Energy and Chemical Change Chapter 10

10.1 Energy Energy - capacity to do work or supply heat. potential energy - stored energy positional - where its at compositional - what it is chemical potential energy - energy stored in chemicals

10.1 Energy kinetic energy - energy of motion heat (q) - energy transferred between two objects of differing temperature. Enthalpy (H) - heat content of a system at constant pressure. b. light, electrical energy, mechanical, ...........

10.1 Temperature verses Heat Heat (q) – type of energy transferred because of a difference in temperature. Can’t be measured directly Flow of energy from a warmer object to a cooler object Temperature – measure of the average kinetic energy of the particles in a sample of matter. Determines the direction of heat transfer

10.1 Temperature verses Heat What contains more heat? a glass of boiling water or an iceberg What does your body sense? temperature or heat

10.1 Energy Law of conservation of energy - in ordinary reactions energy is not created or destroyed but converted from one form to another. Thermochemistry - study of heat changes during chemical and physical reactions.

How does the dam energy work? Mechanical energy of moving water Potential energy of dammed water Mechanical of falling water Mechanical of moving turbine Mechanical energy of spinning in generator Electrical energy of moving electrons

What does it mean to “burn calories”? During exercise your body generate a certain amount of heat (measured in Calories) Occurs from the breakdown of organic molecules

10.2 Units of Energy calorie (cal) - heat needed to raise 1 g of water 1 C. Calorie (Cal) - food Calorie - 1000 cal 1 can of pop contains 140 Calories- This means that 1 can of pop releases 140 kilocalories when it burns completely in the presence of Oxygen. Joule (J) - SI unit of energy 1Cal = 1kcal = 1000 cal = 4186 J = 4.186 kJ

10.2 Heat Capacity Heat Capacity – heat it takes to change a substances temperature by 1 C. Dependant on mass A cup of water has a higher heat capacity than a drop of water Dependant on chemical composition Different substances of the same mass may have different heat capacities

10.3 Specific Heat Specific heat (C) – amount of heat required to raise the temperature of 1 g of a substance 1ºC. Measured in J/(g x ºC) Heat affects the temp of objects with a high specific heat much less than the temperature of those with a low specific heat.

10.3 Specific Heat Which has the highest on the list? Water!! This allows bodies of water to store large quantities of energy. Cities by water have cooler summers, warmer winters, and wind! Land has a much lower specific heat capacity

10.3 Specific Heat How do you calculate the specific heat? q = (m)(c)(∆t) q = energy m = mass c = specific heat capacity ∆t = temperature change = Tfinal - Tinitial

q = (m)(c)(∆t) mass = c = ∆t = How many joules of energy would it take to heat up 68.0 g of aluminum from 2.5ºC to 34.5ºC? q = (m)(c)(∆t) mass = c = ∆t = 68.0 g 0.895 J/gºC 34.5ºC – 2.5ºC = 32.0ºC q = (68.0 g)(.895 J/gºC)(32.0ºC) q = 1950 J

10.3 Finding Specific Heat Solving the energy equation for the specific heat, c, yields Specific heat will have units of J/gºC

10.4 Endothermic and Exothermic Reactions Thermochemistry is concerned with the flow of heat between a chemical system and its surroundings. System - what you are looking at Surroundings - everything else Universe = system + surroundings Endothermic - energy in, system absorbs heat from the surroundings(+ enthalpy change). Exothermic - energy out, system loses heat to surroundings(- enthalpy change).

10.5 Calorimetry Calorimeter – insulated device used to measure heat changes. If the specific heat capacity is a characteristic property of a substance, it can be used to determine the identity of an unknown metal. Calorimetry- The accurate & precise measurement of the heat change of a chemical or physical process. Energy cooling copper Energy warming water =

10.5 Calorimetry Enthalpy (H) – The amount of heat that a substance has at a given temperature and pressure Heat changes are the same as enthalpy changes Change in enthalpy is symbolized as H Therefore H = q H = Endothermic Reaction - H = Exothermic Reaction

10.5 Heat in Chemical Reactions The energy needed to heat up the water can be calculated from the mass of the water, its specific heat, and its temperature change. (or q = (m)(c)(∆t)) This is the same energy that the metal releases. If you know the mass of the metal and its temperature change, you can determine its specific heat.

Sample Lab Problem In the lab, a 25 g sample of a metal that has been sitting in boiling water at 100.0ºC, is placed in a calorimeter containing 65 g of water at 27.5 ºC. The water warms up and reaches a high temperature of 30.0 ºC. What is the specific heat of the metal?

