Presentation on theme: "Thermochemistry Heat and Chemical Change"— Presentation transcript:
1Thermochemistry Heat and Chemical Change Charles Page High SchoolDr. Stephen L. Cotton
2Section 11.1 The Flow of Energy - Heat OBJECTIVES:Explain the relationship between energy and heat.
3Section 11.1 The Flow of Energy - Heat OBJECTIVES:Distinguish between heat capacity and specific heat.
4Energy and HeatThermochemistry - concerned with heat changes that occur during chemical reactionsEnergy - capacity for doing work or supplying heatweightless, odorless, tastelessif within the chemical substances- called chemical potential energy
5Energy and HeatGasoline contains a significant amount of chemical potential energyHeat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.only changes can be detected!flows from warmer cooler object
6Exothermic and Endothermic Processes Essentially all chemical reactions, and changes in physical state, involve either:release of heat, orabsorption of heat
7Exothermic and Endothermic Processes In studying heat changes, think of defining these two parts:the system - the part of the universe on which you focus your attentionthe surroundings - includes everything else in the universe
8Exothermic and Endothermic Processes Together, the system and it’s surroundings constitute the universeThermochemistry is concerned with the flow of heat from the system to it’s surroundings, and vice-versa.Figure 11.3, page 294
9Exothermic and Endothermic Processes The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed.All the energy is accounted for as work, stored energy, or heat.
10Exothermic and Endothermic Processes Fig. 11.3a, p heat flowing into a system from it’s surroundings:defined as positiveq has a positive valuecalled endothermicsystem gains heat as the surroundings cool down
11Exothermic and Endothermic Processes Fig. 11.3b, p heat flowing out of a system into it’s surroundings:defined as negativeq has a negative valuecalled exothermicsystem loses heat as the surroundings heat up
12Exothermic and Endothermic Fig. 11.4, page on the left, the system (the people) gain heat from it’s surroundings (the fire)this is endothermicOn the right, the system (the body) cools as perspiration evaporates, and heat flows to the surroundingsthis is exothermic
13Exothemic and Endothermic Every reaction has an energy change associated with itExothermic reactions release energy, usually in the form of heat.Endothermic reactions absorb energyEnergy is stored in bonds between atoms
14Heat Capacity and Specific Heat A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC.Used except when referring to fooda Calorie, written with a capital C, always refers to the energy in food1 Calorie = 1 kilocalorie = 1000 cal.
15Heat Capacity and Specific Heat The calorie is also related to the joule, the SI unit of heat and energynamed after James Prescott Joule4.184 J = 1 calHeat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 oC
16Heat Capacity and Specific Heat Specific Heat Capacity - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 oC (abbreviated “C”)often called simply “Specific Heat”Note Table 11.2, page 296Water has a HUGE value, compared to other chemicals
17Heat Capacity and Specific Heat For water, C = 4.18 J/(g oC), and also C = 1.00 cal/(g oC)Thus, for water:it takes a long time to heat up, andit takes a long time to cool off!Water is used as a coolant!Note Figure 11.7, page 297
18Heat Capacity and Specific Heat To calculate, use the formula:q = mass (g) x T x Cheat abbreviated as “q”T = change in temperatureC = Specific HeatUnits are either J/(g oC) or cal/(g oC)Sample problem 11-1, page 299
19Section 11.2 Measuring and Expressing Heat Changes OBJECTIVES:Construct equations that show the heat changes for chemical and physical processes.
20Section 11.2 Measuring and Expressing Heat Changes OBJECTIVES:Calculate heat changes in chemical and physical processes.
21CalorimetryCalorimetry - the accurate and precise measurement of heat change for chemical and physical processes.The device used to measure the absorption or release of heat in chemical or physical processes is called a Calorimeter
22CalorimetryFoam cups are excellent heat insulators, and are commonly used as simple calorimetersFig. 11.8, page 300For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system
23Calorimetry Changes in enthalpy = H q = H These terms will be used interchangeably in this textbookThus, q = H = m x C x TH is negative for an exothermic reactionH is positive for an endothermic reaction (Note Table 11.3, p.301)
24CalorimetryCalorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed systemFigure 11.9, page 301Sample 11-2, page 302
26In terms of bonds C O O C O Breaking this bond will require energy. C Making these bonds gives you energy.In this case making the bonds gives you more energy than breaking them.
