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Homework Problems

Homework Problems Chapter 6 Homework Problems: 8, 12, 24, 40, 46, 52, 58, 64, 68, 70, 72, 74, 78, 82, 94, 96. CHAPTER 6 Thermochemistry. Thermochemistry, Work, and Heat Thermochemistry is the study of energy flow in chemical systems. Work (w) is (force) . (displacement)

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Homework Problems

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  1. Homework Problems Chapter 6 Homework Problems: 8, 12, 24, 40, 46, 52, 58, 64, 68, 70, 72, 74, 78, 82, 94, 96

  2. CHAPTER 6 Thermochemistry

  3. Thermochemistry, Work, and Heat Thermochemistry is the study of energy flow in chemical systems. Work (w) is (force) . (displacement) Heat, or thermal energy, (q) is the energy that moves from a hot object to a cold object when placed in contact with each other. T1 > T3 > T2 heat flows from hot to cold block

  4. Energy Energy is the capacity to do work or supply heat. In principal any kind of energy can be converted into an equivalent amount of work or heat. We divide energy into two general types, kinetic (due to motion in a particular direction) and potential (due to position or composition). The total amount of all the different kinds of energy for a system is called the internal energy (E). In a closed system (no energy in or out) energy can be converted from one type to another, but the total amount of energy remains constant. This statement is called conservation of energy.

  5. Units of Energy All forms of energy can be expressed in the same units. To find the MKS unit for energy, it is convenient to use the equation for kinetic energy EK = mv2 So units are (kg) (m/s)2 = kg.m2 = 1 Joule = 1 J 2 s2 Other common units for energy include: calorie (cal) Amount of heat needed to raise the temperature of 1 g of water by 1 C 1 cal = 4.184 J (exact) 1 kcal = 1 food calorie = 1000 cal = 4184. J (exact) Kilowatt hour (kWh) Amount of energy equal to 1000. J/s (1000 Watt = 1 kWatt) for 1 hour. Note that 1 watt = 1 J/s, a unit of power. 1 kWh = 3.6 x 106 J (exact)

  6. System, Surroundings, and Universe The system is a part of the universe that we have separated off for study. The surroundings are everything not included in the system. The universe is everything - system + surroundings. While in principle the surroundings are everything not included in the system, in practice we can usually focus on that part of the surroundings in the immediate vicinity of the system.

  7. State Function A state function is any function whose change in value depends only on the initial and final state of the system, but which is independent of the pathway used to go between them. The value for something that is not a state function depends not only on the initial and final state but also on the pathway used to travel between them. Altitude is a state function, distance traveled is not a state function. In thermodynamics, E is a state function, q and w are not state functions (in fact, they are not functions at all!)

  8. First Law of Thermodynamics The first law of thermodynamics gives a relationship between internal energy, work, and heat. E = q + w E = Ef - Ei = change in internal energy q = heat w = work For now, we limit ourselves to mechanical work, where w = - p V. Note that the first law is based on experimental observation and is not derived from some other law or principle. The first law relates the change in a state function (E) to the sum of two things that are not state functions (q and w).

  9. Mechanical Work Mechanical work is the work associated with the change in the volume of a system. Consider the expansion of a gas

  10. Derivation of the Expression For Mechanical Work w = F . d (from the definition of work) But p = F/A, and so F = p . A So |w| = p . A . d However, from the diagram d = h So |w| = p . A . h But A . d = V where V is the change in volume. So |w| = p . V (absolute value) p d But notice that when a gas expands its volume change is in a direction opposite to that of the applied pressure. This intro-duces a negative sign into the expression for mechanical work. Therefore w = - p . V

  11. Sign Convention for Heat and Work We use the following sign convention for heat and work. q > 0 heat flows from the surroundings into the system (endothermic) q < 0 heat flows from the system to the surroundings (exothermic) w > 0 work is done on the system (system is compressed) w < 0 work is done by the system (system expands) Note that qsurr = - qsyst, and wsurr = - wsyst.

