Thermochemistry PDF

Thermochemistry 2008, Prentice Hall  Heating Your Home most homes burn fossil fuels to generate heat the amount the temperature of your home increases depends on several factors how much fuel is burned the volume of...

Thermochemistry 2008, Prentice Hall  Heating Your Home most homes burn fossil fuels to generate heat the amount the temperature of your home increases depends on several factors how much fuel is burned the volume of the house the amount of heat loss the efficiency of the burning process can you think of any others? Tro, Chemistry: A Molecular Approach 2  Nature of Energy even though Chemistry is the study of matter, energy effects matter energy is anything that has the capacity to do work work is a force acting over a distance Energy = Work = Force x Distance energy can be exchanged between objects through contact collisions Tro, Chemistry: A Molecular Approach 3  Classification of Energy Kinetic energy is energy of motion or energy that is being transferred thermal energy is kinetic Tro, Chemistry: A Molecular Approach 4  Classification of Energy Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object energy stored in the structure of a compound is potential Tro, Chemistry: A Molecular Approach 5  Law of Conservation of Energy energy cannot be created or destroyed  First Law of Thermodynamics energy can be transferred between objects energy can be transformed from one form to another  heat → light → sound Tro, Chemistry: A Molecular Approach 6  Some Forms of Energy Electrical  kinetic energy associated with the flow of electrical charge Heat or Thermal Energy  kinetic energy associated with molecular motion Light or Radiant Energy  kinetic energy associated with energy transitions in an atom Nuclear  potential energy in the nucleus of atoms Chemical  potential energy in the attachment of atoms or because of their position Tro, Chemistry: A Molecular Approach 7  Units of Energy the amount of kinetic energy an object has is directly proportional to its mass and velocity when  the KE =mass ½mvis 2 in kg and speed in m/s, the unit for kinetic energy is 2 2 kg  m s 1 joule of energy is the amount of energy needed to move a 1 kg mass at a speed of 1 m/s 2  1 J = 12 kg  m 8 s  Units of Energy joule (J) is the amount of energy needed to move a 1 kg mass a distance of 1 meter 1 J = 1 N∙m = 1 kg∙m2/s2 calorie (cal) is the amount of energy needed to raise one gram of water by 1°C kcal = energy needed to raise 1000 g of water 1°C food Calories = kcals Energy Conversion Factors 1 calorie (cal) = 4.184 joules (J) (exact) 1 Calorie (Cal) = 1000 calories (cal) 1 kilowatt-hour (kWh) = 3.60 x 106 joules (J) Tro, Chemistry: A Molecular Approach 9  Energy Use Energy Energy Energy Required to Energy used to Used by Raise Required to Run 1 Average Unit Temperature Light 100-W Mile U.S. of 1 g of Bulb for 1 hr Citizen in Water by 1°C (approx) 1 Day joule (J) 4.18 3.60 x 105 4.2 x 105 9.0 x 108 calorie (cal) 1.00 8.60 x 104 1.0 x 105 2.2 x 108 Calorie (Cal) 0.00100 86.0 100. 2.2 x 105 kWh 1.16 x 10-6 0.100 0.12 2.5 x 102 Tro, Chemistry: A Molecular Approach 10  Energy Flow and Conservation of Energy we define the system as the material or process we are studying the energy changes within we define the surroundings as everything else in the universe Conservation of Energy requires that the total energy change in the system and the surrounding must be zero  Energyuniverse = 0 = Energysystem + Energysurroundings   is the symbol that is used to mean change  final amount – initial amount 11  Internal Energy the internal energy is the total amount of kinetic and potential energy a system possesses the change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end a state function is a mathematical function whose result only depends on the initial and final conditions, not on the process used E = Efinal – Einitial Ereaction = Eproducts - Ereactants Tro, Chemistry: A Molecular Approach 12  State Function Tro, Chemistry: A Molecular Approach 13  Energy Diagrams energy diagrams are a Internal Energy “graphical” way of showing final the direction of energy flow energy added during a process E = + if the final condition has a initial larger amount of internal energy than the initial condition, the change in the Internal Energy internal energy will be + initial if the final condition has a energy removed smaller