# Unit 4: THERMOCHEMISTRY AND NUCLEAR CHEMISTRY

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1 Unit 4: THERMOCHEMISTRY AND NUCLEAR CHEMISTRY Chapter 6: Thermochemistry 6.1: The Nature of Energy and Types of Energy Energy (E): - the ability to do work or produce heat. Different Types of Energy: 1. Radiant Energy: - solar energy from the sun. 2. Thermal Energy: - energy associated with the random motion of atoms and molecules. 3. Chemical Energy: - sometimes refer to as Chemical Potential Energy. It is the energy stored in the chemical bonds, and release during chemical change. 4. Potential Energy: - energy of an object due to its position. First Law of Thermodynamics: - states that energy cannot be created or destroyed. It can only be converted from one form to another. Therefore, energy in the universe is a constant. - also known as the Law of Conservation of Energy (ΣE initial = ΣE final ). 6.2: Energy Changes in Chemical Reactions Heat (q): - the transfer of energy between two objects (internal versus surroundings) due to the difference in temperature. Work (w): - when force is applied over a displacement in the same direction (w = F d). - work performed can be equated to energy if no heat is produced (E = w). This is known as the Work Energy Theorem. System: - a part of the entire universe as defined by the problem. Surrounding: - the part of the universe outside the defined system. Open System: - a system where mass and energy can interchange freely with its surrounding. Closed System: - a system where only energy can interchange freely with its surrounding but mass not allowed to enter or escaped the system. Isolated System: - a system mass and energy cannot interchange freely with its surrounding. Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 135.

2 Honour Chemistry Exothermic Process (ΔE < 0): - when energy flows out of the system into the surrounding. (Surrounding gets Warmer.) Potential Energy Diagram for Exothermic Process Potential Energy Reactant(s) Product (s) Energy Output ΔE < 0 (Δq < 0 when w = 0) Surrounding Reaction Pathway Endothermic Process (ΔE > 0): - when energy flows into the system from the surrounding. (Surrounding gets Colder.) Potential Energy Diagram for Endothermic Process Potential Energy Reactant(s) Product (s) ΔE > 0 (Δq > 0 when w = 0) Energy Input Surrounding Reaction Pathway 6.3: Introduction of Thermodynamics Thermodynamics: - the study of the inter action of heat and other kinds of energy. State of a System: - the values of all relevant macroscopic properties like composition, energy, temperature, pressure and volume. State Function: - also refer to as State Property of a system at its present conditions. - energy is a state function because of its independence of pathway, whereas work and heat are not state properties. Pathway: - the specific conditions that dictates how energy is divided as work and heat. - the total energy transferred (ΔE) is independent of the pathway, but the amounts of work and heat involved depends on the pathway. Page 136. Copyrighted by Gabriel Tang B.Ed., B.Sc.

3 Internal Energy (E): - total energy from work and heat within a system. ΔE = q + w ΔE = Change in System s Internal Energy q = heat (q > 0 endothermic; q < 0 exothermic) w = work (w > 0 work done on the system; w < 0 work done by the system) Work as Compression and Expansion - During expansion on the system, w < 0 because the system is pushing out and work is done by the system (energy output to the surrounding). - During compression on the system, w > 0 because the system is being pressed by the surround and work is done on the system (energy input by the surrounding). F w = F Δd (Pressure = Force per unit of Area, P = or F = PA) A w = (PA) Δd (Substitute PA as Force; A Δd = Volume 3 dimensions) w = P ΔV (During Expansion V, and w. Negative is added to PΔV) Assignment 6.1 to 6.3 pg. 198 #1 to 3, 6 to 11 w = P ΔV (1 L atm = J) 6.4: Enthalpy of Chemical Reactions Enthalpy (H): - the amount of internal energy at a specific pressure and volume (H = E + PV). ΔE = q PΔV ΔH PΔV = q PΔV ΔH = q Change in Enthalpy in a Chemical Reaction ΔH = q = H products H reactants ΔH > 0 Endothermic Reaction ΔH < 0 Exothermic Reaction (ΔE = ΔH PΔV Rearrange formula for enthalpy) (Equate ΔE and simplify by cancelling PΔV on both sides) (Change in Enthalpy is Change in Heat of a system at constant pressure and little change in volume.) ΔH = nδh rxn ΔH = Change in Enthalpy moles ΔH rx Molar Enthalpy of Reaction (kj/mol) Potential Energy Diagram for Endothermic Chemical Reactions / Physical Processes Potential Energy Diagram for Exothermic Chemical Reactions / Physical Processes Potential Energy Reactant(s) Product (s) ΔH > 0 (H products > H reactants ) Potential Energy Reactant(s) ΔH < 0 (H products < H reactants ) Product (s) Reaction Pathway Reaction Pathway Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 137.

