Unit I – 1 Thermochemistry PDF

Unit I – 1 1. Thermochemistry Chapter 6 of Chemistry (15th Ed., J. Overby, Mc Graw Hill). Chapter 6 focuses on Thermochemistry, which is the study of how thermal energy changes during chemical reactions Thus, you will learn how to measure or calculate the energy released by chemical reactions or th...

Unit I – 1 1. Thermochemistry Chapter 6 of Chemistry (15th Ed., J. Overby, Mc Graw Hill). Chapter 6 focuses on Thermochemistry, which is the study of how thermal energy changes during chemical reactions Thus, you will learn how to measure or calculate the energy released by chemical reactions or the work they can do under different conditions. 1.1 Energy conversions associated with chemical reactions (Section 6.2) Most chemical reactions absorb or release thermal energy. This is why the most common form of energy transfer you will deal with in this course is heat, the flow of thermal energy associated with temperature differences across the boundaries between two compartments. In order to track the transfer of thermal energy it is necessary to define the boundaries that energy will cross. System: Surroundings: Universe: If the process of interest is a chemical reaction, the reactants consumed and the products formed in that process represent the system while the solvent, the vessel, the bath, the stirrer, etc. represent the surroundings. First law of thermodynamics: Energy cannot be created or destroyed but can only be converted or transferred. The total amount of energy in the universe is constant.  Unit I – 2 Three types of systems can be defined: Open, if it can exchange energy and mass Closed, if it can exchange only energy Isolated, if it cannot exchange mass or energy Units of energy 4.184 𝐽 (𝑒𝑥𝑎𝑐𝑡) = 1 𝑐𝑎𝑙 1000 𝑐𝑎𝑙 = 1 𝐶𝑎𝑙 1 𝑘𝐽 = 1000 𝐽 Thermodynamic quantities consist of 2 parts: A number indicating the magnitude. A sign indicating the direction of the flow from the system’s point of view.  Unit I – 3 1.3 Introduction to Thermodynamics (Section 6.3) Thermochemistry studies the heat exchanges occurring during chemical reactions. To do this, it compares the state of a system before and after a process. The state of a system is defined by all relevant macroscopic properties like energy, pressure, composition, temperature, and volume. These are called state functions because they are determined only by the state of the system regardless of how that condition was achieved. This implies that the change of any state function X during a process is given simply the difference of that function in the final and initial state. ∆𝑋 = 𝑋𝑓 − 𝑋𝑖. Examples: The validity of this principle (state functions depending only on the states of the final and initial states) is key to thermodynamics because thermodynamics is always interested in changes in energy, not in absolute energy values since they are impossible to calculate with any accuracy.  Unit I – 4 1.5 Calorimetry (Section 6.5) Specific Heat and Heat capacity. The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of that substance by 1 °C. As seen in the table below, the specific heat can also be expressed per mol of compound (molar specific heat). Molar specific heats of selected substances Au C (graphite) C (diamond) Fe H2O 0.129 J/g·°C 0.720 J/g·°C 0.502 J/g·°C 0.444 J/g·°C 4.184 J/g·°C 25.40 J/mol 8.65 J/mol of C 6.03 J/mol of C 24.80 J/mol 75.38 J/mol The specific heat is used for pure compounds or for substances with a homogeneous composition. For example, the specific heat of ethanol (a compound), a sample of gasoline (a mixture of compounds), or a sample of brass (a metallic alloy) are fixed values, independent of which portion of the sample was analyzed (every gram of the sample has the same composition). The heat capacity (C) is used for equipment consisting of several parts made of different materials. For example, a constant-pressure calorimeter consists of a container (glass, Styrofoam, or metal), a thermometer (glass), a stirrer (glass, metal), and a lid (plastic). Since the composition of this apparatus is heterogeneous, its ability to absorb heat depends on which part is being considered. For this reason, the heat capacity is referred to the entire apparatus, not to each gram of it. Thus, the heat capacity of an object is the heat required to warm it up by 1 °C. Clearly, changing one part of the apparatus (e.g., the stirrer) will affect the heat capacity of the entire apparatus. Thus, the heat capacity of any object in contact with the system (e.g., the calorimeter where the reaction occurs), must be determined before the thermochemistry of a reaction can be studied (without modifying the elements of the calorimeter. Problem 1.9: Consider an exothermic reaction occurring in an apparatus with a heat capacity (C) of 75.5 𝑘𝐽/°𝐶. The heat released by the reaction is transferred to the apparatus and causes its temperature to increase from 23.12 °C (𝑡𝑖 ) to 27.49 °C (𝑡𝑓 ). Calculate the heat absorbed by the apparatus. Problem 1.10: When 15.0 g of NH4Cl is dissolved in 100. mL of water at 24.0 °C, its temperature drops to 14.5 °C. Calculate the heat lost by the water.  