amount of internal E = ─ final energy than the initial condition, the change in the internal energy will be ─ Tro, Chemistry: A Molecular Approach 14  Energy Flow when energy flows out of a system, it must all flow into Surroundings the surroundings E + when energy flows out of a System system, Esystem is ─ E ─ when energy flows into the surroundings, Esurroundings is + therefore: ─ Esystem= Esurroundings Tro, Chemistry: A Molecular Approach 15  Energy Flow when energy flows into a system, it must all come from Surroundings the surroundings E ─ when energy flows into a System system, Esystem is + E + when energy flows out of the surroundings, Esurroundings is ─ therefore: Esystem= ─ Esurroundings Tro, Chemistry: A Molecular Approach 16  How Is Energy Exchanged? energy is exchanged between the system and surroundings through heat and work  q = heat (thermal) energy  w = work energy  q and w are NOT state functions, their value depends on the process E = q + w system gains heat energy system releases heat energy q (heat) + ─ system releases energy by system gains energy from work w (work) + doing work ─ system gains energy system releases energy E + ─ Tro, Chemistry: A Molecular Approach 17  Energy Exchange energy is exchanged between the system and surroundings through either heat exchange or work being done Tro, Chemistry: A Molecular Approach 18  Heat & Work on a smooth table, most of the kinetic energy is transferred from the first ball to the second – with a small amount lost through friction Tro, Chemistry: A Molecular Approach 19  Heat & Work on a rough table, most of the kinetic energy of the first ball is lost through friction – less than half is transferred to the second Tro, Chemistry: A Molecular Approach 20  Heat Exchange heat is the exchange of thermal energy between the system and surroundings occurs when system and surroundings have a difference in temperature heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature thermal equilibrium Tro, Chemistry: A Molecular Approach 21  Quantity of Heat Energy Absorbed Heat Capacity when a system absorbs heat, its temperature increases the increase in temperature is directly proportional to the amount of heat absorbed the proportionality constant is called the heat capacity, C  units of C are J/°C or J/K q = C x T the heat capacity of an object depends on its mass  200 g of water requires twice as much heat to raise its temperature by 1°C than 100 g of water the heat capacity of an object depends on the type of material  1000 J of heat energy will raise the temperature of 100 g of sand 12°C, but only raise the temperature of 100 g of water by 2.4°C Tro, Chemistry: A Molecular Approach 22  Specific Heat Capacity measure of a substance’s intrinsic ability to absorb heat the specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C  Cs  units are J/(g∙°C) the molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C the rather high specific heat of water allows it to absorb a lot of heat energy without large increases in temperature  keeping ocean shore communities and beaches cool in the summer  allows it to be used as an effective coolant to absorb heat Tro, Chemistry: A Molecular Approach 23  Quantifying Heat Energy the heat capacity of an object is proportional to its mass and the specific heat of the material so we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object Heat = (mass) x (specific heat capacity) x (temp. change) q = (m) x (Cs) x (T) Tro, Chemistry: A Molecular Approach 24  Example 6.2 – How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from -8.0°C to 37.0°C? Sort Given: T1= -8.0°C, T2= 37.0°C, m=3.10 g Information Find: q, J Strategize Concept Plan: Cs m, T q q m  C s  T q = m ∙ Cs ∙ T Relationships: Cs = 0.385 J/g (Table 6.4) Follow the Solution: q m  C s  T Concept Plan to Solve the T T2  T1 T 37.0  C - - 8.0 C    3.10 g  0.385 gJ C  45.0  C  problem 45.0  C 53.7 J Check Check: the unit and sign are correct  Pressure -Volume Work PV work is work that is the result of a volume change against an external pressure when gases expand, V is +, but the system is doing work on the surroundings so w is ─ as long as the external pressure is kept constant ─Work = External Pressure x Change in Volume w = ─PV  to convert the units to joules use 101.3 J = 1 atm∙L Tro, Chemistry: A Molecular Approach 26  Example 6.3 – If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done? Given: V1=0.100 L, V2=1.85 L, P=1.00 atm Find: w, J Concept Plan: P, V w w - P  V Relationships: 101.3 J = 1 atm L Solution: V V2  V1 w  P  V 101.3 J  1.75 atm  L  1 atm  L V 1.85 L - 0.100 L  1.00 atm  1.75 L   1.75 atm  L - 177 J 1.75 L Check: the unit and sign are correct  Exchanging Energy Between System and Surroundings exchange of heat energy q = mass x specific heat x Temperature exchange of work w = −Pressure x Volume Tro, Chemistry: A Molecular Approach 28  Measuring E, Calorimetry at Constant Volume since E = q + w, we can determine E by measuring q and w in practice, it is easiest to do a process in such a way that there is no change in volume, w = 0  at constant volume, Esystem = qsystem in practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction – so instead, we use an insulated, controlled surroundings and measure the temperature change in it the surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water qsurroundings = qcalorimeter = ─qsystem ─Ereaction = qcal = Ccal x T Tro, Chemistry: A Molecular Approach 29  Bomb Calorimeter used to measure E because it is a constant volume system Tro, Chemistry: A Molecular Approach 30  Example 6.4 – When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92°C to 28.33°C. If Ccal = 4.90 kJ/°C, find E for burning 1 mole Given: 1.010 g C12H22O11, T1 = 24.92°C, T2 = 28.33°C, Ccal = 4.90 kJ/°C Erxn, kJ/mol Find: Concept Plan: Ccal, T qcal qcal qrxn qcal C cal T qrxn - qcal qcal = Ccal x T = -qrxn qrxn Relationships: E  MM C12H22O11 = 342.3 g/mol mol C12 H 22O11 Solution: q1.010  C   T 1 mol C12H 22O11 qrxn  16.7 kJ cal gC 12 H 22O11  cal E 06 10 mol 2.95 -3  kJ 342.3 g mol C12 H 22O11 2.5906 10-3 mol  4. T 9028.33 3.41 C 16.7 kJ  C  C  24.92 C T  qrxn 3.41 C  16.7 kJ  qcal - 5.66 103 kJ/mol Check: the units and sign are correct 31  Enthalpy the enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume  H is a state function H = E + PV the enthalpy change, H, of a reaction is the heat evolved in a reaction at constant pressure Hreaction = qreaction at constant pressure usually H and E are similar in value, the difference is largest for reactions that produce or use large quantities of gas Tro, Chemistry: A Molecular Approach 32 Endothermic and Exothermic Reactions when H is ─, heat is being released by the system reactions that release heat are called exothermic reactions when H is +, heat is being absorbed by the system reactions that release heat are called endothermic reactions chemical heat packs contain iron filings that are oxidized in an exothermic reaction ─ your hands get warm because the released heat of the reaction is absorbed by your hands chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process ─ your hands get cold because they are giving away your heat to the reaction 33  Molecular View of Exothermic Reactions in an exothermic reaction, the temperature rises due to release of thermal energy this extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat during the course of a reaction, old bonds are broken and new bonds made the products of the reaction have less chemical potential energy than the reactants the difference in energy is released as heat 34  Molecular View of Endothermic Reactions in an endothermic reaction, the temperature drops due to absorption of thermal energy the required thermal energy comes from the surroundings during the course of a reaction, old bonds are broken and new bonds made the products of the reaction have more chemical potential energy than the reactants to acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products Tro, Chemistry: A Molecular Approach 35  Enthalpy of Reaction the enthalpy change in a chemical reaction is an extensive property  the more reactants you use, the larger the enthalpy change by convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) H = -2044 kJ   2044 kJ 1 mol C 3 H 8 Hreaction for 1 mol C3H8 = -2044 kJ or 1 mol C 3 H 8  2044 kJ  2044 kJ 5 mol O 2 Hreaction for 5 mol O2 = -2044 kJ or 5 mol O 2  2044 kJ Tro, Chemistry: A Molecular Approach 36  Example 6.6 – How much heat is evolved in the complete combustion of 13.2 kg of C 3H8(g)? Given: 13.2 kg C3H8, Find: q, kJ/mol Concept Plan: kg g mol kJ 1000 g 1 mol C3H 8 - 2044 kJ 1 kg 44.09 g 1 mol C3H 8 Relationships: 1 kg = 1000 g, 1 mol C3H8 = -2044 kJ, Molar Mass = 44.09 g/mol