Thermochemistry Lecture Notes PDF

Thermochemistry 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 5 (c) 2014 Lecture presentation adapted from: Chapter 5: Thermochemistry James F. Kirby Quinnipiac University Hamden, CT  Ene...

Thermochemistry 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 5 (c) 2014 Lecture presentation adapted from: Chapter 5: Thermochemistry James F. Kirby Quinnipiac University Hamden, CT  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. This chapter is about thermodynamics, which is the study of energy transformations, and thermochemistry, which applies the field to chemical reactions, specifically. 3  Kinetic & Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition. Kinetic energy is energy an object possesses by virtue of its motion. SI unit of energy is the Joule (Kg.m2s-2 - Note 1 cal = 4.184 J) PE KE 4  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). 5  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. 6  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 7 Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects. 8  Work Changes in HEAT generally accompany chemical reaction, but reactions can also do WORK. – Consider the decomposition of KClO3 in a sealed vessel – The gas formed will expand within the vessel pushing against atmospheric pressure (pressure-volume work) When a gas expands, V is positive and work (w) = force (m*a) x distance (h) w is negative, signifying that energy m=mass leaves the system as work. and pressure (P) = (m*a) a=acceleration When a gas is compressed, V is area negative and w is positive, signifying that energy (as work) enters the system. w (performed by system) = - P * area x h w (performed by system) = - PV 9 Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). E = q + w. 10  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. 11  Internal Energy Change A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is 0.250 L, and the final volume is 0.980 L. If the piston expands against a constant pressure of 1.35 atm, how much work (in J) is done? 1 L.atm = 101.3 J. w = -PV Cylinder expands so w must be -ve V = (Vfinal - Vinitial) = 0.980 L - 0.250 L = 0.730 L w = -PV = -(1.35 atm)*(0.730 L) = - 0.9855 L.atm w = (- 0.9855 L.atm)*(101.3 J/L.atm) = -99.8 J 12  Internal Energy Change Gases A(g) and B(g) are confined in a cylinder-and-piston and react to give a solid product C(s). As the reaction occurs, the system loses 1150 J of heat an the piston moves downward, E = (-1150 J) + (480 J) = -670 J doing 480 J of work on the system. What is the change in the internal energy of the system? The overall effect is a net loss of energy from the system  The reaction is exothermic (q is negative) The piston contracts, indicating pressure decreases (w is positive) 13  State Functions Usually we have no way of knowing the internal energy of a system; – finding that value is simply too complex a problem. 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. 14  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. 15 State Functions A State Function is a thermodynamic quantity whose value depends only on the state at the moment, i. e., the temperature, pressure, volume, etc Whether the battery is shorted out or is discharged by running the fan, its E is the same. – Even though q and w are different in the two cases shown. However, q and w are not state functions. 16  Defining Enthalpy H Internal Energy can be defined as: At constant volume At constant pressure (most reactions) rearrange But Pi = Pf Enthalpy is a state function but Substitute for U Heat (qp) is not. At constant pressure, the change in enthalpy is the heat gained or lost. 17  Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. 18  Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants 19  Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. The instrument used to measure heat flow is called a calorimeter. 20  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, usually given for one mole of the substance. 21  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). 