  12. Use of the First Law A gas is confined inside of a cylinder at an initial pressure and volume pi = 1.00 atm and Vi = 2.000 L. 8000 J of heat is added to the gas under conditions of constant pressure. The final volume of the gas is Vf = 6.000 L. What are q, w, and E for the process?

  13. A gas is confined inside of a cylinder at an initial pressure and volume pi = 1.00 atm and Vi = 2.000 L. 8000 J of heat is added to the gas under conditions of constant pressure. The final volume of the gas is Vf = 6.000 L. What are q, w, and E for the process? From the first law, E = q + w. Heat is added to the system, so q = + 8000. J For constant pressure w = - p V = - (1.00 atm) (6.000 L – 2.000 L) = - 4.00 L.atm 101.3 J = - 405. J * 1 L.atm Finally, E = q + w = 8000. J + (- 405. J) = 7595. J *Note: To find the conversion between L.atm and J we do the following # J = 1 L.atm 1 m3 1.013 x 105 N/m2 = 101.3 N.m = 101.3 J 1000 L 1 atm

  14. Heat Capacity Consider adding a known amount of heat to a substance. One would expect that the temperature of the substance would increase. Heat capacity (C) is a measure of how much heat is required to cause the temperature of a substance to change by a given amount. Heat capacity is defined as C = q/T where q = amount of heat added T = Tf - Ti = change in temperature

  15. Notes on Heat Capacity 1) Since C = q/T, the units for C are (energy)/(temperature). The derived MKS units for C are J/K, though it is often given in J/C. 2) Note that the numerical value for C when expressed in J/K or J/C is identical. That is because we are using the change in temperature. Since the size of a degree Kelvin and a degree Celsius is the same, the numerical value for T is the same whether expressed in K or C. Example: Ti = 15.0 C (288.2 K) and Tf = 21.5 C (294.7 K) T = Tf – Ti = 21.5 C – 15.0 C = 6.5 C = 294.7 K - 288.2 K = 6.5 K 3) C is not a state function, since q is not a state function. However, if we restrict ourselves to processes occurring under conditions of constant pressure then C is a state function. This is sometimes indicated by using the symbol Cp.

  16. Specific Heat and Molar Heat Capacity There are two quantities that are closely related to heat capacity. Specific heat capacity, (specific heat, Cs) - The heat capacity per gram of substance. Molar heat capacity, (Cm) - The heat capacity per mole of substance. The relationships between Cs, Cm, and heat capacity are as follows: Cs = C/m where m= mass of substance Cm = C/n where n = moles of substance So Cs has units of J/g.C and Cm has units of J/mol.C. Note that C is an extensive property, but Cs and Cm are intensive properties.

  17. Finding C, Cs, and Cm We may use the above relationships to find heat capacity and related quantities from experimental data. Example: 1000. J of heat is added to a 20.0 g sample of water (H2O, M = 18.01 g/mol). The initial and final temperatures of the water are Ti = 19.3  and Tf = 31.3 C. What are C, Cs, and Cm?

  18. Example: 1000. J of heat is added to a 20.0 g sample of water (H2O, M = 18.01 g/mol). The initial and final temperatures of the water are Ti = 19.3 C and Tf = 31.3 C. What are C, Cs, and Cm? C = q/T T = Tf - Ti = 31.3 C - 19.3 C = 12.0 C So C = 1000. J = 83.3 J/C 12.0 C Cs = C/m = (83.3 J/C) = 4.17 J/g.C 20.0 g Cm = C/n n = 20.0 g 1 mol = 1.11 mol 18.01 g Cm = (83.3 J/C) = 75.0 J/mol.C 1.11 mol

  19. Specific Heat For Common Substances The above table is at T = 25. C. Note that water has an unusual-ly large value for heat capacity, which acts to moderate temperatures for cities surrounded by large bodies of water.

