Chemistry CH 7

Chemistry CH 7 - Chapter 7 Thermochemistry Why nature gas...

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Unformatted text preview: Chapter 7 Thermochemistry Why nature gas is a better fuel than coal and why the energy value of fats is greater than that of carbohydrates and proteins? Contents 7-1 Getting Started: Some Thermnology 7-2 Heat 7-3 Heats of Reaction and Calorimetry 7-4 Work 7-5 The First Law of Thermodynamics 7-6 Heats of Reaction: U and H 7-7 Indirect Determination of H: Hess's aw 7-8 Standard Enthalpies of Formation 7-9 Fuels as Sources of Energy 7-1 Getting Started: Some Therminology Systems and their surroundings The study of thermochemistry requires the definition of system and surroundings. The system is the particular sample of matter under investigation, and the surroundings include everything else outside the system. Vacuum flask (water Matter vapor) Heat Open a system Heat Heat Closed b system b Heat Isolated c system c a Systems and their Surroundings a. Open system. The beaker of hot coffee transfers energy to the surroundings-it loses heat as it cools. Matter is also transfers in the form of water vapor b. Closed system. The flask of hot coffee transfers energy (heat) To the surroundings as it cools. Because the flask is stoppered, no water vapor escapes and no matter is transferred. c. Isolated system. Hot coffee in an insulated flask approximates an isolated system. No water vapor escapes, and for a time at least, little heat is transferred to the surroundings. Isolated System No energy in or out No matter in or out System System is the part of the universe chosen for study Boundary Isolated System: No interaction across the system boundary. (a system that does not interact with its surroundings) Work: done when a force acts through a distance Wok = force distance = [m(kg) a(m s-2)] d(m) Kinetic energy: The energy of moving object Potential energy: The Energy: The capacity Energy has the to do work potential to do work SI unit of energy: 1 joule(J) = 1 kgm2s-2 1 cal = 4.184J (exactly) (7.2) 1Cal=1000cal=1kcal Kinetic energy = m(kg) [u(m/s)]2 (7.1) 7-2 Heat Heat(q), like work, is energy transferred between a system and its surroundings as a result of a temperature difference. We will discuss the quantity of heat following three questions: Heat capacity Specific heat Law of conservation of energy The quantity of heat required to change the temperature of a system by one degree is called the heat capacity of the system. If the system is a mole of substance, we can use the term molar heat capacity. If it is one gram of substance, we call it specific heat capacity, or more commonly, Specific heat(sp ht). The specific heat of water depends some what on temperature, but, over the range from 0 to 100 C, its value is about: 1.00cal/g C =1cal g-1 C -1 = 4.18J/ g C = 4.18 g-1 C 1 (7.3) Example for calculating a quantity of heat How much heat is required to raise the temperature of 7.35g of water From 21.0 to 98.0 C? (Assume the specific heat of water is 4.18J g-1 C -1 throughout this temperature range. ) Solution The specific heat is the heat capacity of 1.00g water: 4.18J/g water C. The heat capacity of system(7.35g water) is 7.35 g water4.18J/ g water C = 30.7J/ C. The required temperature change in the system is (98.0-21.0) C = 77.0 C. The heat required to produce this temperature change = 30.7 7J/ C 77.0 C = 2.36 103J. Summarized quantity of heat = mass of substance specific heat temperature heat capacity = C (7.4) q = m specific heat T = C T (7.5) T = Tf Ti Tf is the final temperature; Ti is the niitial temperature. If the temperature of a system increases: Tf >Ti, T >0 (the heat is absorbed or gained by the system) If the temperature of a system decreases: Tf <Ti, T <0 (the heat is evolved or lost by the system) Law of conservation of energy In interactions between a system and its surroundings, The total the energy remains constant-energy is neither created nor destroyed. Apllied to the exchange of heat, This means that qsystem + qsurroundings = 0 (7.6) Heat lost by a system is gained by its surroundings, and vice versa qsystem = - qsurroundings (7.7) Experimental determination of specific heats 150.0g Lead 22.0C 28.8 C a b c a: A 150.0-g sample of lead is heated to the temperature of boiling water (100 C). b: A 50.0-g sample of water is added to a thermally insulted beaker, and its temperature is found to be 22.0C. c: The hot lead is dumped into the cold water, and the temperature of the final lead-water mixture is 28.8 C qlead =- qwater (7.8) Example for determining a specific heat from experimental data (use data presented in figure) Solution Note: if you know any foure of the five quantities-q, m, specific heat, T , Table 7.1 Specific heata of several solid elements (in J g-1C -1 ) Metals Pb Cu Fe Al Mg Sp ht Monmetals Sp ht 0.128 Se 0.321 0.385 S 0.706 0.449 P 0.903 1.024 0.777 Metalloids Sp ht Te 0.202 As 0.329 Significance: The relatively high specific heat of aluminum helps to account for its use as "miracle thaw" products designed to thaw frozen foods rapidly. Compounds tend to have higher specific heats. Water has a specific heat that is more than 30 time greater than that of lead. 7 y 3 Heats of reaction and calorimetry Thermal energy Calorimetry Chemical energy Heat of reaction Exothermic and Endothermic reactions 40.5 a: An exothermic reaction. The reactants are mixed At room temperature, but the temperature of the mixture rises to 40. 