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)

   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 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

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

 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 Energies

D

E= +(reactant bonds)-(product bonds) EX: 2N0 2 + H 2 0 →HNO 3 + HNO 2

D E = (4 N-O + 2 H-O) – (5 N-O + 2 N-H) D E =(4(201 KJ/mol) + 2(463 KJ/mol)) – (5(201 KJ/mol) + 2(391 KJ/mol)) D E = -57 KJ/mol (note: some N-O bonds are probably double)

Energy and Changes of State

  Heat of Vaporization (H v ): quantity of heat required to vaporize a substance  Heat of Fusion (H f ): quantity of heat required to melt 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

Energy and Changes of State

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)

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)

Enthalpy

 D

H rxn =

S D

H products –

S D

H reactants



Ex: C

3

H

8

+ 5 O

2

= 3 CO

2

+ 4 H

2

O

 D H rxn =[3(-393.5 KJ/mol)+4(-241.8 KJ/mol)] – [(-103.8 KJ/mol)]  D

H

rxn

=-2043.91 KJ

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 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

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

Entropy

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

       

∆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 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

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