Unit I – 5 Problem 1.11: When 3.50 g of HCl(g) is dissolved in 5.55 mol of water at 20.0 °C, its temperature increases to 37.2 °C. Calculate the heat absorbed by the water. 1.4 Determining the Heat Released/Absorbed by Chemical Reactions (Sect. 6.4) Chemical reactions can be carried out under different experimental conditions. In this course, we will focus on the two most common conditions: constant pressure and constant volume. Constant Pressure Constant Volume Apparatuses that allow measuring the heat transferred by reactions at constant volume are called bomb calorimeters. Bomb calorimeters are usually used to study combustion reactions (which are all exothermic). The sample is placed inside a sealed capsule (closed system) surrounded by O2 to support the combustion and the heat released is absorbed by a water bath in which the bomb is immersed. The temperature increase of the water can be used to calculate the heat released by the reaction using the principles of Calorimetry. A bomb calorimeter  Unit I – 6 Constant-Volume Calorimetry. In a bomb calorimeter, exothermic reactions occur within a sealed steel capsule (the bomb). While the combustion products remain in the capsule, the heat is able to escape the reaction compartment because of the temporary temperature gradient that ensues. The heat is thus transferred to the surrounding water. Based on the principle of conservation of energy, all energy exiting the reaction compartment must enter the surrounding compartment. system surroundings Thus: 𝑞𝑟𝑥𝑛 = −𝑞𝑠𝑢𝑟𝑟 The – sign is placed because the same amount of heat is viewed from two different points of view: heat exiting the system must be negative, but that same amount of heat must be positive since it is described as entering the surroundings. Which elements of the surroundings can absorb that heat in a bomb calorimeter? If you consider the sketch of the bomb calorimeter, the water is the major sink of heat because of its large volume and because of its large specific heat. Other elements that should be considered are the thermometer, the stirrer, the capsule, the ignition wires, and the vessel holding the water. These elements do not change for a given instrument and they are often supplied by the manufacturer with an accurately measured heat capacity (𝐶𝐶𝑎𝑙 ). However, the water is added by the user and its volume can therefore change. Thus the water and the hardware of a bomb calorimeter are usually considered separately. system surroundings Thus: 𝑞𝑟𝑥𝑛 = −(𝑞𝑤𝑎𝑡𝑒𝑟 + 𝑞𝑐𝑎𝑙 ) 𝑞𝑟𝑥𝑛 = −[(𝑠𝑤 ∙ 𝑚𝑤 ∙ ∆𝑡) + (𝐶𝑐𝑎𝑙 ∙ ∆𝑡)] Other times, problems might give the overall heat capacity of the apparatus including the water. In this case, the previous equation is simplified to: 𝑞𝑟𝑥𝑛 = −𝐶𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑐𝑎𝑙 ∙ ∆𝑡  Unit I – 7 Problem: 1.12 When 1.010 g of sucrose (C12H22O11) undergoes combustion in a bomb calorimeter, the temperature rises from 24.922 °C to 28.331 °C. Knowing that the overall heat capacity of the calorimeter, determined in a separate experiment, is 4.901 kJ/°C, find ∆𝑈𝑐𝑜𝑚𝑏 of sucrose in kJ/mol sucrose. [–5662 kJ/mol] Problem 1.13: When 1.550 g of liquid hexane (C6H14) underwent combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. The calorimeter contained 1.000 kg of water and its heat capacity, determined in a separate experiment, was 1.546 kJ/°C. Find ∆𝑈𝑐𝑜𝑚𝑏 of hexane in kJ/mol hexane. [–3905 kJ/mol] Problem 1.14 The combustion of toluene (C7H8) has a ∆𝑈𝑐𝑜𝑚𝑏 of − 3.91 × 103 𝑘𝐽/𝑚𝑜𝑙. When 1.55 g of toluene underwent combustion in a bomb calorimeter, the temperature rose from 23.12 °C to 37.57 °C. Find the heat capacity of the calorimeter knowing that it contained 750. mL of water. [1.414 kJ/°C]  Unit I – 8 Problem 1.15 1.435 𝑔 of naphthalene (𝐶10 𝐻8 ) was burned in a bomb calorimeter. Consequently, the temperature of the water rose from 20.17 𝐶 to 25.84 𝐶. If the mass of water surrounding the calorimeter was exactly 2000. 