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Thermochemistry AP Chemistry Chapters 6 & 10.8 EU 3.C – Chemical and physical transformations may be observed in several ways and typically involve a change in energy. EK 3.C.2 – Net changes in energy for a chemical reaction can be endothermic or exothermic. EU 5.A – Two systems with different temperatures that are in thermal contact will exchange energy. The quantity of thermal energy transferred from one system to another is called heat. EK 5.A.2 – The process of kinetic energy transfer at the particulate scale is referred to in this course as heat transfer, and the spontaneous direction of the transfer is always from a hot to a cold body. EU 5.B – Energy is neither created nor destroyed, but only transformed from one form to another. EK 5.B.1 – Energy is transferred between systems either through heat transfer or through one system doing work on the other system. EK 5.B.2 – When two systems are in contact with each other and are otherwise isolated, the energy that comes out of one system is equal to the energy that goes into the other system. The combined energy of the two systems remains fixed. Energy transfer can occur through either heat exchange or work. The Nature of Energy Thermo 6.1 Enduring Understanding & Essential Knowledge LO 3.11 – The student is able to interpret observations regarding macroscopic energy changes associated with a reaction or process to generate a relevant symbolic and/or graphical representation of the energy change. (See SP 1.5, 4.4; EK 3.C.2) LO 5.3 – The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions. (See SP 7.1; EK 5.A.2) LO 5.4 – The student is able to use conservation of energy to relate the magnitudes of the energy changes occurring in two or more interacting systems, including identification of the systems, the type (heat versus work), or the direction of energy flow. (See SP 1.4, 2.2; EK 5.B.1, 5.B.2) The Nature of Energy Thermo 6.1 Learning Objectives Energy • Energy = the capacity to do work or to produce heat That which is needed to oppose natural attractions. • First Law of Thermodynamics = The total energy of the universe is constant. • Law of Conservation of Energy = Energy can be converted from one form to another but can be neither created nor destroyed. Energy Units • calorie (cal) = the amount of energy (heat) required to raise the temperature of one gram of water 1 oC. 1000 calories = 1 kcal = 1 Calorie Note: “calorie” and “Calorie” (capitalized) are NOT the same thing • Joule (J) 1 cal = 4.184 J Types of Energy • Potential Energy (PE) = energy due to position or composition ie: water behind a dam, energy in bonds • Kinetic Energy (KE) = energy due to the motion of an object depends on the mass (m) of the object and its velocity (v) 1 KE = mv 2 (provided on the AP Equation sheet) 2 System vs. Surroundings • System = the part of the universe we are focusing on • Surroundings = everything else in the universe • Heat either goes from a system into the surroundings (exothermic), or from the surroundings into a system (endothermic) Example 6.1.B For each of the following, define a system and its surroundings and give the direction of energy transfer. • Methane is burning in a Bunsen burner in a laboratory. System = methane & oxygen Surroundings = everything else around the system Direction = system to the surroundings (exothermic) • Water drops, sitting on your skin after swimming, evaporate System = water drops Surroundings = everything else around the system Direction = surroundings (your skin) to the system (endothermic) Example 6.1.B Hydrogen gas and oxygen gas react violently to form water. Which is lower in energy: a) a mixture of hydrogen and oxygen gases b) water Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted. Heat & Work (ΔE = q + w) • Temperature = a measurement of average kinetic energy • Heat (Q or q) = energy that is transferred from a hotter object to a