5C. CaO(s) + H2O(l) Ca(OH)2(s) a 5.8 b: An endothermic reaction. The reactants are mixed At room temperature, but the temperature of the mixture falls to 5.8C. b Ba(OH)2H2O(s)(s) 8 BaCl2 2O(s) + 2NH3(aq) + 8H2O(l) 2H Heat of reaction, qrxn, is the quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system,at constant temperature Exothermic reaction is one that produces a temperature increase In an isolated system or, in a nonisolated system, gives off heat to surroundings. (qrxn< 0) Heats of reaction are experimentally determined in a Calorimeter, a device for measuring quantities of heat. In an endothermic reaction, the corresponding situation is a temperature decrease in an isolated system or a gain of heat from the surroundings by a nonisolated system.(qrxn> 0) Bomb Calorimetry Motorized Electrical leads for igniting sample Thermometer Insulated container O2 inlet Bomb (reaction chamber) Fine wire in contact with sample Cup holding sample Water A bomb calorimeter is ideally suited for measuring the heat evolved in a combustion reaction . The system is isolated from its surroundings. when the combustion reaction occurs, chemical energy is converted to thermal energy, and the temperature of the system rises. The heat of reaction is the quantity of heat that the system would have to lose to its surroundings to be restored to its initial temperature. This quantity Of heat, in turn, is just the negative of the thermal energy gained by the calorimeter and its contents (qcalorim): Note : That the temperature of a reaction mixture usually changes during a reaction, so that we must return the mixture to the initial temperature before we assess how much heat is exchanged with surroundings. qrxn = C qcalorim(where qcalorim = qwater C qbomb C ... ) (7.9) We assemble the calorimeter in exactly the same way each time, a heat capacity of the calorimeter is defined. From qalorim, we then establish qrxn, then we can determine the heat of combution of different substances. qcalorim = heat capacity of calorim T (7.10) Example for using bomb calorimetry data to determine a heat of reaction. The combustion of 1.010 g sucrose, C12H22O11, in a bomb calorimeter causes the temperature to rise from 24.92 to 28.33C. The heat capacity of the calorimeter assembly is 4.90kJ/C. (a) What is the heat of combustion of sucrose, expressed in kilojoules per mole of C12H22O11? Solution (b) Verify the claim of sugar producers that one teaspoon (a) First we can calculate q calorim with equation (7.10). q calorim = 4.90kJ/C (28.33 24.92) C = (4.90 3.14)kJ = 16.7 kJ Now, using equation (7.9), we get qrxn = y qcalorim = 16.7 kJ This is the heat of combustion of the 1.010 g sample. Per gram C12H22O11 qrxn = -16.7kJ = -16.5 kJ/g C12H22O11 1.010 g C12H22O11 qrxn O -16.5kJ Per mol C12H22 = 11 g C12H22O11 = - 5.65 342.3 g C12H22O11 1mol C12H22O11 C12H22O11 10 3 kJ/mol (b) To determine the caloric content of sucrose, we can use heat of combustion per gram of sucrose determined in part(a), together with a factor to convert from kilojoules to kilocalories. (Because 1 cal = 4.184 J, 1 kcal = 4.184 kJ.) ? kcal = 4.8 g C12H22O11 tsp -16.5kJ g C12H22O11 1kcal tsp = -19kcal tsp 1 food Calorie (1 Calorie with a capital C) is actually 1000 cal, or 1 kcal. Therefore, 19 kcal = 19 Calories. The claim is justified. The "Coffee C Cup" Calorimeter Thermometer Glass stirrer Cork stopper Two Styrofoam cups nested together containing reactants in solution. Example for using "Coffee C Cup" calorimetry data to determine a heat of reaction. In the neutralization of a strong acid with a strong base, the essential reaction is the combination of H C (aq) and OH C (aq) to form water(recall page 148). H C (aq) C OH C (aq) H2O(l) two solutions, 100.0mL of 1.00 mol/mL HCl(aq) and 100.0 mL of 1.00 mol/mL NaOH (aq), both initially at 21.1 C, are added to a Styrofoam cup calorimeter and allowed to react. The temperature rises to 27.8 C. Determine the heat of the neutralization reaction, expressed per mole H2O formed. Is the reaction endothermic or exothermic? In addition to assuming that the calorimeter is an isolation system, assume that all there is in the system to absorb heat is 200.0 mL of Water. This assumption ignores the fact that 0.10 mol each of NaCl and H2O are formed in the reaction, that the density of the resulting NaCl (aq) is not quite 1.00 g/mL, And that its specific heat is not quite 4.18 J g C 1C 1. Also, ignore the small heat capacity of the Styrofoam cup itself. Solution Because the reaction is a neutralization reaction, let us call the heat of reaction qneutr. Now, according to equation(7.9), qneutr = C qcalorim, and if we make the assumptions described above, 1.00 g 4.18 J qcalorim = 200.0 mL (27.8 C 21.1 mL g C C) 3 = 5.6 10 J qneutr = C qcalorim = C 5.6 103 J = C 5.6 kJ In 100.0 mL of 1.00 