𝑔 and the heat capacity of the bomb calorimeter was 1.80 𝑘𝐽/(°𝐶), calculate the molar heat of combustion of naphthalene. The specific heat capacity for water is 4.184 𝐽/𝑔 ∙ °𝐶. [−5.14 × 103 𝑘𝐽/𝑚𝑜𝑙] Apparatuses that allow measuring the heat transferred by reactions at constant pressure, ∆𝐻, are called constant A coffee-cup calorimeter pressure calorimeters. They are generally low-tech devices sometimes constructed from two Styrofoam coffee-cups (in this case called coffee-cup calorimeters). Just like in the bomb calorimeter, a thermometer measures the temperature change experienced by the solution and this information is used to deduce the amount of heat the reaction transferred to or absorbed from the water, i.e., ∆𝑯. Note that, unlike bomb calorimeters, Styrofoam coffee-cups are not sealed. This implies that, if they were used to carry out a reaction producing gases, they would escape the vessel together with the thermal energy they have. For example, any hot H2 produced by the reaction: HCl(aq) + Mg(s) → MgCl2(aq) + H2(g) + heat) would escape the system and would remove some heat that would go unaccounted for. For this reason, such calorimeters are not suitable to study reactions that generate gases. Rather, they are generally used to investigate the thermochemistry of aqueous reactions not involving gases.  Unit I – 9 Constant-Pressure Calorimetry. The reactions studied under these conditions are generally aqueous reactions (no gases involved). This implies that the system (the reactants and products) and the surroundings (the water in which they are dissolved) are not physically separated as is the case in bomb calorimeters. Nevertheless, heat exchanges in these reactions follow the same equations that apply to constant-volume calorimetry (Eq. 23 rewritten as Eq. 26 or 27). 𝑞𝑟𝑥𝑛 = −(𝑞𝑠𝑜𝑙 + 𝑞𝑐𝑎𝑙 ) 𝑞𝑟𝑥𝑛 = −[(𝑠𝑠𝑜𝑙 ∙ 𝑚𝑠𝑜𝑙 ∙ ∆𝑡𝑠𝑜𝑙 ) + (𝐶𝑐𝑎𝑙 ∙ ∆𝑡𝑐𝑎𝑙 )] The specific heat of the solution is generally known or, if the reacting solutions are generally diluted, is often approximated to that of water (4.184 J/°C·g). If a Styrofoam coffee-cup is used as a vessel, the heat capacity of the container is generally neglected. Have you noticed how quickly a Styrofoam coffee cup warms up once you put hot water (~75 °C)? This indicates that a very small amount of heat is required to warm up the walls of the container from 25 °C to 75 °C. In this case, the equation above becomes 𝑞𝑟𝑥𝑛 = −𝑠𝑠𝑜𝑙 ∙ 𝑚𝑠𝑜𝑙 ∙ ∆𝑡 Remember that, since these reactions occur at constant pressure, 𝒒𝒓𝒙𝒏 = ∆𝑯. Problem 1.16: 50.0 mL of 0.500 M HCl(aq) were mixed with 50.0 mL of 0.500 M NaOH(aq) in a coffee-cup calorimeter. The initial temperature of both solutions was 22.50 °C and the mixture reached a temperature of 25.84 °C. The density and the specific heat of the solutions can be approximated to those of water. Calculate the enthalpy change associated with the reaction below in kJ/mol of HCl. [–55.9 kJ/mol HCl] HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)  Unit I – 10 Problem 1.17: 75.0 mL of 0.450 M HNO3(aq) were mixed with 75.0 mL of 0.450 M NaOH(aq) in a coffee-cup calorimeter. The initial temperature of both solutions was 20.12 °C and the mixture reached a temperature of 23.19 °C. The density and the specific heat of the solutions can be approximated to those of water. Calculate the enthalpy change associated with the reaction below in kJ/mol of HNO3. [–57.1 kJ/mol HNO3] HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) If the container has a non-negligible and unknown heat capacity, it must be determined by carrying out a reaction whose heat exchange is known. Problem 1.18: 100. mL of 2.00 M HNO3(aq) were mixed with 100. mL of 2.00 M NaOH(aq) in a constant-pressure calorimeter consisting of a glass beaker surrounded by a metal canister. The initial temperature of both solutions was 25.0 °C and the mixture reached a temperature of 36.48 °C. The density and the specific heat of the solutions were 1.00 g/mL and 4.2 J/°C·g. Determine the heat capacity of the calorimeter using any useful data from the previous problem. [155 J/°C]  Unit I – 11 Problem 1.19 A coffee-cup calorimeter is used to determine the heat of the following reaction for the acid-base neutralization CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) ∆𝐻𝑛𝑒𝑢𝑡 = When we add 20.00 mL of 0.625 M NaOH at 21.400 °C to 30.00 mL of 0.500 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 24.347 °C. The heat capacity of the calorimeter has been previously determined to be 27.8 J/°C. Assume that the specific heat of the mixture is the same as that of water, 4.184 J/g°C, and that the density of the mixture is 1.02 g/mL. a) Calculate amount of heat given off in the reaction. [–0.711 kJ] b) Complete the thermochemical equation above. [–56.9 kJ] Problem 1.20 When 1.00 𝐿 of 1.00 𝑀 𝐵𝑎(𝑁𝑂3 )2 solution at 25.0 °𝐶 is mixed with 1.00 𝐿 of 1.00 𝑀 𝑁𝑎2 𝑆𝑂4 solution at 25.0 °𝐶, 𝐵𝑎𝑆𝑂4 forms as a white precipitate and the temperature in the constant pressure calorimeter rises to a final temperature of 28.1 °𝐶. Assume that no heat is absorbed by the calorimeter and that the density and specific heat of the solutions are the same as for water. Write a complete thermochemical equation for this process including its ∆𝐻𝑟𝑥𝑛 value.  