Section9_Thermochemistry2

Section9_Thermochemistry2 - THERMOCHEMISTRY II By the end...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: THERMOCHEMISTRY II By the end of this lecture, you will be able to: (1) Define calorimetry, heat capacity and specific heat capacity. (2) Contrast constant pressure and constant volume calorimetry. (3) Understand standard states and standard enthalpies of formation. (4) Use Hesss Law and standard enthalpies of formation to calculate heats of reaction. (5) Understand how to predict enthalpies of formation using bond dissociation energies. Page 1 of 21 1 calorie = the energy req'd to raise the temp of 1g H2O by 4.183 J 1C (or 1KJ) Calorimetry Calorimetry is the science of measuring heat changes. 1 Calorie = 10^3 Principle of measurement: Energy released by reaction = energy absorbed by calorimeter or Energy required by reaction = energy given up by calorimeter i.e., Ereaction = Ecalorimeter calorie Page 2 of 21 Heat Capacity and Specific Heat Capacity Heat capacity, C, is the energy required to raise the temperature of an object by 1 C. C = heat absorbed____ increase in temperature E = CT Specific heat capacity, c, is the energy required to raise the temperature of 1 kg of a substance by 1 C. c = heat absorbed per kg increase in temperature {J kg C-1 = J kg-1 / C} E = mcT Page 3 of 21 Constant Pressure Calorimetry The "coffee cup calorimeter" E q qP =q+w = E - w = E + PV = H 0 Recall: Enthalpy is heat at constant pressure. Therefore, enthalpy is measured by the CCC. (Note: we cannot measure absolute enthalpies. We can only measure changes in enthalpy.) V will be small (because liquids do not expand appreciably with temperature), but not 0, i.e., a very small amount of work is done. Page 4 of 21 For an exothermic reaction at constant P: energy released = energy absorbed by by reaction solution = specific heat capacity of solution mass of solution increase in T = Hreaction The specific heat capacity of the calorimeter itself (the cup, the stirrer, the thermometer, the lid) is often ignored because it is much less than that of the solution contained in the cup. c m T Page 5 of 21 When 1.0 L of 1.0 M Ba(NO3)2 solution at 25 C and 1.0 L of 1.0 M Na2SO4 are mixed, the temperature of the mixture increases to 28.1 C. Assume the calorimeter absorbs no heat the specific heat capacity of the solution is 4.18 103 J kg-1 C-1 the density of the mixture is 1.0 g L-1. What is the enthalpy change per mole of BaSO4 formed? Vtot = 1 + 1 = 2 L T = Tf - Ti = 28.1-25.0 = 3.1C m = Vtot + p = 2 kg Heat lost by the system = heat taken up by solution = spec heat cap * mass * T = 4.183*10^3 J kg^-1C^-1 * 2Kg * 3.1 C = 25.9 kJ Ba(NO)(aq) + NaSO !TM BaSO(s) + 2NaNO(aq) @ const P therefore qp = H H = -25.9 kJ/mol BaSO system loses energy Page 6 of 21 Constant Volume Calorimetry The "bomb calorimeter" E = q + w q = E - w qV = E 0 w = PV conrtast with qp = H Internal energy is measured. Because V = 0, w = 0, i.e., no work is done by the system. For an exothermic reaction at constant V: energy released = energy absorbed by by reaction calorimeter = heat capacity of calorimeter specific calorimeter not a specific heat capacity, but a heat capacity increase in T = Ereaction Page 7 of 21 In this instance, the calorimeter is massive with an appreciable heat capacity. Note: For both types of calorimeter, the system is the components of the reaction you are studying. The surroundings are the calorimeter and its contents. Thus, an exothermic reaction will warm the calorimeter and an endothermic one will cool it. Page 8 of 21 When 0.5269 g of octane (C8H18) is burned in an excess of O2 in a bomb calorimeter with heat capacity of 11.3 kJ C-1, the temperature of the calorimeter increases by 2.25 C. What is the heat of combustion of octane in kJ mol-1? CH + 25/2 O !TM 8CO +9HO mw(CH) = 8 * 12.011 + 18 * 1.008 = 114.2 g/mol T = Tf - Ti = 2.25C E released by reaction = E absorbed by calorimeter = heat cap of calorimeter * T = 11.3 kJC^-1 * 2.25C = 25.4 kJ n(CH) = 0.5296g / 114.2g/mol = 4.614*10^-3 mol heat of combustion = -25.4 kJ / 4.614*10^-3 mol = 5.5 * 10^3 kJ/mol system must decrease in energy Page 9 of 21 Standard Enthalpies of Formation The standard enthalpy of formation, Hf , is the enthalpy change that results from the formation of 1 mol of a compound from its elements, with all components in their standard states. The degree symbol () indicates that the corresponding process has been performed under standard conditions. Standard conditions: 1 atm 298 K 1.00 M concentration for solutions (Do not confuse this with STP, which is 1 atm and 0 C!) Page 10 of 21 The standard