Ch6.ppt - CHAPTER 6 Thermochemistry 1 Energy Chemistry...

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Unformatted text preview: CHAPTER 6 Thermochemistry 1 Energy & Chemistry Burning peanuts supply sufficient energy to boil a cup of water. 2 Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) • These processes are examples of CONVERSION OF ENERGY Energy & Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy — • light • electrical energy • kinetic and potential energy 3 Potential & Kinetic Energy • Potential energy — energy that is associated with the object’s position. • Kinetic energy — energy associated with motion 4 Potential & Kinetic Energy less stable change in potential energy EQUALS kinetic energy more stable A gravitational system. The state of lower potential energy is more stable. 5 Potential & Kinetic Energy 6 less stable change in potential energy EQUALS kinetic energy more stable A system of two balls attached by a spring. The state of lower potential energy is more stable. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released. Potential & Kinetic Energy less stable change in potential energy EQUALS kinetic energy more stable A system of oppositely charged particles. Again, the state of lower potential energy is more stable. 7 Potential & Kinetic Energy 8 less stable change in potential energy EQUALS 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. Potential Potential Energy Energy on on the the Atomic Atomic Scale Scale • Positive and negative particles (ions) attract one another. • As the particles attract, they have a lower potential energy. NaCl — composed of Na+ and Cl- ions. 9 Potential & Kinetic Energy Kinetic energy — energy of motion. All atoms, molecules, ions have kinetic energy. rotate rotate vibrate vibrate translate translate 10 Law Law of of Conservation Conservation of of Energy Energy 11 The first law of thermodynamics states that there is CONSERVATION OF ENERGY in nature. The total energy is unchanged in any process. Energy can be converted from one form to another but can neither be created nor destroyed. For example, if in a chemical reaction the PE of products is less than that of reactants, the difference must be released as KE. PE PE Energy Energy Change Change in in Chemical Chemical Processes Processes 12 Reactants Reactants Kinetic Kinetic Energy Energy Products Products Potential Energy of system dropped. Kinetic Energy increased. Therefore, you often feel a T increase. 13 Internal Internal Energy Energy (E) (E) • Internal energy of an object contains potential and kinetic energies of all its particles. • it depends on - number of particles - type of particles - temperature - state of matter Heat Heat transfer transfer Heat is a form of energy that transfers between two objects due to a temperature difference. Heat transfers from hotter object to cooler one until thermal equilibrium is established. 14 Units Units of of Energy Energy calorie (cal) 1 calorie = heat required to raise temperature of 1.0 g of water by 1.0 oC. 1000 cal = 1 kcal 15 16 Units of Energy The SI unit is called JOULE (“J”) 1 J = 1 kg · m2/s2 1 kJ = 1000 J 1 cal = 4.184 Joules 17 HEAT CAPACITY Different objects can have different response to a given amount of heat. Which can be heated more quickly? Specific Specific Heat Heat The heat (q) absorbed or lost by an object is related to a) mass of object; b) change in temperature and; c) the material that object is made of. Every material has a property called specific heat. Specific heat shows how much heat is needed to increase the temperature of 1 gram of a substance by 1 Kelvin. 