mol/L HCl, the amount of H C is C 1.00 mol HCl ? Mol HCl H C = 0.1000 L 1 mol H 1 mol HCl 1L = 0.100 mol H C Similarly, in 100.0 mL of 1.00 mol/L NaOH there is 0.100 mol OH C . Thus, the H C and the OH C combine to form 0.100 mol H2O. (The two are in stoichiometric proportions; neither is in excess.) The amount of heat produced per mole of H2O is C 5.6kJ q neutr = = C 56 kJ/mol H2O 0.100 mol H2O 7-4 Work A moving piston does work on the surroundings. The amount of work done is: w = p v Initial state Cross-section Area=A Final state Note: Pressure is a force per unit area (p = F/A), which means that the product of a pressure and an area is a force (p = F A). The product in the expansion, v . It is the "volume" part of pressure-volume work Now we can use equation (7.1) to calculate the work done. Work (w) = force (F) distance (h) = p A h quation (7.11) is the expression we will use to represent a quantity f pressure-volume work. w= pext v wo significant features to note in equation (7.11) are : 1) The negative sign: it is necessary to conform to sign conventions. When a gas expands, v is positive and w is negative, signifying that nergy leaves the system as work; When a gas is compressed, v is egative and w is positive, signifying that energy (as work) enters the ystem. 2) The term pext : it is the external pressure-the pressure against which system expands or the applied pressure that compresses a system. If pressure is stated in atmospheres and volume in liters, the unit of work is the liter-atmosphere, L atm. The SI unit of work is the joule, J. the conversion factor between these two units of work can be obtained from the gas constant, R. 8.3145 m3 Pa mol C 1 K C 1 = 8.3145 m3 N m2 mol C 1 K C 1 = 8.3145 kg m2 s C 2 mol C 1 K C 1 8.3145 J mol C 1 K C 1 = 0.083057 L atm mol C 1 K C 1 and 8.3145 J 0.082057 L atm = 101.33 J/L atm Example for calculating pressure-volume work In last figure, suppose the gas is 0.100 mol He at 298 K. How much work, in, joules, is associated with its expansion at constant temperature ? Solution We are given enough data to calculate the initial and final gas volume ( note that the identity of the gas does not enter into the calculate because we are assuming ideal gas behavior ). Once we have these volumes, we can obtain v. The initial pressure is 2.40 atm. The extermal pressure term in the pressure-volume work is the Final pressure-1.30 atm. Finally, the product C pext v must be multiplied by a factor to convert work in L atm to work in joules. Note: That the ideal gas equation embodies Boyle's law: The volume of a fixed amount of gas at a fixed temperature is inversely proportional to the pressure. Thus, we could simple write that 2.40 atm V final = 1.02 L 1.30 atm = 1.88 L n R t = 0.100 mol 0.0821 L atm mol C 1 K C 1 298 K V initial = P 2.40 atm initial = 1.02 L 0.100 mol 0.0821 L atm mol C 1 K C 1 298 K V final = n R t = Pfinal 1.30 atm = 1.88 L v = V final C V initial = 1.88 L C 1.02 L = 0.86 L w = C pext v = C 1.30 atm 0.86 L 101 J/ 1L atm = C 1.1 102 J The negative value signifies that the expanding gas does work on its surroundings 7-5 The First Law of Thermodynamics Nuclear energy Reactor Hydroelectricity Steam turbine Thermal energy Generator Electrical Mechanical energy energy Friction Motor Thermoelectricity Batery Chemical energy Combustion Schematic diagram of the common energy conversion processes. Internal energy, U, is the total energy (both kinetic and potential) in a system, including translational kinetic energy of molecules, the energy associated with molecular rotation and vibrations, the energy stored in chemical bonds and intermolecular attractions, and the energy associated with electrons in atom. First law of thermodynamics, the law of conservation of energy, dictates the relationship between heat (q), work (w), and changes in internal energy ( U ). U = q + w (7.12) If we consider that an isolated system is unable to exchange either heat or work with its surroundings, then U isolated system = 0, and we can say The energy of an isolated system is constant. In using equation (7.12) we must keep these important points in mind. Any energy entering the system carries a positive sign. Thus, if heat is absorbed by the system, q > 0. If work is done on the system, w > 0 Any energy leaving the system carries a negative sign. Thus, if heat is given off by the system, q < 0. If work is done by the system, w < 0 In general, the internal energy of a system changes as a result of energy entering or leaving the system as heat and/or work. If, on balance, more energy enters the system than leaves, U is positive. If more energy leaves than enters, U is negative. Example for relating U , q, and w Through the first law of thermodynamics A gas, while expanding ( see the figure of 7-4), absorbs 25 J of heat and does 243 J of work. What is U for the gas? Solution The key to problems of this type lies in assigning the correct signs to the quantities of heat and work. Because heat is absorbed by (enters) the system, q is positive. Because work done by the system represents energy leaving the system, w is negative. You may find it useful to