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Thermochemistry

Thermochemistry. The study of heat change in chemical reactions. Thermochemistry is part of a broader subject called thermodynamics , which is the scientific study of the interconversion of heat and other kinds of energy.

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Thermochemistry

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  1. Thermochemistry The study of heat change in chemical reactions

  2. Thermochemistry is part of a broader subject called thermodynamics, which is the scientific study of the interconversion of heat and other kinds of energy. The first law of thermodynamics, which is based on the law of conservation of energy, states that energy can be converted from one form to another, but cannot be created or destroyed.

  3. Types of Energy • Thermal energy - The energy associated with the random motion of atoms and molecules. • Temperature - the average kinetic energy of the particles in a sample of matter • Heat(q)- the total kinetic energy of the particles in a sample of matter

  4. The internal energy of a system has two components: kinetic and potential energy. Kinetic energy - the energy produced by a moving object (various types of molecular motion and the movement of electrons within molecules). Potential energy - energy available by virtue of an object’s position (determined by the attractive and repulsive forces at the atomic level) . It is impossible to measure all these contributions accurately, so we cannot calculate the total energy of a system with any certainty. Changes in energy, on the other hand, can be determined experimentally.

  5. Chemical energy - The energy stored within the structural units of chemical substances; its quantity is determined by the type and arrangement of constituent atoms. • When substances participate in chemical reactions, chemical energy is released, stored, or converted to other forms of energy. • Chemical energy can be considered a form of potential energy because it is associated with the relative positions and arrangements of atoms within a given substance.

  6. Energy can assume many different forms that are interconvertible. When one form of energy disappears, some other form of energy (of equal magnitude) must appear, and vice versa.

  7. Energy Changes in Chemical Reactions Almost all chemical reactions absorb or produce (release) energy, generally in the form of heat. In order to analyze energy changes associated with chemical reactions we must first define the terms system and surroundings.

  8. System - the specific part of the universe that is of interest to us. For chemists, systems usually include substances involved in chemical and physical changes. • Surroundings - the rest of the universe outside the system. • For example, in an acid-base neutralization experiment, the system would be the contents of the beaker, the 50.0 mL of HCl to which 50.0 mL of NaOH are added.

  9. 2H2(g) + O2(g) 2H2O(l) + energy The combustion of hydrogen gas is one of many chemical reactions that release considerable quantities of energy. System - ? Surroundings - ?

  10. Since energy cannot be created or destroyed, any energy lost by the system must be gained by the surroundings. Thus the heat generated by the combustion process is transferred from the system to its surroundings. This is an example of an exothermic process, which is any process which gives off heat - that is, transfers thermal energy to the surroundings.

  11. heat + 2HgO(s) 2Hg(l) + O2(g) The decomposition of mercury (II) oxide is an endothermic process, in which heat has to be supplied to the system by the surroundings. System - ? Surroundings - ?

  12. Calorimetry Our discussion of calorimetry, the measurement of heat changes, will depend on an understanding of specific heat and heat capacity.

  13. Specific Heat and Heat Capacity Specific heat(s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. • Specific heat is an intensive property, its magnitude doesn’t depend on the amount of the substance present • Specific heat has the units J/g.oC or kJ/g.oC

  14. If we know the specific heat and the amount (mass in g) of a substance (m), then the change in the sample’s temperature (DT = Tfinal - Tinitial) will tell us the amount of heat (q) that has been absorbed or released in a particular process. q = msDT The sign convention for q is positive for endothermic processes and negative for exothermic processes.

  15. Example: A 466g sample of water is heated from 8.50 oC to 74.60 oC. Calculate the amount of heat absorbed by the water in kJ. q = msDT q = (466g)(4.184 J/g.oC)(74.60 oC - 8.50 oC) q = 129000 J q = 129 kJ

  16. The heat capacity(C) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius. • Heat capacity is an extensive property (it varies dependent on the amount of substance present). • Heat capacity has the units J/oC or kJ/oC • The relationship between the heat capacity and specific heat of a substance is C = ms

  17. Determine the heat capacity for 60.0g of water. C = ms (60.0g)(4.184 J/g.oC) = 251 J/oC

  18. Calorimeters In a laboratory, heat changes in physical and chemical processes are measured with a calorimeter, a closed container designed specifically for this purpose. • The special design of a calorimeter allows us to assume that no heat is lost to the surroundings during the time it takes to make measurements.

