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Thermodynamics the study of energy changes in reactions Energy Energy is the ability to do work or produce heat. Energy is divided into two basic types: potential energy kinetic energy. Potential Energy Potential energy is stored energy based on: position - gravity-related, like a skateboarder at the top of a half-pipe or stretched rubber band chemical composition food or a battery like the chemicals in Kinetic energy Kinetic energy is energy of motion This includes the constant, random motion of atoms and molecules Is proportional to temperature As temp increases, motion of particles increases →Thermal Expansion Chemical Energy – Form of potential energy based upon chemical composition – Stored in bonds between atoms, molecules, and ionic crystals – Potential to • form new substance • Do work • produce or require heat – Chemical energy of food powers your body – Chemical energy of gas powers your car Thermal Energy: Total internal energy of system: depends on both Heat and Temperature Heat: Form of energy that flows from warmer to cooler object Temperature: measure of the average KE of the particles Energy can be converted between the different forms of energy, but no matter how many times energy is converted the Law of Conservation of Energy States no energy can be created or destroyed Heat (q): energy that moves from hot to cold Will continue to flow until temps are equal Heat is quantified by measuring change in temp, 20°C 10°C Temp does not directly measure heat stored What is final temp? calorie: To measure heat directly, the calorie (cal) was defined as the amount of heat needed to increase the temp of 1 gram of water by 1 °C. Note: The Calories reported on food wrappers are measured as heat needed to increase the temperature of 1000 grams of water by 1 °C, 1 Calorie (Cal) = 1000 calories (cal) = 1 kilocalorie (kcal) Drinking a whole can of Pepsi actually means drinking: 250 Cal = 250,000 calories! Normal diet: 2,000 Cal = 2,000,000 cal Calorimetry: measuring heat energy Calorimeter: device used to measure heat lost or absorbed during chemical reaction, including energy from food Joule: measure energy more accurately Calories are still used in the nutrition field, but in chemistry we use the Joule (J). 1 cal = 4.184 J Ex. 1) If my breakfast contains 230 Calories, how much energy in Joules will it supply? Joule: measure energy more accurately Calories are still used in the nutrition field, but in chemistry we use the Joule (J). 1 cal = 4.184 J Ex. 1) If my breakfast contains 230 Calories, how much energy in Joules will it supply? 230 Cal 1000 cal 1 Cal 4.184 J 1 cal Joule: measure energy more accurately Calories are still used in the nutrition field, but in chemistry we use the Joule (J). 1 cal = 4.18 J Ex. 1) If my breakfast contains 230 Calories, how much energy in Joules will it supply? 230 Cal 1000 cal 1 Cal 4.18 J 1 cal = 961400 J = 9.6 x 105 J Both the calorie and Joule are defined using water But, other substances gain and lose heat at different rates Specific heat (c) – the amount of energy required to raise 1 gram of a any substance by 1°C Liquid water has a relatively high specific heat: loses & gains heat slowly – moderates earth’s climate The better the conductor, the lower the specific heat When a cake has been baked & removed from the oven it would cause severe burns to touch the metal pan, but the top of the cake can be touched to test if the cake is ready…Why? The metal of the pan has a low specific heat b/c it can lose heat quickly This heat would be absorbed by your finger!!! The cake contains a lot of trapped air and other materials with much higher specific heats. These objects lose heat slowly, allowing more time before your finger has absorbed enough heat to cause a burn. Density of a cake is much less than the density of the metal pan, so for the same surface area touched by your finger there are a lot more metal atoms that transfer heat energy than cake molecules. So… the mass also has an affect on the amount of heat transferred Plus… (old stuff) Metals are good conductors of heat and electricity While air is a poor