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Chapter 6 Thermochemistry: Energy Flow and Chemical Change 6-1 Dr. Wolf’s CHM 101 Thermochemistry: Energy Flow and Chemical Change 6.1 Forms of Energy and Their Interconversion 6.2 Enthalpy: Heats of Reaction and Chemical Change 6.3 Calorimetry: Laboratory Measurement of Heats of Reaction 6.4 Stoichiometry of Thermochemical Equations 6.5 Hess’s Law of Heat Summation 6.6 Standard Heats of Reaction (DH0rxn) 6-2 Dr. Wolf’s CHM 101 Thermochemistry: Energy Flow and Chemical Change In the observation and measurement of a change in energy, we talk about the System and the Surroundings. The System is what we are studying, e.g. a reaction taking place in a flask. The Surroundings are the rest of the universe, but generally only the nearby portion that is relevant to the system. 6-3 Dr. Wolf’s CHM 101 A change in energy in the System is always accompanied by an opposite change in energy of the Surroundings. Change in energy, DE = Efinal - Einitial = Eproducts - Ereactants Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. 6-4 Dr. Wolf’s CHM 101 Heat (thermal energy) and Work, Two Forms of D E The symbol for heat is q. The symbol for work is w. They can be positive or negative values. Energy flowing into a system is defined as positive. Energy flowing out of a system is negative. DE = q + w 6-5 In a system transferring energy as heat only. Dr. Wolf’s CHM 101 DE = q + 0 A system losing energy as work only. DE = 0 + w 6-6 Dr. Wolf’s CHM 101 Work is done on the surroundings, so work on the system is negative. The Sign Conventions* for q, w and DE q + w = DE + + + + - 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. 6-7 Dr. Wolf’s CHM 101 Law of Conservation of Energy (First Law of Thermodynamics) The total energy of the universe is constant. Energy transfers can occur between the system and surroundings in the form of heat and/or work. But the energy of the system and the energy of the surroundings remains constant; energy is conserved. DEuniverse = DEsystem + DEsurroundings = 0 Units of Energy Joule (J) Calorie (cal) British Thermal Unit 6-8 Dr. Wolf’s CHM 101 1 J = 1 kg*m2/s2 1 cal = 4.18J 1 Btu = 1055 J (mass x accel x dist) Sample Problem 6.1 PROBLEM: PLAN: System 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 451J of work on the pistons and the system loses 325J to the surroundings as heat, calculate the change in energy (DE) in J, kJ, and kcal. Define system and surrounds, assign signs to q and w and calculate DE. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = - 325J DE = q + w = -776J 6-9 Determining the Change in Internal Energy of a kJ 103J w = - 451J -325J + (-451J) = -776J = 0.776kJ Dr. Wolf’s CHM 101 0.776kJ kcal 4.18kJ = 0.186kcal State functions such as internal energy of a system are dependent only upon the initial state and the final state. It is independent upon the pathway to get between the two states. Two different paths for the energy change of a system. 6-10 Dr. Wolf’s CHM 101 Enthalpy, H In chemistry the work is usually electrical (Ch 21) or pressure-volume (PV) work.....work done by expanding gas (change in volume, DV) DV= Vfinal - Vinitial So this work done on surroundings is negative. DE = q + w DE = q - PDV 6-11 Dr. Wolf’s CHM 101 Pressure-volume work. The Meaning of Enthalpy, H DE = q - PDV qp = DE + PDV (constant pressure) DH = DE + PDV In many reactions little or no work done so DH ~ DE in 1. Reactions that do not involve gases. 2. Reactions in which the number of moles of gas does not change. 3. Reactions in which the number of moles of gas does change but q is >>> PDV. DH is the change in enthalpy - change in internal energy plus PDV work qp = DE + PDV = DH DH = qp , equals the heat gained or lost at constant pressure. 