Thermochemistry 2009, Prentice-Hall, Inc. Topic 9 Thermochemistry Thermochemistry 2009, Prentice-Hall, Inc. Energy Energy is the ability to do work or transfer heat. Energy used to cause an object that has mass to move is called work. Energy used to cause the temperature of an object to rise is called heat. Thermochemistry 2009, Prentice-Hall, Inc. Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition. Thermochemistry 2009, Prentice-Hall, Inc. Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion. 1 2 KE = mv 2 Thermochemistry 2009, Prentice-Hall, Inc. Units of Energy The SI unit of energy is the joule (J). An older, non-SI unit is still in widespread use: the calorie (cal). 1 cal = J 1 J = 1 kg m 2 s2s2 Thermochemistry 2009, Prentice-Hall, Inc. Definitions: System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston). Thermochemistry 2009, Prentice-Hall, Inc. Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects. Thermochemistry 2009, Prentice-Hall, Inc. First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Thermochemistry 2009, Prentice-Hall, Inc. Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. Thermochemistry 2009, Prentice-Hall, Inc. Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = E final E initial Thermochemistry 2009, Prentice-Hall, Inc. Changes in Internal Energy If E > 0, E final > E initial Therefore, the system absorbed energy from the surroundings. This energy change is called endothermic. Thermochemistry 2009, Prentice-Hall, Inc. Changes in Internal Energy If E < 0, E final < E initial Therefore, the system released energy to the surroundings. This energy change is called exothermic. Thermochemistry 2009, Prentice-Hall, Inc. Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is, E = q + w. Thermochemistry 2009, Prentice-Hall, Inc. E, q, w, and Their Signs Thermochemistry 2009, Prentice-Hall, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. Thermochemistry 2009, Prentice-Hall, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic. Thermochemistry 2009, Prentice-Hall, Inc. State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. Thermochemistry 2009, Prentice-Hall, Inc. State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. In the system below, the water could have reached room temperature from either direction. Thermochemistry 2009, Prentice-Hall, Inc. State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so, E depends only on E initial and E final. Thermochemistry 2009, Prentice-Hall, Inc. Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV Thermochemistry 2009, Prentice-Hall, Inc. Enthalpy When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) This can be written H = E + P V Thermochemistry 2009, Prentice-Hall, Inc. Enthalpy Since E = q + w and w = -P V, we can substitute these into the enthalpy expression: H = E + P V H = (q+w) w H = q So, at constant pressure, the change in enthalpy is the heat gained or lost. Thermochemistry 2009, Prentice-Hall, Inc. Endothermic and Exothermic A process is endothermic when H is positive. Thermochemistry 2009, Prentice-Hall, Inc. Endothermicity and Exothermicity A process is endothermic when H is positive. A process is exothermic when H is negative. Thermochemistry 2009, Prentice-Hall, Inc. Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = H products H reactants Thermochemistry 2009, Prentice-Hall, Inc. Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. Thermochemistry 2009, Prentice-Hall, Inc. The Truth about Enthalpy 1.Enthalpy is an extensive property (depends on how much). 2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. 3. H for a reaction depends on the state of the products and the state of the reactants. Thermochemistry 2009, Prentice-Hall, Inc. Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. Thermochemistry 2009, Prentice-Hall, Inc. Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1 C) is its heat capacity. Thermochemistry 2009, Prentice-Hall, Inc. Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K or (1 C). Thermochemistry 2009, Prentice-Hall, Inc. Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass temperature change s = q m Tm T Thermochemistry 2009, Prentice-Hall, Inc. Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter. Thermochemistry 2009, Prentice-Hall, Inc. Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: q = m s T Thermochemistry 2009, Prentice-Hall, Inc. How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = J/g 0 C t = t final t initial = 5 0 C 94 0 C = C q = ms t = 869 g x J/g 0 C x 89 0 C= -34,000 J Thermochemistry 2009, Prentice-Hall, Inc. Bomb Calorimetry Reactions can be carried out in a sealed bomb such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. Thermochemistry 2009, Prentice-Hall, Inc. Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. For most reactions, the difference is very small. Thermochemistry 2009, Prentice-Hall, Inc. Hesss Law H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. However, we can estimate H using published H values and the properties of enthalpy. Thermochemistry 2009, Prentice-Hall, Inc. Hesss Law Hesss law states that [i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. Thermochemistry 2009, Prentice-Hall, Inc. Hesss Law Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products. Thermochemistry 2009, Prentice-Hall, Inc. Enthalpies of Formation An enthalpy of formation, H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Thermochemistry 2009, Prentice-Hall, Inc. Standard Enthalpies of Formation Standard enthalpies of formation, H f , are measured under standard conditions (25 C and 1.00 atm pressure). Thermochemistry 2009, Prentice-Hall, Inc. Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g) H 0 = kJ rxn S (rhombic) + O 2 (g) SO 2 (g) H 0 = kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) H 0 = kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g) H 0 = kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g) H 0 = x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) H 0 = kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l) H 0 = (2x-296.1) = 86.3 kJ rxn Thermochemistry 2009, Prentice-Hall, Inc. Calculation of H Imagine this as occurring in three steps: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) Thermochemistry 2009, Prentice-Hall, Inc. Calculation of H Imagine this as occurring in three steps: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) Thermochemistry 2009, Prentice-Hall, Inc. Calculation of H Imagine this as occurring in three steps: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) Thermochemistry 2009, Prentice-Hall, Inc. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Calculation of H The sum of these equations is: Thermochemistry 2009, Prentice-Hall, Inc. Calculation of H We can use Hesss law in this way: H = n H f products m H f reactants where n and m are the stoichiometric coefficients. Thermochemistry 2009, Prentice-Hall, Inc. H= [3( kJ) + 4( kJ)] [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] [( kJ) + (0 kJ)] = ( kJ) ( kJ) = kJ C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Calculation of H Thermochemistry 2009, Prentice-Hall, Inc. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n H 0 (products) f = m H 0 (reactants) f - H0H0 rxn 6 H 0 (H 2 O) f 12 H 0 (CO 2 ) f = [+] - 2 H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x x187.6 ] [ 2x49.04 ] = kJ kJ 2 mol = kJ/mol C 6 H 6 6.6 Thermochemistry 2009, Prentice-Hall, Inc. The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H 2 (g) H (g) + H 0 = kJ Cl 2 (g) Cl (g) + H 0 = kJ HCl (g) H (g) +Cl (g) H 0 = kJ O 2 (g) O (g) + H 0 = kJ OO N 2 (g) N (g) + H 0 = kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond 9.10 Thermochemistry 2009, Prentice-Hall, Inc. Average bond energy in polyatomic molecules H 2 O (g) H (g) +OH (g) H 0 = 502 kJ OH (g) H (g) +O (g) H 0 = 427 kJ Average OH bond energy = = 464 kJ 9.10 Thermochemistry 2009, Prentice-Hall, Inc. Bond Energies (BE) and Enthalpy changes in reactions H 0 = total energy input total energy released = BE(reactants) BE(products) Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. 9.10 Thermochemistry 2009, Prentice-Hall, Inc. Use bond energies to calculate the enthalpy change for: H 2 (g) + F 2 (g) 2HF (g) H 0 = BE(reactants) BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HH FF Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF H 0 = 2 x = kJ 9.10 Thermochemistry 2009, Prentice-Hall, Inc. 9.3 Lattice energy (E) increases as Q increases and/or as r decreases. cmpd lattice energy MgF 2 MgO LiF LiCl Q= +2,-1 Q= +2,-2 r F < r Cl Electrostatic (Lattice) Energy E = k Q+Q-Q+Q- r Q + is the charge on the cation Q - is the charge on the anion r is the distance between the ions Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. Thermochemistry 2009, Prentice-Hall, Inc. 9.3 Born-Haber Cycle for Determining Lattice Energy H overall = H 1 + H 2 + H 3 + H 4 + H 5 oooooo