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PSU CHEM 110 - Thermochemistry 2

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Sheets Page 1 Lecture 39 Lecture 39: Thermochemistry 2 Read: BLB 5.6–5.7; 8.8 HW: BLB 5:63,67a,b,69,73,75,83,85; 8:65a,67a,c,92a 18:72a,b, 74 Sup 5:1–7; 8:11–13 Know: • Hessʼs Law • heats of formation • enthalpy of reactions • estimating reaction enthalpy from bond energies Check out the grad-u-lator on the Chem110 website FINAL SKILL CHECK TEST DEADLINE: MONDAY, APRIL 27 Need help?? Get help!! TAs in CRC (211 Whitmore) and SI—hours on Chem 110 website; my office hours (Mon 12:30-2 & Tues 10:30-12 in 324 Chem Bldg [or 326 Chem]) Final Exam: MONDAY, May 4, 12:20 pm Concept Final review session with Sheets: Thursday 4/30 @ 6pm in 108 Forum. Please work through the Concept Exam (on lecture note page) before the review session & bring it with you, along with any questions you may have. This review is meant to complement to the review sessions that your TAs will be holding in which they will go over the practice exams.Sheets Page 2 Lecture 39 Enthalpy review • enthalpy is an property (how much stuff you have) • ΔH for a reaction is in magnitude & in sign to ΔH for the reverse reaction • ΔH for a reaction depends on the of the reactants and products (gas, liquid, solid) • ΔH for a sum of steps is the same as ΔH for the overall process (Hessʼs law)—Why is this true? ΔH is a state function— path-independent, history-independent & depends only on current states (T, P, etc.)Sheets Page 3 Lecture 39 Hessʼs law • ΔH for an overall reaction is equal to the sum of the individual steps • a consequence of ΔH being a state functionSheets Page 4 Lecture 39 Example: Given the following information: ΔH 2 SO2(g)+ O2(g) → 2 SO3(g) –196 kJ 2 S(s) + 3 O2(g) → 2 SO3(g) –790 kJ What is ΔHrxn for S(s) + O2(g) → SO2(g) ?Sheets Page 5 Lecture 39 Heat of formation • enthalpy of formation: ΔHf ; heat released or absorbed when combine to form a compound combine elements ⎯⎯⎯⎯→ compound ΔHf • standard enthalpy of formation: ΔHf° ; ΔH to form 1 mole of the compound from its elements which are all in their standard states • standard state: most stable form of element (gas/liquid/solid) at standard temperature and pressure [T = 25°C (298 K); P = 1 atm] • by definition, an element in its standard state has ΔHf° = 0 ΔHf° of O2(g), S(s), C(graphite) = 0 ΔHf° of C(diamond) ≠ 0 • NOTE: ΔHf° values found in BLB Appendix CSheets Page 6 Lecture 39 Heat of formation (cont.) Examples: ΔHf° = –238.6 kJ • reactants = elements in std state NOTE: phase (state) is important (l, s, g) & these reactants might not have whole numbers for coefficients. Why?? Because we are making 1 mole of the product! 1 atm Br2(l) → Br2(g) ΔH°f = 30.71 kJ 25°C C (graphite)2H2(g)1/2 O2(g) CH3OH(l)1 atm25°CSheets Page 7 Lecture 39 Example: For which of the following reactions (at 25°C & 1 atm) does ΔH°rxn = ΔH°f? H2(g) + F2(g) ⎯→ 2HF(g) NO(g)+ 1/2 O2(g) → NO2(g) 2C(graphite) + 3H2(g) + 1/2 O2(g) ⎯→ C2H5OH (l) Pb(s) + Cl2(g) ⎯→PbCl2(s) S(s) + O3(g) ⎯→ SO3(g)Sheets Page 8 Lecture 39 Standard enthalpy of a reaction (ΔH°rxn) • heat of reaction (ΔHrxn) when all reactants and products are in their standard states • obtain ΔH°rxn from Δ H°f ΔH°rxn = Σn ΔHf°(products) − Σm ΔHf°(reactants) n, m are stoichiometric coefficients of each individual product and reactant, respectively • NOTE: this is an application of Hessʼs LawSheets Page 9 Lecture 39 ΔH°rxn example calculate ΔH°rxn C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) substance ΔHf° (kJ/mol) C2H5OH(l) –277.7 O2(g) 0 CO2(g) –393.5 H2O(l) –285.8 !Hrxno = n !Hfo"(products) – m !Hfo"(reactants)Sheets Page 10 Lecture 39 Estimating Δ Hrxn from bond dissociation energies • ΔHrxn ≅ Σn D(broken) − Σ m D(formed) energy is released (ΔH is −) when bonds form n, m = # bonds • helps us understand origins of ΔHrxnSheets Page 11 Lecture 39 Example 2 H2 + O2 → 2 H2O ΔHrxn = ?? • draw Lewis structures of reactants and products Bond D (kJ/mol) H–H 436 O–H 463 O2 495 reactants (broken) products (formed) bond # D bond # DSheets Page 12 Lecture 39 Example: The following reaction is used to produce methanol: CO(g) + 2 H2(g) → CH3OH ΔHrxn = –128 kJ C O H H H CHOHH2 Use the following info to calculate the approximate C O bond enthalpy. Bond D (kJ/mol) C–H 413 C=O 799 H–H 436 C–O 358 O–H 463 reactants (broken) products (formed) bond # D bond # D In BLB Table 8.4 D(C O) = 1072 kJ/molSheets Page 13 Lecture 39 Example: What is ΔH for the following reaction? Is this an exothermic or endothermic reaction? Fe2O3(s) + 3 H2O(l) → 2 Fe(OH)3(s) ΔH°f (kJ/mol) Fe2O3(s) –824.2 H2O(l) –285.8 H2O(g) –241.8 Fe(OH)3(s) –823.0Sheets Page 14 Lecture 39 Before next class: Read: BLB 15.1–15.3 HW: BLB 15:13,14,21 Sup 15:1–4 Know: • concept of equilibrium • equilibrium constant (Keq) Work through the concept final (on lecture notes website). Good luck studying for finals! Please start now. Answers: p. 4: ΔHrxn = –297 kJ p. 9: ΔHrxn = –1366.7 kJ p. 11: ΔHrxn = –485 kJ p. 12: 1060 kJ/mol p 13: ΔHrxn = 35.6 kJ;


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PSU CHEM 110 - Thermochemistry 2

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