6-12 Dr. Wolf’s CHM 101 Change of Enthalpy, D H DH= Hfinal - Hinitial Exothermic / Endothermic If the products have less energy than the reactants, i.e. Hfinal < Hinitial , then DH is negative. Heat has been released. Exothermic processes have a negative DH. Enthalpy diagram for exothermic process. CH4(g) + 2O2(g) 6-13 CO2(g) + 2H2O(g) Dr. Wolf’s CHM 101 Change of Enthalpy, D H DH= Hfinal - Hinitial Exothermic / Endothermic If the products have more energy than the reactants, i.e. Hfinal > Hinitial , then DH is positive. Heat has been absorbed. Endothermic processes have a positive DH. Enthalpy diagram for endothermic process. H2O(s) 6-14 Dr. Wolf’s CHM 101 H2O(l) Sample Problem 6.2 PROBLEM: PLAN: Drawing Enthalpy Diagrams and Determining the Sign of DH 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) 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. H2(g) + 1/2O2(g) (reactants) EXOTHERMIC H2O(l) 6-15 Dr. Wolf’s CHM 101 DH = -285.8kJ (products) (b) The reaction is endothermic. H2O(g) ENDOTHERMIC H2O(l) (products) DH = +40.7kJ (reactants) Some Important Types of Enthalpy Change heat of combustion (DHcomb) C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g) heat of formation (DHf) K(s) + 1/2Br2(l) KBr(s) heat of fusion (DHfus) NaCl(s) NaCl(l) heat of vaporization (DHvap) C6H6(l) 6-16 Dr. Wolf’s CHM 101 C6H6(g) Table 6.3 Heats of Combustion of Some Fats and Carbohydrates Substance DHcomb(kJ/g) Fats vegetable oil -37.0 (x 1/4.18) = -8.85 Cal / g margarine -30.1 butter -30.0 Carbohydrates 6-17 table sugar (sucrose) -16.2 (x 1/4.18) = -3.88 Cal / g brown rice -14.9 maple syrup -10.4 Dr. Wolf’s CHM 101 Components of Internal Energy Contributions to the kinetic energy: • The molecule moving through space, Ek(translation) • The molecule rotating, Ek(rotation) • The bound atoms vibrating, Ek(vibration) • The electrons moving within each atom, Ek(electron) Contributions to the potential energy: • Forces between the bound atoms vibrating, Ep(vibration) • Forces between nucleus and electrons and between electrons in each atom, Ep(atom) • Forces between the protons and neutrons in each nucleus, Ep(nuclei) • Forces between nuclei and shared electron pair in each bond, Ep(bond) 6-18 Dr. Wolf’s CHM 101 Components of internal energy (E) 6-19 Dr. Wolf’s CHM 101 Breaking and Forming Chemical Bonds During a chemical reaction, chemical bonds are broken and new chemical bonds are formed. Breaking each chemical bond requires energy, Endothermic Forming each chemical bond releases energy, Exothermic The DH for a reaction is the sum of the energy required and released during the bond breaking and forming process. The enthalpy for any reverse process has the same numerical value but a change in sign. 6-20 Dr. Wolf’s CHM 101 Specific Heat Capacity When an object absorbs heat, it gets hotter. The more heat it absorbs the hotter it gets, i.e. q ~ DT or q = constant x DT or q /DT = constant This capacity of an object to absorb heat is called its: heat capacity = q / DT (units of J/K) The amount of heat to change the temperature by 1 K A related property is Specific Heat Capacity (c) This is the amount of heat to change the temperature of 1 gram of substance by 1 K c = q / mass x DT (units of J/g-K) Another property is Molar Heat Capacity (C) This is the amount of heat to change the temperature of 1 mole of substance by 1 K C = q / moles x DT (units of J/mol-K) 6-21 Dr. Wolf’s CHM 101 Sample Problem 6.3 PROBLEM: Calculating the Quantity of Heat from the Specific Heat Capacity A layer of copper welded to the bottom of a skillet weighs 125g. How much heat is needed to raise the temperature of the copper layer from 250C to 300.0C? The specific heat capacity (c) of Cu is 0.387J/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: 6-22 q= 0.387J g*K Dr. Wolf’s CHM 101 x 125g x (300-25)0C = 1.33x104J Calorimetry Calorimetry is a way to measure heat from a process by directing the heat into “surroundings” where the increase of heat can be measured. The surroundings will be a container called a calorimeter. The heat lost by the system equals the heat gained by the calorimeter. - q sample = qcalorimeter (minus sign for heat lost) 6-23 Dr. Wolf’s CHM 101 Coffe-cup calorimeter - q sample = qcalorimeter - (mass x heat capacity x DT)samp = (mass x heat capacity x DT)cal. 6-24 Dr. Wolf’s CHM 101 Sample Problem 6.4 PROBLEM: PLAN: Determining the Specific Heat Capacity of a Solid A 25.64g sample of a solid was heated in a test tube to 100.00oC in boiling water and carefully added to a coffee-cup calorimeter containing 50.00g of water. The water temperature increased from 25.10oC to 28.49oC. What is the specific heat capacity of the solid? (Assume all the heat is gained by the water). It is helpful to use a table to summarize the data given. Then work the problem realizing that heat lost by the system must be equal to that gained by the surroundings. SOLUTION: csolid = Mass(g) c(J/g-K) solid 25.64 c H2O 50.00 4.184 25.64g x c x - 50.00g x 4.18J/g-K x 3.39 K Dr. Wolf’s CHM 101 