4 Honour Chemistry Writing ΔH Notations with Chemical Equations / Physical Process: a. Endothermic Reactions / Processes Reactant(s) + Heat Product(s) OR Reactant(s) Product(s) ΔH = + kj Example: Water is vaporized from its liquid state. H 2 O (l) kj H 2 O (g) or H 2 O (l) H 2 O (g) ΔH = kj b. Exothermic Reactions / Processes Reactant(s) Product(s) + Heat OR Reactant(s) Product(s) ΔH = kj Example: Methane undergoes combustion at constant pressure. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) kj or CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) ΔH = 803 kj Example 1: It takes kj to form 84.0 L of NO 2 (g) from its elements at 1.00 atm and 25.0 C. Determine the molar heat of enthalpy for the formation of NO 2 (g). Express the answer in proper ΔH notation. ΔH = kj V = 84.0 L P = 1.00 atm T = 25.0 C = K atm L R = mol K? ΔH rx? PV (1.00 atm) = ( 84.0 L) atm L ( )( K) = mol RT mol K ΔH kj ΔH = nδh rxn ΔH rx = ΔH rx 33.9 kj/mol n mol N 2 (g) + 2 O 2 (g) 2 NO 2 (g) ΔH = 67.8 kj (2 mol of NO 2 in Eq) or ½ N 2 (g) + O 2 (g) NO 2 (g) ΔH = 33.9 kj Example 2: Given that 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O (g) kj, calculate the change in enthalpy when 28.2 g of butane is burned kj ΔH rx = kj/mol 2 mol 28.2 g = mol C 4 H g/mol ΔH =? (There are 2 moles of C 4 H 10 in the chemical equation for 5317 kj.) ΔH = nδh rx ( mol)( kj/mol) ΔH = kj = 1.29 MJ (1 MJ = 1000 kj) Assignment 6.4 pg. 199 #21 to 28 Page 138. Copyrighted by Gabriel Tang B.Ed., B.Sc.

5 6.5 & 12.6: Calorimetry and Phase Changes Energy involved in Physical Change (Temperature or Phase Change): Heating Curve: - a graph of temeparture versus time as a substance is heated from a solid phase to a gaseous phase. - when a substance is undergoing a phase change, its temperature remains at a constant (the plateau on the heating curve) until all molecules aquired enough energy to overcome the intermoelcular forces nexessary. This is commonly refered to as the potential change of a subsatnce. - when a substance is undergoing temperature change within a particular phase, it is refered to as kinetic change (because temeperature is also refered to as the average kinetic energy of a substance). Heating Curve of Water 100 C Temperature 0 C Phase Change Solid / Liquid at Melting Point Water Phase Change Liquid / Gas at Boiling Point Steam Ice Time Molar Enthalpy of Fusion (ΔH fus ): - the amount of heat needed to melt one mole of substance from solid to liquid at its melting point (in kj/mol). Molar Enthalpy of Vaporization (ΔH vap ): - the amount of heat needed to evaporate one mole of substance from liquid to gas at its boiling point (in kj/mol) Molar Enthalpy of Sublimation (ΔH sub ): - the amount of heat needed to sublime one mole of substance from solid to gas (in kj/mol). - because sublimation involves two phase change in one step, the molar enthalpy of sublimation is the sum or the molar ethalphies of fusion and vapourization. ΔH sub = ΔH fus + ΔH vap Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 139.

6 Honour Chemistry Specific Heat (s): - the amount of heat (J or kj) needed to change (1 g or 1 kg) of substance by 1 C or 1 K. - the stronger the intermolecular forces, the higher the specific heat capacity. Heat Capacity (C): - the amount of heat (J or kj) needed to a given amount of substance by 1 C or 1 K. - usually used in a calorimeter (see section 6.5). Physical Potential Change q = nδh fus q = nδh vap q = nδh sub q = Heat Change (J or kj) moles ΔH fus = Molar Enthalpy of Fusion (kj/mol) ΔH vap = Molar Enthalpy of Vaporization (kj/mol) ΔH sub = Molar Enthalpy of Sublimation (kj/mol) ΔH sub = ΔH fus + ΔH vap Physical Kinetic Change q = msδt q = CΔT C = ms q = Heat Change (J or kj) m = mass (g or kg) ΔT = Change in Temperature (in C or K) s = Specific Heat [J/(g C) or kj/(kg C) or J/(g K) or kj/(kg K)] C = Heat Capacity [J/ C or kj/ C or J/K or kj/k)] Physical Thermodynamic Properties of Some Common Substances (at 1.00 atm and K) Substance Melting Boiling Specific Heat ΔH fus ΔH vap Point ( C) Point ( C) [kj/(kg C)] (kj/mol) (kj/mol) Ice H 2 O (s) Water H 2 O (l) Steam H 2 O (g) Ammonia NH 3 (g) Methanol CH 3 OH (l) Ethanol C 2 H 5 OH (l) Aluminum Al (s) Carbon (graphite) C (s) Copper Cu (s) Iron Fe (s) Mercury Hg (l) Example 1: What is the change in enthalpy involved when g of water boils from liquid to gas at 100 C? Since this question involves phase change (vaporization) only, we need to use q = nδh vap. ΔH vap = kj/mol g = mol H 2 O g/mol q =? q = nδh vap q = (2.000 mol)(40.79 kj/mol) q = kj Page 140. Copyrighted by Gabriel Tang B.Ed., B.Sc.