18 19 heat transferred, J Specific Heat = mass, g × T change,°C q C C == m∆T ∆T = Tfinal - Tinitial Note that ∆T in K is the same as ∆T in °C. 20 The higher the specific heat capacity, the more slowly the temperature of material rises in response to heating. 21 Specific heat is an INTENSIVE property. does NOT depend on sample size. Sample Problem 1 Finding the Specific Heat from Quantity of Heat 22 PROBLEM: A sample of copper weighs 125 g. It takes 1.38x104 J to raise the temperature of the copper sample from 250C to 3100C. What is the specific heat of Cu, in J/g*K ? PLAN: Given the mass, heat and change in temperature, we can use q = C x mass x T to find the answer. T in 0C is the same as in K. SOLUTION: T = (310-25)0C = 285 0C The same in K: T = 583 K – 298 K = 285 K q = C x mass x T C = q / (mass x T) = 1.38x104 J (125 g x 285 K) = 0.387 J/gK Specific Specific Heat Heat Capacity Capacity If 25.0 g of Al cool from 266 oC to 37 oC, how many joules of heat energy are lost by the Al? heat = q = (C)(mass)(∆T) C for aluminum = 0.896 J/g•K 23 24 Specific Specific Heat Heat If 25.0 g of Al cool from 266 oC to 37 oC, how many joules of heat energy are lost by the Al? heat = q = (C)(mass)(∆T) where ∆T = Tfinal - Tinitial q = (0.896 J/g•K)(25.0 g)(37 - 266)oC q = - 5130 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al. 25 Specific Specific Heat Heat and and Molar Molar Heat Heat Capacity Capacity specific heat capacity heat capacity per gram = J/g•°C or J/g•K molar heat capacity heat capacity per mole = J/mol•°C or J/mol•K 26 Specific Specific Heat Heat • • A bucket containing 11.5 kg of cold water at 12°C is taken into a house at a warmer temperature and left inside until it has reached thermal equilibrium with its new surroundings. If water absorbed 502 kJ of energy from the surroundings, what is the final temperature of water? Q = 502 kJ m = 11.5 kg c = 4.18 J g-1 °C-1 Tfinal = ? qq ==(C)(mass)(∆T) (C)(mass)(∆T) q ∆T= ∆T= m C ∆ T = 502,000 J / (11,500 g × 4.18 J g-1°C-1 ) = 10.4 °C (to 3 s.f.) So temperature of the water = 12°C + 10.4 °C = 22.4 °C Final T of water ~= 22 °C Heat Heat Exchange Exchange Heat transfer can be used to find specific heat capacity, C When two substances exchange heat, heat lost by hot object = heat gained by cold object (conservation of energy) although they have opposite signs q1 = –q2 27 28 Heat Heat Transfer Transfer • Use heat transfer as a way • • • • to find specific heat capacity, C 55.0 g Fe at 99.8 ˚C are dropped into 225 g water at 21.0 ˚C Water and metal come to 23.1 ˚C C of water = 4.18 J/g•K What is the specific heat capacity of the metal? Heat Heat Transfer Transfer Because of conservation of energy, q(Fe) = –q(H2O) (heat out of Fe = heat into H2O) or q(Fe) + q(H2O) = 0 q(H2O) = (4.184 J/g•K) (225 g)(23.1 ˚C – 21.0 ˚C) q(H2O) = 1977 J q(Fe) = (CFe) (55.0 g)(23.1 ˚C – 99.8 ˚C) q(Fe) = – 4219 • CFe q(Fe) + q(H2O) = – 4219 CFe + 1977 = 0 CFe = 0.469 J/g•K 29 System and Surroundings 30 • SYSTEM – The object under study • SURROUNDINGS – Everything outside the system Universe = System + Surroundings Directionality of Heat Transfer 31 • Heat always transfers from hotter object to cooler one. • EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes down T(surr) goes up Directionality of Heat Transfer • ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up T (surr) goes down 32 Endothermic Endothermic and and Exothermic Exothermic Processes Processes ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. (system gains heat) EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. (system loses heat) 33 34 Endothermic Endothermic or or Exothermic Exothermic ?? vaporization melting solid liquid freezing gas condensation Endothermic (requires heat) Exothermic (releases heat) 35 FIRST FIRST LAW LAW OF OF THERMODYNAMICS THERMODYNAMICS heat transferred ∆E = q + w change in system’s internal energy work done by (or to) the system Energy is conserved! 