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Chapter 8 Thermochemistry

Chapter 8 Thermochemistry. 8.1 Principles of Heat Flow. 8.2 Calorimetry. 8.3 Enthalpy. 8.4 Thermochemical Equations. Hess’s Law. 8.5 Enthalpies of Formation. 8.6 Bond Enthalpy. 8.7 The First Law of Thermodynamics. Thermodynamic s is the study of heat and its transformations.

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Chapter 8 Thermochemistry

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  1. Chapter 8Thermochemistry 8.1 Principles of Heat Flow 8.2 Calorimetry 8.3 Enthalpy 8.4 Thermochemical Equations. Hess’s Law 8.5 Enthalpies of Formation 8.6 Bond Enthalpy 8.7 The First Law of Thermodynamics

  2. 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.

  3. the system A chemical system and its surroundings. the surroundings

  4. Energy is the capacity to do work. energy due to the position of an object or energy due to composition which can be released in a chemical reaction Potential Energy Kinetic Energy energy due to the motion of the object Potential and kinetic energy can be interconverted.

  5. Energy is the capacity to do work. Figure A less stable, Ep= mgh change in potential energyEQUALS kinetic energy more stable, Ek= 1/2 mv2 A gravitational system. The potential energy of a lifted weight is converted to kinetic energy as the weight falls.

  6. Energy is the capacity to do work. Figure B less stable change in potential energyEQUALS kinetic energy more stable A system of two balls attached by a spring. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released.

  7. Energy is the capacity to do work. Figure C less stable change in potential energyEQUALS kinetic energy more stable A system of oppositely charged particles. The potential energy gained when the charges are separated is converted to kinetic energy as the attraction pulls these charges together.

  8. Energy is the capacity to do work. Figure D less stable change in potential energyEQUALS kinetic energy more stable 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.

  9. Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. DE = Efinal - Einitial = Eproducts - Ereactants Internal Energy, E, of the system consists of potential (composition, state of subdivision), and kinetic (vibrational, rotational, and translational) energies.

  10. A system transferring energy as heat only.

  11. Energy, E work done on surroundings DE<0 A system losing energy as work only. Zn(s) + 2H+(aq) + 2Cl-(aq) H2(g) + Zn2+(aq) + 2Cl-(aq)

  12. Changes in Internal Energy • If E < 0, Efinal < Einitial • Therefore, the system released energy to the surroundings. • This energy change is called exergonic, or exothermic.

  13. Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w), or both q, and w. • That is, E = q + w.

  14. The Sign Conventions* for q, w and DE + = DE q w + + + + - depends on sizes of q and w - + depends on sizes of q and w - - - * For q: + means system gains heat; - means system loses heat. * For w: + means word done on system; - means work done by system.

  15. DEuniverse = DEsystem + DEsurroundings Units of Energy Joule (J) 1 J = 1 kg·m2/s2 Calorie (cal) 1 cal = 4.184 J British Thermal Unit 1 Btu = 1055 J

  16. Two different paths for the energy change of a system.

  17. State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. • In the system below, the water could have reached room temperature from either direction. • However, we do know that the internal energy of a system is independent of the path by which the system achieved that state.

  18. State Functions • A state function depends only on the current state of a system. The change in a state function between two states is independent of the pathway between them. • Therefore, internal energy is a state function. • And so, E depends only on Einitial and Efinal.

  19. State Functions • However, q and w are not state functions. • Whether the battery is shorted out (a) or is discharged by running the fan (b), its E is the same. • But q and w are different in the two cases!

  20. PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (DE) in J, kJ, and kcal. kcal kJ -776J -0.776kJ 4.18kJ 103J Sample Problem Determining the Change in Internal Energy of a System PLAN: Define system and surroundings, assign signs to q and w and calculate DE. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = - 325 J w = - 451 J DE = q + w = -325 J + (-451 J) = -776 J = -0.776kJ = -0.185 kcal

  21. Pressure-volume work.

  22. Sample Problem • 1. A system expands against a constant pressure of 1.50 atm from 10.0 L to 25.0 L, while absorbing 150 J of heat. Determine the change in internal energy, ∆E.[1 L·atm = 101.3 J] Plan: Find work done by the system, add to heat change. w = -P∆V = -(1.50 atm)(25.0 L - 10.0 L)(101.3 J/L·atm) = - 2279.25 J = - 2280 J ∆E = q + w = 150 J - 2280 J = -2130 J

  23. 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

  24. 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

  25. 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 = qp • So, at constant pressure the change in enthalpy is the heat gained or lost.

  26. The Truth about Enthalpy • Enthalpy is an extensive property. • H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. • H for a reaction depends on the state of the products and the state of the reactants, that is ∆H is a state function.

  27. The Meaning of Enthalpy w = - PDV DH ≈ DE 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. DH = DE + PDV 3. Reactions in which the number of moles of gas does change but q is >>> (much, much, much greater) PDV qp = DE + PDV = DH When gases are involved, ∆H = ∆E + ∆nRT Where ∆n = moles gaseous products - moles gaseous reactants.