  19. If we know the heat capacity,then the change in the sample’s temperature (DT = Tfinal - Tinitial) will tell us the amount of heat (q) that has been absorbed or released in a particular process. q = CDT

  20. A certain calorimeter has a heat capacity of 1.55 kJ/oC. Calculate the heat absorbed by a calorimeter in kJ that increases in temperature from 23.60 oC to 25.87 oC. q = CDT q = (1.55 kJ/oC)(25.87 oC - 23.60 oC) q = 3.52 kJ

  21. Heat of combustion, the amount of energy given off when a certain amount of a substance is “burned” in the presence of oxygen, is usually measured by placing a known mass of a compound in a steel container called a constant-volume bomb calorimeter. (With this being a steel container, its volume obviously doesn’t change, but the pressure certainly would.) This calorimeter is filled with oxygen before the sample is placed in the cup. The sample is ignited electrically, and the heat produced by the reaction can be accurately determined by measuring the temperature increase in the known amount of surrounding water.

  22. Because no heat enters or leaves the system throughout the process, we can write qrxn = -(qwater + qcalorimeter) where qwater is obtained by q = msDT and qcalorimeter is determined by q = CDT

  23. q = -(qwater + qcalorimeter) qwater = msDT = (2000.g)(4.184J/g.oC)(25.84 oC - 20.17 oC) = 47400 J or 47.4 kJ qcalorimeter = CDT = (1.80 kJ/oC)(25.84 oC - 20.17 oC) = 10.2 kJ q = -(47.4 kJ + 10.2 kJ) = -57.6 kJ A quantity of 1.435g of naphthalene (C10H8), a pungent-smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.17 oC to 25.84 oC. If the mass of water surrounding the calorimeter was exactly 2000g and the heat capacity of the bomb calorimeter was 1.80 kJ/ oC, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion. The molar mass of naphthalene is 128.173g/mol, so the heat of combustion of 1 mol of naphthalene is -57.6 kJ = -5140 kJ__ 1.435 g C10H8 128.173 g C10H8

  24. A simpler device than the constant-volume calorimeter is the constant-pressure calorimeter used to determine the heat changes for noncombustion reactions. An operable constant-pressure calorimeter can be constructed from two Styrofoam coffee cups. This device measures the heat effects of a variety of reactions, such as acid-base neutralization, as well as the heat of solution and heat of dilution.

  25. As the following example show, we can study the heat of acid-base reactions using a constant-pressure calorimeter. 1.00 x 102 mL of 0.500 M HCl is mixed with 1.00 x 102 mL of 0.500 M NaOH in a constant-pressure calorimeter that has a heat capacity of 335 J/oC. The initial temperature of the HCl and NaOH solution is the same, 22.50 oC, and the final temperature of the mixed solution is 24.90 oC. Calculate the heat of neutralization for the following reaction NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) (Assume the densities and specific heats of the solutions are the same as for water.) Assuming that no heat is lost to the surroundings, we write qrxn = -(qsoln + qcalorimeter) qsoln = msDT = (1.00 x 102g + 1.00 x 102g)(4.184 J/g.oC)(24.90 oC - 22.50 oC) = 2010 J or 2.01 kJ qcalorimeter= CDT = (335 J/oC )(24.90 oC - 22.50 oC) = 804 J or .804 kJ qrxn = -(2.01 kJ + .804 kJ) = -2.81 kJ

  26. 2.81 kJ of heat are given off by the neutralization of 1.00 x 102 mL of 0.500 M HCl and 1.00 x 102 mL of 0.500 M NaOH. To calculate the heat of neutralization for this reaction, you need to know the number of kJ released when 1 mol of HCl reacts with 1 mol of NaOH. (Remember, the coefficients in a thermochemical equation represent moles.) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 0.500 M HCl = 0.500 mol HCl = 0.0500 mol HCl 1 L .1 L yield .0500 mol 0.500 M NaOH = 0.500 mol NaOH = 0.0500 mol NaOH H2O 1 L .1 L -2.81 kJ = -56.2 kJ 0.0500 mol 1 mol

  27. Enthalpy To quantify the heat flow into or out of a system, chemists use a property called enthalpy, represented by the symbol H. Enthalpy is an extensive property; its magnitude depends on the amount of the substance present.

