Transcript Thermodynamics

Thermodynamics
Lauren Kanner
Rachel Grotheer
Specific Heat
• In Phase Energy Change
– : q=mcDT q=energy m=mass c=specific heat
constant DT=change in temperature
• Between Phase Energy Change
– : q=mHf or q=mHv Hf=enthalpy of fusion
Hv=enthalpy of vaporization units: J/g, J/mol,
etc
– Melting=-fusion
– Condensation=-vaporization
Calorimetry
• -mmetalcDT = mwatercDT
• -mmetalc(Tf-Ti) = mwaterc(Tf-Ti)
• c for water is 4.18 J/g*C
Breaking and Forming Bonds
• Breaking bonds requires energy
• Forming bonds releases energy
• Exothermic reactions release energy after
forming products
• Endothermic reactions take energy from the
surrounding system to start reaction
Bond Energies
• Bond Energies
 : DE= +(reactant bonds)-(product bonds)
• Ex: CH4 + 2O2 = CO2 + 2 H2O
 DE= (4 C-H + 2 O=O) – (2 C=O + 4 H-O)
 DE= (4(414KJ/mol) + 2(498KJ/mol)) -(2(740KJ/mol)+4(464KJ/mol))
DE= -680KJ
Enthalpy
• Enthalpy Equation
DHrxn=S DHproducts – SDHreactants
– *use table for values
• 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
– Units: KJ/mol
Enthalpy Cont…
• Decreasing enthalpy of a system is
favorable because increased enthalpy
cannot rely on surroundings for energy
 DHrxn=- for exothermic
 DHrxn=+ for endothermic
• Reverse rxns have same number and
different signs
Entropy
• DSsystem = S S*products - S S*reactants
• Ex. CO(g) + 2 H2(g) = CH3OH(l)
– : DSsystem = [CH3OH] – [CO + 2 H2]
– : DSsystem = [1mol (126.8 J/K*mol)] – [1
mol (197.6 J/K*mol) + 2 mol (130.7
J/K*mol)]
– : DSsystem
= -332.2 J/K
– DS units: J/K * mol
Entropy Cont…
• Gas has highest and solids have lowest
entropy
• Solids and liquids have almost same
magnitude of entropy
• 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
Entropy DS
+ increase
+ increase
- decrease
Maybe Never
Enthalpy DH
- decrease
Always Maybe
•++ is spontaneous at high temperatures
•- - is spontaneous at low temperatures
Gibb’s Free Energy
•
DG = DH - T DS
*Favored if negative
• Ex. Cgraphite + 2 H2 = CH4
– DH=-74.8KJ DS=-80.8 J/K
– DG = -74.8 KJ – (298K)(-80.8 J/K)
(1KJ/1000J)
– DG = -74.8 KJ + 24.1 KJ
– DG = -50.7 KJ spontaneous & favored
– DG < 0 spontaneous
– DG > 0 non-spontaneous
More Gibbs Free Energy
• Solve for threshold temp with DG=0 and given DH &
DS
• DG* is standard condition DG=0 is equilibrium
• DG = DG* + RTlnQ
OR
• DG = DG* + RTlnK (at equilibrium)
• Q=reaction quotient
Q=K at equilibrium
• K= e-DG*/RT
• DG < 0 ~ large K
DG=0 ~ K=1
DG>0 ~ small K
Hess’s Law
• Any arithmetic combination of various
reactions that gives a matching net rxn and
enthalpy
• Ex. Want: C + 2H2 --> CH4 DH= ?
– 2H2+ O2 --> 2H2O
H= -572 kJ
– C + O2 --> CO2
H= -394kJ
– CO2 + 2H2O --> CH4 + 2O2 H= +890kJ
C + 2H2 --> CH4
DH= -76kJ
The End
Good Luck on the AP test!!!