Lecture 1A. Thermochemistry PDF

Chemistry 86 Chemistry for Engineers Thermochemistry What is Thermochemistry? It is a branch of thermodynamics that deals with the heat flow that accompanies chemical reactions. What is Thermodynamics? It is the scientific study of the transformations of energy. ...

Chemistry 86 Chemistry for Engineers Thermochemistry What is Thermochemistry? It is a branch of thermodynamics that deals with the heat flow that accompanies chemical reactions. What is Thermodynamics? It is the scientific study of the transformations of energy.  Nature of Energy Even though Chemistry is the study of matter, energy affects 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 4  Classification of Energy Kinetic energy is energy of motion or energy that is being transferred thermal energy is kinetic Tro, Chemistry: A Molecular Approach 5  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 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 Joule is the SI unit of energy. kg ∙m𝟐 1J=1 s𝟐 Watt is the SI unit of power. J 1W=1 s Kilowatt-hour is a measure of power consumption. 1 kWh = 3.6 × 10 𝟔 J Tro, Chemistry: A Molecular Approach 8  Units of Energy 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) Tro, Chemistry: A Molecular Approach 9  Energy Use Energy Required to Energy used Energy Required to Run 1 Raise Unit to Light 100-W Mile Temperature of Bulb for 1 hr 1 g of Water by (approx) 1°C Joule (J) 4.18 3.60 x 105 4.2 x 105 calorie (cal) 1.00 8.60 x 104 1.0 x 105 Calorie (Cal) 0.00100 86.0 100. kWh 1.16 x 10-6 0.100 0.12 Tro, Chemistry: A Molecular Approach 10 The system is the specific part of the universe that is of interest in the study. We define the system as the material or process we are studying the energy changes within. Everything else is considered as the surroundings. open closed isolated Exchange: mass & energy energy nothing Try this! Classify the following as open, closed, or isolated system. a. A chemical reaction taking place in an enclosed flask. b. A cup of tea placed on a table. c. Hot water placed in perfectly insulated closed container. d. A thermos flask containing hot coffee. Tro, Chemistry: A Molecular Approach 12  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 13  There are various forms of energy: chemical, heat, light, mechanical, and electrical. Suggest three ways of interconverting these forms of energy and indicate what energy interconversion occurred. ex. Chemical to heat: burning of wood Tro, Chemistry: A Molecular Approach 14  Energy Flow and Conservation of Energy Conservation of Energy requires that the total energy change in the system and the surrounding must be zero  DEuniverse = 0 = DEsystem + DEsurroundings  D is the symbol that is used to mean change  final amount – initial amount 15  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 DE = Efinal – Einitial DEreaction = Eproducts - Ereactants Tro, Chemistry: A Molecular Approach 16  State Function Tro, Chemistry: A Molecular Approach 17  Energy Diagrams energy diagrams are a Internal Energy “graphical” way of showing final the direction of energy flow energy added during a process DE = + 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 DE = ─ final energy than the initial condition, the change in the internal energy will be ─ Tro, Chemistry: A Molecular Approach 18  Energy Flow when energy flows out of a system, it must all flow into Surroundings the surroundings DE + when energy flows out of a System system, DEsystem is ─ DE ─ when energy flows into the surroundings, DEsurroundings is + therefore: ─ DEsystem= DEsurroundings Energy lost by system = Energy gained by the surroundings Tro, Chemistry: A Molecular Approach 19  Energy Flow when energy flows into a system, it must all come from Surroundings the surroundings DE ─ when energy flows into a system, DEsystem is + System DE + when energy flows out of the surroundings, DEsurroundings is ─ therefore: DEsystem= ─ DEsurroundings Energy gained by system = Energy lost by the surroundings Tro, Chemistry: A Molecular Approach 20  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 DE = 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 DE + ─ Tro, Chemistry: A Molecular Approach 21  Energy Exchange energy is exchanged between the system and surroundings through either heat exchange or work being done Tro, Chemistry: A Molecular Approach 22  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 23  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 24  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 25 Temperature is a measure of the thermal energy (total kinetic energy of the particles of an object). Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature = Thermal Energy 900C 400C greater thermal energy Try this! 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, kJ, and kcal. SOLUTION: q = - 325 J w = - 451 J DU = q + w = -325 J + (-451 J) = -776 J kJ kcal -776 J = -0.776 kJ -0.776 kJ = -0.185 kcal 103J 4.184 kJ  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 DT 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 by 12°C, but only raise the temperature of 100 g of water by 2.4°C Tro, Chemistry: A Molecular Approach 28  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 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 29  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 (DT) Tro, Chemistry: A Molecular Approach 30  Example 1 – 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, DT q q  m  C s  DT Relationships: q = m ∙ Cs ∙ DT Cs = 0.385 J/g Follow the Solution: q  m  C s  DT   Concept DT  T2  T1 Plan to DT  37.0 C - - 8.0C   3.10 g   0.385 gJ C  45.0 C  Solve the  45.0 C  53.7 J problem Check Check: the unit and sign are correct Try this! How much heat is given off when an 869 g iron bar cools from 940C to 50C? c of Fe = 0.449 J/g 0C Dt = tfinal – tinitial = 50C – 940C = -890C q = mcDt = 869 g x 0.449 J/g 0C x –890C = -35,000 J Try this! A 6.22-kg piece of copper metal (c = 0.385 J/g·oC) is heated from 20.5 oC to 324.3 oC. Calculate the heat absorbed (in kJ) by the metal.  Pressure -Volume Work PV work is work that is the result of a volume change against an external pressure when gases expand, DV 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 = ─PDV  to convert the units to Joules use 101.3 J = 1 atm∙L Tro, Chemistry: A Molecular Approach 34 Example 2 – 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, DV w w  - P  DV Relationships: 101.3 J = 1 atm L Solution: DV  V2  V1 w  P  DV 101.3 J  1.75 atm  L  1 atm  L DV  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 Try this! Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and Joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in Joules did he require to take a breath while exercising?  Exchanging Energy Between System and Surroundings exchange of heat energy q = mass x specific heat x DTemperature exchange of work w = −Pressure x DVolume Tro, Chemistry: A Molecular Approach 37  Measuring DE, Calorimetry at Constant Volume since DE = q + w, we can determine DE 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, DEsystem = 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 ─DEreaction = qcal = Ccal x DT Tro, Chemistry: A Molecular Approach 38  Bomb Calorimeter used to measure DE because it is a constant volume system Tro, Chemistry: A Molecular Approach 39 Example 3 – 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 DE for burning 1 mole Given: 1.010 g C12H22O11, T1 = 24.92°C, T2 = 28.33°C, Ccal = 4.90 kJ/°C Find: DErxn, kJ/mol Concept Plan: Ccal, DT qcal qcal qrxn qcal  Ccal  DT qrxn  - qcal qcal = Ccal x DT = -qrxn qrxn Relationships: DE  Molar mass of C12H22O11 = 342.3 g/mol mol C12 H 22O11 Solution: 1 mol C12H 22O11 1.010 g C12H 22O11   2.9506 10-3 mol 342.3 g DT  28.33C  24.92C DT  3.41C Check: 40 Example 3 – 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 DE for