7 Example 2: How much energy is needed to heat g of water from 20.0 C to 80.0 C? Since this question involves temperature (kinetic) change only, we need to use q = mcδt. s = J/(g C) m = g H 2 O ΔT = 80.0 C 20.0 C = 60.0 C q =? q = msδt q = (100.0 g)(4.184 J/(g C))(60.0 C) = J q = J = 25.1 kj Example 3: What is the total energy needed to sublime 40.0 g of solid ammonia to gaseous ammonia? (ΔH fus = 5.66 kj/mol; ΔH vap = kj/mol) Since this question involves phase change (sublimation) only, we need to use q = nδh sub. ΔH sub = ΔH fus + ΔH vap q = nδh sub ΔH sub = 5.66 kj/mol kj/mol q = ( mol)(28.99 kj/mol) ΔH sub = kj/mol 40.0 g q = 68.1 kj = mol NH g/mol q =? Example 4: What is the total energy needed to heat g of water at 80.0 C to steam at 115 C? For this question, we have two kinetic changes (water and steam) and one phase change (vaporization). m = g H 2 O = kg H 2 O g = mol H 2 O g/mol s water = kj/(kg C) ΔT water = C 80.0 C = 20.0 C ΔH vap = kj/mol s steam = 1.99 kj/(kg C) ΔT steam = 115 C 100 C = 15 C q total =? q total = ms w ΔT w + nδh vap + ms s ΔT s (water) (vaporization) (steam) q total = ( kg)(4.184 kj/(kg C))(20.0 C) + (1.000 mol)(40.79 kj/mol) + ( kg)(1.99 kj/(kg C))(15 C) q total = kj kj kj q total = 42.8 kj Molar Enthalpy of Solution (ΔH soln ): - the amount of heat needed to dissolve 1 mole of substance. - it can easily be found using a constant-pressure calorimeter. - for molecular solutes, it involves overcoming the intermolecular forces of the solute and solvent. This is followed by the hydration process as the solute and solvent molecules come together. - for ionic solutes, it first involves overcoming the lattice energy. Finally, the ions and solvent molecules come together during hydration process. Enthalpy of Solution ΔH = nδh soln ΔH = Change in Enthalpy moles ΔH sol Molar Enthalpy of Solution (kj/mol) Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 141.

8 Honour Chemistry Example 5: A cold pack consists of 40.0 g of NH 4 NO 3 is dissolved in water. How much energy is absorbed or released into its surrounding if the ΔH soln is 26.2 kj/mol? ΔH sol kj/mol (ΔH soln > 0; endothermic) (Heat is absorbed from the surrounding) 40.0 g = mol NH 4 NO g/mol ΔH =? ΔH = nδh soln ΔH = ( mol)(26.2 kj/mol) ΔH = 13.1 kj absorbed Example 6: 12.9 kj of heat is released when CaCl 2 is dissolved in water, find the mass of CaCl 2 dissolved if the molar enthalpy of solution of CaCl 2 is 82.8 kj/mol. ΔH sol 82.8 kj/mol (ΔH soln < 0; exothermic) (Heat is released into the surrounding) ΔH = 12.9 kj M = g/mol CaCl 2? m =? Heat of Dilution: - the amount of heat associated during the dilution process of a solution, - a solution that has a endothermic heat of solution (ΔH soln > 0) will have an endothermic heat of dilution. - a solution that has a exothermic heat of solution (ΔH soln < 0) will have an exothermic heat of dilution. (Examples: When NaOH (s) is dissolved to become NaOH (aq), it releases heat. During the dilution process, the solution gets even warmer. This is because more intermolecular forces form between the added water molecules and the ions present. More intermolecular forces or bonds form mean more heat is released. Similarly, when concentrated H 2 SO 4 (aq) is diluted, more intermolecular forces are made and the process releases a lot of heat. Hence, we always add concentrated to water slowly with constant stirring.) Calorimetry: - uses the conservation of energy (Heat Gained = Heat Lost) to measure calories (old unit of heat: 1 cal = J). - physical calorimetry involves the mixing of two systems (one hotter than the other) to reach some final temperature. - the key to do these problems is to identify which system is gaining heat and which one is losing heat. Example 7: Hot water at 90.0 C is poured into 100. g of cold water at 10.0 C. The final temperature of the mixture is 70.0 C. Determine the mass of the hot water. m cold water = 100. g s = J/(g C) ΔT hot water = 90.0 C 70.0 C = 20.0 C ΔT cold water = 70.0 C 10.0 C = 60.0 C m hot water =? ΔH 12.9 kj ΔH = nδh soln = ΔH soln 82.8 kj/mol mol m = nm = ( mol)( g/mol) Heat Lost = Heat Gained (hot water, kinetic) (cold water, kinetic) m hw sδt hw = m cw sδt cw o mcwδtcw 100. g 60.0 C m hw = o ΔT 20.0 C hw m hw = 300. g m = 17.3 g = ( ( )( ) ) Page 142. Copyrighted by Gabriel Tang B.Ed., B.Sc.

9 Example 8: A g of iron metal at 330. C is dropped into a beaker of g of water at 25.0 C. What will be the final temperature? T f =? m Fe = g s Fe = J/(g C) ΔT Fe = 330. C T f m water = g s water = J/(g C) ΔT water = T f 25.0 C Heat Lost = Heat Gained (iron, kinetic) (water, kinetic) m Fe s Fe ΔT Fe = m w s w ΔT w (1500. g)(0.444 J/(g C))(330. C T f ) = (1000. g)(4.184 J/(g C))(T f 25.0 C) T f = 4184T f T f 4184T f = T f = T f = T f = 66.9 C 4850 Energy involved in Chemical Change (Chemical Reaction): Molar Heat of Combustion (ΔH comb ): - the amount of heat released when one mole of reactant is burned with excess oxygen. - the reaction is often exothermic and therefore ΔH comb < 0. Enthalpy of Combustion ΔH = nδh comb ΔH = Change in Enthalpy moles ΔH comb = Molar Heat of Combustion (kj/mol) Schematic of a Bomb Calorimeter - we often use a constant-volume calorimeter (or bomb calorimeter) to determine ΔH comb due to its well-insulated design. It is calibrated for the heat capacity of the calorimeter, C cal, before being use for to calculate ΔH comb of other substances. The sample is measured and burned using an electrical ignition device. Water is commonly used to absorb the heat generated by the reaction. The temperature of the water increases, allowing us to find the amount of heat generated. By applying the law of conservation of energy, we can then calculate the ΔH comb of the sample. Heat Lost = Heat Gained (Combustion Reaction) (water, kinetic) Chemical Combustion Calorimetry n sample ΔH comb = C cal ΔT (if bomb calorimeter is used) or n sample ΔH comb = m w s w ΔT (if the heat absorbed by the calorimeter itself is ignored) Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 143.

