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Dive into the world of Thermochemistry, exploring heat changes in reactions, energy transfer, heat capacity, and more. Learn about endothermic and exothermic processes, calorimetry, enthalpy, and molar heat values.
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Thermochemistry • Thermochemistry is concerned with the heat changes that occur during chemical reactions. • Can deal with gaining or losing heat
Energy • The capacity for doing work or supplying heat. - Energy is only detected because of its effects • Energy stored within the structural units of chemical substances is called Chemical Potential Energy (Gasoline)
Heat (q) • Energy that transfers from one object to another because of a temperature difference between them • Heat is not detectable, only changes caused by heat • Heat always flows from a warmer object to a cooler object
Universe • The system and its surroundings make up the universe • System – part of the universe you focus on • Surroundings – everything else in the universe
Endothermic • A process that absorbs heat from the surroundings • Heat flowing into a system from its surroundings is defined as positive q heat
Exothermic • A process that releases heat to its surroundings • Heat flowing out of the system to its surroundings is defined as negative q heat
Calorie vs. Joule • Calorie – the quantity of heat needed to raise the temperature of 1g of water 1oC • 1 Calorie = 1kilocalorie = 1000 calories • Joule – the quantity of heat needed to raise the temperature of 1g of water 0.2390oC • 1 J = 0.2390 cal 4.184J = 1 cal
Heat Capacity • The amount of heat needed to increase the temperature of an object exactly 1oC • The greater the mass of the object, the greater its heat capacity. • Which has more heat capacity, a drop of water, or an entire pool?
Specific Heat • The amount of heat it takes to raise the temperature of 1g of the substance 1oC • The higher the specific heat, the longer it takes to heat up. q heat C = = m x ΔT mass (g) x change in temp
Calorimetry • The accurate and precise measurement of heat change for chemical and physical processes. • The heat released by the system is equal to the heat absorbed by its surroundings
Enthalpy (H) • For systems at constant pressure, the heat content is the same as a property called enthalpy of the system • If a reaction occurs at constant pressure, enthalpy can be used interchangeably with heat. q = ΔH ΔH = m x C x ΔT
Thermochemical Equation • An equation that includes the heat of change • CaO(s) + H2O(l) Ca(OH)2(s) + 65.2 kJ
Heat of Reaction • The heat change for the equation exactly as written. • Heats of reaction at constant pressure (1atm @ 25o C) reported as ΔH
Cont. • Exothermic • Endothermic CaO(s) + H2O(l) Na2CO3(s) + H2O(g) + CO2(g) ΔH + 129kJ ΔH = -65.2kJ Ca(OH)2(s) 2NaHCO3(s)
Exothermic H2O(l) CaCl2(s) Ca2+(aq) + 2Cl-(aq) ΔHsoln = -82.8kJ/mol
Endothermic H2O(l) NH4NO3(s) NH4+(aq) + NO3-(aq) ΔHsoln = 25.7 kJ/mol
Heat of Combustion • The heat of reaction for the complete burning of one mole of a substance. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 890kJ OR CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = - 890kJ
Molar Heat of Fusion (ΔHfus) • The heat absorbed by one mol of a substance in melting from a solid to a liquid at a constant temperature. H2O(s) H2O(l) ΔHfus = 6.01kJ/mol
Molar Heat of Solidification (ΔHsolid) • The heat lost when one mole of a liquid solidifies at a constant temperature H2O(l) H2O(s) ΔHsolid = -6.01kJ/mol
Molar Heat of Vaporization (ΔHvap) • The amount of heat necessary to vaporize one mole of a given liquid H2O(l) H2O(g) ΔHvap = 40.7kJ/mol
Molar Heat of Condensation (ΔHcond) • The amount of heat released when 1 mol of vapor condenses H2O(g) H2O(l) ΔHcond = -40.7kJ/mol
Molar Heat of Solution (ΔHsoln) • The heat change caused by dissolution of one mole of substance H2O(l) NaOH(s) Na+(aq) + OH-(aq) ΔHsoln = -445.1 kJ/mol
Hess’s Law • Indirect method to measure heat reaction • 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
Cont. Diamond to Graphite a. C(s, graphite) + O2(g) CO2(g) ΔH= -393.5kJ b. C(s, diamond) + O2(g) CO2(g) ΔH= -395.4kJ Write equation a in reverse to cancel out unwanted information c. CO2(g) C(s,graphite) + O2(g)ΔH= 393.5kJ
Cont. • Now, add equation b and c b. C(s, diamond) + O2(g) CO2(g) ΔH= -395.4kJ c. CO2(g) C(s,graphite) + O2(g)ΔH= 393.5kJ
Cont. b. C(s, diamond) + O2(g) CO2(g) ΔH= -395.4kJ c. CO2(g) C(s,graphite) + O2(g)ΔH= 393.5kJ • Now, add equation b and c C (s, diamond) C (s, graphite) ΔH= -1.9kJ
Standard Heat of Formation (ΔHfo) • The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25oC ΔHo = ΔHfo (products) – ΔHfo (reactants)