  20. Constant Volume Processes Consider some general process taking place at constant volume. From the first law E = q + w But for mechanical work, w = - pV so at constant volume, V = 0, so w = 0 Therefore, for a process carried out at constant volume E = qV (V = constant) What does this mean? For a process carried out at constant volume, q is a state function, and so no information is needed concerning path. This makes it far easier to calculate and keep track of heat flow for these kinds of processes.

  21. Enthalpy Most processes in the laboratory are carried out at constant pressure instead of constant volume. It would be nice to have a state function whose change in value was equal to q for constant pressure processes. We define enthalpy, H, as follows H = E + pV Since E, p, and V are state functions, it follows that enthalpy is also a state function.

  22. Constant Pressure Processes Consider some general process taking place at constant pressure. From the definition of enthalpy H = E + pV H = E + (pV) = E + (pfVf - piVi) Now, if pressure is held constant, then pf = pi = p, and so H = E + p(Vf - Vi) = E + pV Now, from the first law E = q + w If we only have mechanical work E = q - p V

  23. If we now substitute into the expression for enthalpy, we get H = E + pV = (q - p V) + p V or (finally!) H = qp (p = constant) What does this mean? For a process carried out at constant pressure, q is a state function, and so no information is needed concerning path. This makes it far easier to calculate and keep track of heat flow for these kinds of processes. To summarize E = qV (constant volume processes) H = qp (constant pressure processes) Since q is something that can be measured experimentally, we now have a way to relate this information to changes in state functions (internal energy or enthalpy).

  24. Calorimetry Calorimetry is the experimental method used to measure the heat produced or taken up in a chemical reaction. The measured value for q can be related to Erxn (for constant volume) or Hrxn (for constant pressure).

  25. Bomb Calorimetry In bomb calorimetry a measured mass of a chemical substance reacts with excess oxygen to form combustion products. If the heat capacity of the calorimeter apparatus is known (and this can be determined experimentally) then q = - C T C = heat capacity of calorimeter T = Tf - Ti = change in temperature The negative sign in the above equation occurs because we are measuring the value of q for the surroundings, and qsyst = - qsurr. Since the combustion process occurs under conditions of constant volume, q = E, the change in internal energy. Example: 1.412 g of carbon (M = 12.01 g/mol) are burned in a bomb calorimeter (C = 10325. J/C). The temperature of the calorimeter changes by 4.48 C. What are E (change in internal energy) and Em (change in internal energy per mole of carbon) for the process.

  26. Example: 1.412 g of carbon (M = 12.01 g/mol) are burned in a bomb calorimeter (C = 10325. J/C). The temperature of the calorimeter changes by 4.48 C. What are E (change in internal energy) and Em (change in internal energy per mole of carbon) for the process. q = - C T = - (10325. J/C) (4.48 C) = - 46300. J For a process carried out at constant volume q = E, so E = - 46300. J Finally, the change in internal energy per mole of carbon is Em = E/n n = 1.412 g 1 mol = 0.1176 mol 12.01 g Em = - 46300. J = - 393000. J/mol = - 393. kJ/mol 0.1176 mol

  27. Constant Pressure Calorimetry Just as bomb calorimetry is carried out under conditions of constant volume, we can also do calorimetry under conditions of constant pressure. In a coffee cup calorimeter a chemical reaction is carried out under conditions of constant pressure. For a reaction in solution qsoln = msoln . Cs,soln . T H = qsyst = - qsoln

  28. Standard Conditions and Standard State It is convenient to report thermodynamic data for a particular set of conditions. These conditions are called the thermodynamic standard conditions, and are usually (but not always) taken to be p = 1.00 atm*, T = 25.0 C = 298.2 K. The standard state for an element is the most stable form of the element for standard conditions. carbon C(s) nitrogen N2(g) bromine Br2() oxygen O2(g) iron Fe(s) argon Ar(g) mercury Hg () chlorine Cl2(g) sulfur S(s) __________________________________________________________ * Technically, standard pressure is now taken to be 1.00 bar = 0.987 atm. This makes only a minor difference in values for thermodynamic quantities.