10 Honour Chemistry Example 9: Octane, C 8 H 18(l) was burned completely to CO 2 (g) and H 2 O (l) in a bomb calorimeter. The following are the observations of the experiment. Mass of C 8 H 18 (l) burned g Initial Temperature of Calorimeter and Water C Final Temperature of Calorimeter and Water C Heat Capacity of Calorimeter kj/ C a. Determine the experimental molar heat of combustion of C 8 H 18 (l). b. The theoretical ΔH comb for C 8 H 18 (g) is kj/mol, calculate the % error of this experiment. a. m octane = g g n octane = g/mol = mol C 8 H 18 C cal = kj/ C ΔT = T f T i = C C ΔT = C ΔH comb =? b. % error = Theoretical Experimental Theoretical Heat Lost = Heat Gained (Combustion Reaction) (water, kinetic) n octane ΔH comb = C cal ΔT o o T kj/ C C ΔH comb = mol C cal Δ = ( )( ) Molar Heat of Reaction (ΔH rxn ): - the amount of heat released when one mole of reactant undergoes various chemical changes. - examples are ΔH comb, ΔH neut (neutralization), ΔH ion (ionization). Constant-Pressure Calorimeter (or Styrofoam Calorimeter) - commonly used to determine ΔH neut, ΔH ion, ΔH fus, ΔH vap, ΔH rxn of non-combustion reaction. First, the n octane ( ) ΔH comb = 5255 kj/mol (negative sign is added because ΔH comb is exothermic) 100% % error = ( kj/mol ) ( 5255 kj/mol ) ( kj/mol) 100% % error = 3.932% (The small % error means the bomb calorimeter was a good heat insulator.) Enthalpy of Chemical Reactions ΔH = nδh rxn ΔH = nδh neut ΔH = nδh ion ΔH = nδh comb ΔH = Change in Enthalpy ΔH rx Molar Heat of Reaction (kj/mol) ΔH io Molar Heat of Ionization (kj/mol) A Simple Styrofoam Calorimeter moles ΔH neut = Molar Heat of Neutralization (kj/mol) ΔH comb = Molar Heat of Combustion (kj/mol) sample s mass is measured. Water is commonly used to absorb or provide the heat for the necessary change. The initial and final temperatures of the water are recorded, allowing us to find the amount of heat change. By applying the law of conservation of energy, we can then calculate the necessary molar enthalpy of change. Page 144. Copyrighted by Gabriel Tang B.Ed., B.Sc.

11 Non-Combustion Calorimetry Heat Gained /Lost = Heat Lost / Gained (Non-Combustion Change) (water, kinetic) n sample ΔH rx m w s w ΔT n sample ΔH fus = m w s w ΔT n sample ΔH vap = m w s w ΔT m sample s sample ΔT sample = m w s w ΔT w Example 10: When ml of HNO 3 (aq) at mol/l is reacted with ml of KOH (aq) at mol/l, the temperature of the final mixture reached C from C. Determine the molar heat of neutralization between HNO 3 (aq) and KOH (aq). Assignment 12.6 pg #69, 71, 72, 78, pg #29 to : Standard Enthalpies of Formation and Reaction (non-combustion chemical change) (physical change - molar heat of fusion) (physical change - molar heat of vapourization) (physical change specific heat) The reaction is acid-base neutralization and will produce water as a result. HNO 3 (aq) + KOH (aq) H 2 O (l) + KNO 3 (aq) Assuming KNO 3 (aq) does not affect the specific heat of water, we would have ml of water produced. Since we have equal moles of acid and base (0.300 mol/l L = mol), the ΔH neut for HNO 3 would be the same as KOH. Heat Lost = Heat Gained n acid = n base = mol (Neutralization) (water, kinetic) m w = kg (500.0 ml is produced) s w = kj/(kg C) ΔT = C C ΔT = 6.50 C n acid ΔH neut = m w s w ΔT m s T ΔH neut = w w Δ ΔH neut = Standard State: - standard conditions of 1 atm and 25 C. It is denote by a superscript o. Standard Molar Enthalpy of Formation (ΔH o f): - the amount of heat required / given off to make 1 mole of compound from its elemental components under standard conditions. - the Molar Heat of Formation of ALL ELEMENTS is 0 kj. - the state of the compound affects the magnitude of H f. (H 2 O (g) has ΔH o f = kj/mol; H 2 O (l) has ΔH o f = kj/mol) (See Appendix 2 on pg. A2 to A5 in the Chang 9 th ed. Chemistry textbook for a list of ΔH f ) n acid o o ( kg)( kj/(kg C)( 6.50 C) ΔH neut = 181 kj/mol ( mol) (negative sign is added due to increased surrounding temperature - an exothermic reaction) Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 145.