colder object If two objects stay in contact, heat will flow from the warmer object to the cooler object until their temperatures are the same. • Work (w) = force acting over a distance System and Energy Endothermic • A system can increase its internal energy, ΔE, in two ways: the system can absorb energy (+q) from the surroundings the surroundings can do work (+w) on the system Exothermic • A system can lose energy in two ways: the system can release energy (-q) to the surroundings the system can do work (-w) on the surroundings Example 6.1.A 1) A system does 20 J of work on the surroundings. The change in energy of -20 J. the system, ΔE, is 2) A system has a change in energy of -25 J. No work is done by or on the 25 system, so J of heat must have been (absorbed / released). 3) The E of a system is 50 J if the system absorbs 30 J of heat and has 20 J of work done on it. 4) The E of a system is -40 J if the system releases 50 J of heat and has 10 J of work done on it. 5) A system has a change in energy of +45 J. If 30 J of work was done by the system, then 75 J of heat must have been (absorbed / released). Exothermic • Exothermic = system releases heat to the surroundings • Example: A + B AB + heat Energy (heat) is a product Energy (heat) was released to make the bond between A & B • Making a bond ALWAYS releases energy • The energy of the reactants is higher than the energy of the products, so energy is lost overall • ΔH has a –q (negative) value Endothermic • Endothermic = system absorbs energy from the surroundings • Example: AB + heat A + B Energy (heat) is a reactant Energy (heat) was absorbed to break the bond between A & B • Breaking a bond ALWAYS requires / uses energy • The energy of the reactants is lower than the energy of the products, so energy was gained overall • ΔH has a +q (positive) value Example 6.1.B 1) A container of melted paraffin wax is allowed to stand at room temperature until the wax solidifies. What is the direction of heat flow as the liquid wax solidifies? 2) When solid barium hydroxide octahydrate is mixed in a beaker with solid ammonium thiocyanate, a reaction occurs. The outside of the beaker quickly becomes very cold. Was the reaction exothermic or endothermic? Example 6.1.C Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. a) Water freezes. Exothermic b) Your hand gets cold when you touch ice. Exothermic c) The ice gets warmer when you touch it. Endothermic d) Water boils in a kettle being heated on a stove. Endothermic e) Water vapor condenses on a cold pipe. Exothermic f) Ice cream melts. Endothermic EU 3.C – Chemical and physical transformations may be observed in several ways and typically involve a change in energy. EK 3.C.2 – Net changes in energy for a chemical reaction can be endothermic or exothermic. EU 5.A – Two systems with different temperatures that are in thermal contact will exchange energy. The quantity of thermal energy transferred from one system to another is called heat. EK 5.A.2 – The process of kinetic energy transfer at the particulate scale is referred to in this course as heat transfer, and the spontaneous direction of the transfer is always from a hot to a cold body. EU 5.B – Energy is neither created nor destroyed, but only transformed from one form to another. EK 5.B.1 – Energy is transferred between systems either through heat transfer or through one system doing work on the other system. EK 5.B.2 – When two systems are in contact with each other and are otherwise isolated, the energy that comes out of one system is equal to the energy that goes into the other system. The combined energy of the two systems remains fixed. Energy transfer can occur through either heat exchange or work. EK 5.B.3 – Chemical systems undergo three main processes that change their energy; heating/cooling, phase transitions, and chemical reactions. EK 5.B.4 – Calorimetry is an experimental technique that is used to measure the change in energy