  28. Enthalpy of a Reaction It is the change in enthalpy, DH, that we measure in a chemical reaction. The change of enthalpy of a reaction, DHrxn commonly represented as just DH, is the difference between the enthalpies of the products and the enthalpies of the reactants: DH = H(products) - H(reactants)

  29. In other words, DH represents the heat given off or absorbed during a reaction. • The enthalpy of a reaction can be positive or negative, depending on the process. • For an endothermic process DH is positive. For an exothermic process DH is negative.

  30. Thermochemical equations Thermochemical equations show the enthalpy changes as well as the mass relationships in a thermochemical reaction.

  31. In a thermochemical equation... • the stoichiometric coefficients always refer to the number of moles of a substance. • When we reverse an equation, we change the roles of reactants and products. Consequently, the magnitude of DH for the equation remains the same, but its sign changes. • If we multiply both sides of a thermochemical equation by a factor n, then DH must also change by the same factor. • When writing thermochemical equations, we must always specify the physical states of all reactants and products, because they help determine the actual enthalpy changes.

  32. H2O(l) H2O(s) DH = -6.01 kJ 2H2O(l) 2H2O(s) DH = -12.02 kJ Examples of thermochemical equations H2O(s) H2O(l) DH = 6.01 kJ Thermochemical equations show the enthalpy changes that occur in a physical or chemical process.

  33. 1 mol H2O 25.0 g H2O x = 1.39 mol H2O 18.02 g H2O H2O(l) H2O(s) DH = -6.01 kJ 1 mol H2O -6.01 kJ 1.39 mol H2O How much energy is absorbed or released when 25.0 g of water is frozen? = -8.35 kJ

  34. It is impossible to determine the absolute enthalpy of a substance. This problem is similar to the one geographers face in expressing the elevations of specific mountains or valleys. Rather than trying to devise some type of “absolute” elevation scale (perhaps based on distance from the center of the Earth?), by common agreement all geographic heights and depths are expressed relative to sea level, an arbitrary reference with a defined elevation of “zero” meters or feet. Similarly, chemists have agreed on an arbitrary reference point for enthalpy.

  35. Enthalpy of Formation The “sea level” reference point for all enthalpy expressions is called the standard enthalpy of formation,(DHof), which is defined as the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. Elements are said to be in the standard state at 1 atm, hence the term “standard enthalpy”. The superscript “o” represents standard-state conditions (1 atm), and the subscript “f” stands for formation. Although the standard state does not specify a temperature, we will always use DHof values measured at 25 oC.

  36. O2 - because the DHof for O2 = 0, and the DHof for O3 = 0 By convention, the standard enthalpy of formation of any element in its most stable form is zero. Using your reference packet, state the most stable (allotrope) form for molecular oxygen State the most stable allotrope for carbon C(graphite) is the most stable allotrope

  37. You can usually get DHf values for compounds from your thermochemistry reference packet. DHf that aren’t listed in the reference packet can be determined in one of two ways, the direct method or the indirect method. • Both of these methods require an understanding of thermochemical equations.

  38. The Direct Method for Determining DHf • Works best for compounds that can be readily synthesized from their elements. Suppose we want to know the enthalpy of formation of carbon dioxide. We must measure the enthalpy of the reaction when carbon (graphite) and molecular oxygen in their standard states are converted to carbon dioxide in its standard state: C(graphite) + O2(g) CO2(g) DHorxn = -393.5 kJ DHorxn = SnDHof(products) - SnDHof(reactants) -393.5 kJ= 1 mol(x) - [1 mol(0 kJ/mol) + 1 mol(0 kJ/mol)] -393.5 kJ= 1(x) x must be equal to -393.5 kJ/mol, therefore, DHof for CO2 is equal to -393.5 kJ/mol.

