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Chemistry-140 Lecture 13 Chapter 6: Enthalpy Changes for Chemical Reactions Chapter Highlights energy transfer, specific heat and heat transfer heat of fusion & heat vapourization exothermic & endothermic reactions first law of thermodynamics enthalpy changes calorimetry Hess’s Law Chemistry-140 Lecture 13 Thermodynamics Thermodynamics: energy and its transformations. Thermochemistry: energy changes and chemical reactions Force: "push" or "pull" exerted on an object. Work: energy required to overcome a force: Work = force x distance. Heat: energy transferred from one object to another because of a temperature difference. Chemistry-140 Lecture 13 Kinetic and Potential Energy Kinetic energy (EK): energy of an object due to its motion. 1 2 EK = mv2 (m = mass, v = velocity) Potential energy (EP): energy stored by an object due to its relative position. Chemistry-140 Lecture 13 Energy Energy can be expressed in a wide variety of units. The joule (J) is the metric unit of energy and is the energy possessed by a 2 kg object moving at a velocity of 1 m/s. 1 J (joule) =1 kg-m2/s2. The calorie (cal) is the energy required to raise the temperature of 1 g of water by 1°C. 1 cal = 4.184 J 1 Cal = 1 kcal = 1000 cal = 4.184 kJ Chemistry-140 Lecture 13 Energy Transfer The environment of a chemical reaction is separated into system and surroundings. System: reactants, products, solvents, etc. in the reaction. Surroundings: the universe, including the vessel. 2 H2(g) + O2(g) 2 H2O(l) + energy Processes that lower a system's internal energy are spontaneous. Processes that increase a system's internal energy are nonspontaneous. Chemistry-140 Lecture 13 Energy Transfer & Specific Heat Heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C, heat transferred = q = C x DT The specific heat of a substance, C, is the amount of heat required to raise the temperature of a 1 g sample of the substance by 1°C, quantity of heat supplied C= mass of object temperature change Chemistry-140 Lecture 13 Energy Transfer & Specific Heat Example 6.2: A lake has a surface area of 2.6 x 106 m2 and an average depth of 10 m. What quantity of heat (kJ) must be transferred to the lake to raise the temperature by 1 oC? (Assume a density of 1.0 g/cm3 the lake water) Chemistry-140 Lecture 13 Answer: Determine the mass of water and calculate the energy required using the concept of specific heat. Volume(H2O) = (2.6 x 106 m2)(10 m) = 2.6 x 107 m3 = 2.6 x 1013 cm3 Mass(H2O) = (2.6 x 1013 cm3)(1.0 g/cm3) = 2.6 x 1013 g Since: quantity of heat supplied C= mass of object temperature change q = m x C x DT = (2.6 x 1013 g)(4.184 J/g-K)(1.0 K) = 1.1 x 1011 kJ Chemistry-140 Lecture 13 Energy and Changes of State Heat transfer: heat is 500 kJ transferred to a substance with no structural change 1100oC Iron, 1.0 kg, 0oC Phase changes or phase transitions: a 500 kJ substance's structure is altered as in melting, freezing, vaporizing, and condensing. Ice, 2.0 kg, 0oC 0oC 0oC Chemistry-140 Lecture 13 Energy and Changes of State Heat of fusion: the enthalpy change associated with melting a substance, DHfus (kJ/mol). Heat of vapourization; the enthalpy associated with vapourizing a substance, DHvap (kJ/mol). Heating a substance is endothermic. BUT... during a phase transition the temperature remains constant Chemistry-140 Lecture 13 Energy