CP Chapter 17 Thermochemistry 2014-2015 Thermochemistry Thermochemistry is the study of energy that occur during chemical and physical changes (changes of state) The Nature of Energy Energy is the ability to do work or produce Heat, q or Q, is ; flows due to temperature differences (always to ) Law of Conservation of Energy Energy CANNOT be created nor ; only converted into different types Kinetic vs. Potential Energy Two main types of energy kinetic and potential o Kinetic energy of o Potential energy due to position or energy in chemical bonds Chemical potential energy - the energy stored in a substance because of its composition Example: gasoline Temperature vs. Heat Temperature is a measure of the in a sample. Temperature is a of heat. Heat is the total energy of molecular motion, dependent upon amount, size, and type of particles. Heat is. Units of Heat calorie - the amount of heat required to raise the temperature of one gram of pure by one degree Celsius Calorie nutritional calorie; o 1 Calorie = 1000 calories = 1 kilocalories (kcal) Joule SI unit of heat o 1 calorie = J Converting Energy Units Calorie/calorie/ kilocalorie calorie/joule Ex 1) A cereal has 155 nutritional Calories per serving. How many calories, kilocalories and Joules is this? Ex 2) A person on a diet consumed 1350 Calories in one day. How many calories, kilocalories and Joules is this? 1
System and Surroundings Universe = system + surroundings System The part of the universe you wish to. In Chemistry this is your chemical /physical process. Surroundings Everything else in the. When heat is transferred it can flow in or out of the system Endothermic vs. Exothermic An Exothermic process is one that (evolved) heat to its surroundings (feels ) o Energy products < Energy reactants An Endothermic process is one that heat from the surroundings (feels ) o Energy products > Energy reactants Exothermic Process E < E Endothermic Process E > E Q and heat flow Exothermic process, heat is, q is. Endothermic process, heat is, q is positive (+) Specific Heat = C p Specific heat of a substance is the amount of heat required to raise the temperature of one gram of that by one degree Celsius. Unit for specific heat is J/g o C Each substance has a specific heat Water = 4.184 J/g o C Gold = 0.129 J/g o C Copper = 0.386 J/g o C The lower the specific heat the lower the amount energy is required to raise its temperature. 2
Calculating Heat Released and Absorbed Q = mc p ΔT o Q = Heat (J) o m is (gram) o ΔT ( ) is temperature change T final -T initial o C p is specific heat at a constant pressure Calculating Heat Ex1:If the temperature of 56.6 g of ethanol increases from 45.0 o C to 80.0 o C, how much heat has been absorbed by the ethanol? Specific heat of ethanol = 2.44 J/g o C Ex2: A 4.00 g sample of a substance was heated from 274 K to 314 K and absorbed 32 J of heat, what is the specific heat of the substance? Ex3: If 98,000 J of energy are added to 6200 g of water, what will the change in temperature of the water be? Specific heat of water = 4.184 J/g o C Calorimetry and Enthalpy Calorimetry Calorimetry is a laboratory used to measure heat flow o Based on o the law of conservation of o Idea that the heat released by the system heat absorbed by the surroundings or vice versa; -q = q Calorimetry Example An ice cube is added to a warm cup of water. o The amount of heat used to melt the ice cube is the same amount of heat lost by the warm water. Calorimeter Insulated device used to measure flow Either measures the heat absorbed released during chemical/physical change 3
Enthalpy and Enthalpy Changes Enthalpy (H) is the content of a system at constant Thermochemistry uses the in enthalpy(δh) to study heat changes At constant pressure, Q = ΔH Heat of reaction (ΔH rxn ) The change of enthalpy in a reaction Endo, Exo, q, and H All chemical and physical changes release or absorb heat Exothermic: heat; Q is negative; H is Endothermic: absorbs ; Q is positive; H is positive Thermochemical Equation equation that includes the physical states of all and products (1 atm, 25 o C) and the energy, ΔH Endo) NH 4 NO 3(s) à NH 4 + (aq) + NO 3 - (aq) ΔH = 27 kj Exo) CaO (s) + H 2 O (l) à Ca(OH) 2(s) ΔH = -65.2 kj Thermochemical Equations for Endothermic Reactions Endothermic : ΔH is positive; heat is on the side of the equation (Heat absorbed) NH 4 NO 3(s) à NH 4 + (aq) + NO 3 - (aq) ΔH = 27 kj NH 4 NO 3(s) + Heat à NH 4 + (aq) + NO 3 - (aq) NH 4 NO 3(s) + 27kJ à NH 4 + (aq) + NO 3 - (aq) Thermochemical Equations for Exothermic Reactions Exothermic : ΔH is negative; heat is on the side of the equation (Heat released) CaO (s) + H 2 O (l) à Ca(OH) 2(s) ΔH = -65.2 kj CaO (s) + H 2 O (l) à Ca(OH) 2(s) + Heat CaO (s) + H 2 O (l) à Ca(OH) 2(s) + 65.2 kj Heat of Combustion Heat of combustion is the enthalpy change for the complete burning of a substance CH 4 (g) + 2O 2 (g) à CO 2 (g) + 2H 2 O(l) ΔH = -891kJ CH 4 (g) + 2O 2 (g) à CO 2 (g) + 2H 2 O(l) + Heat CH 4 (g) + 2O 2 (g) à CO 2 (g) + 2H 2 O(l) + 891kJ Ex 1) A hot piece of metal (at 155 C) with a mass of 4.68 g is placed into 53.9 grams of water at 22 C. The water (Cp =4.184 J/g C) heats up to 37.1 C. What is the specific heat of metal? Ex 2) How much heat is released by the combustion of 250.0 g of octane, C 8 H 18? Octane ΔH comb = -5471 kj/mol 4