Unit I – 12 Thermochemical equations. A balanced equation that includes the corresponding enthalpy change (the heat released or absorbed) associated with the reaction it describes is called a thermochemical equation. Ex.: C2H5OH(l) + 3O2(g) 3 H2O (l) + 2 CO2(g) ∆𝐻 = –1367 kJ/mol Conventions: The enthalpy change refers to the chemical equation as written. Thus the notation “–1367 kJ/mol” refers to 1 mol of reaction, that is to the number of moles of reactants and products that appear in the equation as written. This means: ▪ The consumption of 3 mol of O2 through reaction with C2H5OH releases 1367 kJ of heat or the consumption of 1 mol of O2 through reaction with C2H5OH releases 456 kJ of heat ▪ The production of 3 mol of H2O from the combustion of C2H5OH releases 1367 kJ of heat or the production of 1 mol of H2O from the combustion of C2H5OH releases 456 kJ of heat ▪ The production of 2 mol of CO2 from the combustion of C2H5OH releases 1367 kJ of heat or the production of 1 mol of CO2 from the combustion of C2H5OH releases 384 kJ of heat The coefficients in a thermochemical equation refer to number of moles, not number of molecules. It follows that it is acceptable to use fractional coefficients The physical state of all species must be specified The value of ∆𝐻 does not change significantly with moderate changes of temperature If bearing the same coefficients, the chemical equation for the reverse reaction is associated with a ∆𝐻 equal in magnitude but opposite in sign to the ∆𝐻 for the direct reaction. The notation ∆𝐻 is often followed by a description of the reation it refers to (ex.: ∆𝐻𝑛𝑒𝑢𝑡 , ∆𝐻𝑐𝑜𝑚𝑏 , ∆𝐻𝑑𝑖𝑠𝑠 , ∆𝐻𝑓𝑢𝑠 , ∆𝐻𝑣𝑎𝑝 , ∆𝐻𝑑𝑖𝑙 , ∆𝐻𝑓 , etc.) Problem 1.21 When 2.61 g of dimethyl ether, CH3OCH3, is burned at constant pressure, 82.5 kJ of heat is given off. Complete the following thermochemical equation. [–1460 kJ/mol] CH3OCH3(l) + 3 O2(g) 2 CO2(g) + 3 H2O(l) ∆𝐻𝑐𝑜𝑚𝑏 =  Unit I – 13 Problem 1.22 A liquid propane tank in a home barbeque contains 13.2 kg of propane, C3H8. a) Calculate the heat (in kJ) associated with the complete combustion at constant pressure of all the propane in the tank. [–6.12×105 kJ] C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ∆𝐻 = −2044 𝑘𝐽/𝑚𝑜𝑙 b) Calculate the heat (in kJ) associated with the complete combustion of 47.9 kg of O2 according to the reaction above. [–6.12×105 kJ] Problem 1.23 Ammonia reacts with O2 according to the following reaction: 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) ∆𝐻 = −902.0 𝑘𝐽/𝑚𝑜𝑙 Calculate the heat (in kJ) associated with the complete reaction at constant pressure of 155.0 g of NH3. [−2052 𝑘𝐽]  Unit I – 14 Problem 1.24: Some cold packs take advantage of the fact that dissolving ammonium nitrate in water is an endothermic process. The addition of 5.44 g NH4NO3(s) to 150.0 g of water in a coffee- cup calorimeter (with stirring to dissolve the salt) resulted in a decrease in temperature from 18.6 °C to 16.2 °C. Complete the thermochemical equation below. Assume that the solution (whose mass is 155.4 g) has a specific heat capacity of 4.23 J/g·°C. [23.0 kJ/mol] NH4NO3(s) NH4+(aq) + NO3–(aq) Hdiss. = Problem 1.25: Sucrose (table sugar, C12H22O11) can oxidize to CO2(g) and H2O(l): C12H22O11(s) + 12 O2(g) 12 CO2(g) + 11 H2O(l) Hrxn = –5645 kJ/mol What is the enthalpy change for the oxidation of 5.00 g (1 teaspoonful) of sugar? [–82.5 kJ] Problem 1.26: Almost insoluble AgCl(s) precipitates when aqueous AgNO3 solution is mixed with NaCl solution: Ag+(aq) + Cl–(aq) AgCl(s) Hrxn = When 250 mL of 0.16 M AgNO3 solution and 125 mL of 0.32 M NaCl solution are mixed in a coffee-cup calorimeter of negligible heat capacity, the temperature rises from 21.15 °C to 22.90 °C. Complete the thermochemical equation above. Assume the density of the solution is 1.0 g/mL and its specific heat capacity is 4.2 J/gK. [–68.9 kJ]  Unit I – 15 Problem 1.27 Reaction of 1 mol of H2(g) with O2(g) to form liquid water is represented by the equation. H2(g) + 1/2 O2(g) H2O(l) Hrxn = –285.8 kJ/mol What is the enthalpy change accompanying the decomposition of 12.6 g of liquid water to hydrogen and oxygen gases? [200. kJ] Problem 1.28 Assume you mix 100.0 mL of 0.200 M CsOH solution with 50.0 mL of 0.400 M HCl solution in a coffee cup calorimeter of negligible