12 Honour Chemistry Standard Enthalpy of Formation (Chemical) ΔH = nδh o f ΔH = Change in Enthalpy moles ΔH o f = Standard Molar Enthalpy of Formation (kj/mol) Example 1: Find the standard molar enthalpy of formation for table salt given that its formation reaction, 2 Na (s) + Cl 2 (g) 2 NaCl (s) kj, at standard conditions. ΔH = 822 kj 2 mol of NaCl ΔH o f =? ΔH = nδh o f ΔH o ΔH f = = n 822 kj 2 mol ΔH o f = 411 kj/mol Example 2: What is the amount of heat absorbed / released when 100. g of CO 2 (g) is produced from its elements (CO 2 has ΔH o f = kj/mol)? 100.g 44.01g/mol = mol CO 2 ΔH = nδh o f ΔH = ( mol)( kj/mol) ΔH o f = kj/mol ΔH = 894 kj (894 kj is released) ΔH =? Example 3: Iron (III) oxide, rust, is produced from its elements, iron and oxygen. What is the mass of rust produced when 1.20 MJ is released when iron is reacted with oxygen (ΔH o f = kj/mol for Fe 2 O 3 )? ΔH = 1.20 MJ = kj (exothermic) ΔH o f = kj/mol M = g/mol Fe 2 O 3? m =? ΔH = nδh o f ΔH = o Δ Example 4: Calculate the standard molar enthalpy of formation of silver (I) oxide when 91.2 g of Ag 2 O is produced from its elements and 12.2 kj of heat is released from the process. H f 91.2 g = mol Ag 2 O g/mol ΔH = 12.2 kj (exothermic) ΔH o f =? kj = mol kj/mol m = nm = ( mol)(159.7 g/mol) m = 233 g ΔH = nδh o f ΔH o ΔH f = = n ΔH o f = 31.0 kj/mol 12.2 kj mol Standard Molar Enthalpy of Reaction (ΔH o rxn): - the amount of heat involved when 1 mol of a particular product is produced or 1 mol of a particular reactant is (Note: In hydrocarbon combustion, consumed under standard conditions of 1 atm and 25 C. assume all products are gaseous - it is equal to the difference between of all enthalpies unless otherwise stated.) of products and all enthalpies of reactants. - if the reaction is a combustion, it is called the molar heat of combustion. Page 146. Copyrighted by Gabriel Tang B.Ed., B.Sc.

13 Example 5: Propane is a clean burning fossil fuel that is widely used in outdoor barbecue. a. Calculate the standard molar enthalpy of combustion for propane. (ΔH o f C 3 H 8 = kj/mol; ΔH o f CO 2 = kj/mol; ΔH o f H 2 O (g) = kj/mol) b. Draw its potential energy diagram. c. How much energy will be absorbed or released when 15.0 g or propane is burned? b. Potential Energy Diagram Direct Method to determine Standard Enthalpy of Reaction ΔH o rx ΣH o products ΣH o reactants ΔH o rx Change in Enthalpy of Reaction ΣH o products = Sum of Heat of Products (from all nδh o f of products) ΣH o reactants = Sum of Heat of Reactants (from all nδh o f of reactants) a. We have to first write out a balance equation for the combustion of propane. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) ΔH o f : kj/mol 0 kj/mol kj/mol kj/mol ΔH o rx ΣH o products ΣH o reactants ΔH o rx [3 mol ( kj/mol) + 4 mol ( kj/mol)] [1 mol ( kj/mol) + 5 mol (0 kj/mol)] ΔH o rx [ kj] [ kj] = kj Potential Energy C 3 H 8 (g) + 5 O 2 (g) kj ΔH o rx 2044 kj/mol of C 3 H 8 burned ΔH o rx 2044 kj kj 3 CO 2 (g) + 4 H 2 O (g) c. From part a., for every 1 mol of C 3 H 8 burned, kj is released g = mol C 3 H g/mol ΔH o rx kj/mol ΔH =? ΔH = nδh o rxn ΔH = ( mol)( kj/mol) Reaction Pathway ΔH = 695 kj (695 kj is released) Example 6: Find the amount of heat released when 34.9 g of butane gas is burned at standard conditions. (ΔH o f C 4 H 10 = kj/mol; ΔH o f CO 2 = kj/mol; ΔH o f H 2 O (g) = kj/mol) We have to first write out a balance equation for the combustion of butane. 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O (g) (We have to divide all coefficients by 2 because we are calculating ΔH rxn per mol of butane burned.) 13 C 4 H 10 (g) + O2 (g) 4 CO 2 (g) + 5 H 2 O (g) 2 ΔH f : kj/mol 0 kj/mol kj/mol kj/mol ΔH o rx ΣH o products ΣH o reactants ΔH o 13 rx [4 mol ( kj/mol) + 5 mol ( kj/mol)] [1 mol ( kj/mol) + mol (0 kj/mol)] 2 ΔH o rx [ 2783 kj] [ kj] ΔH rx kj/mol of C 4 H 10 burned 34.9 g = mol C 4 H g/mol ΔH rx kj/mol ΔH =? ΔH = nδh rxn ΔH = ( mol)( kj/mol) = kj ΔH = 1.60 MJ (1.60 MJ is released) Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 147.