36 Two different paths for the energy change of a system. 37 heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM SYSTEM ∆E = q + w w transfer in (+w) w transfer out (-w) 38 The Sign Conventions for q and w ∆E = q + w All signs are from the point of view of the SYSTEM • For q: + means system gains heat; - means system loses heat. • For w: + means work done on system; - means work done by system. Sample Problem 2 PROBLEM: PLAN: Determining the Change in Internal Energy of a System 39 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 (E) in J, and in kJ. Define system and surroundings, assign signs to q and w and calculate E. The answer should be converted from J to kJ and then to kcal. SOLUTION: Heat released by system, q < 0 Work done by system, w < 0 q = - 325 J w = - 451 J E = q + w = -325 J + (-451 J) = -776 J -776J x 1 kJ 10 J 3 = -0.776kJ 40 w=Fxd F=PxA w = P x (A x d) w = - PV W is negative because work done by system Sample Problem 3 PROBLEM: PLAN: PV Work Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. Use the formula w = -PV SOLUTION: V = 64 L – 46 L = 18 L w = - PV = -15 atm x 18 L = -270 L•atm Sign of w is negative, because the gas does work on surroundings while expanding. To convert from L•atm to J, a conversion factor can be used: 1 L·atm = 101.3 J 101.3 J 270 L atm 27350 J 1 L atm 41 6.2. 6.2. ENTHALPY ENTHALPY and and CALORIMETRY CALORIMETRY w = - PV q = ∆E – w = ∆E + P∆V Lets define enthalpy as Then, at constant pressure, H = E + PV H = E + PV H = qp 42 43 ENTHALPY ENTHALPY ∆H = qp Change of enthalpy equals heat transferred at constant Pressure. •Enthalpy is the “heat portion” of energy. • Units of enthalpy: same as for energy (J, kJ) often written in kJ/mol, representing the enthalpy of just 1 mol of system. •The sign of ∆H tells us if the process is endothermic or exothermic. ENTHALPY ENTHALPY ∆ ∆H = H H = Hfinal ­ Hinitial final ­ H initial 44 If ∆H is positive, Hfinal > Hinitial final initial System gained heat Process is ENDOTHERMIC If ∆H is negative, Hfinal < Hinitial final initial System has lost heat Process is EXOTHERMIC In a chemical reaction, ∆H = Hfinal – Hinitial = Hproducts – Hreactants Enthalpy diagrams for exothermic and endothermic processes. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) CH4 + 2O2 heat out Enthalpy, H Enthalpy, H H < 0 H2O(l) Hfinal Exothermic process Hfinal H > 0 CO2 + 2H2O A H2O(g) H2O(g) Hinitial B 45 heat in Hinitial Endothermic process Sample Problem 4 PROBLEM: PLAN: 46 Drawing Enthalpy Diagrams and Determining the Sign of H In each of the following cases, determine the sign of H, state whether the reaction is exothermic or endothermic, and draw the enthalpy diagram. (a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ (b) 40.7kJ + H2O(l) H2O(g) Look at whether heat is written as a “reactant” or as a “product”. If heat is on the right side (“product”), then the system releases heat, and the reaction is exothermic. The opposite is true for an endothermic reaction. SOLUTION: (a) The reaction is exothermic. H2(g) + 1/2O2(g) EXOTHERMIC (reactants) H = -285.8kJ H2O(l) (products) (b) The reaction is endothermic. H2O(g) ENDOTHERMIC H2O(l) (products) H = +40.7kJ (reactants) 47 ConstantPressure Calorimetry How do they experimentally measure heats of reactions? Coffee-cup calorimeter. Sample Problem 5 Determining the Heat of a Reaction 48 PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Assuming the final solution has the mass 75.0 g and the same specfic heat capacity as water: c = 4.18 J/g*K, a) calculate heat gained by solution (qsoln) in J; b) calculate heat released by reaction (qrxn); c) calculate the heat (enthalpy change) of