  29. Enthalpy of Reaction (Hrxn) Because of the importance of enthalpy, we define enthalpy changes for specific kinds of processes. Hrxn (enthalpy of reaction) - The enthalpy change when one mole of a specific reaction is carried out at p = 1.00 atm (); equals qp for the process. Examples: CaCO3(s)  CaO(s) + CO2(g) Hrxn = + 178.3 kJ/mol CaO(s) + CO2(g)  CaCO3(s) Hrxn = - 178.3 kJ/mol 2 CaCO3(s)  2 CaO(s) + 2 CO2(g) Hrxn = + 356.6 kJ/mol HCl(aq) + NaOH(aq)  NaCl(aq) + H2O() Hrxn = - 55.8 kJ/mol Notice that changing the direction of the reaction changes the sign for Hrxn , and multiplying a reaction by a constant multiplies the value for Hrxn .

  30. We can see this as follows: CaCO3(s)  CaO(s) + CO2(g) Hrxn = + 178.3 kJ/mol CaO(s) + CO2(g)  CaCO3(s) Hrxn = ? __________________________________________________________ CaCO3(s)  CaCO3(s) Htotal = 0.0 kJ/mol Carrying out the first process followed by the second process means that our final state is the same as our initial state. There is no change in the system, and so Htotal = 0.0 kJ/mol. Therefore, the change in enthalpy for the second process must be the same as the change in enthalpy for the first process, except for the sign.

  31. Relationship Between Hrxn and Erxn For a process carried out at constant pressure H = E + pV If V > 0 then H > E V = 0 then H = E V < 0 then H < E Now consider a chemical reaction. Since Vg >> Vs, V it follows that V = Vf - Vi  Vg,f - Vg,i If the gases are ideal, then p V = pVg,f - pVg,i = ng,fRT - ng,iRT = ngRT where ng = ng,f - ng,i So Hrxn = Erxn + pV  Erxn + ngRT

  32. Use of Thermodynamic Relationships When ignited magnesium burns with a bright white flame. The balance chemical equation for the process is 2 Mg(s) + O2(g)  2 MgO(s) Hrxn = - 1203.2 kJ/mol Consider carrying out the above process at standard thermodynamic conditions. What are q, w, and E for the process? Also, do you expect Hrxn to be greater than, about the same, or less than Erxn?

  33. When ignited magnesium burns with a bright white flame. The balance chemical equation for the process is 2 Mg(s) + O2(g)  2 MgO(s) Hrxn = - 1203.2 kJ/mol Consider carrying out the above process at standard thermodynamic conditions. What are q, w, and E for the process? Process is carried out at constant pressure, so q = H = - 1203.2 kJ/mol H = E + ngRT , so E = H - ngRT ng = 0 - 1 = -1 (- 1 mol/mol) E = - 1203.2 kJ/mol - (-1)(8.314 x 10-3 kJ/mol.K)(298. K) = - 1200.7 kJ/mol Finally, E = q + w, so w = E - q = (-1200.7 kJ/mol) - (- 1203.2 kJ/mol) = + 2.5 kJ/mol

  34. Enthalpy of Formation (Hf) The formation reaction for a substance is defined as the reaction that produces one mole of a single product out of elements in their standard state. Because of the way we have defined the formation reaction, we may have to use fractional stoichiometric coefficients for some or all of the reactants. The enthalpy change for this reaction is defined as the enthalpy of formation for the substance. H2(g) + 1/2 O2(g)  H2O() Hf(H2O()) = - 285.8 kJ/mol H2(g) + 1/2 O2(g)  H2O(g) Hf(H2O(g)) = - 241.8 kJ/mol C(s) + O2(g)  CO2(g) Hf(CO2(g)) = - 393.5 kJ/mol Pb(s) + C(s) + 3/2 O2(g)  PbCO3(s) Hf(PbCO3(s)) = - 699.1 kJ/mol 3/2 O2(g)  O3(g) Hf(O3(g)) = + 143.0 kJ/mol Example: Write the formation reaction for acetone (CH3COCH3()).