Sample Lab Problem qmetal = qwater First, qmetal = qwater To find the energy of the cooling metal, find the qwater. qwater = (mwater)(cwater)(∆twater) qwater = (65 g)(4.184J/g ºC)(30.0 ºC – 27.5 ºC) qwater = 680 J

Sample Lab Problem qmetal = -qwater = -680 J To find the specific heat of the metal use the equation solved for c. = .39 J/g ºC

10.6 Thermochemical Equations All reactions involve energy. Thermochemical equation - chemical equation with energy included. 2C8H18 + 25O2  16CO2 + 18H2O + 10900kJ Endothermic - energy a reactant Exothermic - energy is a product

10.6 Thermochemical Equations Heat of reaction - energy shown as enthalpy change. 2C8H18 + 25O2  16CO2 + 18H2O H= -10900kJ H = positive(+) is endothermic H = negative(-) is exothermic

10.6 Thermochemical Equations Using the following equation, calculate the kilojoules of heat required to decompose 2.24 mol of NaHCO3 (s) 2NaHCO3  Na2CO3 (s) + H2O (g) + CO2 (g) H = 129 kJ 2.24 mol NaHCO3 x __________ = 129 kJ 145 kJ 2 mol NaHCO3

10.6 Thermochemical Equations Heat of combustion – the heat of reaction for the complete burning of one mole of a substance. Reported as enthalpy changes. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) H= -890.4kJ H = the heat of combustion

10.7 Heat and Changes of State Explain what happens when you drop an ice cube on the floor using the following terms: System, surroundings, heat, temperature. The ice cube is the system. The floor and air surrounding it are the surroundings. The ice absorbs heat from the surroundings and begins to melt. The mixture of ice and water will remain at 0o C until all of the ice is melted.

10.7 Heat and Changes of State Molar Heat of Fusion (Hfus) – the heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature. Molar Heat of Solidification (Hsolid) – the heat lost when one mole of a liquid changes to a solid at a constant temperature. This means that Hfus = Hsolid

10.7 Heat and Changes of State How many grams of ice at 0o C could be melted by the addition of 2.25 kJ of heat? The molar heat of fusion for water is 6.01 kJ/mol

10.7 Heat and Changes of State How much heat is absorbed when 24.8 grams of H2O(l) at 100oC is converted to steam at 1000C? P.276 in book

10.7 Heat and Changes of State So according to what you now know, what is the molar heat of vaporization? Molar heat of condensation? How are these two related?

Heating Curve

10.7 Heat and Changes of State Heat changes can also occur when a substance is dissolved in a solvent. This heat change is called the molar heat of solution. Ex: NaOH (s)  Na (aq) + OH (aq) Hsoln= -445.1kJ/mol Is this exothermic or endothermic? Would it feel hot or cold? Exothermic/ Hot

10.8 Hess’s Law If you add two or more thermochemical equations to give a final equation, then you can also add the heat changes to give the final heat change. Ex: Conversion of diamond to graphite C (s, diamond) + O2  CO2 (g) H= -395.4 kJ CO2 (g)  C (s, graphite) + O2 (g) H=393.5 kJ C (s, diamond)  C(s, graphite) H= When you add these two equations together the CO2 and O2 cancel out. You can add the heat changes to give the final heat change -1.9 kJ

10.8 Hess’s Law Calculate the enthalpy change H (in kJ) for the reaction 2Al (s) + Fe2O3(s)  2Fe (s) + Al2O3 (s) H= ? kJ Use the enthalpy changes for the combustion of aluminum and iron: 2Al (s) + 3/2O2 (g)  Al2O3 (s) H= -1669.8 kJ 2Fe (s) + 3/2O2 (g)  Fe2O3 (s) H= -824.2 kJ

Hess’s Law Calculate the enthalpy change H (in kJ) for the deomposition of hydrogen peroxide 2H2O2(l)  2H2O (l) + O2 (g) H= ? kJ Use the enthalpy changes for the combination of hydrogen with oxygen to make 2 different compounds H2 (g) + O2 (g)  H2O2(l) H= -187.9 kJ H2 (g) + 1/2O2 (g)  H2O (l) H= -285.8 kJ

10.9 Standard Heat of Formation Sometimes it is hard to measure the change for a reaction. If this is the case, you can calculate the enthalpy change of the reaction from standard heats of formation. Standard heat of formation (Hof) is the change in enthalpy that accompanies the formation of one mole of the compound from its elements.

10.9 Standard Heats of Formation Will be using table A.6 to calculate p. 787 Use the following equation… Ho = Hof (products) - Hof(reactants)

Example Problem What is the standard heat of reaction for the reaction of carbon monoxide gas with oxygen to form carbon dioxide? Write a balanced equation 2Co(g) + O2 (g)  2 CO2 (g) Use Table A.6 in Appendix A to find the standard heats of formation Hof O2 (g) = Hof CO (g) = Hof CO2 (g) = 0.0 kJ/mol -110.5 kJ/mol -393.5 kJ/mol

Example Problem 3. Sum the Hof values of the reactants and then the products, taking into account the number of moles of each Reactants: 2 mole CO (g) ________ = 1 mol O2 (g) _________ = Products 2 mol CO2 (g) _________ = -110.5 kJ -221.0 kJ 1 mol CO 0.0 kJ 0.0 kJ 1 mol O2 Total Reactants = -221.0 kJ -393.5 kJ -787.0 kJ 1 mol CO2

Example Problem Ho = Hof (products) - Hof(reactants) Ho = -787.0 kJ – (-221.0 kJ) = -566.0kJ