heat capacity. The following reaction occurs: OH–(aq) + H+(aq) H2O(l) Hneut = The temperature of both solutions before mixing was 22.50 °C, and it rises to 24.28 °C after the acid-base reaction. Complete the thermochemical equation above. Assume the densities of the solutions are all 1.00 g/mL and the specific capacities of the solutions are 4.2 J/gK. [–56 kJ] Problem 1.29 Calcium carbide (CaC2) is manufactured by the reaction of CaO with carbon at high temperature. CaO(s) + 3 C(s) CaC2(s) + CO(g) Hrxn = 464.8 Kj/mol Is this reaction endo- or exothermic? If 6.50 g of C is allowed to react with an excess of CaO, how much heat is absorbed or evolved by the reaction? [83.8kJ]  Unit I – 16 Problem 1.30 Acetic acid (CH3COOH) is made industrially by the reaction of methanol and carbon monoxide: CH3OH(l) + CO(g) CH3COOH(l) Hrxn = –355.9 kJ/mol If you produce 1.00 L of acetic acid by this reaction, what quantity of heat is evolved? (1 mL of CH3 COOH weighs 1.044 g) [–6.187103 kJ] Problem 1.31 Consider the dissolution of CaCl2(s) CaCl2(s) Ca2+(aq) + 2 Cl–(aq) Hdiss = – 81.5 Kj/mol An 8.50 g sample of CaCl2 is dissolved in 125.0 g of water with both substances at 25.0 °C. Calculate the final temperature of the solution, assuming no heat is lost to the surroundings. Also, assume that the heat capacity of the solution is 4.184 J/g°C. [36.2 °C]  Unit I – 17 Problem 1.37 A nutritionist determines the caloric value of a 10.00-g sample of beef fat by burning it in a bomb calorimeter. The calorimeter held 2.500 Kg of water, the heat capacity of the bomb is 1.360 kJ/°C, and the temperature of the calorimeter increased from 25.0 °C to 56.9 °C. a)Calculate the number of joules released per gram of beef fat. [37.7 kJ/g] b) One nutritional Calorie is 4184 joules. What is the caloric value of beef fat, in nutritional Calories/gram? [9.01 Cal] Problem 1.38 The combustion of 0.0222 g of isooctane vapor, C8H18(g), at constant volume raises the temperature of a bomb calorimeter by 0.400 °C. The heat capacity of the calorimeter and water combined is 2.48 kJ/°C. a)Find the molar heat of combustion of gaseous isooctane. [5104 kJ/mol] b)How many grams of C8H18(g) must be burned to obtain 495 kJ of heat? [11.1 g]  Unit I – 18 Mixed problems Problem 1.39 Methylhydrazine is burned with dinitrogen tetroxide in the attitude-control engines of the space shuttles. CH6N2(l) + 5/4 N2O4(l) CO2(g) + 3 H2O(l) + 9/4 N2(g) The two substances ignite instantly on contact, producing a flame temperature of 3000 K. The thermal energy liberated by 0.100 g of CH6N2 at constant temperature (25 °C) and atmospheric pressure is 750. J. a) Calculate H for the reaction as written. [–346 kJ] b) How many kilojoules are liberated when 87.5 g of N2 is produced? [–480. kJ] Problem 1.40 Consider the dissolution of CaCl2(s) CaCl2(s) Ca2+(aq) + 2 Cl–(aq) Hdiss = – 81.5 Kj/mol An 8.50 g sample of CaCl2 is dissolved in 125.0 g of water with both substances at 25.0 °C. Calculate the final temperature of the solution, assuming no heat is lost to the surroundings. Also, assume that the heat capacity of the solution is 4.184 J/g°C. [36.2 °C] Problem 1.41 A hypothetical exothermic reaction has released 3.358 kJ of heat to a calorimeter that contains 50.00 g of water. The temperature of the water and the calorimeter, originally at 4.184 𝐽 22.34 °C, increases to 36.74 °C. Calculate the heat capacity of the calorimeter. 𝑠𝐻2 𝑂 = °𝐶∙𝑔 [24.0 J/°C] Problem 1.42 A hypothetical endothermic reaction has absorbed 3.358 kJ of heat from the same calorimeter used above, containing 50.00 g of water. The temperature of the water and the calorimeter, originally at 22.34 °C, decreased to 7.94 °C. Calculate the heat capacity of the 4.184 𝐽 calorimeter. 𝑠𝐻2 𝑂 = [24.0 J/°C] °𝐶∙𝑔 Problem 1.43 A student wishes to determine the heat capacity of a coffee cup calorimeter. After she mixes 100.0 g of water at 58.5 °C with 100.0 g of water, already in the calorimeter, at 22.8 °C, the final temperature of the water is 39.7 °C. 4.184 𝐽 Calculate the heat capacity of the calorimeter in J/°C. 