  35. Example: Write the formation reaction for acetone (CH3COCH3()). 3 C(s) + 3 H2(g) + ½ O2(g)  CH3COCH3()

  36. Note that based on the above definition it follows that the enthalpy of formation of an element in its standard state is 0.0 kJ/mol. We may see this as follows: The formation reaction for N2(g), by definition (formation of one mole of a single product out of elements in their standard state), is N2(g)  N2(g) But nothing happens in the above process, and so it follows that Hrxn = Hf(N2(g) ) = 0.0 kJ/mol This will be true for any element in its standard (thermodynamically most stable) state.

  37. Enthalpy of Combustion (Hc) The combustion reaction for a substance is defined as the reaction of one mole of a single substance with O2(g) to form combustion products. Because of the way in which we have defined the combustion reaction we may have to use fractional coefficients for some of the reactants and products. The enthalpy change for this reaction is defined as the enthalpy of combustion for the substance. Combustion products for common elements are C  CO2(g) H  H2O() N  N2(g) CH4(g) + 2 O2(g)  CO2(g) + 2 H2O() Hc(CH4(g)) = - 890.3 kJ/mol C6H6() + 15/2 O2(g)  6 CO2(g) + 3 H2O() Hc(C6H6()) = - 3267.4 kJ/mol

  38. Writing Formation and Combustion Reactions We may use the definition of formation reaction and combustion reaction to write own the balanced chemical equations corresponding to these reactions. Example: Hexane (C6H14()) is a hydrocarbon often used as a solvent in organic reactions. Write the formation reaction and the combustion reaction for hexane. CH3CH2CH2CH2CH2CH3

  39. Example: Hexane (C6H14()) is a hydrocarbon often used as a solvent in organic reactions. Write the formation reaction and the combustion reaction for hexane. Formation - One mole of a single product out of elements in their standard state. C6H14() 6 C(s) + 7 H2(g)  C6H14() (balanced) Combustion - One mole of a single reactant, plus oxygen, to form combustion products. C6H14() + ? O2(g)  C6H14() + ? O2(g)  6 CO2(g) + 7 H2O() C6H14() + 19/2 O2(g)  6 CO2(g) + 7 H2O() (balanced)

  40. Enthalpy Change For Phase Transitions There are three phase transitions that can occur by adding heat to a substance. s   fusion (melting) Hfus   g vaporization Hvap s  g sublimation Hsub The enthalpy change when one mole of a substance undergoes the transition at the normal transition temperature is defined as the enthalpy change for the phase transition. It represents the amount of heat required to convert one mole of the substance from the initial phase to the final phase. H2O()  H2O(g) Hvap(H2O) = 40.7 kJ/mol at T = 100.0 C Unlike other processes, the temperature for a phase transition is usually taken to be the temperature for which the two phases exist at equilibrium when p = 1.0 atm.

  41. Hess’ Law We previously noted that the change in the value for a state function depends only on initial and final state and is independent of the path used to travel between the two states. We may put this in a more formal manner in terms of Hess’ law. Hess’ law – The change in value for any state function will be the same for any process or combination of processes that have the same initial and final state. We are particularly interested in applying Hess’ law to chemical reactions.