22  Constant Pressure Calorimetry The heat change for the system can be found by measuring the temperature change for an aqueous reaction in a calorimeter. The specific heat for water is well known (4.184 J/g∙K). We can calculate H (or q) for the reaction with this equation: q = m  Cs  T 23  Bomb Calorimetry Reactions can be carried out in a sealed “bomb”. Because the volume in the bomb calorimeter is constant, what is measured is the internal energy, E, not the H. For most reactions, the difference is very small. 24  Heat Capacity How much heat is released when (a) How much heat is needed to warm 4.50 g of methane gas (16.0 250 g of water (about 1 cup) from g/mol) is burned in a constant- 22°C to 98°C? (b) What is the molar heat capacity of water (Cm)? Specific pressure system. The known heat (Cs)of water = 4.18 J/g.K HCH4 is -890 kJ/mol. How much heat is released? q = Cs* m * T. (-890.0 kJ/mol) = (4.18 J/g.K)(250 g)(76 K)=7.9 x104 J. / = -250 kJ Since 1 mol = 18.0 g H2O Cm = 4.18 J/g.K * 18.0 g/mol = 75.2 J/mol.K 25  Heat Capacity When 4.00 g of methylhydrazine (46.1 g/mol) is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 to 39.50°C. In a separate experiment the heat capacity of the calorimeter was measured to be 7.794 kJ/°C. Calculate the heat of reaction for the T = 139.50 °C - 25.00 °C2 = 14.50 °C combustion of a mole of CH6N2. qrxn = -Ccal * T = -(7.794 kJ/°C)(14.50 °C) = -113.0 kJ H = qP = -113.0 kJ (at constant pressure) 2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l) −113.0 kJ H = qP = 4.00 g ∗ 46.1 g/mol = −1.30x103 kJ/mol 4g of material causes an X K rise in temperature. Previous experiment shows that 1 K rise requires 7.794 kJ 26  Hess’s Law H is 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. Hess’s law: If 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. 27  Hess’s Law Because H is a state function, the total enthalpy change depends only on the initial state (reactants) and the final state (products) of the reaction. 28  Calculation of H We can use Hess’s law with heats of formations (or even bond enthalpies), where n and m are the stoichiometric coefficients. – An enthalpy of formation, Hf, is the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. – Standard enthalpies of formation, ∆Hf°, are measured under standard conditions (25 ºC and 1.00 atm pressure). H = nHf,products – mHf°,reactants 29  Calculation of H C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) Imagine this as occurring in The sum of these equations three steps: is the overall equation! 1. Decomposition of propane to – Use these individual enthalpies its constituent elements: of these steps to estimate the unknown reaction enthalpy C3H8(g)  3 C(graphite) + 4 H2(g) 2. Formation of CO2: C3H8(g)  3 C(graphite) + 4 H2(g) 3 C(graphite) + 3 O2(g) 3 CO2(g) 3 C(graphite) + 3 O2(g) 3 CO2(g) 3. Formation of H2O: 4 H2(g) + 2 O2(g)  4 H2O(l) 4 H2(g) + 2 O2(g)  4 H2O(l) C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) 30  Calculation of H using Values from the Standard Enthalpy Table C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) H = Hproducts - Hreactants H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0 kJ)] H = [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)] H = (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ 31  Heats of Formation Which of these reactions at 25 °C Calculate the standard enthalpy of formation of CaCO3(s) given the enthalpy of reaction below and Hfo represents a standard enthalpy of of CaO and CO2 in Table below. formation? For each that does not, what changes are needed to make it an equation CaCO3(s) → CaO(s) + CO2(g) 𝐻 = 178.1 kJ. whose H is an enthalpy of formation? 𝐻 = 𝐻 CaO + 𝐻 CO2 - 𝑯𝒐𝒇 CaCO3 (a) 2 Na(s)+ O2(g) → Na2O(s) 178.1 = -635.5 kJ - 393.5 kJ - 𝑯𝒐𝒇 CaCO3 YES 𝑯𝒐𝒇 CaCO3 = -1207.1 kJ/mol (b) 2 K(l) + Cl2(g) → 2 KCl(s) NO, potassium is given as a liquid (c) C6H12O6(S) → 6 C(diamond) + 6 H2(g) + 3 O2 (g) NO, a decomposition rxn, need to reverse the reaction 32  Heats of Formation a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and H2O(l). (b) Compare the quantity of heat (a) produced by combustion of 1.00 g propane (44.1 g/mol) with that produced by 1.00 g benzene (78.1 C6H6(l) + O2(g) → 6 CO2(g) + 3 H2O(l) g/mol). 