Solution: 1000 g 1 mol - 2044 kJ 13.2 kg     6.12 10 5 kJ 1kg 44.09 g 1 mol Check: the sign is correct and the value is reasonable 37  Measuring H Calorimetry at Constant Pressure reactions done in aqueous solution are at constant pressure  open to the atmosphere the calorimeter is often nested foam cups containing the solution qreaction = ─ qsolution = ─(masssolution x Cs, solution x T)  Hreaction = qconstant pressure = qreaction  to get Hreaction per mol, divide by the number of moles Tro, Chemistry: A Molecular Approach 38  Example 6.7 – What is Hrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in 100.0 mL of solution changes the temperature from 25.6°C to 32.8°C? Given: 0.158 g Mg, 100.0 mL, Find: q, kJ/mol Concept Plan: kg g mol kJ 1000 g 1 mol C3H 8 - 2044 kJ 1 kg 44.09 g 1 mol C3H 8 Relationships: 1 kg = 1000 g, 1 mol C3H8 = -2044 kJ, Molar Mass = 44.09 g/mol Solution: 1000 g 1 mol - 2044 kJ 13.2 kg     6.12 10 5 kJ 1kg 44.09 g 1 mol Check: the sign is correct and the value is reasonable 39  Example 6.7 – What is Hrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in 100.0 mL of solution to change the temperature from 25.6°C to 32.8°C? Given: 0.158 g Mg, 100.0 mL sol’n, T1 = 25.6°C, T2 = 32.8°C, Cs = 4.18 J/°C, dsoln = 1.00 g/mL Find: Hrxn, J/mol Mg Concept Plan: m, Cs, T qsoln qsoln qrxn qsoln m Cs T qrxn - qsoln Relationships: qrxn qsoln = m x Cs x T = -qrxn H  mol Mg Solution: 1.00 g 100.0 mL  1.00 10 2 g T qsoln m C1 smL qrxn  3.0 103 J 1 mol H   -3 0.158 1.00gMg 2 10 g 4.18  J 6.4 994 7.2  10 C -3  mol 3. 0 10 3 J mol Mg 6.49 94 10 mol 24.31 g g C 5 3  - 4.6 10 J/mol qrxn   q  T 32.8soln   3. 0 10 J  C  2 5.6  C  7.2  C Check: the units and sign are correct 40  Relationships Involving Hrxn when reaction is multiplied by a factor, Hrxn is multiplied by that factor because Hrxn is extensive C(s) + O2(g) → CO2(g) H = -393.5 kJ 2 C(s) + 2 O2(g) → 2 CO2(g) H = 2(-393.5 kJ) = 787.0 kJ if a reaction is reversed, then the sign of H is reversed CO2(g) → C(s) + O2(g) H = +393.5 kJ Tro, Chemistry: A Molecular Approach 41  Relationships Involving Hrxn Hess’s Law if a reaction can be expressed as a series of steps, then the Hrxn for the overall reaction is the sum of the heats of reaction for each step Tro, Chemistry: A Molecular Approach 42  Sample – Hess’s Law Given the following information: 2 NO(g) + O2(g) 2 NO2(g) H° = -173 kJ 2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq) H° = -255 kJ N2(g) + O2(g) 2 NO(g) H° = +181 kJ Calculate the H° for the reaction below: 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) H° = ? [2 NO2(g) 2 NO(g) + O2(g)] x 1.5 H° = 1.5(+173 kJ) [2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq)] x 0.5 H° = 0.5(-255 kJ) [2 NO(g) N2(g) + O2(g)] H° = -181 kJ [3 NO2(g) 3 NO(g) + 1.5 O2(g)] H° = (+259.5 kJ) [1 N2(g) + 2.5 O2(g) + 1 H2O(l) 2 HNO3(aq)] H° = (-128 kJ) [2 NO(g) N2(g) + O2(g)] H° = -181 kJ 3 NO2(g) + Approach Tro, Chemistry: A Molecular H2O(l) 2 HNO3(aq) + NO(g) H° = - 49 kJ  Standard Conditions the standard state is the state of a material at a defined set of conditions  pure gas at exactly 1 atm pressure  pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest  usually 25°C  substance in a solution with concentration 1 M the standard enthalpy change, H°, is the enthalpy change when all reactants and products are in their standard states the standard enthalpy of formation, Hf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements  the elements must be in their standard states  the Hf° for a pure element in its standard state = 0 kJ/mol  by definition Tro, Chemistry: A Molecular Approach 44  Formation Reactions reactions of elements in their standard state to form 1 mole of a pure compound if you are not sure what the standard state of an element is, find the form in Appendix IIB that has a Hf° = 0 since the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions Tro, Chemistry: A Molecular Approach 45  Writing Formation Reactions Write the formation reaction for CO(g) the formation reaction is the reaction between the elements in the compound, which are C and O C + O → CO(g) the elements must be in their standard state  there are several forms of solid C, but the one with Hf° = 0 is graphite  oxygen’s standard state is the diatomic gas C(s, graphite) + O2(g) → CO(g) the equation must be balanced, but the coefficient of the product compound must