conductor of heat and electricity (nonmetals) To calculate the amount of heat gained or lost by an object q = m × c × ∆T q = heat (J) m = mass (g) c = specific heat J g °C ∆T = Temp. final – Temp. initial, in °C If the object is heating up: the final temp is greater than the initial temp, so ∆T is positive and amount of heat gained is positive. +q If the object is cooling down: the final temp is less than the initial temp, so ∆T is negative and the amount of heat lost is negative. -q If ∆T is positive the answer for heat is positive: heat is gained. If ∆T is negative the answer for heat is negative: heat is lost. -q +q Add the following to your notes Specific heats for water: for steam = 1.97 J/goC for liquid = 4.18 J/goC for ice = 2.09 J/goC Specific heats (c) for various metals: aluminum = 0.897 J/goC copper = 0.385 J/goC iron = 0.449 J/goC lead = 0.129 J/goC mercury = 0.140 J/goC Ex. 2) What is the heat absorbed by 100. g of water to raise it from 25 °C to 75 °C? q = m × c × ∆T q = ________ x ________ x (________ - _______) q = _______________ Ex. 2) What is the heat absorbed by 100. g of water to raise it from 25 °C to 75 °C? q = m × c × ∆T q = _100.g__ x _4.18 J/goC_ x (75oC_- 25oC) q = __ ____ __ Ex. 2) What is the heat absorbed by 100. g of water to raise it from 25 °C to 75 °C? q = m × c × ∆T q = _100. g__ x _4.18 J/goC _ x (75oC - 25oC) q = __ 20900 J___ = 2.1 x 104 J or 21000 J Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = ________ x ________ x (________ - _______) q = _______________ Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = 100. x 0.449 J/goC x ( 20. oC – 50. oC) q= Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = 100. g x 0.449 J/goC x ( 20.oC – 50. oC) q = 100.g x 0.449 J/goC x (-30. oC) q= Ex. 3) What is the heat given off by 100. g of iron to cool from 50. °C to 20. °C? q = m × c × ∆T q = 100. x 0.449 x ( 20. – 50.) q = 100. x 0.449 x (-30.) q = -1347 J = -1300 J or -1.3 x 103 J Heat Movement The Law of Conservation of Energy states that heat energy cannot just appear out of nowhere and disappear to nowhere. If an object warms up, that heat had to be absorbed from somewhere and if an object cools down, that heat had to be given off to something. In the Earth’s ecosystem heat is gained by the sun’s radiation and given off to space at night. This is why a cloudy night is typically warmer than a clear night – the heat gets trapped in by the clouds. Heat can move from one object to another by three ways: 1. Radiation 2. Convection 3. Conduction The energy from the sun reaches earth by radiation - electromagnetic waves: The infrared part of the electromagnetic spectrum is the “heat” form of radiation. Ex: heat from fire, light bulb, coil on electric stove Convection is the movement of heat by circulating materials, so anything that is a fluid (gases and liquids) can move heat by convection, often called convection currents. Ex. Most air-conditioning systems, the ocean currents, and the warm and cold “fronts” the weather-person talks about on the news. Conduction is the movement of heat between objects that are touching. Most common form of heat movement when cooking. The food touches a pan that’s touching a burner. As the pan absorbs heat from burner the food absorbs heat from the pan. When you touch a hot pan heat moves from the pan into your hand which then burns you. Often the lasting effects of a burn can be minimized if you can quickly get the heat to move out of your hand by running cold water over your hand for a long time. Heat energy will always naturally move from an object with higher heat to an object with lower heat, regardless of the method of heat transfer The Law of Conservation of Energy also tells us that the amount of heat lost by the hot object must be equal to the amount of heat gained by the cold object. 