  39. The Indirect Method for Determining DHf Many compounds cannot be directly synthesized from their elements. In some cases, the reaction proceeds too slowly, or side reactions produce substances other than the desired compound. In these cases DHof can be determined by an indirect approach, which is based on the law of heat summation (or simply Hess’s law).

  40. Hess’s law can be stated as follows: 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. In other words, if we break down the reaction of interest into a series of reactions for which DHorxn can be measured, we can calculate DHorxn for the overall reaction.

  41. We would represent the synthesis of CH4 from its elements as What is DHof for methane? C(graphite) + 2H2(g) CH4(g) However, this reaction does not take place as written, so we cannot measure the enthalpy change directly. We must use Hess’s law.

  42. The following reactions involving C(graphite), H2(g) and CH4(g) with O2(g) have all been studied and the DHorxn values are accurately known (a) C(graphite) + O2(g) CO2(g) DHorxn = -393.5 kJ (b) 2H2(g) + O2(g) 2H2O(l) DHorxn = -571.6 kJ (c) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DHorxn = -890.4 kJ Remember, we are trying to determine the DHof for C(graphite) + 2H2(g) CH4(g)

  43. C(graphite) + 2H2(g) CH4(g) DHorxn = -74.7 kJ As you can see, all the nonessential species (O2, CO2, and H2O) cancel in this operation. Because the above equation represents the synthesis of 1 mol of CH4 from its elements, we have DHof(CH4) = -74.7 kJ/mol. Since we want to obtain one equation containing only C and H2 as reactants and CH4 as a product, we must reverse (c) to get (d) CO2(g) + 2H2O(l) CH4(g) + 2O2(g) DHorxn = 890.4 kJ The next step is to add equations (a), (b), and (d) (a) C(graphite) + O2(g) CO2(g) DHorxn = -393.5 kJ (b) 2H2(g) + O2(g) 2H2O(l) DHorxn = -571.6 kJ (d) CO2(g) + 2H2O(l) CH4(g) + 2O2(g) DHorxn = 890.4 kJ

  44. The general rule in applying Hess’s law is that we should arrange a series of chemical equations (corresponding to a series of steps) in such a way that, when added together, all species will cancel except for the reactants and products that appear in the overall reaction.

  45. Once we know the standard enthalpies of formation, we can calculate the standard enthalpy of reaction, DHorxn, defined as the enthalpy of a reaction carried out at 1 atm.

  46. Consider this hypothetical reaction aA + bB cC + dD where a, b, c, and d are stoichiometric coefficients.

  47. For this reaction DHorxn is given by DHorxn = [cDHof(C) + dDHof(D)] - [aDHof(A) + bDHof(B)] We can generalize this equation as: DHorxn = SnDHof(products) - SnDHof(reactants) where n denotes the stoichiometric coefficients for the reactants and products, and S (sigma) means “the sum of”.

  48. Calculate the kJ of heat released when 1 mol of liquid pentaborane, B5H9, is exposed to gaseous oxygen producing solid diboron trioxide and liquid water.

  49. (*Note - DHf values vary slightly depending on which reference book you choose, these values are from Raymond Chang Chemistry, 6th Edition) Remember, these are kJ released or absorbed per one mole

  50. aA + bB  cC + dD B5H9(l) + 6O2(g)  2.5 B2O3(s) + 4.5 H2O(l) If it had been written with 2B5H9, you would have to divide the final kJ by 2 to get the amount of heat released when 1 mol of pentaborane reacted with oxygen. Notice, this was balanced in such a way that you only started with 1 mol of B5H9 DH = H(products) - H(reactants) DHorxn = [cDHof(C) + dDHof(D)] - [aDHof(A) + bDHof(B)] DHorxn = [2.5(-1263.6) + 4.5(-285.8)] - [1(73.2) +6(0)] DHorxn = -4518.3 kJ

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