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Unit 5 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. The Nature of Energy • Two forms of energy exist, potential and kinetic. • Potential energy is due to composition or position. • Chemical potential energy is energy stored in a substance because of its composition. • Kinetic energy is energy of motion. Thermochemistry © 2009, Prentice-Hall, Inc. Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant. Thermochemistry © 2009, Prentice-Hall, Inc. Kinetic-Molecular Theory The average kinetic energy of the molecules is proportional to the absolute temperature. Thermochemistry © 2009, Prentice-Hall, Inc. Units of Energy • The SI unit of energy is the joule (J). kg m2 1 J = 1 s2 • An older, non-SI unit is still in widespread use: the calorie (cal). 1 cal = 4.184 J 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. 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. 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 = Efinal − Einitial Thermochemistry © 2009, Prentice-Hall, Inc. Changes in Internal Energy • If E > 0, Efinal > Einitial – 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, Efinal < Einitial – Therefore, the system released energy to the surroundings. – This energy change is called exothermic. 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. 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. 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. Specific Heat • The specific heat of any substance is the amount of heat required to raise one gram of that substance one degree Celsius. Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1oC Depends on both the object’s mass and its chemical composition Thermochemistry © 2009, Prentice-Hall, Inc. Specific Heat (cont.) • Calculating heat absorbed and released – q = c × m × ΔT – q = heat absorbed or released (in Joules) – c = specific heat of substance – m = mass of substance in grams – ΔT = change in temperature in Celsius Thermochemistry © 2009, Prentice-Hall, Inc. Specific Heat (cont.) • Examples: • How much heat does a 20.0 g ice cube absorb as its temperature increases from (-27.0oC) to 0.0oC? Give your answer in both joules and calories. • q = c × m × ΔT • Specific Heat of Ice = 2.03 J/goC • 1 calorie = 4.184 J Thermochemistry © 2009, Prentice-Hall, Inc. Specific Heat (cont.) • Example Cont. q=? c = 2.03 J/goC m = 20.0 grams ΔT = FinalTemp(0.0oC) – InitialTemp (-27.0oC) = Change (27.0oC) q = c × m × ΔT q = (2.03 J/goC)(20.0g)(27.0oC)= 1.10 x 103 J 263 cal Thermochemistry © 2009, Prentice-Hall, Inc. • Example 2: • A 5.00 gram sample of a metal is initially at 55.0 ºC. When the metal is allowed to cool for a certain time, 98.8 Joules of energy are lost and the temperature decreases to 11.0 ºC. What is the specific heat of the metal? What metal is it? • q = c × m × ΔT 0.449 J/g ºC Iron Thermochemistry © 2009, Prentice-Hall, Inc. The liquid is boiling at 100o 120 C; no temperature change (use q = moles x ΔHvap.) The gas temperature is rising from 100 to 120 oC (use The Heat Curve for Water, q =going massfrom x ΔT -20 x C)to 120 oC, The liquid temperature is rising from 0 to 100 oC (use q = mass x ΔT x C) The solid is melting at 0o C; no temperature change (use q = moles x ΔHfus.) The solid temperature is rising from -20 to 0 oC (use q = mass x ΔT x C) Thermochemistry 23 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 Einitial and Efinal. 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 + PV Thermochemistry © 2009, Prentice-Hall, Inc. Enthalpy • Since E = q + w and w = -PV, we can substitute these into the enthalpy expression: H = E + PV 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 = Hproducts − Hreactants 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. Hess’s 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. Hess’s Law Hess’s 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. Hess’s 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, Hf, 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, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). Thermochemistry © 2009, Prentice-Hall, Inc. Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) H0rxn = -393.5 kJ S(rhombic) + O2 (g) CS2(l) + 3O2 (g) SO2 (g) H0rxn = -296.1 kJ CO2 (g) + 2SO2 (g) 0 = -1072 kJ Hrxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. C(graphite) + O2 (g) 2S(rhombic) + 2O2 (g) + CO2(g) + 2SO2 (g) CO2 (g) H0rxn = -393.5 kJ 2SO2 (g) H0rxn = -296.1x2 kJ CS2 (l) + 3O2 (g) 0 = +1072 kJ Hrxn C(graphite) + 2S(rhombic) CS2 (l) Thermochemistry H0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ © 2009, Prentice-Hall, Inc. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) • Imagine this as occurring in three steps: C3H8 (g) 3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g) 3 CO2 (g) 4 H2 (g) + 2 O2 (g) 4 H2O (l) Thermochemistry © 2009, Prentice-Hall, Inc. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) • Imagine this as occurring in three steps: C3H8 (g) 3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g) 3 CO2 (g) 4 H2 (g) + 2 O2 (g) 4 H2O (l) Thermochemistry © 2009, Prentice-Hall, Inc. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) • Imagine this as occurring in three steps: C3H8 (g) 3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g) 3 CO2 (g) 4 H2 (g) + 2 O2 (g) 4 H2O (l) Thermochemistry © 2009, Prentice-Hall, Inc. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) • The sum of these equations is: C3H8 (g) 3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g) 3 CO2 (g) 4 H2 (g) + 2 O2 (g) 4 H2O (l) C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) Thermochemistry © 2009, Prentice-Hall, Inc. Calculation of H We can use Hess’s law in this way: H = nHf°products – mHf° reactants where n and m are the stoichiometric coefficients. Thermochemistry © 2009, Prentice-Hall, Inc. Calculation of H C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l) H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ Thermochemistry © 2009, Prentice-Hall, Inc. Benzene (C6H6) 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 49.04 kJ/mol. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) H0rxn = nH0f (products) - mHf0 (reactants) H0rxn = [ 12H0f (CO2) + 6H0f (H2O)] - [ 2H0f (C6H6)] H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ = - 2973 kJ/mol C6H6 2 mol Thermochemistry © 2009, 6.6 Prentice-Hall, Inc. The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. Bond Energy H0 = 436.4 kJ H2 (g) H (g) + H (g) Cl2 (g) Cl (g) + Cl (g) H0 = 242.7 kJ HCl (g) H (g) + Cl (g) H0 = 431.9 kJ O2 (g) O (g) + O (g) H0 = 498.7 kJ O O N2 (g) N (g) + N (g) H0 = 941.4 kJ N N Bond Energies Single bond < Double bond < Triple bond Thermochemistry © 2009, 9.10 Prentice-Hall, Inc. Average bond energy in polyatomic molecules H2O (g) OH (g) H (g) + OH (g) H0 = 502 kJ H (g) + O (g) H0 = 427 kJ 502 + 427 = 464 kJ Average OH bond energy = 2 Thermochemistry © 2009, 9.10 Prentice-Hall, Inc. Bond Energies (BE) and Enthalpy changes in reactions Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. H0 = total energy input – total energy released = BE(reactants) – BE(products) Thermochemistry © 2009, 9.10 Prentice-Hall, Inc. Use bond energies to calculate the enthalpy change for: H2 (g) + F2 (g) 2HF (g) H0 = BE(reactants) – BE(products) Type of bonds broken H H F F Type of bonds formed H F Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) 1 1 436.4 156.9 436.4 156.9 Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) 2 568.2 1136.4 H0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ Thermochemistry © 2009, 9.10 Prentice-Hall, Inc. Electrostatic (Lattice) Energy Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. Q+Q E=k 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) increases as Q increases and/or as r decreases. cmpd MgF2 MgO LiF LiCl lattice energy 2957 Q= +2,-1 3938 Q= +2,-2 1036 853 rThermochemistry F < r Cl © 2009, 9.3 Prentice-Hall, Inc. Born-Haber Cycle for Determining Lattice Energy o o o o o o Hoverall = H1 + H2 + H3 + H4 + H5 Thermochemistry © 2009, 9.3 Prentice-Hall, Inc.