Presentation on theme: "Chemistry Unit 8 Thermochemistry Chapter 17. 17.1 The Flow of Energy Energy Transformations – Goal 1 Chemical Potential Energy Energy stored in chemical."— Presentation transcript:
17.1 The Flow of Energy Energy Transformations – Goal 1 Chemical Potential Energy Energy stored in chemical bonds Measured in Joules (j) Determined by: –Kinds of Atoms –Arrangement of Atoms Heat (q) Measure of the total amount of energy (E) transferred from an object with high E to an object with low E Also measured in Joules (j) Temperature (T) Measure of average kinetic E of the particles Indicates relative amounts of E Indicates the hotness or coldness of an object Measured in Kelvin, Celsius, or Fahrenheit
Which has more energy? Cup of water at 30°C or tub full of water at 30°C? Cup of water at 100°C or tub full of water at 30°C? What type of energy is stored in the water? –Chemical Potential Energy Locked up (only released by reactions) –Kinetic Energy Easily experienced in our everyday lives
Energy Concepts: System: Substance as Solid, liquid, or gas Surroundings: Environment around Substance Definitions of physical State Changes: Melting Boiling Condensing Freezing Subliming
Heating Curve of Ice/Water Energy and Time Temperature
Endothermic and Exothermic Processes – Goal 2 Law of Conservation of Energy In a chemical reaction, energy is neither created nor destroyed. Enthalpy (H) Total Energy of a system (Mostly Potential Energy & Kinetic Energy) Enthalpy Change (ΔH) Energy change of a process (physical or chemical) ΔH = H (Products) – H (Reactants) If at constant pressure: Heat released or absorbed is Enthalpy Change ΔH = q
Types of Enthalpy Changes Losing energy: Exothermic Products have less energy than the reactants ΔH = small – large = (-) value Δ H = (-) System LOSES energy Surroundings GAIN energy Gaining energy: Endothermic products have more energy then reactants Δ H = large – small = (+) value Δ H = (+) System GAINS energy Surroundings LOSE energy
Examples State Changes Melting? Freezing? Condensing? Evaporation or Boiling? Chemical Reactions Combustion? Heat Pack? Cold Pack?
Heat Capacity & Specific Heat – Goal 3 Heat Capacity Amount of energy needed to raise the temperature of an object Example: »A large body of water (Ocean) has a greater Heat Capacity than a small pond Specific Heat Capacity (c) Amount of energy needed to raise one gram of a substance 1 °C Units: joules per (gram °C) or J/g°C Table p. 296 “Specific” to a substance: Each substance has it’s own unique Specific Heat Capacity.
Heat Equation q = mcΔT q = heat (energy change) m = mass of the substance c = specific heat ΔT = change in temperature ΔT = Final T – Initial T
Sample Problem 17.2 1.The temperature of a piece of copper with a mass of 95.4 g increases from 25.0°C to 48.0°C when the metal absorbs 849 J of heat. What is the specific heat of copper? 2.How much heat is required to raise the temperature of 250.0 g of mercury from 23.6°C to 89.1°C? (the specific heat of mercury is 0.14 J/g°C)
17.2 Measuring and Expressing Enthalpy Changes Calorimetry – Goal 4 Calorimeter An instrument used to measure the heat changes of a reaction or process Qualities: –Well Insulated –Uses a known substance (i.e. water) –Use the specific heat (c) and ∆T to solve for Heat Example: Lab Heat water with a process Measure: 1) mass of water 2) ∆T of water 3) Know c Calculate: 1) Heat absorbed by water Conservation of Energy Heat absorbed by water = Heat lost by Process
Sample Problem 17.3 When 25.0 mL of water containing 0.025 mol HCl at 25˚C is added to 25.0 mL of water containing 0.025 mol NaOH at 25.0˚C in a foam cup calorimeter, a reaction occurs. Calculate the enthalpy change (in kJ) during this reaction if the highest temperature observed is 32.0˚C. Assume the densities of the solutions are 1.00 g/mL and the volume of the final solution is equal to the sum of the volumes of reacting solutions.
Thermochemical Equations – Goal 5 Stoichiometry also works for the energy released or energy absorbed by a reaction. Burning candle wax equation: 2 C 22 H 46 + 67 O 2 → 44 CO 2 + 46 H 2 O (ΔH = - 23,600kJ) What does this mean? MOLE RATIOS! Mole ratio’s related to energy: Coefficients RELATE to energy value!
