Thermodynamics Cheat Sheet
Specific Heat
In Phase Energy Change
–: q=mcT
q=energy m=mass c=specific heat constant T=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
-mmetalcT = mwatercT
-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
: E= +(reactant bonds)-(product bonds)
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
Enthalpy Equation: DHrxn=S DHproducts – SDHreactants
Decreasing enthalpy of a system is favorable because increased enthalpy cannot rely
on surroundings for energy
Hrxn=- for exothermic
Hrxn=+ for endothermic
Reverse rxns have same number and different signs
Ssystem = S*products -  S*reactants
S units: J/K * mol
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G = H - T S*
Favored if negative
G < 0 spontaneous
G > 0 non-spontaneous
Solve for threshold temp with G=0 and given H & S
G* is standard condition G=0 is equilibrium
G = G* + RTlnQ OR G =G* + RTlnK (at equilibrium)
Q=reaction quotient
Q=K at equilibrium
K= e-G*/RT
G < 0 ~ large K G>0 ~ small K
G=0 ~ K=1
++ is spontaneous at high temperatures
- is spontaneous at low temperatures
EntropyS
+ increase
+ increase
- decrease
Maybe
Never
Always
Maybe
Enthalpy H
- decrease