and Changes of State The energy required (q) for the five possible processes is determined by the amount of sample. For warming: q = C x mass x DT, (specific heat, C, is different for each physical state) For the phase transitions: q = (# of moles) x DH. Chemistry-140 Lecture 13 Energy and Changes of State Question (similar to example 6.4): Calculate the enthalpy change upon converting 1.00 mol of ice at -25 oC to water vapour (steam) at 125 oC under a constant pressure of 1 atm. The specific heat of ice, water and steam are 2.09 J/g-K, 4.18 J/g-K and 1.84 J/g-K, respectively. Also for water; DHfus = 6.01 kJ/mol and DHvap = 40.67 kJ/mol. Chemistry-140 Lecture 13 Energy and Changes of State Answer: Divide the total change into calculable segments and calculate the enthalpy change for each segment. Sum to get the total enthalpy change (Hess's Law). Segment 1: Heat ice from -25 oC to 0 oC. Segment 2: Convert ice to water at 0 oC. Segment 3: Heat water from 0 oC to 100 oC. Segment 4: Convert water to steam at 100 oC. Segment 5: Heat steam from 100 oC to 125 oC. Chemistry-140 Lecture 13 Energy and Changes of State Temperature oC Heat liberated 5: Steam 4: Boiling Heat absorbed 3: Water 2: Melting 1: Ice Heat added (kJ) Chemistry-140 Lecture 13 Energy and Changes of State Segment 1: Heating the ice from -25 oC to 0 oC. DH1 = (1.00 mol)(18.0 g/mol)(2.09 J/g-K)(25 K) = 940 J Segment 2: Convert ice to water at 0 oC. DH2 = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ Segment 3: Heating the water from 0 oC to 100 oC. DH3 = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) = 7520 J Chemistry-140 Lecture 13 Energy and Changes of State Segment 4: Convert water to steam at 100oC. DH4 = (1.00 mol)(40.67 kJ/mol) = 40.67 kJ Segment 5: Heating the steam from 100oC to 125oC. DH5 = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = 830 J Total Enthalpy Change: DH = DH1 + DH2 + DH3 + DH4+ DH5 = (0.94 kJ) + (6.01 kJ) + (7.52 kJ) + (40.67 kJ) + (0.83 kJ) = 55.97 kJ Chemistry-140 Lecture 13 The First Law of Thermodynamics Energy is neither created nor destroyed; only exchanged between system and surroundings. Internal energy is the total energy of a system. Only changes in internal energy, DE, can be measured. DE = Efinal - Einitial (+ve = gain by system & -ve = loss to surroundings) DE = q + w work internal heat energy added done (heat added is positive & heat withdrawn is negative) (work done is positive & work done by is negative) Chemistry-140 Lecture 13 The First Law of Thermodynamics Energy transferred from surroundings to system Energy transferred from system to surroundings Chemistry-140 Lecture 13 Heat Transfer Question: The H2(g) and O2(g) in a cylinder are ignited and the system loses 550 J to its surroundings. The expanding gas does 240 J of work on its surroundings. What is the change in internal energy of the system? Chemistry-140 Lecture 13 Heat Transfer Answer: Energy flows from the system, so q = -550 J. Work is done by the system, so w = -240 J Therefore: DE = q + w = (-550 J) + (-240 J) = -790 J Chemistry-140 Lecture 13 Term Test #1 Friday October 12th, 2001 6:30 P.M. Chemistry-140 Lecture 13 A to H Education Gym I to Z Rm 104 Odette Bldg Chemistry-140 Lecture 13 Duration: 75 minutes Contents: SIX “Problems”!! Covers Material From Chapters 4 - 6 A Periodic Table & ALL Required Constants will be Supplied Chemistry-140 