represent the values of q and w, with their correct signs, within parentheses. Then complete the algebra. U = q + w = ( + 25 J) + ( C 243 J) = 25 J C 243 J = C 218 J Functions of state: a state function. any property that has a unique value for a specified state of a system. For example A sample of pure water at 20 C (293.15 K) and under one standard atmosphere of pressure is in a specified state. The density of water in this state is 0.99820 g/mL. We can establish that this density is a unique value- a function of state-in this way: Obtain three different samples of water: (a) one purified by extensive distillation of groundwater; (b) one synthesized by burning pure H2 (g) in pure O2 (g); (c) one prepared by driving off the water of hydration from CuSO45 H2O and condensing the gaseous water to a liquid. The densities of the three different samples for the state that we specified will all the same- 0.99820 g/mL. Thus, the value of a function of state depends on the state of the system, and not on how that state was established. The internal energy of a system is a function of state. Its value can' t be established. The difference in internal energy between the two states has a unique value, U = U 1 - U 2. As a further illustration, consider the scheme outlined here and illustrated by the diagram below. Imagine that a system changes from state 1 to state 2 and then back to state 1. U State 1 (U ) State 1 (U ) U State 2 (U ) 1 2 1 U2 Step 1: U = State 2 Step 2: C U = Internal energy U1 C U2 U2 C U1 U1 State 1 U 0verall= U 2 C U 1 C U 1 C U 2 = 0 Because U has a unique value in each state, U also has a unique value; it is U 2 C U 1. The change in internal energy when the system is returned From state 2 to state 1 is U = U 1 C U2 Thus, the overall change in internal energy is U ( C U )= (U 2 C U 1) C (U 1 C U 2) = 0 This means that the internal energy returns to its initial value of U 1, which it must do, since it is a function of state. It is important to note here that when we reverse the direction of change, we change the sign of U Path-Dependent Function: Unlike internal energy and changes in internal energy, heat (q), and work (w) are not functions of state. Their values depend on the path followed when a system undergoes a change. Recall the figure in 7-4.Think of the 0.100 mol of He at 298 K and under a pressure of 2.40 atm as state 1, and under a pressure of 1.30 atm as state 2. Now, suppose we first reduced the external pressure on the gas from 2.40atm to 1.80 atm ( at which point, the gas volume would be 1.36 L). Then, in a second stage, we could reduce the pressure to 1.30 atm, arriving at state 2. We calculated the amount of work done by the gas in a singlestage expansion in example of 7-4; it was w = C 1.1 102 J. The amount of work done in the two-stage process is the sum of two terms: the pressure-volume work for each stage of the expansion. w= = = = C 1.80 atm ( 1.36 L C 1.02 L) C 1.30 atm ( 1.88 L C 1.36 L) C 0.61 L atm C 0.68 L atm C 1.3 L atm 101 J/1 L atm C 1.3 102 J The value of U is the same for the single-and two-stage expansion processes because internal energy is a function of state. However, we see that slightly more work is done in the two-stage expansion. work is not a function of state; it is path dependent. Note: If w differs in the two different expansion processes, q must also differ, and in such a way that q + w for the one-stage and two-stage expansions are the same. This then makes q + w = U a unique quantity, as required by the first law of thermodynamics. 7-6 Heats of Reaction: U and H Think of the reactants in a chemical reaction as the initial state of a system and the products as the final state. reactants (initial state ) Ui U = Uf C products (final state) Uf Ui According to the first law of thermodynamics, we can also say that U = q C w We have previously identified a heat of reaction as qrxn, and so we can write U = qrxn C w A combustion reaction carried out in a bomb calorimeter. The volume is constant, V = 0. That is, w = -p V = 0. qv is the heat of reaction for a constant-volume reaction, we see that U = q v U = qrxn C w = qrxn C 0 = qrxn = q v (7.13) The heat of reaction measured in aabomb calorimeter isin a bombU . How does the heat of reaction measured equal to calorimeter compare with the heat of reaction if the reaction is carried out in some other way? Such as in beakers, flasks, and other containers open to the atmosphere and under the atmosphere and under the constant pressure of the atmosphere Two different paths leading to the same internal energy change in a system. Initial state U Initial state Ui i w qV Internal energy qp Final state U = q v (a) Uf Final state Uf U = qp C w (b) (a) The volume of the system retains constant and no internal energy is converted into work. Think of burning gasoline in a bomb calorimeter (b) The system does work, so some of the internal energy change is used to do work. Think of burning gasoline in an automobile engine. From the figure and the first law of thermodynamics, we see that for the same reaction at constant pressure U = qp y w , which means U = qV = qp y w . Thus, unless w = 