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Unit 4 Thermodynamics. By: Elliot Prizant and Zak Abecassis. 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. Specific Heat Capacity (C).

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unit 4 thermodynamics

Unit 4 Thermodynamics

By: Elliot Prizant and Zak Abecassis

thermodynamics
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.
specific heat capacity c
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)

  • C(J/g(K) = ___q(J)__

[m(g)] x [ΔT(K)]

slide4
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

heat transferred q
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
bond energies
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
bond energies1
Bond Energies

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)

energy and changes of state
Energy and Changes of State
  • 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
calorimetry
Calorimetry
  • mmetalcΔT = mwatercΔT
  • mmetalc(Tf-Ti) = mwaterc(Tf-Ti)
  • c for water is 4.18 J/g*C
  • q reaction = -(q water + q bomb)
enthalpy
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)
enthalpy1
Enthalpy
    • DHrxn=S DHproducts – SDHreactants
  • Ex: C3H8 + 5 O2 = 3 CO2 + 4 H2O
    • DHrxn=[3(-393.5 KJ/mol)+4(-241.8 KJ/mol)] – [(-103.8 KJ/mol)]
    • DHrxn=-2043.91 KJ
hess s law
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
hess s law1
Hess’s Law
  • 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
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
spontaneity
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
gibbs free energy
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
gibbs free energy1
Gibbs Free Energy
  • 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
the end
THE END

Good Luck

Just remember you put the Thermo in Thermodynamics

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

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