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CAS CH 131 (61)
Lecture

# Enthalpy and Entropy

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School
Boston University
Department
Chemistry
Course
CAS CH 131
Professor
Andrei Lapets
Semester
Fall

Description
Finding Enthalpy and Intro to Entropy • Enthalpy 1. thermodynamic quantity used to describe heat changes taking place at constant pressure 2. H (enthalpy) is a state function 3. ΔH = H final H initial 1. ΔH > 0  endothermic reaction 1. heat absorbed by the system from the example 2. ΔH < 0  exothermic reaction 1. heat released by the system to the surroundings 3. ΔH = 0  no heat exchange 4. H O 2 H(s) 2 (l) 1. ΔH = 6.01 kJ/mol 5. H O 2 H(l) 2 (g) 1. ΔH  40.66 kJ/mol 6. 2H O 22H(l) 2 (g) 1. ΔH  81.32 kJ/mol 7. CH + 2O  CO + 2H O 4 2 2 2 1. ΔH = -890 kJ/mol 8. CO + 22 O  CH2+ 2O 4 2 1. ΔH = 890 kJ/mol 4. Hess Law 1. finds enthalpyΔH 2. indirect method 3. when reactants become products, the change in enthalpy is the same whether the reaction happens in one or multiple steps 4. (1) take 2 or more chemical reactions 5. (2) add them 6. (3) add the H terms 7. (4) find solution 8. Ex. 2C + H  C 2 2 2 1. Given: C + O  CO 2ΔH = -293.5 kJ/mol 2. Given: H + 025O  H O,2H = -285.8 kJ/mol 3. Given: 2C H + 2O 2 4CO +22H O,ΔH 2 -2598.2 kJ/mol 4. 4CO + 22 O  2C2H + 5O ,2H 2 2598.2 kJ/mol 5. 2H + O  2H O,ΔH = -571.6 kJ/mol 2 2 2 6. 4C + 4O  4C2 ,ΔH = -2574 kJ/mol 7. 4C + 2H  2C H ,ΔH = 453.2 kJ/mol 2 2 2 8. 2C + H  C2H ,ΔH2= 226.6 kJ/mol 9. Ex. C H (6H)4+ H O 2 C H 2 +22H O 6 4 2 2 1. Given: C H (6H)6 C H 2 + H 6ΔH4= 277 kJ2mol 2. Given: H O 2H 2 + 0.2O ,ΔH = -9426 kJ/mol 3. Given: H + 025O  H O,2H = -286 4. ΔH = -204 kJ/mol 5. Standard EnthalpyΔH o 1. measuring changes using standard states 2. standard states for solids and liquids 1. 1 atm 2. defined temperature (usually 25 C) o 3. dissolved solution 1. 1 atm 2. 1 M 4. elements at 1 atm and 298.15 K have 0 enthalpy 5. most stable state = 0 1. O = 2, but O =/= 3 6. Ex. C + O  C2 2 o o 1. findΔH reactiondΔH final 2. ΔH oreactionΣΔH pro0ucts – Σ ΔH reactantf o o o o 3. ΔH reactoon(1 mol)ΔH CO –f[1 m2lΔH C + 1 mol Δf O ] = 1 f 2 molΔH CO f -392.5 kJ/mol 7. Ex. 2B H + 12O  5B O + 9H O 5 9 2 2 3 2 1. find heat released per gram, have a gram of B H 5 9 2. ΔH oreaction[(5 mol)(-1263.6) + (9 mol)(-285.8)] – [(2 mol)(73.2) + (12 mol)(0)] = -9036.6 kJ/mol 3. 1 g B H 51 9o
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