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Thermochemistry
 

Thermochemistry

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    Thermochemistry Thermochemistry Presentation Transcript

    • 6.1 Basic principles 6.2 measurement of heat flow– Calorimetry (Laboratory Measurement of Heats of Reaction) 6.3 Energy and Changes of State 6.4 The first law of thermodynamics 6.5 Enthalpy change for chemical reaction 6.6 Hess’s Law 6.7 Standard enthalpy of formation 6.8 Product or reactant-favored reaction and thermochemistry
    • 6.1 Some basic principles State A – Fuel in the tank change in potential energy EQUALS kinetic energy State B - Fuel burned and exhaust produced A system of fuel and exhaust. A fuel is higher in chemical potential energy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car.
    • A chemical system and its surroundings When a chemical reaction takes place, we consider the substances involved. Therefore, the reactants and products are the system.
    • Thermodynamics is the study of heat and its transformations. Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. Fundamental premise When energy is transferred from one object to another, it appears as work and/or as heat. For our work we must define a system to study; everything else then becomes the surroundings. The system is composed of particles with their own internal energies (E or U). Therefore the system has an internal energy. When a change occurs, the internal energy changes. State properties State property of system is described by giving its composition, temperature, and pressure. State property depends on the state of the system, not on the way the system reaches the state.
    • A system transferring energy as heat only When q is negative (-), the heat When q is positive (+), the heat flows out of the system into flows into the system from surrounding. surrounding.
    • A system losing energy as work only Zn(s) + 2H+(aq) + 2Cl-(aq) Energy, E work done on surroundings DE<0 H2(g) + Zn2+(aq) + 2Cl-(aq) When w is negative (-), When w is positive (+), work done by system. work done on system.
    • The Sign Conventions* for q, w and ∆E q + w = ∆E + + + + - depends on sizes of q and w - + depends on sizes of q and w - - - • For q (heat): + means system gains heat, Endothermic; - means system loses heat. Exothermic. • For w (work): + means work done on system; - means work done by system.
    • Magnitude of heat In any process, we are interested in the direction of heat flow and heat magnitude. We express heat, q, in the unit of joules (SI unit) and kilojoules. The joules is named for James Joule who carried out the precise thermodynamic measurement. Traditionally, chemists use the calorie as an energy unit. Calorie is the amount of heat needed to raise 1.00 g English physicist water 1 °C. 1 cal = 4.184 J 1 kcal = 4.184 kJ
    • 6.2 Specific Heat Capacity and Heat Transfer It is important to discuss the magnitude of heat flow in chemical reactions of phase changes. The equation, q = C × ∆T, express the relationship between the magnitude of heat flow and temperature change. ∆T = Tfinal - Tinitial The quantity C is known as heat capacity of the system, having a unit J/°C. Finding the Quantity of Heat from Specific Heat Capacity PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25 0C to 300 0C? The specific heat capacity (c) of Cu is 0.387 J/g*K. PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = c x mass x DT to find the answer. DT in 0C is the same as for K. SOLUTION: 0.387 J q= x 125 g x (300-25) 0C = 1.33x104 J g*K
    • Coffee-cup calorimeter • The heat given out by a reaction is absorbed by water. • The mass of water can be determined. • The heat capacity of water is 4.18 J/g • °C. • The temperature change can be measured by the thermometer. • Heat flow can be calculated for the reaction. • Equation, q = mass × c × ∆T, express the relationship of heat flow and temperature change. • The heat flow for the reaction is equal in magnitude, but opposite in sign to that measured by calorimeter
    • Determining the Heat of a Reaction PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00 0C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.21 0C. Calculate qsoln (in J). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.18 J/g*K) PLAN: 1. We need to determine the limiting reactant from the net ionic equation. 2. The moles of NaOH and HCl as well as the total volume can be calculated. 3. From the volume we use density to find the mass of the water formed. 4. At this point, qsoln can be calculated using the eqaution, q = mass × c × ∆T. • The heat divided by the M of water will give us the heat per mole of water formed.