  28. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g) Enthalpy, H Enthalpy, H heat out heat in Enthalpy diagrams for exothermic and endothermic processes. CH4 + 2O2 H2O(g) Hfinal Hinitial DH < 0 DH > 0 CO2 + 2H2O H2O(l) Hfinal Hinitial A Exothermic process B Endothermic process

  29. C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g) K(s) + 1/2Br2(l) KBr(s) NaCl(s) NaCl(l) C6H6(l) C6H6(g) Some Important Types of Enthalpy Change heat of combustion (DHcomb) heat of formation (DHf) heat of fusion (DHfus) heat of vaporization (DHvap)

  30. 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 250C to 300.0C? The specific heat capacity (Sp) of Cu is 0.387 J/g·˚C. 0.387 J g·˚C Sample Problem Finding the Quantity of Heat from Specific Heat Capacity PLAN: Given the mass, specific heat capacity and change in temperature, we can use q = Sp x mass x DT to find the answer. DT in 0C is the same as for K. SOLUTION: = 1.33x104 J q = x 125 g x (300-25)0C

  31. Coffee-cup calorimeter.

  32. PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and DHrxn (in kJ/mol). (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 Sp = 4.184 J/g·˚C) Sample Problem Determining the Heat of a Reaction PLAN: We need to determine the limiting reactant from the net ionic equation. The moles of NaOH and HCl as well as the total volume can be calculated. From the volume we use density to find the mass of the water formed. At this point qsoln can be calculated using the mass, c, and DT. The heat divided by the M of water will give us the heat per mole of water formed.

  33. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + OH-(aq) H2O(l) 0.500 M x 0.0500 L = 0.0250 mol OH- For NaOH For HCl 0.500 M x 0.0250 L = 0.0125 mol H+ Sample Problem Determining the Heat of a Reaction continued SOLUTION: 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 x 1.00 g/mL = 75.0 g of water q = - mass x specific heat x DT = - 75.0 g x 4.184 J/g·0C x (27.21-25.00)0C = - 693.50 J (-693.50 J/0.0125 mol H2O)(kJ/103 J) = -55.5 kJ/ mol H2O formed

  34. Figure 5.10 A bomb calorimeter

  35. PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories 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/˚C). The temperature increases 4.9370C. Is the manufacturer’s claim correct? 40.24 kJ kcal 4.18 kJ Sample Problem Calculating the Heat of Combustion PLAN: Energy conservation; qsample + qcal = 0 => - q sample = qcalorimeter SOLUTION: qcalorimeter = heat capacity x DT = 8.151 kJ/˚C x 4.937 ˚C = 40.24 kJ = 9.63 kcal or Calories The manufacturer’s claim is true.

  36. AMOUNT (mol) of compound A AMOUNT (mol) of compound B HEAT (kJ) gained or lost Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction. molar ratio from balanced equation DHrxn (kJ/mol)

  37. 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) DHrxn = 1676 kJ If aluminum is produced this way, how many grams of aluminum can form when 1.000x103 kJ of heat is transferred? 26.98 g Al 2 mol Al 1 mol Al 1676 kJ Sample Problem Using the Heat of Reaction (DHrxn) to Find Amounts PLAN: SOLUTION: heat(kJ) 1.000x103 kJ x 1676kJ=2mol Al mol of Al = 32.20 g Al x M g of Al

  38. 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) DH = ? Given the following information, calculate the unknown DH: Equation A: CO(g) + 1/2O2(g) CO2(g) DHA = -283.0 kJ Equation B: N2(g) + O2(g) 2NO(g) DHB = 180.6 kJ CO(g) + 1/2O2(g) CO2(g) DHA = -283.0 kJ NO(g) 1/2N2(g) + 1/2O2(g) CO(g) + NO(g) CO2(g) + 1/2N2(g) Sample Problem Using Hess’s Law to Calculate an Unknown DH 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. DHB = -90.3 kJ DHrxn = -373.3 kJ

  39. PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DH0f. (a) Ag(s) + 1/2Cl2(g) AgCl(s) (b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s) (c)1/2H2(g) + C(graphite) + 1/2N2(g) HCN(g) Sample Problem Writing Formation Equations (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: DH0f = -127.0 kJ DH0f = -1206.9 kJ DH0f = 135 kJ

  40. decomposition formation The general process for determining DH0rxn from DH0f values. Elements -DH0f DH0f Enthalpy, H Reactants Hinitial DH0rxn Products Hfinal DH0rxn = SmDH0f(products) - SnDH0f(reactants)

  41. 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 DH0f values. Sample Problem Calculating the Heat of Reaction from Heats of Formation PLAN: Look up the DH0f values and use Hess’s Law to find DHrxn. SOLUTION: DHrxn = SmDH0f (products) - SnDH0f (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)] DHrxn = -906 kJ

  42. State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

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