0, qV and qp must be different. So U is a function of state and q and w are not. We can use the relationship between qV and qp to devise another state function that represents the heat flow for a process at constant pressure. We begin by writing qV = qp y w Now, using U = qV , w = -p V and rearranging terms, we obtain U = qp y w qp = U y p V Enthalpy H : a state function, is the sum of the internal energy and the pressure volume product of a system: H=U C PV. Enthalpy change H for a process between initial and final state is H = Hf y Hi = (Uf y PfVf) y (Ui y PiVi) H = = (Uf y Ui) y (PfVf y PiVi) H = U y p V If Pi = Pf , the work limited to pressure volume work, the enthalpy change is H = U y p V And the heat flow for the process under these conditions is H =qp (7.14) Enthalpy (H ) and internal energy (U ) changes in a chemical reaction H and U are related by the expression U = H y p V (7.15) In order to assess just how significant pressure volume work is, let's consider the following reaction 2CO(g) y O2(g) 2CO2(g) The reaction is also illustrated in the figure Comparing heats of reaction at constant volume and constant pressure for the reaction 2CO(g) y O2(g) 2CO2(g) (a) No work is performed: qV = U = y 563.5KJ (a) The surroundings do work on the qp system: U = qp y p V qV = qp y p V qp = H = y 566.0KJ If we measure the heat of this reaction under constant pressure conditions at a constant temperature of 298 K, we get y 566.0 KJ, indicating that 566.0 KJ of energy has left the system as heat: H = -566.0KJ. To evaluate the pressure- volume work, we begin by writing: P V = P(Vf - Vi) Then we can use the ideal gas equation to write this alternate expression PV = RT (nf - ni) Here, nf is the number of moles of gas in the products (2 mol CO2) and ni is the number of moles of gas in the reactants(2 mol CO + 1 mol O2). Thus , P V = 0.0083145 KJ mol-1 K-1 298K [2 y (2 y 1)] mol = y 2.5 KJ The change in internal energy is U = H y P V = 566.0 KJ- ( 2.5 KJ ) =563.5 KJ This calculation shows that the pV term is quite small compared to H and that U and H are almost the same. Now let's consider the combustion of sucrose. C12H22O11(s) y 12O2 (g) 12 CO2(g) y 11H2O (l) H = y 5.65 103 KJ (7.16) There is no change in volume in the combustion of sucrose: qp = qV That is 1 mol C12H22O11(s) reacts with 12 mol O2 (g) to produce 12 mol CO2(g), 11H2O (l), and y 5.65 103 KJ of evolved heat In summary, in most reactions, the heat of reaction we measure is H. In some reactions, notably combustion reactions, we measure U (that is, qv). In reaction(7.16): U = H, but this is not always the case. Where it is not, we can obtained a value of H from U by the method we illustrated in our discussion of expression(7.15), but even in those cases, H and U will be nearly equal. In this text, treat all heats of reactions as H values unless there is an indication to the contrary. Example shows how enthalpy changes can provide conversion factors for problem solving. Stoichiometric Calculation Involving Quantities of Heat. How much heat is associated with the complete combustion of 1.00 kg of sucrose, C12H22O11? Solution First, express the quantity of sucrose in moles. 1000 g C12H22O11 ? mol = 1.00 kg C12H22O11 = C12H22O11 1 kg C12H22O11 342.3 g C12H22O11 = 2.92 mol C12H22O11 1 mol Now, formulate a conversion factor based on the information in equation (7.16)---that is, -5.65 103 kJ of heat is associated with the combustion of 1 mol C12H22O11 -5.65 103 kJ ? KJ = 2.92 mol C12H22O11 1 mol C12H22O11 = -1.65 104 kJ The negative sign denotes that heat is given off in the combustion. The heat required to vaporize the fixed quantity of liquid is called the enthalpy(or heat) of vaporization. Usually the fixed quantity of liquid chosen is one mole, and we can call this quantity the molar enthalpy of vaporization. For example: H2O(l) H2O(g) H = 44.0 KJ at 298 K We described the melting of a solid in a similar fashion, on page 223. The energy requirement in this case is called the enthalpy (or heat) of fusion. For~the,.melting of one mole of ice, we can write H2O(s) H2O(l) H = 6.01 KJ at 273.15 K Next will give a example Example for enthalpy changes accompanying of changes in states of matter Calculate H for the process in which 50.0 g of water is converted from liquid at 10.0 C to vapor at 25.0 C? Solution The key to this calculate is to view the process as proceeding in two steps: raising the temperature of liquid water from 10.0 to 25.0 C. the total enthalpy change is the sum of the changes in each step. For a process at constant pressure, H = qp , so we need to calculate the heat absorbed in each step. Heating water from 10.0 to 25.0 C we determine this heat requirement in the same way as shown in example 7-1; that is, we apply equation(7.5) 4.18 J ? KJ = 50.0 g H2O g H2O C = 3.14 KJ (25.0 C 10.0) C 1KJ 1000J Vaporizing water at 25.0 C. For this part of the calculation, we need to express the quantity of water on a mole basis, so that we can then use the molar enthalpy of vaporization at 25.0 C: 44.0 KJ/mol. 1mol H2O 44.0KJ ? KJ = 50.0 g H2O 1mol H2O 18.02 g H2O = 122 KJ Total enthalpy change H = 3.14 KJ y 122 KJ = 125 