  42. Hess’ Law For a Chemical Reaction Let us use Hess’ law to find the value for (Hrxn) for the following chemical reaction: CaCO3(s)  CaO(s) + CO2(g) Hrxn We may obtain this same reaction as follows: Ca(s) + ½ O2(g)  CaO(s) Hf(CaO(s)) C(s) + O2(g)  CO2(g) Hf(CO2(g)) CaCO3(s)  Ca(s) + C(s) + 3/2 O2(g) - Hf(CaCO3(s)) CaCO3(s)  CaO(s) + CO2(g) Hrxn So by Hess’ law, Hrxn = Hf(CaO(s)) + Hf(CO2(g)) - Hf(CaCO3(s)) We can get Hrxn using only data on enthalpies of formation!

  43. General Method For Finding Hrxn Based on the same procedure used in the previous example the following general relationship can be derived Hrxn = [  Hf(products) ] - [  Hf(reactants) ] Notice what this means. If we have a table for formation enthalpies we can find the value for Hrxn for any chemical reaction. In fact, the same general procedure can be used to find the values for the change in any state function. Also note that when we use the superscript  this indicates not only standard conditions but also standard concentrations for reactants and products. gases p = 1.0 atm solutes [M] = 1.0 mol/L solids, liquids, solvents must be present in the system

  44. Example: Find Hrxn for the conversion of acetylene into dichloroethane C2H2(g) + 2 HCl(g)  CH2ClCH2Cl()

  45. Example: Find Hrxn for the conversion of acetylene into dichloroethane C2H2(g) + 2 HCl(g)  CH2ClCH2Cl() Hrxn = [Hf(CH2ClCH2Cl()) ] - [Hf(C2H2(g)) + 2 Hf(HCl(g))] = [ (- 165.2 kJ/mol) ] - [ (226.7 kJ/mol) + 2 ( - 92.3 kJ/mol) ] = - 207.3 kJ/mol Thermodynamic data are given in Appendix II-B.

  46. Other Uses of Hess’ Law Hess’ law can be used to find Hrxn for any process that can be written as a combination of other chemical reactions whose values for Hrxn are known. Example (6.71): Consider the following formation reaction 5 C(s) + 6 H2(g)  C5H12() Hrxn = ? Using the following information find Hrxn for this reaction (1) C5H12() + 8 O2(g)  5 CO2(g) + 6 H2O(g) Hrxn = - 3505.8 kJ/mol (2) C(s) + O2(g)  CO2(g) Hrxn = - 393.5 kJ/mol (3) 2 H2(g) + O2(g)  2 H2O(g) Hrxn = - 483.5 kJ/mol Note all of the above are combustion reactions, which are particularly easy to carry out experimentally.

  47. Example (6.71): Consider the following formation reaction 5 C(s) + 6 H2(g)  C5H12() Hrxn = ? Using the following information find Hrxn for this reaction (1) C5H12() + 8 O2(g)  5 CO2(g) + 6 H2O(g) Hrxn = - 3505.8 kJ/mol (2) C(s) + O2(g)  CO2(g) Hrxn = - 393.5 kJ/mol (3) 2 H2(g) + O2(g)  2 H2O(g) Hrxn = - 483.5 kJ/mol reverse 1st reaction 5 CO2(g) + 6 H2O(g)  C5H12() + 8 O2(g) Hrxn = + 3505.8 kJ/mol 5 times the 2nd reaction 5 C(s) + 5 O2(g)  5 CO2(g) Hrxn = - 1967.5 kJ/mol 3 times 3rd reaction 6 H2(g) + 3 O2(g)  6 H2O(g) Hrxn = - 1450.5 kJ/mol 5 C(s) + 6 H2(g)  C5H12() Hrxn = + 87.8 kJ/mol

  48. End of Chapter 6 “In this house we obey the laws of thermodynamics!” - Homer Simpson “...the Dutch physicist Heike Kamerlingh Onnes gave H the name enthalpy, from the Greek  (in) and  (heat), or from the single Greek word  (enthalpos), to warm within.” K. J. Laidler, The World of Physical Chemistry “[Thermodynamics] is the only physical theory of universal content which, within the framework of the applicability of its basic concepts, I am convinced will never be overthrown.” Albert Einstein

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