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Chapter 17 - Thermochemistry Heat and Chemical Change Energy and Heat Thermochemistry - concerned with heat changes that occur during chemical reactions. 2 Energy and Heat Energy - capacity for doing work or supplying heat • weightless, odorless, tasteless • if within the chemical substances - called chemical potential energy 3 Energy and Heat Gasoline contains a significant amount of chemical potential energy 4 Energy and Heat - 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 Heat 5 Exothermic and Endothermic Processes Essentially all chemical reactions, and changes in physical state, involve either: • release of heat, or • absorption of heat Exothermic and Endothermic In studying heat changes, think of defining these two parts: • the system - the part of the universe on which you focus your attention • the surroundings - includes everything else in the universe Exothermic and Endothermic Together, the system and it’s surroundings constitute the universe Thermochemistry is concerned with the flow of heat from the system to it’s surroundings, and vice-versa. • Figure 17.2, page 506 Law of Conservation of Energy 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. Endothermic Fig. 17.2a, p.506 - heat flowing into a system from it’s surroundings: • defined as positive • q has a positive value • called endothermic –system gains heat as the surroundings cool down Endothermic Exothermic Fig. 17.2b, p.506 - heat flowing out of a system into it’s surroundings: • defined as negative • q has a negative value • called exothermic –system loses heat as the surroundings heat up Exothermic Exothermic and Endothermic Every reaction has an energy change associated with it Exothermic reactions release energy, usually in the form of heat. Endothermic reactions absorb energy Energy is stored in bonds between atoms 14 Heat 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. 15 Heat Capacity and Specific Heat • Used except when referring to food • a Calorie, written with a capital C, always refers to the energy in food • 1 Calorie = 1 kilocalorie = 1000 cal. 16 Heat Capacity and Specific Heat The calorie is also related to the joule, the SI unit of heat and energy • named after James Prescott Joule • 4.184 J = 1 cal 17 Heat Capacity and Specific Heat Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 oC 18 Heat 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 17.1, page 508 19 Heat Capacity and Specific Heat Water has a HUGE value, compared to other chemicals 20 Heat Capacity and Specific Heat 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, and • it takes a long time to cool off! Water is used as a coolant! For 21 Heat Capacity and Specific Heat q = mass (g) x T x C heat abbreviated as “q” T = change in temperature C = Specific Heat Units are either J/(g oC) or cal/(g oC) 22 Example When 435J of heat is added to 3.4 g of olive oil at 21 °C, the temperature increases to 85 °C. What is the specific heat of olive oil? q = m x T x C C= q m x T C= C= q m x T 435J 3.4g x (85°C - 21°C) C = 2 J/g°C Calorimetry Calorimetry - 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 Calorimetry Foam cups are excellent heat insulators, and are commonly used as simple calorimeters For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system. Calorimeter Calorimetry in enthalpy = H q = H These terms will be used interchangeably in this textbook. Changes Calorimetry Thus, q = H = m x C x T H is negative for an exothermic reaction. H is positive for an endothermic reaction. Calorimetry Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system Bomb Calorimeter Calorimeter Problem 50ml of HCl (in H2O) Ti= 22.5°C Calorimeter Problem Add 50ml of NaOH (in H2O) Ti= 22.5°C Calorimeter Problem Chemical Reaction Calorimeter Problem 100ml of HCl and NaOH (in H2O) Tf= 26°C Calorimeter Problem 1ml of H2O = 1g 100ml of H2O = 100g m = 100g Ti= 22.5°C Tf= 26°C CH20= 4.18J/g°C Calorimeter Problem H = m x C x T H = m x C x (Tf –Ti) = 100g x 4.18J/g°C x (26°C - 22.5°C) H = 1460J = 1.46kJ Energy C + O2 CO2 + 395 kJ C + O2 395kJ C O2 Reactants Products 38 In terms of bonds O O C O C O Breaking this bond will require energy. O C O O C O Making these bonds gives you energy. In this case making the bonds gives you more energy than breaking them. 39 Exothermic The products are lower in energy than the reactants Releases energy 40 Energy CaCO CaO CaCO CaO + CO+2 CO2 3 + 176 3 kJ CaO + CO2 176 kJ CaCO3 Reactants Products 41 Endothermic The products are higher in energy than the reactants Absorbs energy 42 Chemistry Happens in MOLES An equation that includes energy is called a thermochemical equation CH4+2O2CO2+2H2O + 802.2 kJ 43 CH4+2O2CO2+2H2O + 802.2kJ 1 mole of CH4 releases 802.2kJ of energy. When you make 802.2 kJ you also make 2 moles of water. 