𝑠𝐻2 𝑂 = ; [47.0 J/°C] °𝐶∙𝑔  Unit I – 19 Problem 1.44 25.0 g of water at 83.5 °C were added to 34.2 g of water at 15.0 °C. Determine the temperature of the final mixture assuming that the heat capacity of the container was negligible. [43.9 °C] Problem 1.45 A sheet of gold weighing 10.0 g and at a temperature of 18.0 °C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 °C. What is the final temperature of the combined metals? (Assume that all heat lost by the gold sheet is absorbed by the iron sheet with no losses). [50.8 °C] 0.129 𝐽 0.444 𝐽 𝑠𝐴𝑢 = ; 𝑠𝐹𝑒 = °𝐶∙𝑔 °𝐶∙𝑔 Problem 1.46 A coffee cup calorimeter is used to determine the specific heat of a metallic sample. The calorimeter is filled with 50.0 mL of water at 25.0 °C (𝑑𝐻2 𝑂 = 0.997 g/mL). A 36.5-g sample of the metallic material is taken from water boiling at 100.0 °C and placed in the calorimeter. The equilibrium temperature of the water and sample is 32.5 °C. The heat capacity of the calorimeter is known to be 1.87 J/°C. Calculate the specific heat of the metallic sample. [0.66 J/g°C] Problem 1.47 A 5.1-gram piece of gold jewelry is removed from water at 100.0 °C and placed in a coffee-cup calorimeter containing 16.9 g of water at 22.5 °C. The equilibrium temperature of the water and jewelry is 23.2 °C. The heat capacity of the calorimeter is 1.54 J/°C. What is the specific heat of this piece of jewelry? The specific heat of pure gold is 0.129 J/°Cg. Is the jewelry pure gold? [Yes, it is gold] Problem 1.48 A 44.0-g sample of an unknown metal at 99.0 °C was placed in a constant- pressure calorimeter containing 80.0 g of water at 24.0 °C. The final temperature of the system was found to be 28.4 °C. The heat capacity of the calorimeter is 12.4 J/°C. Calculate the specific heat of the metal. [0.49 J/g°C]  Unit I – 20 1.6 Standard Enthalpy of Formation and Reaction (Section 6.6) Many of the heat values we have calculated in the past problems have been determined experimentally. In theory, since ∆𝐻 = 𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 , they could also have been calculated if we knew the enthalpy of the reactants and products. However, there is no way to determine the absolute 𝐻 value of a substance. In this situation is useful to set an arbitrary but consistent reference against which eth enthalpies of all compounds should be compared. This is the same approach adopted by geographers when they decided that all altitudes would be measured from the sea level to which it was assigned an altitude of 0 m. This standard reference has three parts: 1. the standard state ° 2. the standard enthalpy change (∆𝐻𝑟𝑥𝑛 ) 3. the standard enthalpy of formation (∆𝐻𝑓° ) also called standard heat of formation 1. Standard State: ▪ For a gas: the standard state of a gas is the pure gas at 1 atm. ▪ For a liquid or solid: the standard state of a liquid or solid is the pure substance in its most stable allotropic form at 1 atm and at the temperature of interest. Although the enthalpy change of a reaction depends on the temperature at which it takes place, the standard state does not imply a specific temperature. Rather, there is a different value of ° ∆𝐻 ° at each temperature (for example ∆𝐻380 indicates the enthalpy change at standard conditions and 380 K). However, textbooks refer consistently refer to a temperature of 25 °C unless otherwise stated. 60 C(graphite) + 60 O2(g) → 60 CO2(g) ∆𝐻 = −23610 𝑘𝐽/𝑚𝑜𝑙 60 C(diamond) + 60 O2(g) → 60 CO2(g) ∆𝐻 = −23732 𝑘𝐽/𝑚𝑜𝑙 C60(fullerene) + 60 O2(g) → 60 CO2(g) ∆𝐻 = −25930 𝑘𝐽/𝑚𝑜𝑙 H 60 CO2  Unit I – 21 ▪ For a substance in solution: The standard state for a substance in solution is a concentration of 1 M. ° 2. Standard Enthalpy Change of a reaction (∆𝐻𝑟𝑥𝑛 ): ▪ This is the change in enthalpy for a process when all reactants and products are in their standard states. This condition is indicated with the symbol “°” (pron. naught). 𝑘𝐽 CaCO3(s, calcite) CaO(s) + CO2(g, 1 atm) 𝐻° = 178 𝑚𝑜𝑙 3. Standard Enthalpy of Formation (∆𝐻𝑓° ): ▪ For a pure compound: the standard enthalpy of formation of a pure compound is the heat associated with the formation of 1 mol of compound from its constituent elements in their standard states. ▪ For a pure element in its standard state: ∆𝐻𝑓° = 0 𝑘𝐽 A reaction of formation is one in which a compound is formed from its elements. Its equation is always written for the formation of 1 mol of the compound of interest. Ex: C(graphite) + O2(g) CO2(g) ∆𝐻𝑓° (CO2, g) = –393.5 kJ/mol Na(s) + ½ Cl2(g) NaCl(s) ∆𝐻𝑓° (NaCl, s) = –411.12 kJ/mol ∆𝐻𝑓° (C2H5OH, l) = –277.0 kJ/mol ∆𝐻𝑓° (NH3, g) = –46.3 kJ/mol  Unit I – 22 How to predict the Standard Enthalpy Change of a reaction (∆𝑯°𝒓𝒙𝒏 ) The Indirect Method This method is based on Hess’s law of heat summation: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. This is because H is a state function and ∆𝐻 ° depends only on the initial and final state of the system (i.e., on the nature of the reactants and products). Thus, if we can break down a reaction with unknown ∆𝐻 ° in a series of reactions with known ∆𝐻 ° , we can add up the known ∆𝐻 ° values to calculate the unknown ∆𝐻 ° data. When adding reactions the following rules apply: 1. If a reaction is reversed, the sign of H is also reversed. N2(g) + O2(g) → 