14 Honour Chemistry Example 7: When g of liquid heptane is burned in the reaction vessel of a calorimeter, 1.50 L of water around the vessel increased its temperature from 20.0 C to 85.0 C. Ignoring the metallic material of the calorimeter, a. determine the experimental standard molar enthalpy of combustion heptane. b. find the theoretical standard molar enthalpy of combustion of heptane. (ΔH f heptane = kj/mol) c. explain why the experimental ΔH rxn is different than its theoretical counterpart. a. We use the conservation of heat to calculate experimental ΔH rxn g g/mol mol C 7 H 16 m water = 1.50 kg (1 kg = 1 L of water) ΔT = 85.0 C 20.0 C = 65.0 C s water = kj / (kg C) ΔH rx? nδh rx m w s w ΔT o m s T ΔH rx w w Δ = ( 1.50 kg) kj/ ( kg C) n mol ΔH rx kj/mol (released) o ( )( 65.0 C) ( ) Experimental ΔH rx 4.08 MJ/mol of C 7 H 16 burned b. To find theoretical ΔH rxn for the combustion of heptane, we have to use the direct method. C 7 H 16 (l) + 11 O 2 (g) 7 CO 2 (g) + 8 H 2 O (g) ΔH f : kj/mol 0 kj/mol kj/mol kj/mol ΔH rx ΣH products ΣH reactants ΔH rx [7 mol ( kj/mol) + 8 mol ( kj/mol)] [1 mol ( kj/mol)] ΔH rx [ kj] [ kj] = kj Theoretical ΔH rx 4.46 MJ/mol of C 7 H 16 burned c. Some of the possible reasons why experimental ΔH rxn ( 4.08 MJ) is different than the theoretical ΔH rxn ( 4.46 MJ) Some of the heat released by the reaction is absorbed by the metal calorimeter itself. Thus, the temperature gained by the water is not an exact reflection of the energy lost by the combustion. The calorimeter is not a closed system. Heat might escape into the surrounding. Even if the system is closed, the buildup of gases from the reaction would increase pressure and volume. Hence, some of the energy produced from the reaction is used to do work by the system. Thereby, lowering the heat available to warm the water. Page 148. Copyrighted by Gabriel Tang B.Ed., B.Sc.

15 Example 8: HCOOH (g) were completely burned to CO 2 (g) and H 2 O (l) in a calorimeter. The following are the observation of the experiment. Mass of HCOOH (g) burned Initial Temperature of Calorimeter and Water Final Temperature of Calorimeter and Water Heat Capacity of Calorimeter and Water 9.22 g 21.5 C 37.3 C 3.20 kj/ C a. Determine the experimental molar enthalpy of formation of HCOOH (l) assuming standard conditions. (ΔH o f CO 2 = kj/mol; ΔH o f H 2 O (l) = kj/mol) b. If the theoretical ΔH f for HCOOH (g) is 363 kj/mol, calculate the % error of this experiment. a. We use the conservation of heat to calculate experimental ΔH rxn g nδh rx C cal ΔT g/mol o o C T ΔH rx cal Δ = ( 3.20 kj/ C)( 15.8 C) mol HCOOH n ( mol) C cal = 3.20 kj/ C ΔH rx kj/mol (released) ΔT = 37.3 C 21.5 C = 15.8 C ΔH rx? Next, we use the direct method to find the ΔH f of HCOOH. Experimental ΔH rx 252 kj/mol of HCOOH burned HCOOH (l) + ½ O 2 (g) CO 2 (g) + H 2 O (l) ΔH f :? kj/mol 0 kj/mol kj/mol kj/mol ΔH rx ΣH products ΣH reactants kj = [1 mol ( kj/mol) + 1 mol ( kj/mol)] [1 mol (ΔH f )] kj = [ kj] [1 mol (ΔH f )] ΔH f = kj kj = kj Experimental ΔH f of HCOOH = 427 kj/mol Theoretical Experimental b. % error = 100% Theroretical % error = ( 363 kj ) ( 427 kj ) ( 363 kj) 100% % error = 17.6% Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 149.

16 Honour Chemistry Hess s Law: - the indirect method of obtaining overall ΔH rxn of a net reaction by the addition of ΔH rxn of a series of reactions. - when adding reactions, compare the reactants and products of the overall net reaction with the intermediate (step) reactions given. Decide on the intermediate reactions that need to be reversed and / or multiply by a coefficient, such that when added, the intermediate products will cancel out perfectly yielding the overall net reaction. - if a particular reaction needs to be reversed (flipped), the sign of the ΔH for that reaction will also need to be reversed. - if a coefficient is used to multiply a particular reaction, the ΔH for that reaction will also have to multiply by that same coefficient. (Check out Hess s Law Animation at Example 9: Calculate ΔH rxn for the reaction N 2 (g) + 2 O 2 (g) 2 NO 2 (g), when the following reactions are given. N 2 (g) + O 2 (g) 2 NO (g) ΔH rx 180 kj 2 NO 2 (g) 2 NO (g) + O 2 (g) ΔH rx 112 kj Note that 2 NO 2 in the net reaction is on the product side, whereas 2 NO 2 in the second reaction is on the reactant side. Hence, we need to reverse the second reaction and its sign of the ΔH rxn. N 2 (g) + O 2 (g) 2 NO (g) ΔH rx 180 kj (Flipped) 2 NO (g) + O 2 (g) 2 NO 2 (g) ΔH rx 112 kj N 2 (g) + 2 O 2 (g) 2 NO 2 (g) ΔH rx + 68 kj Example 10: Determine the ΔH rxn for the reaction S (s) + O 2 (g) SO 2 (g), when the following reactions are given. S (s) O2 (g) SO 3 (g) ΔH rx kj 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) ΔH rx kj a. SO 2 in the net reaction is on the product side, whereas 2 SO 2 in the second reaction is on the reactant side. Hence, we need to reverse the second reaction and its sign of the ΔH rxn. b. There is only 1 SO 2 in the net reaction, whereas there are 2 SO 2 in the second reaction. Therefore the second reaction and its ΔH rxn need to be multiply by the coefficient of ½. S (s) O2 (g) SO 3 (g) ΔH rx kj (Flipped and ½) ½ (2 SO 3 (g) 2 SO 2 (g) + O 2 (g) ) ΔH rx ½( kj) S (s) O2 (g) SO 3 (g) ΔH rx kj SO 3 (g) SO 2 (g) + ½ O 2 (g) ΔH rx 99.1 kj S (s) + O 2 (g) SO 2 (g) ΔH rx kj Page 150. Copyrighted by Gabriel Tang B.Ed., B.Sc.