burning 1 mole Given: 1.010 g C12H22O11, T1 = 24.92°C, T2 = 28.33°C, Ccal = 4.90 kJ/°C Find: DErxn, kJ/mol Concept Plan: Ccal, DT qcal qcal qrxn qcal  Ccal  DT qrxn  - qcal qcal = Ccal x DT = -qrxn qrxn Relationships: DE  Molar mass of C12H22O11 = 342.3 g/mol mol C12 H 22O11 Solution: qcal  Ccal  DT qrxn  16.7 kJ DE    4.90 kJC  3.41C  16.7 kJ mol C12 H 22O11 2.5906 10-3 mol qrxn  qcal  16.7 kJ  - 5.66 103 kJ/mol Check: the units and sign are correct 41 Try this! A 0.88 g gummy bear is burned in a bomb calorimeter. The temperature started at 21.5 °C and leveled off at 24.2 °C. The manufacturer of the bomb calorimeter determined the heat capacity of the calorimeter to be 11.4 kJ/°C. Calculate the heat of combustion per gram of gummy bear.  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, DH, of a reaction is the heat evolved in a reaction at constant pressure DHreaction = qreaction at constant pressure usually DH and DE are similar in value, the difference is largest for reactions that produce or use large quantities of gas Tro, Chemistry: A Molecular Approach 43 Endothermic and Exothermic Reactions when DH is ─, heat is being released by the system reactions that release heat are called exothermic reactions when DH 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 44  Endothermic and Exothermic Reactions Give two examples of exothermic processes and another two for endothermic processes that you can observe in your homes. Tro, Chemistry: A Molecular Approach 45 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts > Hreactants Hproducts < Hreactants DH > 0 DH < 0  Measuring DH 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 DT)  DHreaction = qconstant pressure = qreaction  to get DHreaction per mol, divide by the number of moles Tro, Chemistry: A Molecular Approach 47  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) DH = -2044 kJ  2044 kJ 1 mol C3H8 DHreaction for 1 mol C3H8 = -2044 kJ 1 mol C3H8 or  2044 kJ  2044 kJ 5 mol O 2 DHreaction for 5 mol O2 = -2044 kJ or 5 mol O 2  2044 kJ Tro, Chemistry: A Molecular Approach 48  Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) DH = 6.01 kJ  Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = -6.01 kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ  Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) DH = 6.01 kJ H2O (l) H2O (g) DH = 44.0 kJ How much heat is evolved when 266 g of white phosphorus (P 4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ 1 mol P4 266 g P4 x x 3013 kJ = 6470 kJ 123.9 g P4 1 mol P4  Example 4 – How much heat is evolved in the complete combustion of 13.2 kg of C3H8(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 105 kJ 1kg 44.09 g 1 mol Check: the sign is correct and the value is reasonable 53  Example 5 – What is DHrxn/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: DHrxn, J/mol Mg Concept Plan: m, Cs, DT qsoln qsoln qrxn qsoln  m  Cs  DT qrxn  - qsoln qrxn Relationships: qsoln = m x Cs x DT = -qrxn DH  mol Mg Solution: 1.00 g 100.0 mL   1.00 10 2 g 1 mL 1 mol 0.158 g Mg   6.4994  10-3 mol 24.31 g DT  32.8C  25.6C  7.2C Check: 54  Example 5 – What is DHrxn/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: DHrxn, J/mol Mg Concept Plan: m, Cs, DT qsoln qsoln qrxn qsoln  m  Cs  DT qrxn  - qsoln qrxn Relationships: qsoln = m x Cs x DT = -qrxn DH  mol Mg Solution: qsoln  m  Cs  DT qrxn  3.0 103 J DH   mol Mg 6.4994 10-3 mol  1.00 10 2 g  4.18 gJ C  7.2C  3.0 103 J  - 4.6 105 J/mol qrxn  qsoln  3.0 103 J Check: the units and sign are correct 55 Try this! How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ 1 mol P4 266 g P4 x x 3013 kJ = 6470 kJ 123.9 g P4 1 mol P4 Try this! 1) Given the thermochemical equation N2(g) + 3 H2(g) → 2 NH3(g) ΔH = −91.8 kJ how much energy is given off when 222.4 g of N2 reacts? 2) Determine the amount of heat (in kJ) given off when 1.26 x 104 g of NO2 are produced according to the equation: 2NO(g) + O2(g) → 2NO2(g) ΔH = -114.6 kJ/mol