Lecture 14 Chapter 6: Enthalpy Changes for Chemical Reactions Chapter Highlights energy transfer, specific heat and heat transfer heat of fusion & heat vapourization exothermic & endothermic reactions first law of thermodynamics enthalpy changes calorimetry Hess’s Law Chemistry-140 Lecture 14 The First Law of Thermodynamics Energy is neither created nor destroyed; only exchanged between system and surroundings. Internal energy is the total energy of a system. Only changes in internal energy, DE, can be measured. DE = Efinal - Einitial (+ve = gain by system & -ve = loss to surroundings) DE = q + w work internal heat energy added done (heat added is positive & heat withdrawn is negative) (work done is positive & work done by is negative) Chemistry-140 Lecture 14 The First Law of Thermodynamics Energy transferred from surroundings to system Energy transferred from system to surroundings Chemistry-140 Lecture 14 Heat & Enthalpy Changes Enthalpy: amount of heat energy possessed by a substance. Enthalpy change corresponds to the heat change of the system at constant pressure: DH = qp DH = Hfinal - Hinitial Endothermic reaction: heat is added (positive value). Exothermic reaction: heat is withdrawn (negative value). Chemistry-140 Lecture 14 Enthalpies of Reaction Enthalpy change: the sum of the absolute enthalpies of the products minus the absolute enthalpies of the reactants. DH = H(products) - H(reactants) 2 H2(g) + O2(g) 2 H2(g) + O2(g) 2 H2O(l) + energy 2 H2O(l) DH = -483.6 kJ Chemistry-140 Lecture 14 Enthalpies of Reaction Enthalpy is an extensive property. The magnitude of DH is proportional to the amount of reactant consumed. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DH = -802 kJ 2 CH4(g) + 4 O2(g) 2 CO2(g) + 4 H2O(g) DH = -1604 kJ Chemistry-140 Lecture 14 Enthalpies of Reaction The enthalpy change is equal in magnitude but opposite in sign to DH for the reverse reaction. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DH = -802 kJ CO2(g) + 2 H2O(g) CH4(g) + 2 O2(g) DH = +802 kJ Chemistry-140 Lecture 14 Enthalpies of Reaction The enthalpy change depends on the states of the reactants and the products. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DH = -802 kJ CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DH = -890 kJ 2 H2O(g) 2 H2O(l) DH = -88 kJ Chemistry-140 Lecture 14 Enthalpies of Reaction Question: Ammonium nitrate can decompose by the reaction: NH4NO3(s) N2O(g) + 2 H2O(g) DH = -37.0 kJ Calculate the quantity of heat produced when 2.50 g of NH4NO3 decomposes at constant pressure. Chemistry-140 Lecture 14 Enthalpies of Reaction Answer: We know the heat produced from 1 mole of NH4NO3. Calculate moles in 2.50 g and determine the heat produced by that amount. 1 mol NH 4 NO3 heat = (2.50 g) 80.06 g NH 4 NO3 = -1.16 kJ - 37.0 kJ 1 mol NH 4 NO3 Chemistry-140 Lecture 14 Calorimetry Calorimetry: The measurement of heat flow. Measurements are made using a calorimeter. Recall that: The heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C, heat transferred = q = C x DT In constant-pressure calorimetry, the pressure remains constant because the apparatus is open to the atmosphere. At constant pressure, the heat change of the reaction is the enthalpy change, DH. Chemistry-140 Lecture 14 Calorimetry Question: When 50 mL of 1.0 M HCl(aq) and 50 mL of 1.0 M NaOH(aq) are mixed in a constant-pressure