20°C 10°C 15°C Hot becomes cooler, and cold becomes warmer. -qlost = qgained -- (mlost × clost × ∆Tlost ) = mgained × cgained × ∆Tgained Phases or States of Matter: Solid, Liquid, Gas Sometimes when an object gains or loses heat, it is not just the temperature that changes but the phase of the object can change. The phase depends on how much heat energy is in the object, often measured by the temperature. The solid will have lower temperature than the liquid form, which will have lower temperature than the gas (vapor) form. Energy in states of matter Hi energy & velocity - Low attractive forces & density Hi temp Gas Energy Released Exothermic Liquid Solid Energy In Endothermic Low energy & velocity - Hi attractive forces & density Low temp Phase changes occur when sufficient heat energy is added or removed +q: heat is added or absorbed – Endothermic –q: heat is lost or released – Exothermic Phase changes occur when sufficient heat energy is added or removed to change the object’s phase. Melting is the solid phase changing to the liquid phase Freezing is the liquid phase turning to a solid phase. Melting and freezing both occur at the melting point ~ the temp at which a solid becomes a liquid. If heat is being absorbed (+q), then melting is occurring. If heat is given off (-q), then freezing is happening. Boiling (vaporizing) is the liquid phase changing to the gas phase, and Condensation is the gas phase changing to the liquid phase. Boiling and condensing occur at the boiling point ~ the temp at which a liquid becomes a gas. If heat is being absorbed (+q), then boiling is occurring. If heat is given off (-q), then condensing is occurring. Sometimes liquids evaporate instead of boiling. Evaporation is boiling, but occurs below the boiling point. Evaporation can only happen at the surface of the liquid (boiling happens everywhere inside) and is a cooling process. There are two other phase change possibilities, but they are rare in everyday life. Sublimation is the solid phase changing directly to the gas phase (no liquid in-between) Deposition is the gas phase changing directly to the solid phase. Ex. Dry ice, CO2 A +q would indicate sublimation A –q would indicate deposition. From this heat curve (for water) we can see how the temperatures change and the phases change. Notice that the temperature does not change during a phase change (the two flat lines). The heat equation q = m × c × ∆T will not work during a phase change, so new equations are needed during a phase change bc heat is still needed For melting or freezing q = m × Hf q = heat (add “-“ sign for freezing), (J) m = mass, (g) Hf = heat of fusion, J g For boiling/condensing q = m × Hv q = heat (add “-“ sign for condensing), (J) m = mass, (g) J Hv = heat of vaporization, g Ex. 4) Calculate the heat needed for the heat curve shown below for 150. g of water. Hfwater = 334 J/g Hvwater = 2260 J/g c ice = 2.09 J/g °C c water = 4.18 J/g °C c steam = 1.97 J/g °C Step 1: Ice from -40 °C to 0 °C. q1 = m × c × ∆T q1 = ________ x ________ x (________ - _______) q1 = _______________ Step 1: Ice from -40.0 °C to 0 °C. q1 = m × c × ∆T q1 = 150. g x 2.09 J/goC x ( 0oC – (-40.0oC)) q1 = 12400 J Step 2: Turning ice into liquid water q2 = m × Hf q2 = ___________ x ___________ q2 = _______________ Step 2: Turning ice into liquid water q2 = m × Hf q2 = 150. g x 334 J/g q2 = 50100 J Step 3: Water from 0 °C to 100 °C. q3 = m × c × ∆T q3 = ________ x ________ x (________ - _______) q3 = _______________ Step 3: Water from 0 °C to 100 °C. q3 = m × c × ∆T q3 = 150. g x 4.18 J/goC x ( 100oC – 0oC) q3 = 62700 J Step 4: Turning liquid water into water vapor (steam). q4 = m × Hv q4 = ___________ x ___________ q4 = _______________ Step 4: Turning liquid water into water vapor (steam). q4 = m × Hv q4 = 150. g x 2260 J/g q4 = 339000 J Step 5: Water vapor from 100 °C to 140 °C q5 = m × c × ∆T q5 = ________ x ________ x (________ - _______) q5 = _______________ Step 5: Water vapor from 100 °C to 140.0 °C q5 = m × c × ∆T q5 = 150. g x 1.97 J/goC x ( 140.0oC – 100oC) q5 = 12200 J Total heat involved = q1 + q2 + q3 + q4 + q5 Total heat involved = 12400 J + 50100 J + 62700 J + 339000 J + 12200 J Total heat = __________ Total heat involved = 12400 J + 50100 J + 62700 J + 339000 J + 12200 J Total heat = 476400 J (with proper addition sig figs = 476000J) Heat & Chemical Reactions Thermochemical Equations: A balanced chemical equation that includes all physical states (s, l, g) and change in enthalpy (H) (heat