Principles of Reactivity: Energy and Chemical Reactions Chapter 6 Chapter 6 1 1 1 1

Energy: Some Basics From Physics: Force – a kind of push or pull on an object. Energy – the capacity to do work. Work – force applied over a distance w = F  d Heat – energy transferred from a warmer object to a cooler object. Chapter 6

Energy: Some Basics Kinetic and Potential Energy Kinetic Energy (Thermal Energy) – energy due to motion. Chapter 6

Energy: Some Basics Kinetic and Potential Energy Potential Energy (Stored Energy) – the energy an object possesses due to its position. Potential energy can be converted into kinetic energy. Example: a ball of clay dropping off a building. Chapter 6

First Law of Thermodynamics Energy: Some Basics First Law of Thermodynamics “The total amount of energy in the universe is fixed.” Also referred to as the “Law of Conservation of Energy” Chapter 6

Energy: Some Basics Temperature and Heat Temperature is a measure of heat energy Heat is not the same as temperature. The more thermal energy a substance has the greater its molecular motion (kinetic energy). The total thermal energy in an object is the sum of the energies of all the “bodies” in the object. Chapter 6

Universe = System + Surroundings Energy: Some Basics Systems and Surroundings System – portion of the universe we wish to study. Surroundings – everything else. Universe = System + Surroundings Chapter 6

Energy: Some Basics Directionality of Heat Heat energy always flows from the hot object to the cold object. - this flow continues until the two objects are at the same temperature (thermal equilibrium). Chapter 6

Energy: Some Basics Directionality of Heat Exothermic – Heat is transferred from the system to the surroundings (object will feel “hot”). Endothermic – Heat is transferred to the system from the surroundings (object will fell “cold”). Chapter 6

Energy: Some Basics Energy Units SI Unit for energy is the joule, J: A more traditional unit is the Calorie Calorie (cal) – amount of energy required to raise 1.0 g of water 1oC. 1cal = 4.184J Chapter 6

Specific Heat Capacity The amount of heat transferred is dependant on three quantities: Quantity of material Size of temperature change Identity of the material Chapter 6

Specific Heat Capacity q = energy c = specific heat capacity DT = temperature change Chapter 6

Specific Heat Capacity exothermic -DT -q endothermic +DT +q Chapter 6

Specific Heat Capacity Specific heat capacity can be either per gram (J/g(oC) or per mole (J/mol(oC). The smaller a substances specific heat capacity, the better a thermal conductor it is. Chapter 6

Energy and Changes of State Chapter 6

Energy and Changes of State In the previous slide there is a continuous, steady application of energy. The sections that show increasing temperature are the result of the particular phase being warmed. q = cm(DT) The “flat” sections occur when all the applied energy is used to change the phase of the substance. Fusion – solid  liquid Vaporization – liquid  gas Chapter 6

Energy and Changes of State The energy required to change the phase of a substance is unique and is described in a physical constant. Solid  Liquid Heat of Fusion (water, 333J/g) Liquid  Gas Heat of Vaporization (water, 2256J/g) These constants can be used to determine the energy used in melting or vaporizing a substance. q = (Heat of Fusion)(mass of sample) q = (Heat of Vapor.)(mass of sample) Chapter 6

Energy and Changes of State q = cm(DT) q = (Heat of Vapor.)(mass) Chapter 6

First Law of Thermodynamics Internal Energy Internal Energy – sum of all kinetic and potential energy in an object. It is very hard to determine an objects internal energy, but it is possible to determine the change in energy (DE). Change in internal energy, DE = Efinal - Einitial A positive DE means Efinal > Einitial or the system gained energy from the surroundings (endothermic) A negative DE means Efinal < Einitial or the system lost energy to the surroundings (exothermic) Chapter 6

First Law of Thermodynamics Relating DE to Heat and Work DE = q + w q = heat w = work Both heat energy and work can change a systems internal energy. Chapter 6

First Law of Thermodynamics State Functions State function – a process that is determined by its initial and final conditions. Chapter 6

First Law of Thermodynamics State Functions State function – a process that is determined by its initial and final conditions. “A process that is not path dependant.” Work (w) and heat (q) are not state functions. Energy change (DE) is a state function. Chapter 6

First Law of Thermodynamics Enthalpy (H) - Heat transferred between the system and surroundings carried out under constant pressure. DE = q + w Most reactions occur under constant pressure, so DE = q + (-P(DV)) If volume is also constant, DV = 0 DE = qp So, Energy change is due to heat transfer, DE = DH = qp Chapter 6

DH = Hfinal - Hinitial = qP Enthalpy Enthalpy Change (DH) – The heat evolved or absorbed in a reaction at constant pressure DH = Hfinal - Hinitial = qP Chapter 6

Enthalpy Enthalpy Change (DH) – The heat evolved or absorbed in a reaction at constant pressure H and DH are state functions, depending only on the initial and final states. Chapter 6

Enthalpies of Reaction 2 H2(g) + O2(g)  2 H2O(g) DH = -483.6 J Chapter 6

Enthalpies of Reaction For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ 2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) DH = -1604 kJ Chapter 6