calorimeter, the temperature increases from 21.0 to 27.5 oC. Calculate DH for this reaction assuming a specific heat of 4.18 J/g-oC and a density of 1.0 g/mL for the final solution. Chemistry-140 Lecture 14 Calorimetry Answer: Recall, C = q m x DT q = C x m x DT Since the total solution is 100 mL and the density is 1.0 g/mL, then m = (100 mL) (1.0 g/mL) = 100 g DT = (27.5 - 21.0 oC) = 6.5 oC and C = 4.18 J/g-oC Chemistry-140 Lecture 14 Calorimetry Answer: then qp = (4.18 J/g-oC) (100 g) (6.5 oC) = -2700 J = -2.7 kJ Since moles in solution were (0.050 L)(1.0 M ) = 0.050 mol - 2.7 kJ DH = = -54 kJ/mol 0.050 L Chemistry-140 Lecture 14 Bomb Calorimetry In bomb calorimetry, the apparatus is sealed and the experiment is a constant-volume process. The heat change of the reaction is the internal energy change. DE = qevolved = -Ccalorimeter x DT Chemistry-140 Lecture 14 Bomb Calorimetry Chemistry-140 Lecture 14 A Student “Bomb” Calorimetry Thermometer Cardboard or Styrofoam Lid Nested Styrofoam Cups Exothermic Reaction Occurs in Solution Chemistry-140 Lecture 14 Bomb Calorimetry Question: Hydrazine, N2H4, and its derivatives are widely used as rocket fuels. N2H4(l) + O2(g) N2(g) + 2 H2O(g) When 1.00 g of hydrazine is burned in a bomb calorimeter, the temperature of the calorimeter increases by 3.51 oC. If the calorimeter has a heat capacity of 5.510 kJ/oC, what is the quantity of heat evolved? Chemistry-140 Lecture 14 Bomb Calorimetry Answer: Recall that: DE = qevolved = -Ccalorimeter x DT qevolved = (-5.510 kJ/oC) x (3.51 oC) qevolved = -19.3 kJ Calculate this on a per mole basis: qevolved - 19.3 kJ 32.0 g N 2 H 4 = -618 kJ/mol = 1.00 g N 2 H 4 1 mol N 2 H 4 Chemistry-140 Lecture 14 Term Test #1 Friday October 12th, 2001 6:30 P.M. Chemistry-140 Lecture 14 A to H Education Gym I to Z Rm 104 Odette Bldg Chemistry-140 Lecture 14 Duration: 75 minutes Contents: SIX “Problems”!! Covers Material From Chapters 4 - 6 A Simple Periodic Table & ALL Required Constants will be Supplied Chemistry-140 Lecture 16 Chapter 6: Enthalpy Changes for Chemical Reactions Chapter Highlights energy transfer, specific heat and heat transfer heat of fusion & heat vapourization exothermic & endothermic reactions first law of thermodynamics enthalpy changes calorimetry Hess’s Law Chemistry-140 Lecture 16 Hess’s Law If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy changes for each step. For example: (1) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DH = -802 kJ (2) 2 H2O(g) 2 H2O(l) DH = -88 kJ Chemistry-140 Lecture 16 Hess’s Law Add: (1) + (2) (1) + (2) CH4(g) + 2 O2(g) + 2 H2O(g) CO2(g) + 2 H2O(l) + 2 H2O(g) DH = (-802 - 88) kJ = -890 kJ Net Equation: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DH = -890 kJ Chemistry-140 Lecture 16 Hess’s Law Question: Calculate DH for the reaction: 2 C(s) + H2(g) C2H2(g) Chemistry-140 Lecture 16 Hess’s Law Question: Given the following: (1) C2H2(g) + 5 O2(g) 2 2 CO2(g) + H2O(l) DH = -1299.6 kJ (2) C(s) + O2(g) CO2(g) DH = -393.5 kJ (3) 1 H2(g) + O2(g) 2 H2O(l) DH = -285.9 kJ Chemistry-140 Lecture 16 Hess’s Law Answer: Step 1: Target equation has C2H2(g) as a product so we reverse the first equation and change the sign on DH: -(1) 2 CO2(g) + H2O(l) C2H2(g) + 5 O2(g) 2 DH = +1299.6 kJ Chemistry-140 Lecture 16 Hess’s Law Step 2: Target equation has 2 C(s) as a reactant so multiply the second equation x 2: 2 x (2) 2 