KJ Standard States and Standard Enthalpy Changes The standard state of a solid or liquid substance is the pure Element or compound at a pressure of 1 bar (105 Pa)* and at the temperature of the standard state is the pure gas behaving as an (hypothetical): ideal gas at a pressure of 1 bar and the temperature of interest. The values given in this text are all for 298.15 K (25 C) unless otherwise stated. we say that the standard enthalpy change is the enthalpy change in a reaction in in which the reactants and products are in their standard states. This so-called standard enthalpy of reaction is denoted with a superscript degree symbol, H . Enthalpy diagram Enthalpy Products Enthalpy Products H > 0 H < 0 Reactants Endothermic reactants Reactants Exothermic reactants Horizontal lines represent absolute values of enthalpy. The higher a horizontal line, the greater the value of H that it represents. Vertical lines or arrows represent changes in enthalpy(). Arrows pointing up signify increases in enthalpy-endothermic reactions. Arrows pointing down signify decrease in enthalpyexothermic reactions. 7-7 Indirect Determination of H: Hess's Law The following features of enthalpy change (H) make us can calculate large numbers of heats of reaction . 1. H is an Extensive Property. Consider the standard enthalpy change in the formation of NO(g) from its elements at 25 C. N2(g) y O2 (g) 2 NO(g) H = 180.50 KJ To express the enthalpy change in terms of one mole of NO(g), we divide all coefficients and the H value by two. 1/2N2(g) y 1/2O2 (g) NO(g) H = 180.50 KJ = 90.25 KJ 2. H Changes Sign When a Process Is Reversed. As we learned on page 234, if we reverse a process, the change in a function of state reverses sign. Thus, H for the decomposition of one mole of NO(g) is - H for the formation of one mole of NO(g). NO(g) 1/2N2(g) y 1/2O2 (g) H = y 90.25 KJ 3. Hess's Law of Constant Heat Summation. To describe the standard enthalpy change for the formation of NO2(g) from N2(g) and O2(g). N2(g) y O2 (g) NO2(g) H=? We can think of the reaction as proceeding in two steps: First we form NO(g) from N2(g) and O2(g), and then NO2(g) from NO(g) and O2(g). When we add the equations for these two steps, together with their individual and distinctive H values, we get the overall equation and H value that we are seeking. 1/2N2(g) y 1/2O2 (g) NO(g) y 1/2O2 (g) 1/2N2(g) y O2 (g) NO(g) H = y 90.25 KJ NO2(g) H = y 57.07 KJ NO2(g) H = y 33.18KJ Note that in summing the two equations, we canceled out NO(g), a species that would have appeared on both sides of the overall equation. An enthalpy diagram illustrating Hess's law NO(g) C O2(g) H = y 57.07 KJ Enthalpy H = y 90.25 KJ NO2(g) H = y 33.18 KJ N2(g) C O2(g) Whether the reaction occurs through a single step (blue arrow) or in two steps (black arrows), the enthalpy change is H = 33.18 kJ for the overall reaction: 1/2N2(g) y O2 (g) NO2(g) Suppose we want the standard enthalpy change for the reaction 3 C(graphite) + 4 H2(g) C3H8(g) H = ? (7.18) We use the following standard heats of combustion to calculate H for reaction (7.18). H combustion: C3H8(g) = C 2219.9 KJ/mol C3H8(g) C(graphite) = C 393.5 KJ/mol C(graphite) H2(g) = C 285.8 KJ/mol H2(g) Note: H is an extensive property. In a chemical equation, the stoichiometric coefficients specify the amounts involved, and the unit kJ suffices for H . When H is not accompanied by an equation, the amount involved must somehow be specified, such as per mole of C3H8(g) in the expression H comb = -2219.9 kJ/mol C3H8(g). Example for Applying Hess's Law. Use the heat of combustion data listed above to determine for reaction (7.18): 3 C(graphite) + 4 H2(g) C3H8(g) H = ? Solution To determine an enthalpy change with Hess's law, we need to combine the appropriate chemical equations. A good starting point is to write chemical equations for the given combustion reactions, based on one mole of the indicated reactant. Recall from section 3-4 that the products of the combustion of carbonhydrogen-oxygen compounds are CO2(g) and H20(l). a) C3H8(g) + 5 O2(g) b) C(graphite) + O2(g) a) H2(g) y 1/2O2 (g) 3 CO2(g) + 4 H2O(1) H = y 2219.9 KJ CO2(g) H = y 393.5 KJ H2O (g) H = y 285.8 KJ Because our objective in reaction (7.18) is to produce C3H8(g), the next step is to find a reaction in which C3H8(g) is formed--the reverse of reaction (a). y (a) 3 CO2(g) + 4 H2O(1) C3H8(g) + 5 O2(g) H = y ( y 2219.9 KJ) = y 2219.9KJ Now, we turn our attention to the reactants, C(graphite) and H2(g). To get the proper number of moles of each, we must multiply equation (b) by 3 and equation (c) by 4. 