of a chemical system. Enthalpy & Calorimetry Thermo 6.2 Essential Knowledge LO 3.11 – The student is able to interpret observations regarding macroscopic energy changes associated with a reaction or process to generate a relevant symbolic and/or graphical representation of the energy change. (See SP 1.5, 4.4; EK 3.C.2) LO 5.3 – The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions. (See SP 7.1; EK 5.A.2) Enthalpy & Calorimetry LO 5.5 – The student is able to use conservation of energy to relate the magnitudes of the energy changes when two nonreacting substances are mixed or brought into contact with one another. (See SP 2.2; EK 5.B.1 & 5.B.2) Thermo 6.2 LO 5.6 – The student is able to use calculations or estimations to relate: energy changes associated with heating/cooling a substance to the heat capacity, energy changes associated with a phase transition to the enthalpy of fusion/vaporization, energy changes associated with a chemical reaction to the enthalpy of the reaction, and energy changes to PV work. (See SP 2.2 & 2.3; EK 5.B.3) Learning Objectives LO 5.7 – The student is able to design and/or interpret the results of an experiment in which calorimetry is used to determine the change in enthalpy of a chemical process (heating/cooling, phase transition, or chemical reaction) at constant pressure. (See SP 4.2 & 5.1; EK 5.B.4) Enthalpy • Enthalpy (ΔH) = Heat = the total energy of a system when it is at constant pressure ΔH = Hfinal – Hinitial • For a chemical reaction, the initial state refers to the reactants and the final state refers to the products, so we can write: ΔHreaction = Hproducts – Hreactants • Heat of Reaction = ΔHreaction = ΔHrxn = ΔH (when written with an equation) Enthalpy Signs • If ΔH is negative (-), the reaction is exothermic – ΔH (negative) when Hproducts < Hreactants Energy is leaving the system and goes into the surroundings • If ΔH is positive (+), the reaction is endothermic ΔH (positive) when Hproducts > Hreactants Note: “+” sign is not required when writing ΔH , but a “-” sign is Energy is being absorbed by the system from the surroundings Different Forms of Enthalpy/Heat • ΔH = Heat/Enthalpy The units of ΔH are usually kilojoules (kJ) • ΔHrxn = ΔHreaction = ΔH (with an equation) = Enthalpy/Heat of Reaction • ΔHo = ΔH at standard conditions Thermochemical reactions are generally written at standard thermochemical temperature and pressure, 25 oC and 1 atm • ΔHf = Enthalpy/Heat of Formation Thermochemical Equation Example N2 (g) + 3 H2 (g) 2 NH3 (g) ΔHo = – 92.38 kJ Things to note: • All physical states (solid/liquid/gas) of the reactants and products are cited in this equation. • How can you tell that this reaction is exothermic? • ΔHo represents the energy (Heat of Reaction) released when 1 mol N2 reacts with 3 mol H2 to form 2 mol NH3 • Look at the coefficients in the balanced equation Example 6.2.A N2 (g) + 3 H2 (g) 2 NH3 (g) ΔHo = –92.38 kJ What is the enthalpy of this reaction given only 1.5 mol of H2? According to the equation, the reaction of 3 mol H2 has a ΔHo of –92.38 kJ Conversion Factor: 3 mol H2 = –92.38 kJ 1.5 mol H2 –92.38 kJ 3 mol H2 = – 46.19 kJ = – 46 kJ Heats of Combustion Heat of Combustion = the amount of energy released during a combustion reaction Substance HEATS OF COMBUSTION AT 25 oC H Formula Substance (kJ/mol) Formula H (kJ/mol) Hydrogen H2 (g) -286 Glucose C6H12O6 (s) -2808 Carbon C (s), graphite -394 Octane C8H18 (l) -5471 Methane CH4 (g) -890 Sucrose C12H22O11 (s) -5645 Acetylene C2H2 (g) -1300 Ethanol C2H5OH (g) -1368 Propane C3H8 (g) -2220 Example 6.2.A C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure. According to the equation, the reaction of 1 mol C3H8 has a ΔH of –2221 kJ 5.00 g C3H8 1 mol C3H8 –2221 kJ = – 252 kJ 44.094 g C3H8 1 mol C3H8 Specific Heat (q = mcΔT) q (or Q) m C (or c) ΔT Represents Heat or energy Mass Specific Heat Change (Δ) in temperature Definition The transfer of energy released or