be 1  use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient C(s, graphite) + ½ O2(g) → CO(g) 46  Calculating Standard Enthalpy Change for a Reaction any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products the H° for the reaction is then the sum of the Hf° for the component reactions H°reaction =  n Hf°(products) -  n Hf°(reactants)  means sum n is the coefficient of the reaction Tro, Chemistry: A Molecular Approach 47  The Combustion of CH4 Tro, Chemistry: A Molecular Approach 48  Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) 1. Write formation reactions for each compound and determine the Hf° for each 2 C(s, gr) + H2(g) C2H2(g) Hf° = +227.4 kJ/mol C(s, gr) + O2(g) CO2(g) Hf° = -393.5 kJ/mol H2(g) + ½ O2(g) H2O(l) Hf° = -285.8 kJ/mol Tro, Chemistry: A Molecular Approach 49  Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) 2. Arrange equations so they add up to desired reaction 2 C2H2(g) 4 C(s) + 2 H2(g) H° = 2(-227.4) kJ 4 C(s) + 4 O2(g) 4CO2(g) H° = 4(-393.5) kJ 2 H2(g) + O2(g) 2 H2O(l) H° = 2(-285.8) kJ 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) H = -2600.4 kJ Tro, Chemistry: A Molecular Approach 50  Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) H°reaction = n Hf°(products) - n Hf°(reactants) Hrxn = [(4 HCO2 + 2 HH2O) – (2 HC2H2 + 5 HO2)] Hrxn = [(4 (-393.5) + 2 (-285.8)) – (2 (+227.4) + 5 (0))] Hrxn = -2600.4 kJ Tro, Chemistry: A Molecular Approach 51  Example 6.11 – How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? Given: 1.0 x 1011 kJ Find: mass octane, kg Concept Plan: Write the balanced equation per mole of octane Hf°’s Hrxn° H rxn nH f products - nH f reactants kJ mol C8H18 g C8H18 kg C8H18 from 114.2 g 1 kg above 1 mol 1000 g Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g Solution: C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g) 11 1 mol C8H18 H rxn 114.22 nH f products - ngH f reactants1Material kg Hf°, kJ/mol - 1.0 10 kJ    8H f kJ - 5074.1 CO 2 1 9mol Look up the H f HC  18fHf1000 2 O8H H ° H18H C 8C g 25 (l) 8 18 2 H f O 2   -250.1 for each material 8 393.5 in 2.3 106 kg C8 H18 kJ Appendix 8 kJ    O250  9 241.IIB 2(g) .1 kJ  25 2 00 kJ   CO2(g) -393.5  5074.1 kJ H2O(g) -241.8 Check: the units and sign are correct the large value is expected 52  Energy Use and the Environment in the U.S., each person uses over 105 kWh of energy per year most comes from the combustion of fossil fuels  combustible materials that originate from ancient life C(s) + O2(g) → CO2(g) H°rxn = -393.5 kJ CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g) H°rxn = -802.3 kJ C8H18(g) +12.5 O2(g) → 8 CO2(g) + 9 H2O(g) H°rxn = -5074.1 kJ fossil fuels cannot be replenished at current rates of consumption, oil and natural gas supplies will be depleted in 50 – 100 yrs. Tro, Chemistry: A Molecular Approach 53  Energy Consumption the increase in energy consumption in the US the distribution of energy consumption in the US Tro, Chemistry: A Molecular Approach 54  The Effect of Combustion Products on Our Environment because of additives and impurities in the fossil fuel, incomplete combustion and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy therefore fossil fuel emissions contribute to air pollution, acid rain, and global warming Tro, Chemistry: A Molecular Approach 55  Global Warming CO2 is a greenhouse gas  it allows light from the sun to reach the earth, but does not allow the heat (infrared light) reflected off the earth to escape into outer space  it acts like a blanket CO2 levels in the atmosphere have been steadily increasing current observations suggest that the average global air temperature has risen 0.6°C in the past 100 yrs. atmospheric models suggest that the warming effect could worsen if CO2 levels are not curbed some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitats Tro, Chemistry: A Molecular Approach 56  CO2 Levels Tro, Chemistry: A Molecular Approach 57  Renewable Energy our greatest unlimited supply of energy is the sun new technologies are being developed to capture the energy of sunlight parabolic troughs, solar power towers, and dish engines concentrate the sun’s light to generate electricity solar energy used to decompose water into H2(g) and O2(g); the H2 can then be used by fuel cells to generate electricity H2(g) + ½ O2(g) → H2O(l) H°rxn = -285.8 kJ hydroelectric power wind power Tro, Chemistry: A Molecular Approach 58