3 (b): 3C(graphite) + 3O2(g) 4 : 4H2(g) y 2O2 (g) CO2(g) 4H2O (g) H =3 ( y 393.5 KJ) = y 1181 KJ H = 4( y 285.8 KJ) = y 1143 KJ Here is the overall change we have described: 3 mol C(graphite) and 4 mol H2(g)have been consumed, and 1 mol C3H8(g) has been produced. This is exactly what is required in Equation (7.18). We can now combine the three modified equations. C (a): 3 CO2(g) + 4 H2O(1) C3H8(g) + 5 O2(g) H = y 2219.9KJ 3 (b): 3C(graphite) + 3O2(g) CO2(g) H = y 1181 KJ 4 : 4H2(g) y 2O2 (g) 4H2O (g) H = y 1143 KJ 3 C(graphite) + 4 H2(g) C3H8(g) H = y 104 KJ 7-8 Standard Enthalpies of Formation Using Hess's law allows us to determine H values for reactions without actually performing them, as long as we have sufficient data about the steps that make up the reactions. We do not always have all of the information necessary to do this. Fortunately, there is another way to determine the heat of reaction without actually performing an experiment. Hf("standard delta H of formation"; also called heat of formation) values are tabulated for a large number of compounds. (Appendix C in your textbook contains Hf values.) Hf is defined as the heat of reaction for the production of one mole of a substance from its constituent elements, each in their standard states. For instance, the Hf of liquid water is the H for the following reaction (at 298 K). Up until this point you have been taught to balance equations using only whole numbers. In this case, however, the coefficient for oxygen gas is a fraction. This is necessary in order to balance the equation and have it produce only 1 mol of water. (Remember that production of one mole is part of the definition of Hf ) Example Standard enthalpy of formation of formaldehyde, HCHO(g) 0 Enthalpy of formation H2(g) + C(graphite) + O2(g) Enthalpy of formation Hf = C 108.6 KJ HCHO(g) The formation of HCHO(g) from its elements in their standard states is an exothermic reaction. The heat evolved per mole of HCHO(g) formed is the standard enthalpy (heat) of formation. Relating a standard enthalpy of formation to a chemical equation The enthalpy of formation of formaldehyde at 298 K is Hf = C 108.6 KJ/mol HCHO(g). Solutio n The equation must be witten for the formation of one mole of gaseous HCHO. The most stable forms of the elements at 298 K and 1 bar are gaseous H2 and O2 and solid carbon in the for of graphite. Note that we need one fractional coefficient in this equation. H2(g) + C(graphite) + O2(g) HCHO(g) Hf = C 108.6 KJ Standard Enthalpy of Reaction Let us use Hess's law to calculate the standard enthalpy of reaction for the decomposition of sodium bicarbonate, a minor reaction that occurs when baking soda is used in baking. 2 NaHCO3(s) NaCO3(s) + H2O(l) + CO2(g) H = ? (7.19) Computing heats of reaction from standard enthalpies of formation 2 Na(s) + H2(g) + 2C(graphite) + 3O2(g) Decomposition Formation NaCO3(s) + CO2(g) + H2O(l) Overall 2 NaHCO3 is the sun of the enthalpy changes for the two steps shown. enthalpy is a state function, hence H for the overall reaction 2 NaHCO3(s) NaCO3(s) + H2O(l) + CO2(g) From Hess's law, we see that the following four equation yield equation (7.19) when added together. a) 2 NaHCO3(s) 2 Na(s) + H2(g) + 2C(graphite) + 3O2(g) H = C 2 Hf [NaHCO3(s)] b) 2 Na(s) + C(graphite) + 3/2 O2(g) NaCO3(s) H = Hf [Na2CO3(s)] c) H2(g) + O2(g) H2O(l) H = Hf [H2O(l)] d) C(graphite) + O2(g) CO2 H = Hf [CO2(g)] 2 NaHCO3(s) NaCO3(s) + H2O(l) + CO2(g) H = ? Then we can express the value of H for the decomposition reaction as H = Hf [Na2CO3(s)] + Hf [H2O(l)] + Hf [CO2(g)] y 2 Hf [NaHCO3(s)] (7.20) Imagine the decomposition of sodium bicarbonate taking place in two steps. In the first step, suppose a vessel contains 2 mol NaHCO3, which is allowed to decompose into 2 mol Na(s), 2 mol C(graphite), 1 mol H2(g), and 3 mol O2(g), as in equation (a) above. In the second step, combine the 2 mol Na(s), 2 mol C(graphite), 1 mol H2(g), and 3 mol O2(g) to form the products according to equations (b), (c) and (d) above. The enthalpy change for the overall reaction is the sum of the standard enthalpy changes of the individual steps. H = Hdecomposition + Hformation Hdecomposition = C 2 Hf [NaHCO3(s)] Hformation = Hf [Na2CO3(s)] + Hf [H2O(l)] + Hf [CO2(g)] So that H = Hf [Na2CO3(s)] + Hf [H2O(l)] + Hf [CO2(g)] y 2 Hf [NaHCO3(s)] Equation (7.20) is a specific application of the following more general relationship for a standard enthalpy of reaction. H = p Hf(products) y p Hf(reactants) (7.21) The symbol (Greek, sigma) means "the sum of." The terms that are added together are the products of the standard enthalpies of formation (Hf) and their stoichiometric coefficients, . One sum is required for the reaction products, and another for the initial reactants. The enthalpy change of the reaction is the sum of terms for the products minus the sum of terms for the reactants. Diagrammatic representation of equation Elements Formation Decomposition Enthalpy Enthalpy Products Overall( H> 0 ) Reactants Endothermic reaction Elements Formation Decomposition Reactants Overall( H<0 ) Products Exothermic reaction The state function basis for equation(7.21) is shown in figure and is applied in Example Next page. Calculating H from Tabulated Values of Hf . Let us Example 1 apply equation 7.21 to calculate the standard enthalpy of combustion of ethane, C2H6(g), a component of natural gas. Solution The reaction is C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(l) The relationship we need is equation (7.21). The data we substitute into the relation are from Table 7.2. H = {2 mol CO2 Hf [CO2(g)] + 3 mol H2O Hf [H2O(l)]} y {1 mol C2H6 Hf [C2H6(g)] y 7/2 mol O2 Hf [O2(g)]} = 2 mol CO2 ( y 393.5 KJ/mol CO2) y 3 mol H2O ( y 285.8 KJ/mol H2O) y 1 mol C2H6 ( y 84.7 KJ/mol C2H6) y 7/2 mol O2 0 KJ/mol O2 = y 787.0 KJ y 857.4 KJ y 84.7 KJ = y 1559.7 KJ Example 2 Calculating an unknown Hf value. Use the data here and in Table 7.2 to calculate Hf of benzene, C6H6(l). 