C(s) + 2 O2(g) 2 CO2(g) DH = -787.0 kJ Chemistry-140 Lecture 16 Hess’s Law Step 3: Target equation has H2(g) as a reactant so we keep the third equation as is: (3) 1 H2(g) + O (g) 2 2 H2O(l) DH = -285.9 kJ Chemistry-140 Lecture 16 Hess’s Law Step 4: Add the three equations from steps 1,2 and 3: -(1) 2 x (2) (3) 2 CO2(g) + H2O(l) 2 C(s) + 2 O2(g) H2(g) + 1 O (g) 2 2 C2H2(g) + 5 O2(g) 2 2 CO2(g) H2O(l) Total 2 CO2(g) + H2O(l) + 2 C(s) + 2 O2(g) + H2(g) + 1 O2(g) 2 5 C2H2(g) + O2(g) + 2 CO2(g) + H2O(l) 2 Chemistry-140 Lecture 16 Hess’s Law Net 2 C(s) + H2(g) C2H2(g) -(1) 2 x (2) (3) DH = +1299.6 kJ DH = -787.0 kJ DH = -285.9 kJ Total DH = +226.7 kJ Chemistry-140 Lecture 16 Enthalpies of Formation We can calculate enthalpy associated with many changes. Vapourization (DHvap for converting liquids to gas) Fusion (DHfus for melting solids) Combustion (DHcom for combusting in oxygen) Enthalpy of the reaction that forms a substance from its constituent elements is called the enthalpy of formation, DHf, or the heat of formation. 2 C(s) + H2(g) C2H2(g) DHf = +226.7 kJ Chemistry-140 Lecture 16 Enthalpies of Formation Standard state: the state that is most stable for the substance at the temperature of interest, usually 298 K and 1 atm. C(s) diamond graphite fullerene Standard enthalpy of formation, DH°f, is the enthalpy of the reaction that forms the substance from its constituent elements in their standard states. 2 C(s) + H2(g) C2H2(g) DHof = +226.7 kJ Chemistry-140 Lecture 16 Enthalpies of Formation Enthalpies of reaction can be calculated by applying Hess's law to the enthalpies of formation of the participants: DH°rxn = S [nDH°f(products)] - S [mDH°f(reactants)] Chemistry-140 Lecture 16 Enthalpies of Formation The standard enthalpy of formation for ethanol, C2H5OH, is: 1 2 C(graphite) + 3 H2(g) + O2(g) 2 C2H5OH(l) DH°f = -277.7 kJ By definition, DH°f = 0 for the standard state of an element. Thus, DH°f = 0 for C(graphite) DH°f = 0 for H2(g) DH°f = 0 for O2(g) Chemistry-140 Lecture 16 Enthalpies of Formation Question: Calculate DH for the combustion of propane, C3H8(g) using tables of standard enthalpies of formation. TABLE 6.2 Selected Standard Enthalpies of Formation o Substance H D f (kJ/mol) C3H8(g) -103.8 CO2(g) -393.5 H2O(l) -285.8 Chemistry-140 Lecture 16 Enthalpies of Formation Answer: C3H8(g) + 5 O2(g) DH°rxn = 3 CO2(g) + 4 H2O(l) S [nDH°f(products)] - S [mDH°f(reactants)] DH°rxn = [3DH°f(CO2) + 4DH°f(H2O)] - [DH°f(C3H8) + 5DH°f(O2)] = [3(-393.5 kJ/mol) + 4(-285.8 kJ/mol)] [(-103.8 kJ/mol) + 5(0.0 kJ/mol)] = (-2324 kJ) - (-103.8) = -2220 kJ/mol Chemistry-140 Lecture 16 Enthalpies of Formation Question: The standard enthalpy change for the reaction CaCO3(s) CaO(s) + CO2(g) is 178.1 kJ. Calculate the standard enthalpy of formation of CaCO3(s) from the standard enthalpies of formation of CaO(s) and CO2(g); -635.1 kJ and 393.5 kJ respectively. Chemistry-140 Lecture 16 Enthalpies of Formation Answer: DH°rxn = S [nDH°f(products)] - S [mDH°f(reactants)] DH°rxn = [DH°f(CaO) + DH°f(CO2)] - [DH°f(CaCO3)] 178.1 kJ = [(-635.1 kJ/mol) + (-393.5 kJ/mol)] - [DH°f(CaCO3)] [DH°f(CaCO3)] = -635.1 kJ - 393.5 kJ - 178.1 kJ [DH°f(CaCO3)] = -1206.7 kJ/mol Chemistry-140 Lecture 14 Textbook Questions From Chapter #6 Specific Heat: 15, 18, 24 Changes of State: 30 Enthalpy: 34, 38 Hess’s Law: 40, 42, 44, 48, 54 Calorimetry: 62 General & Conceptual 70, 78