17 Example 3: Find the ΔH rxn for the overall reaction of 2 N 2 (g) + 5 O 2 (g) 2 N 2 O 5 (g), when the following reactions are given. H 2 (g) + ½ O 2 (g) H 2 O (l) ΔH rx kj N 2 O 5 (g) + H 2 O (l) 2 HNO 3 (l) ΔH rx 76.6 kj ½ N 2 (g) O2 (g) + ½ H 2 (g) HNO 3 (l) ΔH rx kj a. 2 N 2 O 5 in the net reaction is on the product side, whereas N 2 O 5 in the second reaction is on the reactant side. Hence, we need to reverse the second reaction and its sign of the ΔH rxn. b. There are 2 N 2 O 5 in the net reaction, whereas there is only 1 N 2 O 5 in the second reaction. Therefore the second reaction and its ΔH rxn need to be multiply by the coefficient of 2. c. There are 2 N 2 in the next reaction on the reactant side. Since ½ N 2 is on the reactant side of the third reaction, we need to multiply the third reaction and its ΔH rxn by the coefficient of 4. d. In order for H 2 O to cancel from the first and second reactions, we have to multiple the first reaction by 2 and flipped. This is because H 2 O in the second reaction has also flipped and has been multiplied by 2. (Flipped and 2) 2 (H 2 O (l) H 2 (g) + ½ O 2 (g) ) ΔH rx 2( kj) (Flipped and 2) 2 (2 HNO 3 (l) N 2 O 5 (g) + H 2 O (l)) ΔH rx 2( kj) ( 4) 4 (½ N 2 (g) O2 (g) + ½ H 2 (g) HNO 3 (l) ) ΔH rx 4( kj) 2 H 2 O (l) 2 H 2 (g) + O 2 (g) ΔH rx kj 4 HNO 3 (l) 2 N 2 O 5 (g) + 2 H 2 O (l) ΔH rx kj 2 N 2 (g) + 6 O 2 (g) + 2 H 2 (g) 4 HNO 3 (l) ΔH rx kj 2 N 2 (g) + 5 O 2 (g) 2 N 2 O 5 (g) ΔH rx kj Assignment 6.6 pg #39 to 42, 45 to 49, 51, 52, 54 to 56, 58, 60 to 64, 67, 68, 74, 76, 80, : Present Sources of Energy and New Energy Sources Fossil Fuel: - hydrocarbon fuels that came from fossils of decayed organisms. 1. Natural Gas: - fossil fuel that consists of mainly small alkanes (80% methane, 10% ethane, 4% propane, 2% butane, 4% nitrogen). - usually burns efficiently (complete combustion). Complete Combustion: - where the products of combustion are carbon dioxide and water vapour only. -characterized by a blue flame. Example: Propane burns completely. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 151.

18 Honour Chemistry Incomplete Combustion: - where the main product of combustion is carbon monoxide, along with carbon dioxide and water vapour. - happens when carbon particles started to form during combustion and deposited as soot as they cooled, or when there is insufficient oxygen. - characterized by a yellow flame. Example: Incomplete combustion of Propane. C 3 H 8 (g) + 4 O 2 (g) 2 CO (g) + CO 2 (g) + 4 H 2 O (g) 2. Petroleum (Crude Oil): - fossil fuels that consist mainly of heavier alkanes along with small amounts of aromatic hydrocarbons, and organic compounds that contain sulfur, oxygen and nitrogen. - gasoline is composed of 40% of crude oil, whereas natural gas is composed of only 10%. Fractional Distillation: - a method of heating crude oil in a tall column to separate its different components by their different boiling points. - lighter alkanes in the natural gas will rise up to the top of the column because of their low boiling points. - the heavier, fuel and lubricating oils will boil off at the bottom of the column due to their high boiling points. Petroleum Refining: - a process to isolate different types of fuel from crude oil using fractional distillation or cracking. Cracking: - a chemical process whereby bigger alkanes are broken up into smaller ones using a catalyst and heat. - since gasoline and natural gas only consists of 50% of crude oil, cracking is necessary to convert heavier fuel to more common fuel used in today s world. Example: The Cracking of Hexadecane. C 16 H H 2 catalyst and heat C 8 H 18 + C 8 H 18 Page 152. Copyrighted by Gabriel Tang B.Ed., B.Sc.