44 Thermochemical Equations A heat of reaction is the heat change for the equation, exactly as written • The physical state of reactants and products must also be given. • Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 oC 45 CH4+2O2 CO2 + 2 H2O + 802.2 kJ If 10. 3 grams of CH4 are burned completely, how much heat will be produced? 10.3g CH4 1 mol CH4 802.2 kJ 16.05g CH4 1mol CH4 ΔH = 514 kJ 46 CH4+2O2 CO2+2H2O + 802.2 kJ How many liters of O2 at STP would be required to produce 23 kJ of heat? 22.4L x 2 = 44.8L produces 802.2kJ 23kJ = 0.029 0.029 x 44.8L = 802.2kJ VO2 = 1.30 L 47 CH4+2O2 CO2+2H2O+802.2 kJ How many grams of water would be produced with 506 kJ of heat? 2 mol of H2O produced with 802.2kJ 2 mol of H2O is 36g 506kJ x 36g = 802.2kJ 22.71g of H2O 48 Heat of Combustion The heat from the reaction that completely burns 1 mole of a substance Note Table 17.2 page 517 49 Summary, so far... Enthalpy The heat content a substance has at a given temperature and pressure Can’t be measured directly because there is no set starting point The reactants start with a heat content The products end up with a heat content So we can measure how much enthalpy changes 51 Enthalpy Symbol is H Change in enthalpy is H (delta H) If heat is released, the heat content of the products is lower H is negative (exothermic) If heat is absorbed, the heat content of the products is higher H is positive (endothermic) 52 Energy Change is down H is <0 Reactants Products 53 Energy Change is up H is > 0 Reactants Products 54 Heat of Reaction The heat that is released or absorbed in a chemical reaction Equivalent to H C + O2(g) CO2(g) + 393.5 kJ C + O2(g) CO2(g) H = -393.5 kJ 55 Heat of Reaction In thermochemical equation, it is important to indicate the physical state H2(g) + 1/2O2 (g) H2O(g) H = -241.8 kJ H2(g) + 1/2O2 (g) H2O(l) H = -285.8 kJ 56 Section 17.3 Heat in Changes of State 57 Heats 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 liquid 58 Heats of Fusion and Solidification Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies 59 Heats of Fusion and Solidification Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies • Thus, Hfus = -Hsolid Note Table 17.3, page 522 Sample Problem 17-4, page 421 60 Sample Problem How many kJs of heat are required to melt a 10.0g popsicle at 0°C? Assume the popsicle has the same molar mass and heat of fusion as water. It takes 6.01kJ to melt 1 mol of ice. There are 18g in 1 mol of ice. 10g x 6.01kJ/mol = 3.34 kJ 18g/mol Heats 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 17.3, page 521 62 Heats of Vaporization and Condensation Condensation is the opposite of vaporization. Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses Hvap = - Hcond 63 ΔHfusion ΔHvaporization Heats of Vaporization and Condensation The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerous • You can receive a scalding burn from steam when the heat of condensation is released! 65 Heats of Vaporization and Condensation H20(l) H20(g) m = 63.7g of liquid Hvap = ? 63.7g x 40.7kJ/mol 18g/mol = 144kJ 66 Heat of Solution Heat 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 substance Sodium hydroxide provides a good example of an exothermic molar heat of solution: 67 NaOH(s) H2O(l) Na1+(aq) + OH1-(aq) Hsoln = - 445.1 kJ/mol The heat is released as the ions separate and interact with water, releasing 445.1 kJ of heat as Hsoln thus becoming so hot it steams! Sample Problem 17-7, page 526 68 Section 17.4 Calculating Heat Changes 69 Hess’s Law If 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 summation Example shown on page 529 for graphite and diamonds 70 Diamond to Graphite C(s,diamond) C(s,graphite) C(s,diamond) + O2 CO2 ΔH=-393.5kJ C(s,graphite) + O2 CO2 ΔH=-395.4kJ CO2 C(s,graphite) + O2 ΔH=395.45kJ C(s,diamond) + O2 CO2 ΔH=-393.5kJ CO2 C(s,graphite) + O2 ΔH=395.45kJ + C(s,diam) C(s,graph) ΔH=-1.9kJ Why Does It Work? If you turn an equation around, you change the sign: H2(g)+1/2 O2(g)H2O(g) H=-285.5 kJ H2O(g)H2(g)+1/2 O2(g)H =+285.5 kJ If you multiply the equation by a number, you multiply the heat by that number: 2H2O(g)H2(g)+O2(g) H =+571.0 kJ Why does it work? You make the products, so you need their heats of formation You “unmake” the products so you have to subtract their heats. How do you get good at this? 74 Standard Heats of Formation H for a reaction that produces 1 mol of a compound from its elements at standard conditions Standard conditions: 25°C and 1 atm. The Symbol is ΔHf 0 75 Standard Heats of Formation The standard heat of formation of an element = 0 This includes the diatomics O2 N2 Cl2 etc…… 76 What good are they? Table 17.4, page 530 has standard heats of formation The heat of a reaction can be calculated by: • subtracting the heats of formation of the reactants from the products Ho = (H 0f Products) - (H 0f Reactants) 77 CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) 0 H f CH4 (g) = - 74.86 kJ/mol 0 H f O2(g) = 0 kJ/mol 0 CO (g) = - 393.5 kJ/mol H f 2 0 H2O(g) = - 241.8 kJ/mol H f H= [-393.5+2(-241.8)]-[-74.68+2 (0)] H= - 802.4 kJ 78 Examples Sample Problem 17-8, page 531 79