absorbed by the reaction The amount of energy needed to change the temperature of 1 gram of substance by 1 degree Celsius ΔT = Tf – Ti (Final temperature minus initial temperature) Unit kJ or J or kcal (Cal) or cal g goC Sign (-) = heat is released (+) = heat is absorbed (+) (+) J • Specific Heat Capacity = Specific Heat oC (-) = cooling down (+) = heating up Specific Heat & Heat Capacity Specific Heat (AKA Specific Heat Capacity)= the amount of energy needed to raise the temperature of one gram of substance by one degree Celsius. • Specific heats are different for the same substance in different phases • ie: ice water, liquid water, and gaseous water (steam) are all H2O but they have different specific heats. • Substances with lower specific heats are (easier / more difficult) to heat up or cool down than substances with higher specific heats. Heat Capacity = the amount of energy needed to raise the temperature of an entire sample of substance by one degree Celsius. • Units: J/oC Example 6.2.B 74.8 J of heat is required to raise the temperature of 18.69 g of silver from 10.0C to 27.0C. a) What is the heat capacity (J/C) of the silver sample? Note: This is a useful value only for this specific sample of silver. b) What is the specific heat capacity (J/g·C) of silver? Note: This is a useful value for any sample of silver that is heated or cooled. This is equivalent to the 4.184 J·g-1·°C-1 that we use for water. This value is also called the specific heat. Example 6.2.C #1 Water has a specific heat capacity of 4.184 J/g·°C. This means it takes 4.184 J to heat 1.00 gram of water 1.00°C. a) How much energy will it take to heat 10.0 grams of water 1°C? b) How much energy is needed to heat 30.0 g H2O from 10.0 °C to 50.0 °C? Example 6.2.C #2 A pot of water (2.5 Liters of water) initially at 25.0C is heated to boiling (100.°C). a) How much energy (in J) is needed to heat the water? (The density of water is 1 g/mL.) b) What would this amount of heat be in kJ? Example 6.2.C #3 What amount of heat is released when 175 g of water cools from 100.°C to room temperature, 20.0 °C? Example 6.2.C #4 We don’t always have to warm up or cool down water. The specific heat capacity of copper metal is 0.39 J/g·°C. How much energy would it take to heat up a 5.20 g sample of copper from 20.0 °C to 100.°C? Calorimetry • Calorimeter = a device used to determine the heat from a chemical reaction • The calorimeter is insulated so the energy gained and lost are kept in a specific container. • –Qexo = Qendo • The energy gained and the energy lost within a calorimeter are equal to one another in value (but not in sign). Example 6.2.C #5 If 300. J of heat energy were used to heat up a 5.00 gram sample of copper metal and a 5.00 gram sample of water both starting at 10.0°C, calculate the final temperature of each sample. Example 6.2.C #6 Suppose we mix 90.0 grams of hot water (90.0°C) with 10.0 grams of cold water (10.0°C). Let x = the final temperature. C = 4.184 J/g·C a) Set up an expression for the energy released (q) by the hot water (qhot = mhotCThot) b) Set up an expression for the energy absorbed (q) by the cold water (qcold = mcoldCTcold) c) Knowing that the heat absorbed = heat released, combine the two expressions and solve for x. Example 6.2.C #7 175 grams of hot aluminum (100.°C) is dropped into an insulated cup that contains 40.0 mL of ice cold water (0.0°C). Determine the final temperature. a) Set up an expression for the heat lost by the aluminum (C=0.900 J/g·°C) b) Set up an expression for the heat gained by the cold water. c) Put the two expressions together and solve for x. EU 2.A – Matter can be described by its physical properties. The physical properties of a substance generally depend on the spacing between the particles (atoms, molecules, ions) that make up the substance and the forces of attraction among them. EK 2.A.1 – The different properties of solids and liquids can be explained by differences