- 99 Views
- Uploaded on
- Presentation posted in: General

Unit 4 Thermodynamics

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Unit 4 Thermodynamics

By: Elliot Prizant and Zak Abecassis

- 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.

- 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)

- C(J/g(K) = ___q(J)__
[m(g)] x [ΔT(K)]

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 capacityof 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

- 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

- 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

DE= +(reactant bonds)-(product bonds)

EX: 2N02 + H20 →HNO3 + HNO2

DE = (4 N-O + 2 H-O) – (5 N-O + 2 N-H)

DE =(4(201 KJ/mol) + 2(463 KJ/mol)) – (5(201 KJ/mol) + 2(391 KJ/mol))

DE = -57 KJ/mol

(note: some N-O bonds are probably double)

- Heat of Fusion (Hf): quantity of heat required to melt a substance
- Heat of Vaporization (Hv): 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=mHf or q=mHv Hf=enthalpy of fusion Hv=enthalpy of vaporization units: J/g, J/mol

- mmetalcΔT = mwatercΔT
- mmetalc(Tf-Ti) = mwaterc(Tf-Ti)
- c for water is 4.18 J/g*C
- q reaction = -(q water + q bomb)

- 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)

- DHrxn=S DHproducts – SDHreactants

- DHrxn=[3(-393.5 KJ/mol)+4(-241.8 KJ/mol)] – [(-103.8 KJ/mol)]
- DHrxn=-2043.91 KJ

- 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

- PbS (s) + 3/2 O2 → PbO (s) + SO2 (g)
- ΔH = -413.7 kJ
- PbO (s) + C (s) → Pb (s) + CO (g)
- ΔH = +106.8 kJ
- PbS (s) + C (s) + 3/2 O2 (g) → Pb (s) + CO (g) + SO2 (g) ΔH = - 325.3 kJ

- 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

- 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

- ∆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

- When ∆G˚ < 0 and K > 1 Products are favored
- When ∆G˚ = 0 and K = 1 It is at equilibrium (Rare)
- When ∆G˚ > 0 and K < 1 reactants are favored

Good Luck

Just remember you put the Thermo in Thermodynamics

Sadi Carnot (1796-1832): the "father" of thermodynamics