reaction (Hrxn) in kJ/mol. PLAN: a) qsoln can be calculated using the mass, c, and T. b) qrxn can be found from conservation of energy. c) we need to find the number of moles of product (water) produced, which can be done from determining the limiting reactant. Then, the heat qrxn is divided by the number of moles of water to give us the heat of reaction per mole of water formed. Sample Problem 5 Determining the Heat of a Reaction continued SOLUTION: a) qsoln = mass x specific heat x T = 75.0 g x 4.18 J/g*0C x (27.21-25.00)0C b) = 693 J qrxn + qsoln = 0 qrxn = - qsoln = - 693 J c) HCl(aq) + NaOH(aq) = - 0.693 kJ NaCl(aq) + H2O(l) For NaOH: 0.500 M x 0.0500 L = 0.0250 mol NaOH For HCl: 0.500 M x 0.0250 L = 0.0125 mol HCl HCl is the limiting reactant. 0.0125 mol of H2O will form during the rxn. Hrxn = (-0.693 kJ/0.0125 mol H2O) = -55.4 kJ/ mol 49 CALORIMETRY CALORIMETRY 50 Measuring Heats of Reaction Constant Volume “Bomb” Calorimeter • Burn combustible sample. • Measure heat evolved in a reaction. • Derive ∆H for reaction. Fig. 6.6 51 Calorimetry Some heat from reaction warms water qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Total heat evolved = qtotal = qwater + qbomb Measuring Measuring Heats Heats of of Reaction Reaction CALORIMETRY CALORIMETRY Calculate heat of combustion of 1 mol of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O • Burn 1.00 g of octane • Temp rises from 25.00 to 33.20 oC • Calorimeter contains 1200 g water • Heat capacity of bomb = 837 J/K 52 Measuring Measuring Heats Heats of of Reaction Reaction CALORIMETRY CALORIMETRY Step 1 Calculate heat transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calculate heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = ­ 48.0 kJ Step 4 Heat of combustion per mole of octane = = (­ 48.0 kJ/g)(114 g/mol) = ­ 5472 kJ/mol 53 Using Using Enthalpy Enthalpy 1) Enthalpy change ( 1) Enthalpy change ( H Hoo) depends on physical states ) depends on physical states of reactants and products. of reactants and products. H2(g) + 1/2 O2(g) --> H2O(g) H2(g) + 1/2 O2(g) --> H2O(l) ∆H˚ = -242 kJ ∆H˚ = -286 kJ 54 Using Using Enthalpy Enthalpy oo 2) If a reaction is reversed, H is also 2) 2) If a reaction is reversed, H is also 2) reversed. reversed. H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ H2O(g) ---> H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ 55 Using Using Enthalpy Enthalpy 3) If the coefficients of a reaction are multiplied 3) 3) If the coefficients of a reaction are multiplied 3) by an integer, H by an integer, Hoo is multiplied by that same is multiplied by that same integer. integer. H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 2 H2(g) + O2(g) --> 2 H2O(g) ∆H˚ = -484 kJ 56 57 Sample Problem 6 PROBLEM: Using Enthalpy Given the thermochemical equation: N2(g) + O2(g) 2NO(g) H = 180 kJ find the enthalphy change for equations: a) 2NO(g) N2(g) + O2(g) b) 6NO(g) 3N2(g) + 3O2(g) H = ? H = SOLUTION: a) Reverse the original equation; H switches sign. 2NO(g) N2(g) + O2(g) H = 180 kJ b) Triple all coefficients; H triples. 6NO(g) 3N2(g) + 3O2(g) H = 540 kJ Using Using Enthalpy Enthalpy oo 4) The value of H given with a balanced 4) The 4) The value of H given with a balanced 4) The chemical equation assumes the moles of chemical equation assumes the moles of reactants and products are exactly equal to the reactants and products are exactly equal to the balancing coefficients. balancing coefficients. H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 242 kJ released when 1 mol H2 reacts with ½ mol O2 58 Sample Problem 7 PROBLEM: Using the Heat of Reaction (H) Thermal decomposition of aluminum oxide can be represented by Al2O3(s) 2Al(s) + 3/2O2(g) Hrxn = 1676 kJ How much heat is required to produce 12.5 g of Al? PLAN: Hrxn, which is equal to 1676 kJ, is given as heat of the reaction assuming 1 mole of reactant decomposes. Therefore, this Hrxn is per 1 mole of a reactant, and units should be 1676 kJ/mol. We need to find how many moles of Al2O3 are reacting, and combine it with Hrxn. SOLUTION: 59 12.5 g 1 mol Al 1 mol Al2O3 0.239 mol Al2O3 27.0 g Al 2 mol Al 1676 kJ 0.239 mol 401 kJ 1 mol 6.3. 