27Exothermic The products are lower in energy than the reactants Releases energy
28CaCO3 + 176 kJ ® CaO + CO2 CaCO3 ® CaO + CO2 Energy Reactants Products
29Endothermic The products are higher in energy than the reactants Absorbs energyNote Figure 11.11, page 303
30Chemistry Happens in MOLES An equation that includes energy is called a thermochemical equationCH4 + 2O2 ® CO2 + 2H2O kJ1 mole of CH4 releases kJ of energy.When you make kJ you also make 2 moles of water
31Thermochemical Equations A heat of reaction is the heat change for the equation, exactly as writtenThe physical state of reactants and products must also be given.Standard conditions for the reaction is kPa (1 atm.) and 25 oC
32CH4 + 2 O2 ® CO2 + 2 H2O kJIf grams of CH4 are burned completely, how much heat will be produced?1 mol CH4802.2 kJ10. 3 g CH416.05 g CH41 mol CH4= 514 kJ
33How many grams of water would be produced with 506 kJ of heat? CH4 + 2 O2 ® CO2 + 2 H2O kJHow many liters of O2 at STP would be required to produce 23 kJ of heat?How many grams of water would be produced with 506 kJ of heat?
35EnthalpyThe heat content a substance has at a given temperature and pressureCan’t be measured directly because there is no set starting pointThe reactants start with a heat contentThe products end up with a heat contentSo we can measure how much enthalpy changes
36Enthalpy Symbol is H Change in enthalpy is DH (delta H) If heat is released, the heat content of the products is lowerDH is negative (exothermic)If heat is absorbed, the heat content of the products is higherDH is positive (endothermic)
39Heat of ReactionThe heat that is released or absorbed in a chemical reactionEquivalent to DHC + O2(g) ® CO2(g) kJC + O2(g) ® CO2(g) DH = kJIn thermochemical equation, it is important to indicate the physical stateH2(g) + 1/2O2 (g)® H2O(g) DH = kJH2(g) + 1/2O2 (g)® H2O(l) DH = kJ
40Heat of CombustionThe heat from the reaction that completely burns 1 mole of a substanceNote Table 11.4, page 305
41Section 11.3 Heat in Changes of State OBJECTIVES:Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.
42Section 11.3 Heat in Changes of State OBJECTIVES:Calculate heat changes that occur during melting, freezing, boiling, and condensing.
43Heats of Fusion and Solidification Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquidMolar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies
44Heats of Fusion and Solidification Heat absorbed by a melting solid is equal to heat lost when a liquid solidifiesThus, Hfus = -HsolidNote Table 11.5, page 308Sample Problem 11-4, page 309
45Heats of Vaporization and Condensation When liquids absorb heat at their boiling points, they become vapors.Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.Table 11.5, page 308
46Heats of Vaporization and Condensation Condensation is the opposite of vaporization.Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condensesHvap = - Hcond
47Heats of Vaporization and Condensation Note Figure 11.5, page 310The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerousYou can receive a scalding burn from steam when the heat of condensation is released!
48Heats of Vaporization and Condensation H20(g) H20(l) Hcond = kJ/molSample Problem 11-5, page 311
49Heat of SolutionHeat changes can also occur when a solute dissolves in a solvent.Molar Heat of Solution (Hsoln) - heat change caused by dissolution of one mole of substanceSodium hydroxide provides a good example of an exothermic molar heat of solution:
50Heat of Solution NaOH(s) Na1+(aq) + OH1-(aq) Hsoln = - 445.1 kJ/mol The heat is released as the ions separate and interact with water, releasing kJ of heat as Hsoln thus becoming so hot it steams!Sample Problem 11-6, page 313H2O(l)
51Section 11.4 Calculating Heat Changes OBJECTIVES:Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.
52Section 11.4 Calculating Heat Changes OBJECTIVES:Calculate heat changes using standard heats of formation.
53Hess’s LawIf you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.Called Hess’s law of heat summationExample shown on page 314 for graphite and diamonds
54Why Does It Work? If you turn an equation around, you change the sign: If H2(g) + 1/2 O2(g)® H2O(g) DH= kJthen, H2O(g) ® H2(g) + 1/2 O2(g) DH = kJalso,If you multiply the equation by a number, you multiply the heat by that number:2 H2O(g) ® 2 H2(g) + O2(g) DH = kJ
55Why does it work?You make the products, so you need their heats of formationYou “unmake” the products so you have to subtract their heats.How do you get good at this?
56Standard Heats of Formation The DH for a reaction that produces 1 mol of a compound from its elements at standard conditionsStandard conditions: 25°C and 1 atm.Symbol isThe standard heat of formation of an element = 0This includes the diatomics
57What good are they? DHo = ( Products) - ( Reactants) Table 11.6, page 316 has standard heats of formationThe heat of a reaction can be calculated by:subtracting the heats of formation of the reactants from the productsDHo= (Products) -(Reactants)