ENTHALPY CHANGE CHAPTER 4
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1 ENTHALPY CHANGE CHAPTER 4
2 ENTHALPY Is the total energy of a system. E k = Kinetic energy. Vibrational Rotational Translational E due to motion H = E k + E p E P = Potential energy Attractive force b/w nucleons & electrons E stored by the bonds between atoms & molecules. H is the change in total energy of the system during a rxn or phase change. Most of the E exchanged is in the form of heat! Therefore we use the change in heat as a indicator of the change in enthalpy expressed per mol! Q mol = H at constant T & P
3 H PHASE CHANGES Change in E p Energy breaks intermolecular bonds Energy increases the speed of molecules Change in E k
4 HOW TO CALCULATE H MELTING ICE VAPOUR Q ice = mc T (c=2.05 J/g ) + At the melting point H = 6.01kJ/mol until all the ice is melted! + Q water = mc T (c=4.19 J/g ) + At the boiling point H = kj/mol until all the water is vaporised! + Q vapour = mc T (c=1.41 J/g )
5 Gas Exothermic Process Terminology for phase changes Endothermic Process Solid Melting (fusion) Freezing (solidification) Liquid Liquid
6 Temperature ( C) HEAT CURVE OF A PURE SUBSTANCE solid & liquid ΔH fus Heat of fusion c (l) = H (l) T (l) ΔH (l) liquid & gas ΔT (l) ΔH vap Heat of vaporization Specific heat capacity of the liquid can be determined from the inverse slope of the curve Energy absorbed by substance (joules / gram)
7 Temperature ( C) REVERSING A HEAT CURVE If instead we cool a hot gaseous substance. liquid & gas boiling point condensation point. heat of vaporization --> heat of condensation. ΔH cond Heat of condensation Energy (joules / gram) released melting point freezing point. heat of vaporisation heat of solidification. solid & liquid ΔH s Heat of solidification
8 IMPORTANT PTS FROM A HEAT CURVE: The melting/freezing point temperature (1 st plateau) The boiling point temperature/vaporization/condensation/ (2 nd plateau) The heat of fusion (ΔH fus ) or solidification (ΔH s ) (change in x-axis) The heat of vaporization (ΔH vap ) or condensation (ΔH cond ) (x-axis) The specific heat capacity (c) of the solid or liquid substance. Calculated from the H/g and T from the inverse slope Since Q=mc T & Q/m = H(J/g) from the graph so c = H T
9 Temperature ( C) HEAT CURVE OF A MIXTURE The mixture melts more gradually than pure matter. The heat curve of a mixture of substances. The mixture melts over a range of temperatures, and each component boils at a different temperature. We can use the different boiling temperatures of the mixtures components to separate them by capturing the vapours separately a process called fractional distillation Energy absorbed by substance (joules / gram)
10 Temperature ( C) SAMPLE PROBLEM IMAGINARY HEAT CURVE Use the heat curve to determine: a) The melting point b) The boiling point c) The specific heat capacity of liquid d) the heat of fusion of e) the heat of vaporization of Energy absorbed by substance (joules / gram)
11 Temperature ( C) SAMPLE PROBLEM IMAGINARY HEAT CURVE Use the heat curve to determine: a) The melting point b) The boiling point c) The specific heat capacity of liquid d) the heat of fusion of e) the heat of vaporization of this substance ΔH fus = ( )= 200 J/g Heat of fusion ΔT (l) = (70-(-20)) =90 ΔH (l) = ( ) =400J/g 400 J/g c (l) = = 4.44 J/g 90 ΔH vap =( )=400 J/g Heat of vaporization Energy absorbed by substance (joules / gram) Answers: a) 20 C b) +70 C c) 4.44 J/g C d) 200 J/g* e) 400 J/g* *Heat of fusion and vaporization are given in J/g. To convert to the more standard form ( J/mol) we could multiply this answer by the molar mass.
