl 2 (g) (hint: balance first!!) H rxn? N-Cl = 200 kj/mol N N = 945 kj/mol Cl-Cl = 243 kj/mol Practice Problems Which amount of energy is higher? Energy that is released when 55g of Al is cooled from 5 C to -25 C Energy that is absorbed when 64g of Cu is heated from 298K to 315K? 8750 J of heat are applied to a 170 g sample of metal, causing a 56 o C increase in its temperature. What is the specific heat of the metal? Which metal is it? 33 Practice Problems Practice: Worked Example 8.10, page 320. Identify how to set up the following problems: Calculate the H o of reaction for: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (l) H o f C 3 H 8 (g): -103.95 kj/mol; H o f CO 2 (g): -393.5 kj/mol; H o f H 2 O(l): -285.8 kj/mol 8750 J of heat are applied to a 170 g sample of metal, causing a 56 o C increase in its temperature. What is the specific heat of the metal? Which metal is it? Practice Problems C 2 H 4 (g ) + 6F 2 (g) 2CF 4 (g) + 4HF(g) H o =? H 2 (g) + F 2 (g) 2HF (g) H o = -537 kj C (s) + 2F 2 (g) CF 4 (g) H o = -680 kj 2C (s) + 2H 2 (g) C 2 H 4 (g) H o = 52.3 kj Use average bond energies to determine the enthalpy of the following reaction (from Table 7.1). CH 4 (g) + Cl 2 (g) CH 3 Cl (g) + HCl (g) (BE C-Cl = 328 kj/mol)