2 NO(g) ∆𝐻° = 180 𝐾𝑗/𝑚𝑜𝑙 2 NO(g) → N2(g) + O2(g) ∆𝐻° = −180 𝐾𝑗/𝑚𝑜l 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 𝐾𝑗 6 NO(g) → 3 N2(g) + 3 O2(g) ∆𝐻° = −540 𝑚𝑜𝑙 [3 × (−180 𝐾𝑗/𝑚𝑜l)] Problem 1.49. Calculate the enthalpy of formation of methane, CH4, from its elements according to the equation C(s) + 2 H2(g) → CH4(g)∆𝐻𝑓° = ? from the following data: C(s) + O2(g) → CO2(g) ∆𝐻° = – 393.5 𝐾𝑗/𝑚𝑜𝑙 H2(g) + ½ O2(g) → H2O(l) ∆𝐻° = – 285.8 𝐾𝑗/𝑚𝑜𝑙 CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆𝐻° = – 890.3 𝐾𝑗/𝑚𝑜𝑙  Unit I – 23 Problem 1.50. From the following enthalpies of reaction: 4 HCl(g) + O2(g) 2 H2O(l) + 2 Cl2(g) H° = –202.4 Kj/mol ½ H2(g) + ½ F2(g) HF(l) H° = –600.0 Kj/mol H2(g) + ½ O2(g) H2O(l) H° = –285.8 Kj/mol Find H°rxn for 2 HCl(g) + F2(g) 2 HF(l) + Cl2(g) Problem 1.51. From the following enthalpies of reaction: CaCO3(s) CaO(s) + CO2(g) H° = –178.1 kJ/mol CaO(s) + H2O(l) Ca(OH)2(s) H° = –65.3 kJ/mol Ca(OH)2(s) Ca2+(aq) + 2 OH–(aq) H° = –16.2 kJ/mol calculate H°rxn for Ca2+(aq) + 2 OH–(aq) + CO2(g) CaCO3(s) + H2O(l)  Unit I – 24 Problem 1.52. From the following enthalpies of reaction: S(s) + O2(g) SO2(g) H° = –296.8 kJ/mol S(s) + 3/2 O2(g) SO3(g) H° = –395.6 kJ/mol determine the standard enthalpy change for the decomposition reaction 2 SO3(g) 2 SO2(g) + O2(g) Problem 1.53. 2 H2(g) + O2(g) 2 H2O(l) H° = –571.6 Kj/mol C3H4 (g) + 4 O2(g) 3 CO2(g) + 2 H2O(l) H° = –1937 Kj/mol C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) H° = –2220. Kj/mol determine the standard heat change of the hydrogenation reaction C3H4(g) + 2 H2(g) C3H8(g) Textbook problems: 6.61-6.64,6.84.  Unit I – 25 Direct Method 𝐻𝑟𝑥𝑛 ° = ∑ 𝐻 ° (products) − ∑ 𝐻 ° (reactants) but 𝐻 ° values are unknown! Workaround using the known heat of formation of products and reactants: Example: CaCO3(s, calcite) CaO(s) + CO2(g, 1 atm) 𝐻𝑟𝑥𝑛 ° =? 𝐻𝑟𝑥𝑛 ° = ∑ 𝑛𝑖 𝐻𝑓° (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝑛𝑖 𝐻𝑓° (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) valid if the reaction is carried at standard state and at temperature relatively close to 25 °C (common tables refer to a temperature of 25 °C).  Unit I – 26 Problem 1.47. Nitroglycerin is a powerful explosive that forms four different gases when detonated (which is why it has explosive power): 4 C3H5(NO3)3(l) 6 N2(g) + O2(g) + 12 CO2(g) + 10 H2O(g) Calculate the standard enthalpy change at 25 °C (𝐻𝑟𝑥𝑛 ° ) when 10.0 g of nitroglycerin is detonated. For nitroglycerin, 𝐻𝑓 at 25 °C is –364 kJ/mol. [–62.57 kJ] ° Problem 1.48. Calculate the standard molar enthalpy change of combustion 𝐻𝑐𝑜𝑚𝑏 ° of liquid benzene (C6H6) at 25 °C. [–3267 kJ] C6H6(l) + 15/2 O2(g) 6 CO2(g) + 3 H2O(l) Textbook problems: 6.53,6.54,6.56,6.57,6.58..  Unit I – 27 Calculation of The Enthalpy Change of a Reaction Through Bond Enthalpy ⬚ We have seen that the enthalpy change associated to a reaction (∆𝐻𝑟𝑥𝑛 ) can be determined: experimentally, by carrying out the reaction under constant pressure and measuring the heat the system has gained from or released to the surroundings; theoretically, by breaking down the reaction into a number of reactions whose ∆𝐻 values are known and applying Hess’s law to them; theoretically, by breaking down the reaction into a number of reactions of formation whose ∆𝐻° values can be easily found in tables (such as the TSTV distributed on Lea). As demonstrated in class, the application of Hess’s law to these reactions of formation can be summarized in the equation: ° ∆𝐻𝑟𝑥𝑛 = ∑ 𝑛𝑖 ∆𝐻𝑓° (products) − ∑ 𝑛𝑖 ∆𝐻𝑓° (reactants) Since the ∆𝐻𝑓° values refer to the formation of compounds at standard state, this approach is only applicable to the calculation of the enthalpy change of reactions carried out under standard state (i.e., starting from reactants at std state and ending with products at std state). Another approach to the calculation of ∆𝐻𝑟𝑥𝑛 is through the bond enthalpy values. Bond enthalpy is the energy that must be spent (positive value!) to break 1 mol of a particular bond (6.022×1023 bonds) in a gaseous molecule. Page 3 of the TSTV contains a list of bond enthalpy values for the most common bonds. Importantly, these are average values. The reason for this feature can be easily understood if we consider the bonds in the molecules below. H 4 C sp H C O O H H H H 1 2 3 1 2 3 4 5 F Csp Csp C 3 3 2 H H O C 2 Csp O 3 C sp3 C Csp O3 H sp sp F H H H H H A B In molecule A, breaking the C−H bonds involving C1, C2, C3 and C4 does not cost the same amount of energy. C2, C3 and C4 have different hybridization and this has an impact on the strength of the bond and therefore on the energy required to break them. Even when the hybridization