𝐻 = [6 𝐻 (CO2) + 3 𝐻 (H2O)] – [𝐻 (C6H6)+ 𝐻 (O2)] = [6(-393.5 kJ) + 3(-285.8 kJ)] - [(49.0 kJ)+ (0 kJ)] = (-2361 - 857.4 - 49.02) kJ = -3267 kJ 𝐻 per g C6H6 = (3267 kJ/mol)/(78.1g/mol) = -41.8 kJ/g (b) Water H2O(l) -285.8 Note 𝐻 C3H8(g) = -2220 kJ/mol or -50.3 kJ/g Oxygen O2 (g) 0 Propane C3H8(g) -227.0 33  Hess’s Law What is the enthalpy of reaction of C and O2 leading to CO. The enthalpy of reaction for the combustion of C to CO2 is -393.5 kJ/mol. The enthalpy for the combustion of CO to CO2 is -283.0 kJ/mol. Known 1: C(s) + O2(g) → CO2(g) H = -393.5 kJ Known 2: CO(g) + O2(g) → CO2(g) H = -283.0 kJ Unknown: C(s) + O2(g) → CO(g) H = ? Reverse 2 and combine with 1 C(s) + O2(g) → CO2(g) H = -393.5 kJ CO2(g) → CO(g) + O2(g) H = +283.0 kJ Known 1: C(s) + O2(g) → CO2(g) Hf o = -393.5 kJ C(s) + O2(g) → CO(g) H = -393.5 kJ +283.0 kJ Known 2: CO(g) + O2(g) → CO2(g) Hf = -283.0 kJ o H = -110.5 kJ 34  Hess’s Law Calculate H for the reaction below given the following information. 2 C (s) + H2(g) → C2H2(g) Hf o= ? Reverse 1 and combine with 2 & 3 (1) C2H2(g) + O2(g) → 2 CO2 (g) + H2O(l) Hf o = -1299.6 kJ (2) C(s) + O2(g) → CO2 (g) Hf o = -393.5 kJ (3) H2(g) + O2(g) → H2O(l) Hf o = -285.8 kJ 2 CO2 (g) + H2O(l) → C2H2(g) + O2(g) H = + 1299.6 kJ C(s) + O2(g) → CO2 (g) H = -393.5 kJ H2(g) + O2(g) → H2O(l) H = -285.8 kJ 2 C (s) + H2(g) → C2H2(g) H = + 1299.6 kJ -393.5 kJ -285.8 kJ H = 226.8 kJ 35  Average Bond Enthalpies Average bond enthalpies are positive, because bond breaking is an endothermic process. Note that these are averages over many different compounds; not every bond in nature for a pair of atoms has exactly the same bond energy. 36  Using Bond Enthalpies to Estimate Enthalpy of Reaction To estimate H for a reaction we can use the bond enthalpies of bonds broken and the new bonds formed. Energy is added to break bonds and released when making bonds. In other words, Hrxn = (bond enthalpies of all bonds broken) − (bond enthalpies of all bonds formed). 37  Bond Energies Calculate the overall reaction enthalpy for the reaction below using the table right. In this example, one C-H bond and one Cl-Cl bond are broken; one C-Cl and one H-Cl bond are formed. CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) H = [H (C-H) + H (Cl-Cl)] − [H (C-Cl) + H (H-Cl)] = [(413 kJ/mol) + (242 kJ/mol)] − [(328 kJ/mol) + (431 kJ/mol)] = (655 kJ/mol) − (759 kJ/mol) = −104 kJ/mol (exothermic) 38  Bond Enthalpy and Bond Length We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases. 39  Phase Changes Conversion from one state of matter to another is called a phase change. Energy is either added or released in a phase change. Phase changes include: – melting/freezing – vaporizing/condensing – subliming/depositing 40  Energy Change & Change of State The heat of fusion is the energy required to change a solid at its melting point to a liquid. The heat of vaporization is the energy required to change a liquid at its boiling point to a gas. The heat of sublimation is the energy required to change a solid directly to a gas. 41  Heating Curves Temperature & heat added is plotted in a heating curve. Within a phase, heat is the product of specific heat, sample mass, and temperature change. The temperature of the For the phase changes, the substance does not rise product of mass and the heat of during a phase change. fusion of vaporization is heat. 42  Phase Changes & H Calculate the enthalpy change upon converting 5 separate steps to calculate: 1.00 mol of ice at -25 °C to steam at 125 °C under a constant pressure of 1 atm. s  ∆HA→B = (1.00 mol)(18.0 g/mol)(2.03 J/g.K) (25 K) = 914 J = 0.91 kJ ∆HB→C = Hfus = 6.01 kJ/mol ∆HC→D = (1.00 mol)(18.0 g/mol)(4.18 J/g.K) (100 K) = 7520 J = 7.52 kJ ∆HD→E = Hvap = 40.67 kJ/mol ∆HE→F = (1.00 mol)(18.0 g/mol)(1.84 J/g.K) (25 K) = 830 J = 0.83 kJ/mol ∆HA→F = 0.91 kJ + 6.01 kJ + 7.52 kJ + 40.7 kJ + 0.83 kJ The specific heats of ice, liquid water, and = 56.0 kJ/mol steam are 2.03, 4.18, and 1.84 J/g.K, respectively. For H2O, Hfus = 6.01 kJ/mol and Hvap = 40.67 kJ/mol. 43