19 Reforming: - a chemical process where smaller alkanes are combined together and hydrogen is removed to form heavier alkanes or changed unbranched alkanes into branched alkanes. - branched alkanes are easier to burn and has a higher octane value in gasoline. (isooctane or 2,2,4-trimethylpentane has the best octane rating assigned as 100) 3. Coal: - a carbon-based mineral consists of very dense hydrocarbon ring compounds with high molar masses. - leaves a lot of soot and burns incompletely. - usually contains 7% sulfur and when combusted with oxygen gives off SO 2 and SO 3, which is the main source of air pollution and acid rain. Greenhouse Effect: - the emission of greenhouses gases that traps more of the sun s radiant (heat) energy in the atmosphere than it occurs naturally. Greenhouses Gases: - man-made and naturally occur gases that contribute to the Greenhouse Effect. 1. Carbon dioxide (CO 2 ) is released to the atmosphere when solid waste, fossil fuels (oil, natural gas, and coal), and wood and wood products are burned. 2. Methane (CH 4 ) is emitted during the production and transport of coal, natural gas, and oil. Methane emissions also result from the decomposition of organic wastes in municipal solid waste landfills, and the raising of livestock. 3. Nitrous oxide (N 2 O) is emitted during agricultural and industrial activities, as well as during combustion of solid waste and fossil fuels. Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 153.

20 Honour Chemistry 4. Hydrofluorocarbons (HFCs), Perfluorocarbons (PFCs), and Sulfur Hexafluoride (SF 6 ) are very powerful greenhouse gases that are not naturally occurring that are generated in a variety of industrial processes. Each greenhouse gas differs in its ability to absorb heat in the atmosphere. HFCs and PFCs are the most heat-absorbent. Methane traps over 21 times more heat per molecule than carbon dioxide, and nitrous oxide absorbs 270 times more heat per molecule than carbon dioxide. Often, estimates of greenhouse gas emissions are presented in units of millions of metric tons of carbon equivalents (MMTCE), which weights each gas by its GWP value, or Global Warming Potential. (Information from US. EPA) - Automobiles and Major Transportations account for 34% of CO 2 emissions globally (Power Plants contributes 33%; Major Industries and Home Heating contribute the remaining 33%). - Presently 89% of Energy Productions involve the burning of Fossil Fuels (Coal, Petroleum, Natural Gas and Biomass). - Heat and Electricity generated from combustion of fossil fuel is at most 30% efficient. (Data from University of Michigan: The Environmental Effect of Using Fossil Fuel: (Greenhouse Effect) 1. Global Warming: - the warming of global temperature due to an increased of greenhouse gases in the atmosphere. 2. Rise of Water Level: - low-lying islands and coastal area are endangered as polar icecaps melt due to the rise of temperature as a result of the greenhouse effect. 3. Unpredicted and Erratic Climate: - greenhouse effect is related to droughts and dry whether in many parts of the world. 4. Deforestation: - another cause of an increased in CO 2 level in the atmosphere. As forests disappeared, there is a lack of plants to absorb carbon dioxide using photosynthesis. - also causes mud and landslides, demineralization of the soil, lost animal habitats and extinction, destruction of entire ecosystems. Plants that may have important medicinal values can also be destroyed. Alternate Energy Sources without the Emission of Greenhouse Gas 1. Solar Energy: - the most efficient energy source where energy from the sun is converted directly to electricity through the use of photovoltaic cells (solar panels) or heat using high efficient insulated glass and an effective water heating system. - technology exists but fairly expensive; requires many solar panels to generate adequate amount of electricity. 2. Wind Power: - the use of wind turbines to generate electricity. - very efficient and extremely economical, but location specific and not very reliable when there is no wind. - can disrupt migratory routes of birds (they get caught in the turbine), aesthetic problems for the various landscapes. Page 154. Copyrighted by Gabriel Tang B.Ed., B.Sc.

21 3. Geothermal Power: - the use of underground steam to generate electricity. - very efficient and reliable, but location specific. - geothermal power is widely use in Iceland where it is sitting on the Atlantic ridge and there are lots of hot springs. 4. Tidal Power: - the use of tidal current to generate electricity. - very efficient and somewhat reliable, but location specific. - tidal power involves placing electric turbines at a narrow mouth of a channel where normal tides can cause bigger current and quick rise in water levels. It is being used in the Bay of Fundy at Nova Scotia, Canada and Kvalsund at the Arctic tip of Norway. - tidal power can sometimes disrupt migratory routes of marine species. 5. Hydroelectricity: - the use of dam and reservoir to turn electric turbines as water falls from a higher level to the spillway (potential energy converted to kinetic energy to electricity). - very efficient and no emission of greenhouse gas. - location specific and very expensive to built. The reservoir flooding can destroy ecological habitats and force migrations of people from towns and villages (Aswan Dam in Egypt and the Three Gorges Dam in China displaced thousands of people and submerged ancient cities). The presence of the dam can disrupt aquatic migratory routes as well. - dams have a limited life span (the collection of silt and mud at the bottom of the dam has to be clear periodically to maintain the structural integrity of the dam). Dams can burst during earthquakes or poor maintenance. Flash flooding of towns along spillway is always a danger. 6. Hydrogen Fuel: - burning hydrogen to form water and generate heat and electricity. - very efficient and zero pollution. - hydrogen is very explosive and technologies are still needed for supplying and storing hydrogen safely in automobiles and homes. Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 155.

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