Thermochemistry Part 2

Transcription

1 Thermochemistry Part 2 Read: BLB 5.6 7; 8.8 HW: BLB 5:63, 67a,b, 69, 73, 75, 83, 85; BLB 8:65a, 67a,c, 92a BLB 18:72ab, 74 Supplemental 5:1 7; Supplemental 8:11 13 Know: Hess!s Law heats of formation enthalpy of reactions estimating reaction enthalpy from bond energies Check out the grade finder on Angel; what do you need to get on your final?? FINAL DEADLINE for credit on skill check tests: (you must get 100% on any test to receive credit for that test): Final Exam: HESS S LAW The sum of the!h values for each step is the same as!h for the overall process. This is true because!h is a state function. C!H rxn A A B!H 1 B C!H 2 A+B "# B + C!H 1!H 2 B!H 1 +!H 2 =!H rxn Dr. L. S. Van Der Sluys Page 1 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 2 Ch. 5 Part 2

2 Example: Given the following information A H 2 (g) + F 2 (g) # 2HF(g)!H A = 537kJ B 2H 2 (g) + O 2 (g) # 2 H 2 O(g)!H B = 572kJ Determine!H for the reaction: C 2F 2 (g) + 2H 2 O(g)# 4HF(g) + O 2 (g)!h C =? IDEA: find combinations of reactions such that then n A + m B = C n!h A + m!h B =!H C Given the following information:!h 2SO 2 (g)+ O 2 (g) # 2SO 3 (g) $196kJ 2S(s) + 3 O 2 (g) # 2SO 3 (g) $790kJ What is!h rxn for the following reaction? A. 986 kj B. 594 kj C kj D. 297 kj S(s) + O 2 (g) # SO 2 (g) Here: 2xA 1xB _ 2H 2 (g) + 2F 2 (g) # 4HF(g)!H = 2( 537kJ) 2 H 2 O(g) # 2H 2 (g) + O 2 (g)!h = ( 572kJ) 2F 2 (g) + 2H 2 O(g)# 4HF(g) + O 2 (g)!h C = 2( 537kJ) 1( 572kJ) = 502kJ Dr. L. S. Van Der Sluys Page 3 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 4 Ch. 5 Part 2

3 Heat of Formation!H f enthalpy of formation heat given off (or absorbed) when elements combine to give one mole or one molecule of a compound combine Elements Compounds!H f!h f standard enthalpy of formation enthalpy of formation when all substances are in their Standard State DEFINITION OF STANDARD STATE 1. P = 1 atm 2. T = 25 C (298K) 3. element is in its most stable state (gas/liquid/solid) For an element in its standard state:!h f = (by definition) Standard States of the Elements (Most stable phase under standard conditions of 298K and 1 atm) 1. Metals: all are at 298K and 1 atm except one (Which one?) 2. Semi metals (metalloids): all are at 298K and 1 atm 3. Nonmetals at 298K and 1 atm i: Noble gases (Group 8): atomic ii: Diatomics: H 2, N 2, O 2, Group 7 (F 2, Cl 2, Br 2, I 2 ) H 2, N 2, O 2, F 2, Cl 2, are Br 2 is a I 2 is a iii: all other non-metals are C (graphite), S 8, P(s), Se Dr. L. S. Van Der Sluys Page 5 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 6 Ch. 5 Part 2

4 Are these Elements in their standard states? O 2 (g) N 3 $ (s) F 2 (g) N 2 (g) O 3 (g) C(diamond) C(graphite) Fe(s) Br 2 (g) H(g) YES NO Dr. L. S. Van Der Sluys Page 7 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 8 Ch. 5 Part 2

5 For which of the following reactions (at 25 C and 1 atm) is!h rxn =!H f? Answer A:!H rxn =!H f Answer B:!H rxn!!h f 1) H 2 (g) + F 2 (g) "# 2HF(g) 2) NO(g)+ 1/2 O 2 (g) # NO 2 (g) 3) 2C(graphite)+3H 2 (g)+1/2o 2 (g)"#c 2 H 5 OH(l) 4) Pb(s) + Cl 2 (g) "#PbCl 2 (s) 5) S(s) + O 3 (g) "# SO 3 (g) 6) Br 2 (l) "# Br 2 (g) We know that!h rxn is the Standard Enthalpy of reaction, or Heat of reaction; all elements must be in their. We know that!h f is the Heat of formation; formation of (how much?) from when all reactants are in their How are they related? Obtaining!H rxn from!h f :!H rxn = %n!h f (prod) $ %m!h f (react) where n and m are stoichiometric coefficients of products and reactants. (This is an application of Hess s Law.) Dr. L. S. Van Der Sluys Page 9 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 10 Ch. 5 Part 2

6 Using the following information, what is!h rxn for the reaction: C 2 H 5 OH(l) + 3O 2 (g)#2co 2 (g) + 3H 2 O(l)!H f (in kj/mole): a) kj b) kj c) kj d) kj If a piece of fruit contains 16.0 g of fructose, and!h rxn = $2803 kj, how many food calories does it contribute to the body? 1 cal = 4.184J 1kcal = 1 Cal (= 1 food calorie) Combustion! C 6 H 12 O 6 (s) +6 # NOTE: Appendix C in the text has a more extensive table of!h f Dr. L. S. Van Der Sluys Page 11 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 12 Ch. 5 Part 2

7 Changes in Enthalpy with Breaking and Formation of Bonds Bond Properties Review COVALENT BOND LENGTHS and ENERGIES Bond length: distance between nuclei bond Bond energy Bond length pm kj/mol C"C C=C C&C more electrons shared, shorter bond length! Bond Enthalpies (bond energy) can be used to estimate!h rxn.! The overall reaction has two steps; breaking the original bonds, and forming new ones. bond Bond energy kj/mol Bond length pm C"H C"Cl C"Br Shorter bond length, stronger the bond Dr. L. S. Van Der Sluys Page 13 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 14 Ch. 5 Part 2

8 Estimating!Hrxn Bond energies provide estimates of reaction enthalpies:!hrxn ' %ndbroken $ %mdformed n, m = # of bonds Energy is given off ($) when bonds form. Helps understand origins of!hrxn Dr. L. S. Van Der Sluys Page 15 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 16 Ch. 5 Part 2

9 Example: Predict the heat of reaction for the decomposition of hydrogen peroxide. 2 H 2 O 2 # 2 H 2 O + O 2!H rxn =? Draw Lewis structures of reactants and products: Reactants (broken) Products (formed) bond # D bond # D Peroxide Decomposition: Elephant Toothpaste Demonstration 2 H 2 O 2 (l) # 2 H 2 O(l) + O 2 (g)!h rxn =? Add KI to catalyze the reaction: I - (aq) + H 2 O 2 (l) # H 2 O(l) + IO - (aq) IO - (aq) + H 2 O 2 (l) # H 2 O(l) + O 2 (g)+ I - (aq) 2 H 2 O 2 (l) # 2 H 2 O(l) + O 2 (g) Can also catalyze the reaction with MnO 2 Compare to Actual value of!h rxn = -196 kj/mole Dr. L. S. Van Der Sluys Page 17 Ch. 5 Part 2 Dr. L. S. Van Der Sluys Page 18 Ch. 5 Part 2