CHAPTER 17: THERMOCHEMISTRY Mrs. Brayfield
REVIEW What is the law of conservation of energy? It states that energy cannot be created or destroyed So the energy of any process is the same
THERMOCHEMISTRY Thermochemistry is the study of energy changes that occur during chemical reactions and changes in state Every chemical bond has energy stored in it This is called chemical potential energy
HEAT TRANSFER Let s dissolve some anhydrous CuSO 4 and then some NH 4 NO 3 in water What happened? Remember that heat is different from temperature: Temperature is the average kinetic energy Heat is the energy transferred from one object to another
HEAT What is the flow of heat? Heat flows from an object of higher temperature to one of lower temperature When we study heat, we look at the flow of heat from the system to the surroundings The system is where we focus our attention The surroundings is everything else
HEAT FLOW Endothermic Endothermic processes are a process that absorbs heat from the surroundings This process has a positive heat flow because heat flows INTO the system At room temperature this process feels cool to the touch Exothermic Exothermic processes are a process that releases heat to its surroundings This process has a negative heat flow because heat flows OUT of the system At room temperature this process feels hot to the touch
ENDOTHERMIC VS. EXOTHERMIC Endothermic Exothermic
ENDOTHERMIC VS. EXOTHERMIC Classify the following as exothermic or endothermic: Melting ice cubes Endothermic A candle flame Exothermic Burning sugar Exothermic Cooking an egg Endothermic
ENDOTHERMIC VS. EXOTHERMIC Classify the following as exothermic or endothermic: Conversion of frost to water vapor Endothermic Combustion of hydrogen gas Making ice cream Exothermic Melting solid salts Endothermic Exothermic
Given the following foods, which has the most Calories?
ENERGY UNITS When we burn calories we actually break down sugars and fats which releases heat The heat is produced from the chemical reactions in our bodies A calorie is the quantity of heat needed to raise the temperature of 1g of pure water 1 C
CALORIES VS. CALORIES The dietary Calorie is NOT the food calorie The dietary Calorie is always written with a capital C 1 Calorie = 1 kilocalorie = 1000 calories So from the previous slide: An orange has 62000 calories A cupcake has 241000 calories A can of soda has 152000 calories
ENERGY UNITS The SI unit of energy (standard unit) is the joule (J) You can convert between joules (J) and calories (cal) by: 1 J = 0.2390 cal 4.184 J = 1 cal
HEAT CAPACITY AND SPECIFIC HEAT Heat capacity is the amount of heat required to increase the temperature of an object by 1 C As the size increases, the heat capacity also increases Specific heat is the amount of heat required to raise the temperature of 1g of the substance by 1 C
SPECIFIC HEAT On a sunny day does asphalt or a puddle get warmer? The asphalt does This means that the asphalt has a lower specific heat The higher the specific heat of a substance, the harder it is to heat it up or cool it down
CALCULATING SPECIFIC HEAT We can calculate specific heat using the equation: C = Where: q m T = heat mass g changein temp( ) q is heat (in either J or cal) m is mass (in g) ΔT is change in temperature (in C) C is the specific heat of a substance (in either J/(g C) or cal/(g/ C)
PHASE CHANGE DIAGRAM Remember this?
PHASE CHANGE DIAGRAM We can only use the specific heat equation if we are only in ONE PHASE (no phase changes)
CALCULATING SPECIFIC HEAT The temperature of a 95.4g piece of copper increases from 25.0 C to 48.0 C when the copper absorbs 849J of heat. What is the specific heat of copper? Givens: m=95.4g, ΔT=(48-25 C), q=849j C = q m T = 849J 95.4g (48.0 25.0) C = 849J 95.4g 23 = 0.387 J g
SPECIFIC HEAT Now, the specific heat value for any given substance (at a particular phase) DOES NOT CHANGE ever! So, we can arrange the specific heat equation to find how much heat the system produced by looking at the temperature of the surroundings: q = m C T
EXAMPLE How much heat is required to raise the temperature of 250.0g of mercury by 52 C? Givens: m = 250g, ΔT = 52 C But we don t know C of mercury! But we do See table 17.1 on pg. 559 of your book C of mercury = 0.14 J/(g C) q = mc T = 250.0g 0.14 q = 1820J J g 52
CALORIMETRY Calorimetry is where we measure the heat flow We can use a calorimeter to insulate the system and some surroundings (not all of the surroundings) For example, we can take a heated piece of copper, put it in water that is in a calorimeter, and measure the temperature change Then using the specific heat equation, we can measure how much heat was given off of the copper
CALORIMETRY In a calorimeter, we need to look at the water s temperature because we can t directly measure the temperature of the copper So we look at the SURROUNDINGS
CALORIMETRY If we look at the surroundings, we can get the equation: q surr = m C T However, we want to know how much heat the copper RELEASED! Well, energy must be conserved in a process: q sys = q surr = m C T
CALORIMETRY EXAMPLE If we added a 25.0g piece of iron (originally at 85.0 C) to 75.0g of water, originally at 20.0 C. If the water heats up to 22.25 C, what is the specific heat of the iron? q water = mc T = 75g 4.18 J g (22.25 20.0) q water = 705.4J q Fe = q H2 O = 705.4J C = q m T = 705.4J 25g ( 22.25 85.0) = 0.450 J g
THERMOCHEMICAL EQUATIONS In a chemical equation, the heat term can be written as a reactant or a product It is a reactant if the reaction is endothermic (requiring heat) and a product if the reaction is exothermic (releasing heat) A thermochemical equation is a chemical reaction where the heat is in the equation
THERMOCHEMICAL EQUATION The heat, or enthalpy, is the total energy for the system (or surroundings) Enthalpy = ΔH = q For example: CaO s + H 2 O l Ca OH 2 s H = 65.2kJ can be written as: CaO s + H 2 O l Ca OH 2 s + 65.2kJ
ENTHALPY If ΔH < 0 then the process is exothermic This is because q is flowing out of the system (it has negative heat flow) If ΔH > 0 then the process is endothermic This is because q is flowing into the system (it has positive heat flow)
EXAMPLE Calculate the amount of heat (in kj) required to decompose 2.24mol NaHCO 3. 2NaHCO 3 s + 85kJ Na 2 CO 3 s + H 2 O l + CO 2 (g) H = 2.24 mol NaHCO 3 85kJ 2 mol NaHCO 3 H = 95kJ
HEAT IN CHANGES OF STATE Since we can t use our heat equation, we need to find another way to calculate the heat during a phase change We do this using the heat of fusion, heat of solidification, heat of vaporization, and heat of condensation
HEAT OF FUSION The heat of fusion (ΔH fus ) is the heat absorbed by one mole of a solid as it melts to a liquid at constant temperature The heat of solidification (ΔH solid ) is the heat lost by one mole of a liquid as it solidifies at constant temperature The quantity of heat absorbed is the same as the quantity of heat released, so ΔH fus = ΔH solid
HEAT OF VAPORIZATION The heat of vaporization (ΔH vap ) is the amount of heat required to vaporize one mole of a liquid at constant temperature The heat of condensation (ΔH cond ) is the heat lost by one mole of a gas as it condenses at constant temperature The quantity of heat absorbed is the same as the quantity of heat released, so ΔH vap = ΔH cond
HEAT IN CHANGES OF STATE We use the chart (pg. 571) to find these values They are constant for each substance
HEATING CURVE