    • Determining the Heat of a Reaction SOLUTION: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + OH-(aq) H2O(l) For NaOH 0.500 M × 0.0500 L = 0.0250 mol OH- For HCl 0.500 M × 0.0250 L = 0.0125 mol H+ HCl is the limiting reactant. 0.0125 mol of H2O will form during the rxn. Total volume after mixing = 0.0750 L 0.0750 L x 103 mL/L × 1.00 g/mL = 75.0 g of water Q = mass x specific heat x DT = 75.0 g × 4.18 J/g* 0C × (27.21-25.00) °C = 693 J
    • Take-home message: The schematic diagram of A bomb calorimeter 6.6 Calorimetery • The heat given out by a reaction is absorbed by water. • The mass of water can be determined. • The heat capacity of water is 4.18 J/g • °C. • The temperature change can be measured by the thermometer. • Heat flow can be calculated for the reaction. • Equation, q = mass × c × ∆T, express the relationship of heat flow and temperature change. • The heat flow for the reaction is equal in magnitude, but opposite in sign to that measured by calorimeter
    • Calculating the Heat of Combustion PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 KiloCalories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2 (the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases 4.937 0C. Is the manufacturer’s claim correct? PLAN: - q sample = qcalorimeter SOLUTION: = heat capacity × ∆T qcalorimeter = 8.151 kJ/K × 4.937 K = 40.24 kJ kcal 40.24 kJ × = 9.63 Kilocalories 4.18 kJ The manufacturer’s claim is true.
    • 6.3 Energy and changes of state A cooling curve for the conversion of gaseous water to ice. Five stages – vapor cools, vapor condenses (constant temperature), liquid water cools, liquid water freezes (constant tempterature), solid water cools
    • Quantitative Aspects of Phase Changes Within a phase, a change in heat is accompanied by a change in temperature which is associated with a change in average Ek as the most probable speed of the molecules changes. q = (amount)(molar heat capacity)(∆T) During a phase change, a change in heat occurs at a constant temperature, which is associated with a change in Ep, as the average distance between molecules changes. q = (amount)(enthalpy of phase change)
    • 6.4 The first law of thermodynamics The Meaning of Enthalpy w = - P∆V ∆H ≈ ∆ E in H = E + PV 1. Reactions that do not involve gases. Where H is enthalpy 2. Reactions in which the number of moles of gas does not change. ∆H = DE + P ∆ V 3. Reactions in which the number of moles of qp = ∆ E + P ∆ V = ∆ H gas does change but q is >>> P ∆ V.
    • If a chemical reaction occurs under a constant pressure, the difference in enthalpy between product and reactant equals the heat flow for the reaction. Qreaction at a constant pressure = ∆E + P∆V = ∆H Enthalpy diagrams for exothermic and endothermic processes. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g) CH4 + 2O2 H2O(g) Hinitial Hfinal Enthalpy, H Enthalpy, H ∆H < 0 heat out ∆H > 0 heat in CO2 + 2H2O H2O(l) Hfinal Hinitial A Exothermic process B Endothermic process q < 0, Hproduct < H reactant q > 0, Hproduct > H reactant
    • Drawing Enthalpy Diagrams and Determining the Sign of ∆H PROBLEM: In each of the following cases, determine the sign of DH, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram. (a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ (b) 40.7kJ + H2O(l) H2O(g) PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction SOLUTION: (a) The reaction is exothermic. (b) The reaction is endothermic. H2(g) + 1/2O2(g) (reactants) H2O(g) (products) EXOTHERMIC ∆H = -285.8kJ ENDOTHERMIC ∆H = +40.7kJ H2O(l) (products) H2O(l) (reactants)