state of a substance is the state in which it is found under standard conditions. e.g., Substance Water Mercury Gold Argon Standard state H2O(l) Hg(l) Au(s) Ar(g) The enthalpy of formation of an element in its standard state is defined to be zero. Hf(element) = 0 Page 11 of 21 Consider the enthalpy of formation of nitrogen dioxide: N2(g) + O2(g) NO2(g) Hf = 34 kJ mol-1 One mole of the product is formed. The reactants and products are in their standard states. Some standard enthalpies of formation Hf (kJ mol-1) Compound - 46 NH3(g) - 286 H2O(l) - 394 CO2(g) - 239 CH3OH(l) NO2(g) 34 Page 12 of 21 How is this useful? Combining Hesss Law and standard enthalpies of formation allows us to calculate enthalpies for other reactions. e.g., What is the enthalpy of combustion of methane (Hreaction)? CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) Remember: Hesss Law allows us to choose any convenient pathway! Well choose: C(s) + 2 H2(g) + 2 O2(g) CO2(g) + 2 H2O(l) (1) Break down CH4 into its elements. CH4(g) C(s) + 2 H2(g) Page 13 of 21 H1 This is just the reverse of the formation reaction: C(s) + 2 H2(g) CH4(g) Hf = -75 kJ mol1 So, all we have to do is reverse the sign of Hf to get H1 = 75 kJ mol-1. (2) Allow H2 and O2 to react to give water. H2(g) + O2(g) H2O(l) Hf = -286 kJ mol-1 But we need 2 moles of H2O for our overall reaction: 2 H2(g) + O2(g) 2 H2O(l) H2 = -572 kJ mol-1 Page 14 of 21 (3) Allow C and O2 to react to give CO2. C(s) + O2(g) CO2(g) Hf = -394 kJ mol-1 = H3 Hreaction = H1 + H2 + H3 = 75 - 572 - 394 kJ mol-1 = -891 kJ mol-1 Note that there is no H term associated with O2(g) because it is an element in its standard state. Generally speaking: Hreaction = nPHf(product) - nRHf(reactant) where nP and nR are the numbers of moles or products and reactants respectively. Page 15 of 21 Calculate the enthalpy of synthesis of diborane from its elements 2 B(s) + 3 H2(g) B2H6(g) given the following data: Reaction 2 B(s) + 3/2 O2(g) B2O3(s) B2H6(g) + 3 O2(g) B2O3(s) + 3 H2O(g) H2(g) + O2(g) H2O(l) H2O(l) H2O(g) 1 Assemble the reactants H (kJ) -1273 -2035 -286 44 2 Assemble the products 3H2O(g) + B2O3 ---> B2H2 + 3O2 -H3 Add 1 + 2 2B + 3/2O2 --> B2O3 H1 2B + 3H2 + 3O2 --> B2H6 + 3O2 + B2O3 3* H2 + 1/2O2 --> H2O(l) 3H2 + 3H2O(g) + B2O3 + 3H2O(l) _____________________________________ Fix the H2O state problem 2B + 3H2 + 3O2 ---> B2O3 + 3H2O(l) 3 (H2O (l) ---> H2O(g)) 3H4 Add 3 + 4 reactants 2B + 3H2 + 3H2O(g) ---> B2H6 + 3H2O(l) + 3H2O(l) 3H2O(g) Hrxn = H1 + 3H2 - H3 + 3H4 = -1273 + 3(-286) - (-2035) + 3(44) = + 36 kJ 3 4 Page 16 of 21 What is the heat of combustion of methanol (CH3OH) in kJ mol-1? -239 0 -394 -286 2CH3OH + 3O2 -------> 2CO2 + 4H2O np Hf(products) = 2(-394) + 4(-286) = -1.932*10^3 kJ np Hf (reactants) = 2(-239) + 3(0) = - 478 kJ Hrxn = -1932*10^3 - -478 = -1.454*10^3 kJ H(combustion of methanol) = Hrxn/2 = -727 kJ/mol Hf Page 17 of 21 Bond dissociation energies If we dont know Hf for a compound, we can often estimate it by considering bond energies. Recall: Bond breaking requires energy Bond formation releases energy The stronger the bond, the more energy is involved. The energy required to break 1 mole of bonds is known as the bond dissociation energy, D. This energy is released when 1 mol of the bonds are formed. Page 18 of 21 Consider: H2(g) + F2(g) 2 HF(g) Hreaction can be estimated by the sum of the energies of the bonds broken minus the sum of the energies of the bonds formed. Hreaction = D(H-H) + D(F-F) - 2 D(H-F) = 1 mol 432 kJ mol-1 + 1 mol 154 kJ mol-1 - 2 mol 565 kJ mol-1 = - 544 kJ -544/2 = -272 kJ/mol estimated value where D(X-Y) represents the bond energy per mole of X-Y bonds. Note: D always has a positive sign. Hf for HF = - 271 kJ mol-1 Page 19 of 21 tabulated value In this reaction, 2 mol of HF are formed, so Hreaction = 2 mol - 271 kJ mol-1 = - 542 kJ mol-1 So, the "bond method" gives a pretty good estimate. (Breaks down for complex molecules, however.) Generally speaking: H = D(bonds broken) - D(bonds formed) energy released Energy Required Page 20 of 21 Calculate H for the reaction given below using the tabulated bond dissociation energies (kJ mol-1). CH4(g) + 2Cl2(g) + 2F2(g) CF2Cl2(g) + 2HF(g) + 2HCl(g) D(C-H) = 413 D(Cl-Cl) = 239 D(F-F) = 154 D(C-F) = 485 D(C-Cl) = 339 D(H-F) = 565 D(H-Cl) = 427 reactants --------> atoms -------> Products Bond bond breaking forming (requires E) (releases E) 1) Bonds 4 * 2 * 2 * broken: C-H = 4*413 Cl-Cl = 2*239 F-F = 2*154 = 1652 = 478 = 308 _____ 2438 2) Bonds 2 * 2 * 2 * 2 * Formed C-F C-Cl H-F H-Cl 3632kJ/mol -1194kJ/mol Page 21 of 21 ...
View Full Document

Ask a homework question - tutors are online