12 WORKSHEET DUE NEXT CLASS
13 Today: 1. Expressing H as a simplified graph 2. Expressing H within a rxn
14 The bonds in C + O 2 contain more energy than the bonds in CO 2. Energy is released (exothermic)( H is neg) Written as: C (s) + O 2 (g) CO 2 (g) kj/mol or C (s) + O 2 (g) CO 2 (g) H= kj/mol
15 The bonds in SnO 2 contain less energy than the bonds in Sn + O 2. Energy is absorbed (endothermic)( H is pos) Written as: SnO 2 (g) kj/mol Sn (s) + O 2 (g) or SnO 2 (g) Sn (s) + O 2 (g) H= kj/mol
16 EXAMPLES Determine the sign of H; state exothermic or endothermic & draw energy diagram for each of the following: 1. H 2(g) + ½ O 2(g) H 2 0 (g) kJ/mol kJ/mol + H 2 0 (l) H 2 0 (g) 3. When nitroglycerine decomposes, it causes a violent explosion, releasing 5.72 x 10 3 kj of heat for each mole. C 3 H 5 (NO 3 ) 3 3CO 2 + 5/2 H 2 O + ¼O 2 + 3/2N 2
17 1. H 2 (g) + ½ O 2 (g) H 2 0(g) kJ H = KJ IS NEGATIVE REACTION IS EXOTHERMIC Graph H 2(g) + ½ O 2(g) H = kj H (kj) H 2 0 (g) Reaction path
18 kJ +H 2 0(l) H 2 0(g) H IS POSITIVE REACTION IS ENDOTHERMIC Graph H 2 0 (g) H (kj) H = 40.7 kj H 2 0 (l) Reaction path
19 3. C 3 H 5 (NO 3 ) 3 3CO H 2 O + ¼O N 2 H IS NEGATIVE REACTION IS EXOTHERMIC C 3 H 5 (NO 3 ) 3 H = x 10 3 kj/mol H (kj) 3CO 2 + 5/2 H 2 O + ¼O 2 + 3/2N 2 Reaction path
20 Expressing H as a graph and calin 4 different ways 1. Q = mc T & H = Q Q = energy expressed in Joules. n H = energy expressed in Joules/mol. 2. Using table with H f H rxn = H f products H f (reactants) 3. Using bond energies H rxn = σ E of broken bonds (σ E of bonds formed) 4. Hess s Law (adding known equations)
21 Using a calorimeter chemist have formed for: Heat of formation ( H f ) = energy stored in a substance during its Heat of fusion ( H fus ) Heat of vaporisation ( H vap ) Heat of dissolution ( H d ) Heat of combustion ( H comb ) Heat of a certain reaction ( H rxn ) etc formation from elements melting boiling Textbook appendix 8
22 HAND-OUT BLUE TABLE! Please note! The heat of formation ( H f ) is the energy stored in a substance during its formation from elements. So substances in their elemental form have a H f = 0. H f of O 2 = 0. H f of Ag = 0. The solid metal not the ion! H f of CO= kj/mol because it is made from C + O 2
23 How do you calculate H rxn from H f? H rxn = H f products H f (reactants) Ex 1. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) kj/mol H rxn = H f products H f (reactants) = ( (-285.8))-((-74.4) + 2(0)) = (-965.1) (-74.4) = kj/mol of methane Exothermic! Graph H f Bond energies Hess s Law
24 Ex 2. H 2 (g) + Cl 2 (g) 2HCl (g) H rxn = kj/mol H rxn = H f products H f (reactants) = (2(-92.3))-(0 + 0) = (-184.6) (0) = kj/mol of hydrogen Exothermic!
25 STOP WORKSHEET USE TABLE OF H f TO SOLVE
26 How do you calculate H rxn from bond energies? Find the energy difference in making and breaking bonds. Based on average bond enthalpies. Provides a rough estimate! Graph H f Bond energies Hess s Law H rxn = σ E of broken bonds (σ E of bonds formed)
27 Ex 3. H 2 (g) + Cl 2 (g) 2HCl (g) H rxn = σ E of broken bonds (σ E of bonds formed) = ( )-(2(432)) = (679) (864) = -185 kj/mol of hydrogen 2x P 419 Bond H-H H-O H-F H-Cl C-H C-C C=C C-O O=O C=O Cl-Cl N=N N=O Energy 436 kj/mol 460 kj/mol 570 kj/mol 432 kj/mol 413 kj/mol 347 kj/mol 607 kj/mol 358 kj/mol 498 kj/mol 745 kj/mol 243 kj/mol 418 kj/mol 631 kj/mol
28 WHY DO AN ESTIMATE???? Can t find H f Dealing with a macromolecule
29 Use bond energies to determine the enthalpy of the following rxns. 2H 2 + O 2 2 H 2 O H rxn = broken (formed) = (2(436) + 498) (4(460)) = (1370) (1840) = -470 kj/mol H rxn = broken (formed) = (5(413) ) (4(413)+607+2(460)) = (3230) (3179) = 51 kj/mol Lewis dot diagrams help to determine what bonds are present. Bond H-H H-O H-F H-Cl C-H C-C C=C C-O O=O C=O Cl-Cl N=N N=O Energy 436 kj/mol 460 kj/mol 570 kj/mol 432 kj/mol 413 kj/mol 347 kj/mol 607 kj/mol 358 kj/mol 498 kj/mol 745 kj/mol 243 kj/mol 418 kj/mol 631 kj/mol
30 STOP - PRACTICE
31 S (s) O 2(g) SO 3(g) SOME REACTIONS OCCUR IN MULTIPLE STEPS. S (s) + O 2(g) SO 2(g) H = kJ SO 2(g) + ½ O 2(g) SO 3(g) H = -99.2kJ EACH STEP HAS AN ENTHALPY VALUE.