Tfinal 100.00 28.49 25.10 -71.51K = - 50.00g x 25.64g x -71.51K 6-25 Tinitial = 28.49 DT -71.51 3.39 4.18J/g-K x 3.39 K 0.387J/g-K A bomb calorimeter - q sample = qcalorimeter qcalorimeter = heat capacity x DT 6-26 Dr. Wolf’s CHM 101 Sample Problem 6.5 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 Consummer Affairs places one serving in a bomb calorimater and burns it in O2(the heat capacity of the calorimeter = 8.15kJ/K). The temperature increases 4.9370C. Is the manufacturer’s claim correct? PROBLEM: PLAN: Calculating the Heat of Combustion - q sample = qcalorimeter SOLUTION: qcalorimeter = heat capacity x DT = 8.151kJ/K x 4.937K = 40.24kJ 40.24kJ kcal = 9.62kcal or Calories 4.18kJ The manufacturer’s claim is true. 6-27 Dr. Wolf’s CHM 101 AMOUNT (mol) of compound A Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction. AMOUNT (mol) of compound B molar ratio from balanced equation HEAT (kJ) DHrxn (kJ/mol) 6-28 Dr. Wolf’s CHM 101 gained or lost Sample Problem 6.6 PROBLEM: Using the Heat of Reaction (DHrxn) to Find Amounts 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 = 1676kJ If aluminum is produced this way, how many grams of aluminum can form when 1.000x103kJ of heat is transferred? PLAN: SOLUTION: heat(kJ) 1676kJ=2mol Al mol of Al xM g of Al 6-29 Dr. Wolf’s CHM 101 1.000x103kJ x 2mol Al 26.98g Al 1676kJ 1mol Al = 32.20g Al Hess’s Law Since enthalpy is a state function, the enthalpy for an overall process is equal to the sum of the enthalpies of all individual steps. Procedure: Identify the target equation whose DH is unknown. Manipulate contributing equations by multiplying coefficients and/or reversing to get the correct amounts of compounds as is target equation. Remembering to change the sign of DH when reversing equations and multiplying DH when multiplying coefficients. The sum of all equations must be the same as the target equation. The sum of the individual DH values will then equal the target DH . 6-30 Dr. Wolf’s CHM 101 Sample Problem 6.7 PROBLEM: Using Hess’s Law to Calculate an Unknown DH Two gaseous pollutants that form 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) Equation B: N2(g) + O2(g) PLAN: 2NO(g) DHB = 180.6kJ 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) NO(g) CO(g) + NO(g) 6-31 CO2(g) DHA = -283.0kJ Dr. Wolf’s CHM 101 CO2(g) DHA = -283.0kJ 1/2N2(g) + 1/2O2(g) DHB = -90.6kJ CO2(g) + 1/2N2(g) DHrxn = -373.6kJ Formation Equations The chemical equation showing the formation of 1 mole of a compound from the elements in their standard state. The enthalpy change for the formation is called the Standard Heat of Formation, DHof . Element in their standard state have a DHof = 0 Most compounds have a negative DHof , i.e. The reaction is exothermic........the compound is lower in energy than the elements from which it was formed. 6-32 Dr. Wolf’s CHM 101 Table 6.5 Selected Standard Heats of Formation at 250C(298K) Formula calcium Ca(s) CaO(s) CaCO3(s) DH0f(kJ/mol) 0 -635.1 -1206.9 carbon C(graphite) C(diamond) CO(g) CO2(g) CH4(g) CH3OH(l) HCN(g) CSs(l) chlorine Cl(g) 6-33 0 1.9 -110.5 -393.5 -74.9 -238.6 135 87.9 121.0 Dr. Wolf’s CHM 101 Formula DH0f(kJ/mol) Formula DH0f(kJ/mol) 0 -92.3 silver Ag(s) AgCl(s) hydrogen H(g) H2(g) 218 0 sodium nitrogen N2(g) NH3(g) NO(g) 0 -45.9 90.3 oxygen O2(g) O3(g) H2O(g) 0 143 -241.8 H2O(l) -285.8 Cl2(g) HCl(g) Na(s) Na(g) NaCl(s) 0 -127.0 0 107.8 -411.1 sulfur S8(rhombic) 0 S8(monoclinic) 2 SO2(g) -296.8 SO3(g) -396.0 Sample Problem 6.8 PROBLEM: Writing Formation Equations Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DH0f. (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) (b) Ca(s) + C(graphite) + 3/2O2(g) (c) 1/2H2(g) + C(graphite) + 1/2N2(g) 6-34 Dr. Wolf’s CHM 101 DH0f = -127.0kJ AgCl(s) CaCO3(s) HCN(g) DH0f = -1206.9kJ DH0f = 135kJ The general process for determining DH0rxn from DH0f values Using Hess’s Law to Calculate DH0rxn 6-35 Dr. Wolf’s CHM 101 Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation Nitric acid, whose worldwide annual production is about 8 billion kg, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: PROBLEM: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Calculate DH0rxn from DH0f 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)] = (4mol)(90.3kJ/mol) + (6mol)(-241.8kJ/mol) [(4mol)(-45.9kJ/mol) + (5mol)(0kJ/mol)] DHrxn = -906kJ 6-36 Dr. Wolf’s CHM 101 End of Chapter 6 6-37 Dr. Wolf’s CHM 101