    • 6.5 Enthalpy change for chemical reactions Thermal chemical reaction – shows the enthalpy relationship between reactants and products Rules of thermochemistry v The magnitude of ∆H is directly proportional to the amount of reactants or products. v ∆H for a reaction is equal in magnitude but opposite in sign for the reverse reaction. v The value of for a reaction is the same whether it occurs in one step or multi-steps. v Hess law ∆H = ∆H1 + ∆H2 + ∙∙∙∙ Some Important Types of Enthalpy Change Heat of combustion (∆Hcomb) C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g) Heat of formation (∆Hf) K(s) + 1/2Br2(l) KBr(s) Heat of fusion (∆Hfus) NaCl(s) NaCl(l) Heat of vaporization (∆Hvap) C6H6(l) C6H6(g)
    • Using the Heat of Reaction (∆Hrxn) to Find Amounts PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by Al2O3(s) 2Al(s) + 3/2O2(g) ∆Hrxn = 1676 kJ If aluminum is produced this way, how many grams of aluminum can form when 1.000x103 kJ of heat is transferred? PLAN: SOLUTION: Heat (kJ) 1.000x103 kJ x 2 mol Al 26.98 g Al 1676 kJ = 2 mol Al 1676 kJ 1 mol Al mol of Al X M = 32.20 g Al g of Al
    • 6.7 Hess’s Law --- To Calculate an Unknown ∆H PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation: CO(g) + NO(g) CO2(g) + 1/2N2(g) ∆ H = ? Given the following information, calculate the unknown ∆ H: Equation A: CO(g) + 1/2O2(g) CO2(g) ∆ HA = -283.0 kJ Equation B: N2(g) + O2(g) 2NO(g) ∆ HB = 180.6 kJ PLAN: Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation. SOLUTION: Multiply Equation B by 1/2 and reverse it. CO(g) + 1/2O2(g) CO2(g) ∆ HA = -283.0 kJ NO(g) 1/2N2(g) + 1/2O2(g) ∆ HB = -90.3 kJ CO(g) + NO(g) CO2(g) + 1/2N2(g) ∆ Hrxn = -373.3 kJ
    • 6.8 Standard enthalpies of Formation PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DH 0f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. PLAN: Use the table of heats of formation for values. SOLUTION: (a) Ag(s) + 1/2Cl2(g) AgCl(s) ∆H 0f = -127.0 kJ (b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s) ∆H 0f = -1206.9 kJ (c) 1/2H2(g) + C(graphite) + 1/2N2(g) HCN(g) ∆H 0f = 135 kJ
    • The general process for determining ∆H 0rxn from ∆H0f values. Elements decomposition Enthalpy, H -∆H0f ∆H0f formation Reactants Hinitial ∆H0rxn Products Hfinal ∆H0rxn = Σ m∆H0f(products) - Σ n∆H0f(reactants)
    • Calculating the Heat of Reaction from Heats of Formation PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Calculate DH0rxn from DH 0f values. PLAN: Look up the DH0f values and use Hess’s Law to find DHrxn. SOLUTION: DHrxn = S mDH0f (products) - S nDH0f (reactants) DHrxn = [4(DH0f NO(g) + 6(DH0f H2O(g)] - [4(DH0f NH3(g) + 5(DH0f O2(g)] = (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) - [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)] ∆Hrxn = -906 kJ Lectrue 3
    • Specific Heat Capacities of Some Elements, Compounds, and Materials Substance Specific Heat Substance Specific Heat Capacity (J/g*K) Capacity (J/g*K) Elements Materials aluminum, Al 0.900 wood 1.76 graphite,C 0.711 cement 0.88 iron, Fe 0.450 glass 0.84 copper, Cu 0.387 granite 0.79 gold, Au 0.129 steel 0.45 Compounds water, H2O(l) 4.184 ethyl alcohol, C2H5OH(l) 2.46 ethylene glycol, (CH2OH)2(l) 2.42 carbon tetrachloride, CCl4(l) 0.864
    • Selected Standard Heats of Formation at 250C(298K) Formula ∆H0f(kJ/mol) Formula ∆H0f(kJ/mol) Formula ∆H0f(kJ/mol) calcium silver Cl2(g) 0 Ca(s) 0 Ag(s) 0 HCl(g) -92.3 CaO(s) -635.1 AgCl(s) -127.0 CaCO3(s) -1206.9 hydrogen H(g) 218 sodium carbon H2(g) 0 Na(s) 0 C(graphite) 0 C(diamond) 1.9 nitrogen Na(g) 107.8 CO(g) -110.5 N2(g) 0 NaCl(s) -411.1 CO2(g) -393.5 NH3(g) -45.9 sulfur CH4(g) -74.9 NO(g) 90.3 S8(rhombic) 0 CH3OH(l) -238.6 S8(monoclinic) 2 oxygen HCN(g) 135 SO2(g) -296.8 O2(g) 0 CSs(l) 87.9 O3(g) 143 SO3(g) -396.0 chlorine H2O(g) -241.8 Cl(g) 121.0 H2O(l) -285.8