Changes of State (Phase Changes) Phase Change A phase change is going from one of matter to another (Physical change) o Gas to liquid o Liquid to solid Changes of state either or (evolve/liberate) energy Endothermic phase changes (Absorb energy) o Melting (solid to liquid) o Evaporation/ (liquid to gas) o Sublimation ( to gas) Exothermic phase changes (Release energy) o (gas to liquid) o Freezing (liquid to solid) o Deposition ( to solid) Temperature and Phase Change During a phase change, as energy is added or removed, there is temperature change. Latent Heat Since there is no change, we cannot use Q = mc P ΔT to calculate enthalpy change. Instead we use latent (hidden) heat, which is the quantity of heat released or absorbed during a phase change. Units are usually Joule/gram or kilojoule/mole Changes of State Latent Heat of Vaporization is the heat required to one mole of a liquid o Going from liquid to gas o For water, ΔH vap = 40.7 kj/mol Latent Heat of Condensation is the heat required to one mole of a gas o Going from gas to liquid o For water, ΔH cond = -40.7 kj/mol Thermochemical Equations for Vaporization and Condensation The reverse process of vaporization is condensation o The process has an opposite sign o ΔH vap = - ΔH cond o H 2 O (l) à H 2 O (g) ΔH vap = 40.7 kj/mol o H 2 O (g) à H 2 O (l) ΔH cond = -40.7 kj/mol Changes of State Latent Heat of Fusion is the heat required to one mole of a solid substance. o Going from solid to liquid o For water, ΔH fus = 6.01 kj/mol (Endothermic) 5
Latent Heat of Solidification is the heat required to one mole of a liquid substance. o Going from liquid to solid o For water, ΔH solid = -6.01 kj/mol (Exothermic) Thermochemical Equations for Fusion and Solidification The reverse process of fusion is solidification. o The reverse process has an sign o ΔHfus = - ΔHsolid o H 2 O (s) à H 2 O (l) ΔH fus = 6.01 kj/mol o H 2 O (l) à H 2 O (s) ΔH solid = -6.01 kj/mol Heating Curve Explanation of Heating Curve When there is a temperature change use Q=m C p ΔT to calculate the amount of energy used. When the heating curve is flat (no temperature change), there is a phase change. Use the latent heat to calculate the amount of energy used. Calculate the heat required to melt 45.6 g of water at its melting point. Latent heat of fusion ( H fus ) = 6.01 kj/mol Calculate the heat evolved to condense 75.4 g of water at its boiling point Latent heat of vaporization ( H vap ) = 40.7 kj/mol 6
What mass of ammonia (NH 3 ) must be vaporizes that absorbs 345 kj of heat? Latent heat of vaporization ( H vap ) = 23.3 kj/mol Hess s Law - Calculating Enthalpy Change (Heat of Reaction) Hess s Law states that you can two or more equations to produce a equation, the reaction enthalpy of unknown is the of the related reactions Ex: H 2 (g) + ½ O 2 (g) à H 2 O(g) DH = -242 KJ H 2 O(g) à H 2 O(l) DH = -44 KJ H 2 (g) + ½ O 2 (g) à H 2 O(l) DH = -286 KJ General principles for combining thermochemical equations. 1. If a reaction is reversed, the sign of ΔH is reversed 2. If the coefficients are multiplied, then the ΔH is multiplied by the same factor. Calculate the enthalpy change for: Ex1: 2S (s) + 3O 2(g) à 2SO 3(g) H =? S (s) + O 2(g) à SO 2(g) H = -297 kj 2SO 3(g) à 2SO 2(g) + O 2(g) H = 198 kj Ex2: 2NO (g) + O 2(g) à 2NO 2(g) H =? N 2(g) + O 2(g) -à 2NO (g) H = 180.6 kj N 2(g) + 2O 2(g) à 2NO 2(g) H = 66.4 kj Ex3: 2C (s) + O 2(g) à 2CO (g) H =? C (s) + O 2(g) à CO 2(g) H = 393.5 kj 2CO + O 2 à 2CO 2 H = -566kJ Ex4: C (s) + 2H 2(g) à CH 4(g) H =? C (s) + O 2(g) à CO 2(g) H =-393.5 kj 2H 2(g) + O 2(g) à 2H 2 O (l) H =-571.6 kj + 2O 2(g) à CO 2(g) + 2H 2 O (l) H = -890.8 kj CH 4(g) 7
Standard Heat of Formation Standard heat of formation is defined as the in that accompanies the of one mole of the compound at its state from its in their standard state. Standard state is Every free element in its standard state is assigned a H f of. Free elements include elements and any molecule (H 2 O 2 N 2 Cl 2 Br 2 I 2 F 2 ) Heat of Formation Equation Ex: Ca(s) + C(s) + 3/2 O 2 (g) à CaCO 3 (s) Practice: Write a formation equation for 1 mole of Na 2 CO 3 (s) ΔH rxn = Use the standard enthalpies of formation to calculate Hrxn Ex1: CH 4(g) +2O 2(g) à CO 2(g) + 2H 2 O (l) H f (CO 2(g) ) = -394kJ H f (H 2 O (l) ) = -286 kj H f (CH 4(g) ) = -75kJ H f (O 2(g) ) = 0 kj Use Table C-13 to calculate Hrxn for the following reactions Ex2: 2HF (g) à H 2(g) + F 2(g) Ex3: 2H 2 S (g) + 3O 2(g) à 2H 2 O (l) + 2SO 2(g) Ex4: 4Fe (s) + 3O 2(g) à 2Fe 2 O 3(s) Ex5: 2NO (g) + O 2(g) à 2NO 2(g) Ex6: 2CO (g) + O 2(g) à 2CO 2(g) 8