in their structures, both at the particulate level and in their supramolecular structures. EU 2.C – The strong electrostatic forces of attraction holding atoms together in a unit are called chemical bonds. EK 2.C.2 – Ionic bonding results from the net attraction between oppositely charged ions, closely packed together in a crystal lattice. EU 5.B – Energy is neither created nor destroyed, but only transformed from one form to another. EK 5.B.3 – Chemical systems undergo three main processes that change their energy; heating/cooling, phase transitions, and chemical reactions. EU 5.D – Electrostatic forces exist between molecules as well as between atoms or ions, and breaking the resultant intermolecular interactions requires energy. EK 5.D.2 – At the particulate scale, chemical processes can be distinguished from physical processes because chemical bonds can be distinguished from intermolecular interactions Vapor Pressure & Changes of State Thermo 10.8 Essential Knowledge LO 2.1 – Students can predict properties of substances based on their chemical formulas and provide explanations of their properties based on particle views. (See SP 6.4; EK 2.A-2.D) LO 2.19 – The student can create visual representations of ionic substances that connect the microscopic structure to macroscopic properties, and/or use representations to connect the microscopic structure to macroscopic properties (e.g., boiling point, solubility, hardness, brittleness, low volatility, lack of malleability, ductility, or conductivity. (See SP 1.1, 1.4 & 7.1; EK 2.C.2, 2.D.1, 2.D.2) LO 5.6 – The student is able to use calculations or estimations to relate: energy changes associated with heating/cooling a substance to the heat capacity, energy changes associated with a phase transition to the enthalpy of fusion/vaporization, energy changes associated with a chemical reaction to the enthalpy of the reaction, and energy changes to PV work. (See SP 2.2 & 2.3; EK 5.B.3) LO 5.10 – The student can support the claim about whether a process is a chemical or physical change (or may be classified as both) based on whether the process involves changes in intramolecular versus intermolecular interactions (See SP 5.1; EK 5.D.2) Vapor Pressure & Changes of State Thermo 10.8 Learning Objectives Phase Changes Endothermic • Exothermic phase changes: Deposition Condensation Freezing Exothermic • Endothermic phase changes: Sublimation Vaporization Melting Vaporization • Vaporization = the phase change from a liquid to a gas • (Equilibrium) Vapor Pressure = pressure of the vapor present at equilibrium In a closed container, a liquid will vaporize. As moles of vapor molecules increases, some vapor will condense. Equilibrium occurs when ratecondensation = ratevaporization Brain Check • What is the vapor pressure of water at 100°C? How do you know? • The vapor pressure of water at 100oC is 1 atm. You know this because atmospheric pressure is 1 atm and this is the temperature at which we observe water to boil. Factors That Affect Vapor Pressure • Liquids in which the intermolecular forces are strong have relatively low vapor pressures. • Vapor pressure increases significantly with temperature. Boiling vs. Vaporization • (Equilibrium) Vapor Pressure = pressure of the vapor present at equilibrium • Boiling = when the vapor pressure of a liquid is equal to the atmospheric pressure Boiling and Freezing Points • Boiling = when the vapor pressure of a liquid is equal to the atmospheric pressure Normal Boiling Point = the temperature at which the vapor pressure of the liquid is exactly 1 atm. • Freezing Point (AKA Melting Point) = the temperature at which a substance goes from a liquid to a solid (or from a solid to a liquid) Normal Melting Point = the temperature at which the solid and liquid states have the same vapor pressure under conditions where the total pressure is 1 atm. Energy of Phase Changes • When we write an equation, there is usually energy