Download presentation
Presentation is loading. Please wait.
Published byValentine Tyler Peters Modified over 9 years ago
1
Unit 4 Thermodynamics By: Elliot Prizant and Zak Abecassis By: Elliot Prizant and Zak Abecassis
2
Thermodynamics The study of the laws that govern the conversion of energy from one form to another, the direction in which heat will flow, and the availability of energy to do work.
3
Specific Heat Capacity (C) This is the heat required to produce a given temperature change per gram of material. Specific heat capacity (C) = Quantity of heat supplied (mass of object) (temperature change) (mass of object) (temperature change) C(J/g(K) = ___q(J)__ [m(g)] x [ΔT(K)] [m(g)] x [ΔT(K)]
4
The average specific heat capacity of a human body is approximately 3500 J.kg-1.K-1. A certain Japanese sumo wrestler has a body mass of 200 kg. What is the wrestler's heat capacity? The heat capacity of a body is obtained by multiplying its specific heat capacity by its mass in kilograms. In this case, the heat capacity of the wrestler is: 3500 (J.kg-1.K-1) x 200 (kg) = 700000 (7.0x105) J.K-1
5
Heat Transferred (q) q = mCΔT Units: mass in grams, C is the specific heat capacity in J/g(K), ΔT is the temperature final minus initial in Kelvins. +q means heat is transferred to a substance. -q means heat is transferred from the system = q transferred into the surroundings
6
Bond Energies q = Σ Reactants Bond Energy – Σ Products Bond Energy Potential energy of the elemental state is ZERO Breaking bonds requires energy to be added to the system Forming bonds requires energy to be released from the system
7
Bond Energies E= +(reactant bonds)-(product bonds) EX: 2N0 2 + H 2 0 →HNO 3 + HNO 2 E = (4 N-O + 2 H-O) – (5 N-O + 2 N-H) E =(4(201 KJ/mol) + 2(463 KJ/mol)) – (5(201 KJ/mol) + 2(391 KJ/mol)) E = -57 KJ/mol (note: some N-O bonds are probably double)
8
Energy and Changes of State Heat of Fusion (H f ): quantity of heat required to melt a substance Heat of Vaporization (H v ): quantity of heat required to vaporize a substance Takes more energy to vaporize than to melt (need to break more bonds to get gas) : q=mH f or q=mH v Hf=enthalpy of fusion Hv=enthalpy of vaporization units: J/g, J/mol
9
Energy and Changes of State
10
Calorimetry m metal cΔT = m water cΔT m metal c(Tf-Ti) = m water c(Tf-Ti) c for water is 4.18 J/g*C q reaction = -(q water + q bomb)
11
Enthalpy Enthalpy (H): Heat transferred into or out of a system at a constant pressure. ΔH = Σ Product Potential Energy - Σ Reactant Potential Energy Units: Joules (J) or Kilojoules (KJ). Negative ΔH is Exothermic/Positive ΔH is Endothermic Enthalpy change for a reaction = ΔHºrxn = Σ ΔHºf products - Σ ΔHºf reactants q = n(ΔH) or q = m(ΔH)
12
Enthalpy H rxn = H products – H reactants Ex: C 3 H 8 + 5 O 2 = 3 CO 2 + 4 H 2 O H rxn =[3(-393.5 KJ/mol)+4(-241.8 KJ/mol)] – [(-103.8 KJ/mol)] H rxn =-2043.91 KJ
13
Hess’s Law Hess’s Law: If a reaction is the sum of two or more other reactions, then ΔH for the overall process must the sum of ΔH values of the constituent reactions. Reversing an equation, causes the sign on ∆H to change. Multiplying an equation, multiply ∆H by same number
14
Hess’s Law PbS (s) + 3/2 O 2 → PbO (s) + SO 2 (g) ΔH = -413.7 kJ PbO (s) + C (s) → Pb (s) + CO (g) ΔH = +106.8 kJ PbS (s) + C (s) + 3/2 O 2 (g) → Pb (s) + CO (g) + SO 2 (g) ΔH = - 325.3 kJ
15
Entropy Entropy (S): A measure of randomness in a system In order of greatest to least random Gases>Liquids>Solids. ∆S = Σ S (products( – Σ S (reactants) Units: J/K(mol) Increasing # of moles of gas in reaction from reactants to products makes more entropy and vice versa Product-favored reactions have higher entropy and vice versa
16
Entropy Entropy
17
Spontaneity If a reaction is spontaneous, then the reaction is product favored. Is the reaction spontaneous? T =∆H / ∆S ∆S 0 ∆S 0 ∆H > 0 Never Spontaneous Maybe (at high temps) ∆H < 0 Maybe (at low temps) Always Spontaneous
18
Gibbs Free Energy ∆G = ∆H - T∆S Units: T is in Kelvin, S is in J/(mol*K) A reaction is SPONTANEOUS when ∆G is NEGATIVE. A reaction is NOT SPONTANEOUS when ∆G is POSITIVE. A reaction is at Equilibrium when ∆G is zero. ∆G = ∆G˚ + RT lnQ ∆G˚ = -RT lnK (at equilibrium) R=8.314 J/(mol*K) T=Kelvin K is the Thermodynamic Equilibrium Constant
19
Gibbs Free Energy When ∆G˚ 1 Products are favored When ∆G˚ = 0 and K = 1 It is at equilibrium (Rare) When ∆G˚ > 0 and K 0 and K < 1 reactants are favored
20
THE END Sadi Carnot (1796-1832): the "father" of thermodynamics Sadi Carnot (1796-1832): the "father" of thermodynamics Good Luck Just remember you put the Thermo in Thermodynamics
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.