Enthalpies of Reaction For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH: CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) DH = +802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ Chapter 6

Enthalpies of Reaction For a reaction Enthalpy is an extensive property (magnitude DH is directly proportional to amount): When we reverse a reaction, we change the sign of DH: Change in enthalpy depends on state: 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 Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ a) Is this reaction endothermic or exothermic? Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ a) Is this reaction endothermic or exothermic? Exothermic, this is indicated by the negative DH. Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Chapter 6

Enthalpies of Reaction 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O2 at constant pressure? Chapter 6

Enthalpies of Reaction 2 MgO(s)  2 Mg(s) + O2(g) DH = 1205 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O2 at constant pressure? Chapter 6

Enthalpies of Reaction 2 MgO(s)  2 Mg(s) + O2(g) DH = 1205 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O2 at constant pressure? Chapter 6

Enthalpies of Reaction 2 MgO(s)  2 Mg(s) + O2(g) DH = 1205 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O2 at constant pressure? Chapter 6

Calorimetry Constant-Pressure Calorimetry Chapter 6

qsystem = -qsurroundings qsystem = -(qwater + qcalorimeter) Calorimetry Constant-Pressure Calorimetry Atmospheric pressure is constant! DH = qP qsystem = -qsurroundings The surroundings are composed of the water in the calorimeter and the calorimeter. qsystem = -(qwater + qcalorimeter) Chapter 6

Calorimetry Constant-Pressure Calorimetry Atmospheric pressure is constant! DH = qP qsystem = -qsurroundings The surroundings are composed of the water in the calorimeter and the calorimeter. For most calculations, the qcalorimeter can be ignored. qsystem = - qwater csystemmsystem DTsystem = - cwatermwater DTwater Chapter 6

Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry) Chapter 6

qrxn = -Ccalorimeter(DT) Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry) Special calorimetry for combustion reactions Substance of interest is placed in a “bomb” and filled to a high pressure of oxygen The sealed bomb is ignited and the heat from the reaction is transferred to the water This calculation must take into account the heat capacity of the calorimeter (this is grouped together with the heat capacity of water). qrxn = -Ccalorimeter(DT) Chapter 6

Calorimetry NH4NO3(s)  NH4+(aq) + NO3-(aq) DTwater = 16.9oC – 22.0oC = -5.1oC mwater = 60.0g cwater = 4.184J/goC msample = 4.25g qsample = -qwater qsample = -cwatermwater DTwater qsample = -(4.184J/goC)(60.0g)(-5.1oC) qsample = 1280.3J - Now calculate DH in kJ/mol Chapter 6

Calorimetry NH4NO3(s)  NH4+(aq) + NO3-(aq) DTwater = 16.9oC – 22.0oC = -5.1oC mwater = 60.0g cwater = 4.184J/goC msample = 4.25g qsample = 1280.3J moles NH4NO3 = 4.25g/80.032g/mol = 0.0529 mol DH = qsample/moles DH = 1280.3J/0.0529mol DH = 24.2 kJ/mol Chapter 6

Calorimetry 2 C8H18 + 25O2  16 CO2 + 18 H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -Ccal (DTwater) qrxn = -11.66kJ/oC(7.42oC) qrxn = -86.52kJ Chapter 6

Calorimetry 2 C8H18 + 25O2  16 CO2 + 18 H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -86.52kJ DHcombustion(in kJ/g) DHcombustion = -86.52kJ/1.80g = Chapter 6

Calorimetry 2 C8H18 + 25O2  16 CO2 + 18 H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -86.52kJ DHcombustion(in kJ/g) DHcombustion = -86.52kJ/1.80g = -48.1 kJ/g DHcombustion(in kJ/mol) DHcombustion = -86.52kJ/0.01577mol = Chapter 6

Calorimetry 2 C8H18 + 25O2  16 CO2 + 18 H2O DTwater = 28.78oC – 21.36oC = 7.42oC Ccal = 11.66kJ/oC msample = 1.80g qrxn = -86.52kJ DHcombustion(in kJ/g) DHcombustion = -86.52kJ/1.80g = -48.1 kJ/g DHcombustion(in kJ/mol) DHcombustion = -86.52kJ/0.01577mol = -5485 kJ/mol Chapter 6

Hess’s Law Hess’s law - if a reaction is carried out in a series of steps, H for the overall reaction is the sum of H’s for each individual step. For example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ 2H2O(g)  2H2O(l) H = -88 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ Chapter 6

Enthalpies of Formation (Heat of Formation) There are many type of H, depending on what you want to know Hvapor – enthalpy of vaporization (liquid  gas) Hfusion – enthalpy of fusion (solid  liquid) Hcombustion – enthalpy of combustion (energy from burning a substance) Chapter 6

Enthalpies of Formation (Heat of Formation) A fundamental H is the Standard Enthalpy of Formation ( ) Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure. “The standard enthalpy of formation of the most stable form on any element is zero” Chapter 6

Enthalpies of Formation Chapter 6

Enthalpies of Formation Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction: Chapter 6

Homework Problems 4, 14, 20, 24, 28, 36, 40, 44, 46, 52, 54, 56a Chapter 6 54 54 54 54