involved. Consider the evaporation of water: H2O(l) + energy H2O(g) • This energy is called the heat of vaporization with the symbol, Hvap. • The equation represents one mole of water being vaporized, so the energy is called the molar heat of vaporization and is reported in kJ/mol. • If “energy / heat” is on the reactants side, the reaction is endothermic. • If “energy / heat” is on the products side, the reaction is exothermic. Example 10.8.A Write an equation for the following changes and include energy: a) The melting of ice: b) The condensation of steam: c) The sublimation of dry ice, CO2(s): d) The freezing of liquid water: Latent Heat (q = mΔHfus) (q = mΔHvap) Represent s q (or Q) m ΔHfus ΔHvap Heat or energy Mass Heat of Fusion Heat of Vaporization The amount of energy needed to melt or freeze (l ↔ s) one gram of substance The amount of energy needed to vaporize or condense (l ↔ g) one gram of substance J/g (kJ/g) (cal/g) J/g (kJ/g) (cal/g) Sometimes “g” is replaced with “mol” and then it becomes the Molar Heat of Fusion Sometimes “g” is replaced with “mol” and then it becomes the Molar Heat of Vaporization (+) = melting (–) = freezing (+) = vaporization (–) = condensation Definition Unit kJ or J or kcal (Cal) or cal g Sign (+) = endo (–) = exo (+) ΔHfus as a Conversion Factor a) When 1.00 gram of H2O(s) melts, 334.6 Joules of energy is absorbed. Calculate the amount of energy (in kJ/mol) required to melt 1.00 mole of ice. (This is the molar heat of fusion for H2O.) b) Calculate the amount of heat needed to melt 35.2 grams of ice. c) What mass of ice can be melted with 975 J of energy? Example 10.8.B How many kilocalories are given off when 50.0 g of water at 0 oC freezes? (The accepted value of the heat of fusion of water, Hfus, is 6.03 kJ/mol.) 1) What information is provided by the problem? m = 50.0 g T = 0 oC (T does not change) Water freezes (l s), so ΔHfus = - 6.01 kJ/mol 2) What equation will be used to solve this problem? q = m ΔHfus Example 10.8.B (Continued) How many kilocalories are given off when 50.0 g of water at 0 oC freezes? (The accepted value of the heat of fusion of water, Hfus, is 6.03 kJ/mol.) 3) Isolate the unknown variable from the equation. (Not necessary because the equation already has “q” isolated.) 4) Plug in the numbers for all the known variables and solve. q = 50.0 g H2O - 6.03 kJ 1 mol H2O 1 mol H2O 18.02 g H2O 1 kcal 4.184 kJ = -3.99 kcal Answer: 3.99 kcal are given off. * Note: Although the value for q is negative, the question asks how many kilocalories are released. You cannot release a negative amount of energy. The negative sign for “q” indicates that the process is exothermic, so energy is being released, rather than absorbed. Example 10.8.C The accepted value for Hfus is 6.03 kJ/mol. The accepted value for Hvap is 40.67 kJ/mol. 1) Calculate the amount of heat energy needed to boil 40.0 grams of water. 2) What mass of water can be vaporized with 55.0 kJ of energy. 3) How much heat is needed to vaporize 1.00 gram of H2O? Example 10.8.C (Continued) Hfus = 6.03 kJ/mol Hvap = 40.67 kJ/mol 4) 100. grams of steam condenses to liquid water. What amount of energy is released? 5) 775 J of energy is added to liquid water at 100°C. What mass of water is vaporized? Heating Curves (General) • Energy can either be used for a temperature change or for a phase change, but not both simultaneously. • Horizontal lines indicate phase changes (temperature remains constant). These are labeled as the normal boiling point and the normal melting point • Each diagonal lines indicates a temperature changes for a different phase (solid, liquid or gas) Heating Curve for Water • Normal Freezing/Melting Point = 0 oC (@ 1 atm) • Normal Boiling Point = 100 oC (@ 1 atm) • Density Liquid Water = 1.00 g/mL Ice = 0.917 g/mL Cooling Curve for Water Things to note: • From left to right, all processes on this curve are exothermic. • This curve is the mirror image of the heating