is the same (C1, C2), the structural surroundings are not the same. C1 is connected to two F atoms which, being strongly electronegative, withdraw electron density from the C atom and, to a certain extent, from the H atom. Thus, this bond cannot have the same bond enthalpy of C2 where this effect is not present. The same can be said for the C−O bonds in molecule B. Here, C2, C3, C4 and C5 have  Unit I – 28 either a different structural environment or different hybridization. Therefore the enthalpy of their bonds with O can be expected to differ to some extent. For this reason, tables of bond enthalpy provide average energy values. Only when only one bond of a given type can exist (only one Br−Br bond, one H−H bond, one H−F bond or one N≡N bond exists) the bond enthalpy has more significant figures since they have are not average values. How to use bond enthalpy values A reaction can be imagined as a process in which the atoms in the reactant species completely dissociate and then recombine forming the different bonds present in the products. Although this is not the true mechanism of reactions, the calculation of ∆𝐻𝑟𝑥𝑛 (a state function) is independent of the path through which the reaction occurs so only the bonds in the reactants and in the products need to be considered and not the way in which they are broken and formed. Remember that energy spent to break bonds must enter the system and therefore must be accounted as positive while energy released when bonds are formed exits the system and therefore must be accounted as negative. To calculate the ∆𝐻𝑟𝑥𝑛 from bond enthalpy values, it is necessary to know how many bonds of each type are broken. For this reason, it is more convenient to draw the reactants and products using their Lewis structures. The energies absorbed and released, each with their appropriate sign, are then added up to give the ∆𝐻 for the reaction of interest. Below is an example (problem 9.76 of Overby). a) Calculate the ∆𝑯 for the combustion of ethane from the average bond enthalpies: C2H6(g) + O2(g) → CO2(g) + H2O(g) The equation must first be balanced so that the correct number of bonds involved can be calculated: 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) The Lewis structures are then drawn to clearly display the bonds involved. H H 2 H C C H + 7 O O 4 O C O + 6 H O H H H 12 mol C−H bonds 414 𝑘𝐽 2 mol C−C bonds 347 𝑘𝐽 ( × )+( × )+ 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 C−H bonds 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 C−C bonds 7 mol O = O bonds 498.7 𝑘𝐽 8 mol C = O bonds 799 𝑘𝐽 ( × )−( × )− 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 O = O bonds 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 C = O bonds 12 mol O−H bonds 460 𝑘𝐽 𝟐𝟒𝟓𝟗 𝒌𝑱 ( × )=− 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 O−H bonds 𝒎𝒐𝒍 𝒓𝒙𝒏  Unit I – 29 b) Calculate the ∆𝑯 for the combustion of ethane from the standard enthalpies of formation of the reactants and the products: c) 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) 𝐻𝑟𝑥𝑛 ° = ∑ 𝑛𝑖 𝐻𝑓° (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝑛𝑖 𝐻𝑓° (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) 2 mol C2 H6 (g) – 84.86 𝑘𝐽 7 mol O2 (g) 0 𝑘𝐽 ( × )+( × ) mol rxn mol C2 H6 (g) mol rxn mol O2 (g) 4 mol CO2 (g) – 393.5 𝑘𝐽 6 mol H2 O(g) −241.8 𝑘𝐽 𝟐𝟖𝟓𝟓. 𝟏 𝒌𝑱 −( × )−( × )=− mol rxn mol CO2 (g) mol rxn mol H2 O(g) 𝒎𝒐𝒍 𝒓𝒙𝒏 As you might have noticed, a discrepancy exists between the values obtained through the two approaches. The most accurate value is the one obtained from the enthalpies of formation because the latter have been directly (experimentally) or indirectly (through application of Hess’s law) obtained for a specific compound of interest. On the contrary, bond enthalpy values are average values of many different experimental values. N.B. The approach based on the bond enthalpy values can be applied only to reactions involving gases. The reason for this limitation lies in the fact that solid, liquid, and aqueous species, require extra energy to break several types of intermolecular interactions (dispersion forces, dipole-dipole interaction, dipole-ion interactions, H-bond, etc.). This energy is not accounted by the bond enthalpy approach. In order to compensate for this weakness, we would have to determine the extra energy required to overcome all these intermolecular interactions as each reactant in converted to a gas and the extra energy released when each products is converted to the condensed phase which appears in the equation. Since these energy values are rarely available, other more convenient approaches are used in these cases.