6.3. Hess’s Hess’s Law Law 60 Consider the formation of water H2(g) + 1/2 O2(g) ­­> H2O(g) + 241.8 kJ Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ 61 Hess’s Hess’s Law Law Making liquid H2O from H2 + O2 involves two exothermic steps. H2 + O2 gas H2O vapor Liquid H2O Hess’s Law Making H2O(l) from H2 involves two steps. H2(g) + 1/2 O2(g) ­­­> H2O(g) + 242 kJ H2O(g) ­­­> H2O(l) + 44 kJ ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­ H2(g) + 1/2 O2(g) ­­> H2O(l) + 286 kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns. 62 Hess’s Law 63 The change in enthalpy (H) is the same whether the reaction takes place in one step or a series of steps, and is the same no matter which path is followed. 64 Hess’s Law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. 65 ∆∆HH along along one one path path == ∆∆HH along along another another path path • This equation is valid because ∆H is a STATE FUNCTION (depends only on the initial and final states of the system, but does not depend on the pathway). • Other example of state function: change of altitude during a trip to Denver. • NOT a state function: mileage during a trip to Denver. Using Hess’s Law How to find H of a reaction using H’s of some other reactions? •• Work backward from the required reaction, using Work backward from the required reaction, using “target” reactants and products to decide how to “target” reactants and products to decide how to manipulate other reactions. manipulate other reactions. •• Reverse available reactions and/or multiply them Reverse available reactions and/or multiply them by appropriate numeric factors, and then add them by appropriate numeric factors, and then add them together. together. 66 Sample Problem 8 Using Hess’s Law to Calculate an Unknown H 67 PROBLEM: Given the following information: Equation A: CO(g) + 1/2O2(g) Equation B: N2(g) + O2(g) CO2(g) HA = -283.0 kJ 2NO(g) HB = 180.6 kJ calculate the unknown H of the following equation: CO(g) + NO(g) PLAN: CO2(g) + 1/2N2(g) H = ? 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) CO2(g) HA = -283.0 kJ 1/2N2(g) + 1/2O2(g) CO2(g) + 1/2N2(g) HB = -90.3 kJ Hrxn = -373.3 kJ Sample Problem 9 Using Hess’s Law to Calculate an Unknown H 68 PROBLEM: Calculate the H for the following reaction: 3C (s) + 4H2 (g) C3H8 (g) Given the following information: C (s) + O2 (g) CO2 (g) H = ­394 kJ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = ­2220 kJ H2 (g) + 1/2O2 (g) H2O (l) H = ­286 kJ PLAN: Identify the target reactants and products in given equations, then manipulate them by reversal and/or multiplication by factors. SOLUTION: Step 1. Only reaction #1 has C(s), but we need 3 C(s). Therefore, we will multiply it by 3. 3C (s) + 3O2 (g) 3CO2 (g) H = ­1182 kJ 69 Sample Problem 9 3C (s) + 4H2 (g) C3H8 (g) 3 C (s) + O 3 2 (g) CO 3 2 (g) H = ­394 kJ ­1182 kJ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = ­2220 kJ H2 (g) + 1/2O2 (g) H2O (l) H = ­286 kJ Step 2. To get C3H8 on the product side of the reaction, we need to reverse reaction #2. 3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ Step 3: Only reaction #3 has H2(g) but we need 4H2(g). Therefore, we will multiply by 4. 4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ 70 Sample Problem 9 Step 4: Add the equations together a...
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