curve of water. • The normal freezing point is still 0 oC. • The normal boiling point is still 100 oC. Brain Check 1) Explain why the Hvap is larger than the Hfus . It takes a lot more energy to break the intermolecular forces between the liquid molecules to completely break apart to form a gas. 2) As intermolecular forces increase, what happens to each of the following: Boiling point Viscosity Surface tension Enthalpy of fusion Freezing point Vapor pressure Heat of vaporization increases increases increases increases increases decreases increases No AP Learning Objectives Hess’ Law Thermo 6.3 Hess’ Law When going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Hess’ Law = If you add two or more thermochemical equations to get a final equation, then you can also add the heats of reaction to give the final heat of reaction. For Example: Equation 1 Equation 2 + Equation 3 New Equation ΔH1o ΔH2o ΔH3o ΔH1o + ΔH2o + ΔH3o Hess’ Law Visualization Rule #1 For Manipulating Equations When an equation is reversed (with the reactants and products switched), the sign of ΔHo must also be reversed. C (s, graphite) + O2 (g) CO2 (g) ΔHo = –393.5 kJ When reversed becomes: CO2 (g) C (s, graphite) + O2 (g) ΔHo = 393.5 kJ Rule #2 For Manipulating Equations If all the coefficients of an equation are multiplied or divided by a common factor, the value of ΔHo must be likewise changed. N2 (g) + 3 H2 (g) 2 NH3 (g) ΔHo = –92.38 kJ When multiplied by 3 becomes: 3 N2 (g) + 9 H2 (g) 6 NH3 (g) ΔHo = –277.14 kJ Rule #3 For Manipulating Equations Cancel (cross out) identical formulas from both sides of an equation in identical physical states. It is okay if the formulas have different coefficients. However, you can only cancel out the smaller coefficient and you must modify the larger coefficient accordingly. C (s graphite) + O2 (g) CO2 (g) C (s diamond) + O2 (g) CO2 (g) C (s graphite) C (s diamond) ΔH = 393.5 kJ ΔH = –395.4 kJ ΔH = –1.9 kJ Example 6.3.A 1) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g) CO2(g) 2) CO(g) + ½ O2(g) ΔHo = –26.7 kJ ΔHo = –283.0 kJ Calculate the ΔHo for the following reaction: 2 Fe (s) + 3/2 O2 (g) Fe2O3 (s) ΔHo = ? This is the “target equation.” Note that thermochemical equations can have fractions for coefficients. Example 6.3.A (Continued #1) Step 1: Look for a reactant/product from the target equation that shows up in only ONE of the given equations & compare its position in all the given equations. 1) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g) ΔHo = –26.7 kJ CO2(g) 2) CO(g) + ½ O2(g) ΔHo = –283.0 kJ 2 Fe (s) + 3/2 O2 (g) Fe2O3 (s) ΔHo = ? We can focus on any of these chemicals. Let’s choose to focus on Fe (s) first. Problem: To match the target equation correctly, Equation #1 must have 2 Fe on the left, but it has 2 Fe on the right. Example 6.3.A (Continued #2) Problem: To match the target equation correctly, Equation #1 must have 2 Fe on the left, but it has 2 Fe on the right. Solution: Reverse Equation #1 and the ΔHo sign. 1) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g) 1*) 2 Fe(s) + 3 CO2(g) Fe2O3(s)+ 3 CO(g) ΔHo = –26.7 kJ ΔHo = 26.7 kJ Example 6.3.A (Continued #3) Step 2: If possible, repeat Step 1 until each equation has been modified. Focus on another reactant/product that shows up only once in the target equation. For this example, we will focus on O2 next. 1) Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g) CO2(g) 2) CO(g) + ½ O2(g) 2 Fe (s) + 3/2 O2 (g) Fe2O3 (s) ΔHo = –26.7 kJ ΔHo = –283.0 kJ ΔHo = ? Problem: We need 3/2 O2 on the left, but we only have ½ O2 on the left in equation #2. Example 6.3.A (Continued #4) Problem: We need 3/2 O2 on the left, but we only have ½ O2 on the left in equation #2. Solution: Multiply Equation #2 by 3. 