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
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Exothermic A process that releases heat to its surroundings Heat flowing out of the system to its surroundings is defined as negative q heat
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Calorie vs. Joule Calorie – the quantity of heat needed to raise the temperature of 1g of water 1 o C 1 Calorie = 1kilocalorie = 1000 calories Joule – the quantity of heat needed to raise the temperature of 1g of water o C 1 J = cal 4.184J = 1 cal
Heat Capacity The amount of heat needed to increase the temperature of an object exactly 1 o C 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 1 o C The higher the specific heat, the longer it takes to heat up. q m x ΔT C == heat 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
Calorimeters
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) + H 2 O(l) Ca(OH) 2 (s) kJ
Heat of Reaction The heat change for the equation exactly as written. Heats of reaction at constant pressure 25 o C) reported as Δ H
Cont. Exothermic Endothermic CaO(s) + H 2 O(l) Ca(OH) 2 (s) ΔH = -65.2kJ ΔH + 129kJ Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g) 2NaHCO 3 (s)
Exothermic CaCl 2 (s) Ca 2+ (aq) + 2Cl - (aq) Δ H soln = -82.8kJ/mol H 2 O(l)
Endothermic NH 4 NO 3 (s)NH 4 +(aq) + NO 3 - (aq) ΔH soln = 25.7 kJ/mol H 2 O(l)
Heat of Combustion The heat of reaction for the complete burning of one mole of a substance. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) + 890kJ CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) OR ΔH = - 890kJ
Molar Heat of Fusion ( Δ H fus ) The heat absorbed by one mol of a substance in melting from a solid to a liquid at a constant temperature. H 2 O(s) H 2 O(l) ΔH fus = 6.01kJ/mol
Molar Heat of Solidification ( Δ H solid ) The heat lost when one mole of a liquid solidifies at a constant temperature H 2 O(l) H 2 O(s) ΔH solid = -6.01kJ/mol
Molar Heat of Vaporization ( Δ H vap ) The amount of heat necessary to vaporize one mole of a given liquid H 2 O(l) H 2 O(g) ΔH vap = 40.7kJ/mol
Molar Heat of Condensation ( Δ H cond ) The amount of heat released when 1 mol of vapor condenses H 2 O(g) H 2 O(l) ΔH cond = -40.7kJ/mol
Molar Heat of Solution ( Δ H soln ) The heat change caused by dissolution of one mole of substance NaOH(s)Na + (aq) + OH - (aq) ΔH soln = kJ/mol H 2 O(l)
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) + O 2 (g)CO 2 (g) ΔH= kJ b. C(s, diamond) + O 2 (g)CO 2 (g) ΔH= kJ Write equation a in reverse to cancel out unwanted information c. CO 2 (g) C(s,graphite) + O 2 (g) ΔH= 393.5kJ
Cont. Now, add equation b and c b. C(s, diamond) + O 2 (g)CO 2 (g) ΔH= kJ c. CO 2 (g) C(s,graphite) + O 2 (g) ΔH= 393.5kJ
Cont. b. C(s, diamond) + O 2 (g)CO 2 (g) Δ H= kJ c. CO 2 (g) C(s,graphite) + O 2 (g) Δ H= 393.5kJ Now, add equation b and c C (s, diamond) C (s, graphite) ΔH= -1.9kJ
Standard Heat of Formation ( Δ H f o ) 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 25 o C ΔH o = ΔH f o (products) – ΔH f o (reactants)