32 Hess Law of Constant Heat Summation: The enthalpy of a rxn is the same whether it happens in one step or in a series of steps. Therefore enthalpy values of known rxns can be added to give the enthalpy of a needed rxn. + S (s) + O 2(g) SO 2(g) SO 2(g) + ½ O 2(g) SO 3(g) H = kJ/mol H = -99.2kJ/mol S (s) O 2(g) SO 3(g) H = -396kJ/mol
33 Rules for rearranging equations: Step 1: Rewrite so that the reactants & products are on the desired side. (Switch sign of H when reversing an equation) Step 2: Adjust for the # of moles by multiplying the entire equation. (You also multiply the H value!)
34 You are given. C (s) + O 2(g) CO 2(g) ΔH = kj/mol CO (g) + ½ O 2(g) CO 2(g) ΔH = kj/mol Use the above rxns to create the desired reaction & calculate the enthalpy C (s) + ½O O 2(g) 2 CO 2(g) ΔH = kj/mol + CO 2(g) CO (g) + ½ O 2(g) ΔH = +283 kj/mol We want C (s) + ½ O 2(g) CO (g) ΔH =? H = kj/mol
35 GIVEN: 2 Al (s) + 3/2 O 2(g) Al 2 O 3(s) ΔH = kj/mol 2 Fe (s) + 3/2 O 2(g) Fe 2 O 3(s) ΔH = kj/mol Calculate the unknown ΔH for the reaction below. Fe 2 O 3(s) + 2 Al (s) Al 2 O 3(s) +2 Fe (s)
36 2Al (s) + 3/2O 2(g) Al 2 O 3(s) ΔH = kj/mol + Fe 2 O 3(s) 2 Fe (s) + 3/2 O 2(g) ΔH = +838 kj/mol Fe 2 O 3(s) + 2 Al (s) Al 2 O 3(s) +2 Fe (s) ΔH = -837 kj/mol
37 Ex. The formation of nitric acid is given by the following equation: 12NH O 2 8HNO 3 + 4NO + 14H 2 O Given the equations below, calculate the ΔH for nitric acid. NH 3 + 5/4 O 2 NO + 3/2H 2 O NO + ½ O 2 NO 2 H = -293 kj/mol H = -59 kj/mol NO 2 + 1/3H 2 O 2/3HNO 3 + 1/3NO H = -45 kj/mol
38 12 NH 3 + 5/4 15 O 2 12 NO /2H 2 O H = -293 kj x12 H = kj 12 NO 2 + 1/3H 4 2 O 2/3HNO /3NO 4 H H = kj kj x12 12 NO + ½ 6 O 2 12 NO 2 H H = kj kj x NH O 2 8HNO 3 + 4NO + 14H 2 O ΔH = - 4,764 kj/mol
39 We have found four ways to calculate enthalpy, H: Q = mc T & H = Q Q = energy expressed in Joules. n H = energy expressed in Joules/mol. H rxn = H f products H f (reactants) H rxn = σ E of broken bonds (σ E of bonds formed) Energy is needed to break bonds but is released when bonds are formed. Therefore the equation is energy absorbed energy released. Hess s Law H rxn = positive # means more E is absorbed endothermic rxn H rxn = negative # means more E is released exothermic rxn
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