2) CO(g) + ½ O2(g) CO2(g) 2) 3 [CO(g) + ½ O2(g) CO2(g) ] 2*) 3 CO(g) + 3/2 O2(g) 3 CO2 (g) ΔHo = –283.0 kJ ΔHo = 3(–283.0 kJ) ΔHo = –849.0 kJ Example 6.3.A (Continued #5) Step 3: If there is an equation that was not modified because none of its reactants or products appear in the Target Equation, modify the last equation so that it is canceled out by the reactants/products in the previously modified equations. • Note that an original equation does not have to be modified if the reactant/product that you are focusing on appears in the Target Equation with the same coefficient and on the same side of the equation. • This step does not apply to this example because we modified all the equations that were originally given. Example 6.3.A (Continued #6) Step 4: Cancel (cross out) identical formulas in identical physical states from both sides of the modified equations (or unmodified equations, if modification was not necessary). (Rule #3 for Modifying Equations) 1*) 2 Fe(s) + 3 CO2(g) Fe2O3(s) + 3 CO(g) 2*) 3 CO(g) + 3/2 O2(g) 3 CO2(g) 2 Fe(s) + 3/2 O2(g) Fe2O3(s) ΔHo = 26.7 kJ ΔHo = –849.0 kJ ΔHo = –822.3 kJ EU 5.C – Breaking bonds requires energy, and making bonds releases energy. EK 5.C.2 – The net energy change during a reaction is the sum of the energy required to break the bonds in the reactant molecules and the energy released in forming the bonds of the product molecules. The net change in energy may be positive for endothermic reactions where energy is required, or negative for exothermic reactions where energy is released. Standard Enthalpies of Formation Thermo 6.4 Essential Knowledge LO 5.8 – The student is able to draw qualitative and quantitative connections between the reaction enthalpy and the energies involved in the breaking and formation of chemical bonds. (See SP 2.3, 7.1 & 7.2; EK 5.C.2) Standard Enthalpies of Formation Thermo 6.4 Learning Objectives Heats of Formation (ΔHf) • ΔHo = (sum of ΔHfo of all the products) – (sum of ΔHfo of all the reactants) • Standard heat (enthalpy) of formation (ΔHfo ) = the change in enthalpy when one mole of substance is formed at standard thermochemical conditions from its elements and in its standard form (solid/liquid/gas). Standard States for a compound = 1 atm (gas), 1 M (soln) Standard States for a gas = 1 atm & 25 oC • Tables for heats of formation are provided. • You are not required to memorize the different heats of formation, except for pure elements in their standard states (because ΔHf under these conditions is zero) Sample ΔHf Table • ΔHfo for all pure elements in their standard states = 0 (zero) kJ/mol • Most ΔHfo are exothermic, but some are endothermic. • Different ΔHfo for the same compound in different states (gas, liquid or solid) Substance ΔHfo (kJ/mol) Substance ΔHfo (kJ/mol) Al (s) 0 Cl2 (g) 0 BaCO3 (s) - 1219 H2O (l) - 286 CO (g) - 110.5 H2O (g) - 241.8 CO2 (g) - 393.5 NaHCO3 (s) - 947.7 CS2 (g) + 117 Na2CO3 (s) - 1131 Heats of Formation (ΔHf) Visualization • CH4(g) + 2O2(g) CO2(g) + 2H2O(l) • ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ Example 6.6.B 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g) Some chefs keep baking soda, NaHCO3 , handy to put out grease fires. When thrown on the fire, baking soda partly smothers the fire and the heat decomposes it to give CO2 , which further smothers the flame. The equation for the decomposition of NaHCO3 is shown above. What is the ΔHo of this reaction? Example 6.6.B (Continued) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g) ΔHo = (sum of ΔHfo of all the products) – (sum of ΔHfo of all the reactants) Use a reference table to find the heats of formation for each of the reactants and products ΔHfo of NaHCO3 (s) = – 947.7 kJ/mol ΔHfo of Na2CO3 (s) = – 1131 kJ/mol ΔHfo of H2O (l) = – 286 kJ/mol ΔHfo of CO2 (g) = – 394 kJ/mol Example 6.6.B (Continued) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g) ΔHo = (ΣΔHfo products) – (ΣΔHfo reactants) ΔHo = (ΔHfo of Na2CO3 + ΔHfo of H2O + ΔHfo of CO2) – (ΔHfo of NaHCO3) ΔHo = [(-1131) + (-286) + (-394)] – [2 x (-947.7)] The ΔHfo of NaHCO3 must be multiplied by 2 because it has a coefficient of 2. ΔHo = 84 kJ