content). Heat released 4 Fe(s) + 3O2(g) → 2Fe2O3(s) + 1625 kJ Heat can be written as either a reactant or product Every chemical reaction involves a change in energy: heat and/or light . If a chemical reaction feels cold: absorbing heat. endothermic reaction – energy in heat energy is written on the reactant side q is positive (+q) 27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq) Heat Added If a chemical reaction feels warm, giving off heat. exothermic reaction - energy out heat energy is written on the product side q is negative (-q) Heat released 4 Fe(s) + 3O2(g) → 2Fe2O3(s) + 1625 kJ Ex. 5) If 515 kJ were needed in the dissolving process, a) How many moles of ammonium nitrate were originally used? b) How many grams would that be? 27 kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq) Ex. 6) How many Joules of heat are released when 77 grams of iron reacts with excess oxygen? 4 Fe(s) + 3O2(g) → 2Fe2O3(s) + 1625 kJ Endothermic (a reactant) solid + energy (heat) → liquid Heat flows into the system from the surroundings. This is why your skin gets cold when you hold ice; the heat is flowing out of your skin to melt the ice. System: ice Surrounding: your skin Exothermic (a product) Iron + oxygen → iron(III) oxide + energy (heat) heat is released from the system and is absorbed by the surrounding environment which is why it’s used in instant hot pads. System: hot pad Surrounding: your skin Enthalpy (H): total heat energy of system ∆H = change in enthalpy Exothermic ∆H is negative Energy of system decreased Endothermic ∆H is positive Energy of system increased To separate water into hydrogen and oxygen (electrolysis) electricity is often used. 2H2O 2H2 + O2 Is electrolysis of water endothermic or exothermic? Where does the energy “go” in the equation? To separate water into hydrogen and oxygen (electrolysis) electricity is often used. Is electrolysis of water endothermic or exothermic? Where does the energy “go” in the equation? ΔH + 2H2O 2H2 + O2 The enthalpy of reaction (∆H) can be calculated to determine if heat is released or gained. If ΔH is +, then heat is absorbed, so it is an endothermic reaction If ΔH is -, then heat is released, so it is an exothermic reaction ΔH = ∑∆nHfo(products) - ∑∆nHfo(reactants) ∆Hfo = standard enthalpy (heat) of formation from the formation of 1 mole of compound Ex. 7) Find the ∆Hrxn for liquid ethanol, C2H5OH, undergoing combustion: C2H5OH + 3O2 → 2CO2 + 3H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH -277.69 kJ/mol O2 0 kJ/mol CO2 -393.509 kJ/mol H2O -285.830 kJ/mol Ex. Find the ∆Hrxn for liquid ethanol, C2H5OH, undergoing combustion: C2H5OH + 3O2 → 2CO2 + 3H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH -277.69 kJ/mol O2 0 kJ/mol CO2 -393.509 kJ/mol H2O -285.830 kJ/mol ∆Hfo(reactants): ∆Hfo(products): ΔH = ∆Hfo(products) - ∆Hfo(reactants) Ex. Find the ∆Hrxn for liquid ethanol, C2H5OH, undergoing combustion: C2H5OH + 3O2 → 2CO2 + 3H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH -277.69 kJ/mol O2 0 kJ/mol CO2 -393.509 kJ/mol H2O -285.830 kJ/mol Per mole ∆Hfo(reactants): (-277.69) + 3(0) = -277.69 kJ ∆Hfo(products): 2(-393.509) +3(-285.830) = -787.018 + (-857.49) = -1644.508 kJ ΔH = ∆Hfo(products) - ∆Hfo(reactants) = (-1644.508) – (-277.69) = -1366.818 kJ Ex. Find the ∆Hrxn for liquid ethanol, C2H5OH, undergoing combustion: C2H5OH + 3O2 → 2CO2 + 3H2O Is this an endothermic or exothermic reaction? Thermodynamic values table Compound ∆Hfo C2H5OH -277.69 kJ/mol O2 0 kJ/mol CO2 -393.509 kJ/mol H2O -285.830 kJ/mol Per mole ∆Hfo(reactants): (-277.69) + 3(0) = -277.69 kJ ∆Hfo(products): 2(-393.509) +3(-285.830) = -787.018 + (-857.49) = -1644.508 kJ ΔH = ∆Hfo(products) - ∆Hfo(reactants) = (-1644.508) – (-277.69) = -1366.818 kJ Neg = Exothermic An additional way to show a change in enthalpy without a calculation is with an enthalpy diagram. What is the key to knowing if the diagram shows exothermic or endothermic? Energy in or out An additional way to show a change in enthalpy without a calculation is with an enthalpy diagram. Energy out Energy in What is the key to knowing if the diagram shows exothermic or endothermic? Energy in or out