2 ways to write out a thermochemical equation: Energy in the equation: 2 C 22 H 46 + 67 O 2 → 44 CO 2 + 46 H 2 O + 23,600kJ 2 NaHCO 3 + 129kJ → Na 2 CO 3 + H 2 O + CO 2 Energy as ΔH 2 C 22 H 46 + 67 O 2 → 44 CO 2 + 46 H 2 O (ΔH = - 23,600kJ) 2 NaHCO 3 → Na 2 CO 3 + H 2 O + CO 2 (ΔH = +129kJ)
2 C 22 H 46 + 67 O 2 → 44 CO 2 + 46 H 2 O (ΔH = - 23,600kJ) Example Problems: 1)How much energy is released when 50.0g of candle wax burns? 2)How much energy is released when 1mol of candle wax burns? 3) How much energy is released when 1g of candle wax burns?
Heat of Combustion The amount of energy released by the complete combustion (burning) of a substance. Can either be: –Energy per gram (J/g) or –Energy per mol (J/mol) What is the Heat of Combustion for Candle wax? (J/g) (J/mol) 2 C 22 H 46 + 67 O 2 → 44 CO 2 + 46 H 2 O (ΔH = - 23,600kJ)
17.3 Heat in Changes of State – Goal 6 Heat can change 2 things for a substance Change Temperature ( q = mcΔT ) Change State (new) State Changes require Energy Change from Solid to Liquid »Heat of Fusion »Opposite? »Heat of solidification (opposite value of fusion) Change from Liquid to Gas »Heat of Vaporization »Opposite? »Heat of Condensation (opposite value of Vapor)
Heat of Fusion Symbol: ΔH fus Energy required to change 1g of solid to liquid Equation ΔH fus = q/m or q = (m)(ΔH fus ) Water: ΔH fus = 333J/g ΔH fus = +333J/g (melting) ΔH sol = -333J/g (Solidification)
Heat of Vaporization Symbol: ΔH vap Energy required to change 1g of Liquid to gas Equation ΔH vap = q/m or q = (m)(ΔH vap ) Water: ΔH vap = 2260J/g ΔH vap = +2260J/g (Vaporizing) ΔH con = -2260J/g (Condensation)
Temperature Change q = mcΔT State Change q = mΔH vap State Change q = mΔH fus
Temperature & State Change Summary Heat energy to change Temperature Substance must stay as “one” state Each state has its own unique (c) q = mcΔT Heat energy to change State q = m ΔH fus q = m ΔH vap If both (or more) occur: Add together for total heat required
Example Problems 1.How much heat energy is required to completely freeze 50.0g of water that is at (+20˚C)? 2.How much heat energy is required to change 125g of ice at -12°C to a gas at 108°C?
17.4 Hess’s Law – Goal 7 Thermochemical Review Heat of Reaction Heat released or absorbed for a known chemical reaction Heat (ΔH) depends on the Balanced Equation (ΔH)=(-): Heat was released (ΔH)=(+): Heat was absorbed Hess’s Law explains how to _________the Heat of Reaction from other known reactions
17.4 Hess’s Law – Goal 7 Thermochemical Review Heat of Reaction Heat released or absorbed for a known chemical reaction Heat (ΔH) depends on the Balanced Equation (ΔH)=(-): Heat was released (ΔH)=(+): Heat was absorbed Hess’s Law explains how to “calculate” the Heat of Reaction from other known reactions
Hess’s Law: Concepts Equations can be added together to obtain a new equation. The heats of reaction (ΔH) must also be added to obtain the new equations heat of reaction (ΔH). Known equations may be reversed or multiplied by a factor so that they will “add” together for the needed result.
Hess’s Law: Rules If an Equation is reversed, then its ΔH is opposite (change the sign). Some equations may need to be multiplied by a factor, if so then their ΔH also need to be multiplied by the factor.
Hess’s Law: Example What is the Heat of Reaction for: C diamond C graphite You are given: C graphite + O 2 CO 2 ΔH = -393.5kJ C diamond + O 2 CO 2 ΔH = -395.4kJ Reverse CO 2 C graphite + O 2 ΔH = +393.5kJ C diamond + O 2 CO 2 ΔH = -395.4kJ Add C diamond C graphite ΔH =
Hess’s Law: Example If: PCl 5 PCl 3 + Cl 2 ∆H= 87.9 kJ 2P + 3Cl 2 2PCl 3 ∆H= -574kJ What is the ∆H of: 2P + 5Cl 2 2PCl 5