12 CO2(g) C 6 H2O(l) Hf = y 6535 2 C6H6(l) C 15 O2 KJ Solution To organize the data needed in the calculation, let's begin by writing the chemical equation for the reaction, with Hf data listed under the chemical formulas. 2 C6H6(l) C 15 O2 12 CO2(g) C 6 H2O(l) Hf = y 6535 KJ Hf , KJ/mol ? 0 y 393.5 y 285.5 Now, we can substitute known data into expression (7.21) and rearrange the equation to obtain a lone term on the left: Hf ?[C6H6(1)]. The remainder of the problem simply involves numerical calculations. Hf = 12 mol CO2 ( y 393.5 KJ/mol CO2) y 6 mol H2O ( y 285.8 KJ/mol H2O) y 2 mol C6H6 [ Hf C2H6 (l)] = y 6535 KJ { C 4722 KJ C 1715 KJ} C 6535 KJ [ Hf C2H6 (l)] = 2 mol C6H6 = 49 KJ/mol C6H6 (l) Ionic Reactions in Solutions Consider the neutralization of a strong acid by a strong base. We can write H C (aq) C OH C (aq) H2O(l) H = C 55.8 KJ (7.22) The ion we arbitrarily choose for our zero is H+(aq). Now let us see how we can use expression (7.21) and data from equation (7.22) to determine the enthalpy of formation of OH C (aq) H = 1 mol H2O Hf [H2O(l)] y 1 mol H C Hf [H y (aq)] y 1 mol OH C Hf [OH y (aq)] = y 55.8 KJ Hf [OH y (aq)] = 55.8 KJ C (1 mol H2O Hf[H2O(l)] ) C (1 mol H C Hf[H y (aq)]) 1 mol OH C Hf [OH y 55.8 KJ C 285.8 KJ C 0 KJ (aq)] = = C 230.0 KJ/mol OH C 1 mol OH C Table 7.3 lists data for several common ions in aqueous solution. Enthalpies of formation in solution depend on the solute concentration. These data are representative for dilute aqueous solutions (about 1 M), the type of solution that we normally deal with. Example Calculating the Enthalpy Change in an Ionic Reaction. Given Hf[BaSO4(s)] = C 1473 kJ/mol, what is the standard enthalpy change for the precipitation of barium. Solution First, let's write the net ionic equation for the reaction and introduce the relevant data. 2 Ba 2 C (aq) C SO4 2 C Hf , KJ/mol y 537.6 y 909.3 BaSO4(s) y 1473 Hf = ? Then we can substitute data into equation (7.21) H = 1 mol BaSO4 Hf [BaSO4(s)] y 1 mol Ba C Hf [Ba y (aq)] y 1 mol SO42 C Hf [SO42 C (aq)] = 1 mol BaSO4 ( y 1473 KJ/mol BaSO4) y 1 mol Ba C ( y 573.6 KJ/mol Ba C ) y 1 mol SO42 C ( y 909.3 KJ/mol SO42 C ) = y 1473 KJ y 573.6 KJ y 909.3 KJ = C 26 KJ 7-9 Fuels as Sources of Energy The reactions involved in producing energy from food and those involved in the combustion of fossil fuels are remarkably similar. Both ultimately convert carbon-based compounds to carbon dioxide and water and produce energy in the process. !) The digestion process breaks down the food we eat into "blood sugar" (glucose), which combines with oxygen in what is essentially a controlled combustion process. (Without the flame, of course!) C6H12O6(s) C O2(g) CO2(g) C H2O(g) H = y 2816 KJ You may recognize this as the reverse of the equation for photosynthesis. CO2(g) C H2O(g) C6H12O6(s) C O2(g) H = y 2816 KJ This is the process by which plants consume the water and CO2 from the atmosphere and produce oxygen and more plant mass. The positive H indicates that energy must be consumed for this process to occur. The necessary energy comes from the sun. The negative sign on the H indicates that heat (energy) is given off in this reaction. This process produces the heat necessary to maintain body temperature and function. Any such energy produced in the body and not expended is stored as fat for future energy production. The energy unit Calorie is still commonly used to indicate the fuel content of foods. 1 calorie = 4.184 joules The calorie reported on food packaging are actually kilocalories. Among proteins, carbohydrates, and fats (the principal constituents of a human diet), fats provide the most energy per gram consumed. In other words, fats have a higher fuel value than do either proteins or carbohydrates. Below is a table of various foods and their fuel values. Fuel value: the energy released when 1 g of a substance is combusted. Fossil fuels such as natural gas, petroleum, and coal can also be characterized by fuel value. Here is a table of fuels--most of them petroleum--with their percentage compositions and fuel values. Fossil fuels: Coal, oil and natural gas, which are presently major of our sources of energy. Natural gas : A naturally occurring mixture of gaseous hydrocarbon, compounds composed of hydrogen and carbon. Petroleum: A naturally occurring combustible liquid composed of hundreds of hydrocarbon, and other organic compounds Petroleum: A naturally occurring solid containing hydrocarbons, of high molecular weight, as well as compounds containing sulfur, oxygen, and nitrogen. ...
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This note was uploaded on 05/07/2010 for the course CHEM 401 taught by Professor Chemistry during the Spring '10 term at Uni Potsdam.

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Chemistry CH 7 - Chapter 7 Thermochemistry Why nature gas...

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