1. THERMODYNAMICS!!!!
Nick Fox
Dan Voicu

2. Changes of State
Order of phase changes based on temperature (lower-higher)
Solid
Melting/solidification
Liquid
Boiling/Condensation
Gas
For more info on the changes of state look at Review of Section 2

3. Specific Heat
Heat capacity: amount of heat required to raise its temperature by 1K
Molar heat capacity Cm: heat capacity of one mole of a substance
Specific heat: the heat capacity of one gram of a substance
Formula= (quantity of heat transferred)/(grams of substance) X (temperature change)
CS= q/m*∆T
Water Specific heat= 4.18 J/g*K

4. Heating Curves This is the heating curve of water (H2O)
1st plateau: 0 °C = K
2nd plateau:
100 °C = K

5. Calorimetry
Calorimetry: experiments measuring the amount of heat transferred between the system and the surroundings
Calorimeter: measures temperature change accompanying a process

6. Covalent Bond Energies
Covalent Bond: when two atoms share a pair of valence electrons
The energy that is released when covalent bonds are formed=∆H (kJ)
Forming bonds releases energy which makes ∆H negative
Breaking bonds takes energy in which makes ∆H positive

7. Enthalpy of Formation/Reaction
∆Hf
It is the enthalpy change for the reaction where the substance is formed from the reactants
To solve the Enthalpy of a reaction you would use this forumla:
∆Horxn= ∑n∆Hof (products) – ∑m∆Hof (reactants)

8. Entropy and Spontaneity
Three Types of Motion
Translational motion is the movement of an entire molecule in one direction
Vibrational motion is the periodic movement of atoms in a molecule
Rotational motion is the spinning movement of a molecule about an axis

9. Entropy and Spontaneity (Cont.)
Three Main Factors Affecting Entropy
Temperature
Increasing the temperature increases the kinetic energy of the molecules, therefore the there is more movement of the molecules, increasing the randomness/disorder; decreasing the temperature has the opposite effect
Volume
When the molecules occupy a greater volume, there is more disorder to the system because they are more spread out and have more motional energy; less volume means more order because molecules are less spread out and have less motional energy
Number or independently moving particles
The more molecules there are present in the system, the greater the entropy because there is more motional energy present in a greater number of molecules
The Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is zero. S(0 K) = 0

10. Entropy and Spontaneity (Cont.)
One way for a reaction to be spontaneous is for the change in entropy to be positive (and the change in enthalpy to be negative)
∆S° = ∑nS°(products) - ∑mS°(reactants) where n and m are coefficients in the chemical equation
We also can look at the change in entropy of the surroundings
∆Ssurr = -q /T = - ∆H/T
∆S°univ = ∆S°sys + ∆S°surr
∆S°univ will be positive for any spontaneous reaction

11. Gibb’s Free Energy and Spontaneity
∆G = ∆H – T∆S, under standard conditions ∆G° = ∆H° – T∆S° (standard temperature is 298K)
If Gibb’s free energy is negative, the reaction is spontaneous in the forward direction
If Gibb’s free energy is zero, the reaction is in equilibrium
If Gibb’s free energy is positive, the forward reaction is nonspontaneous, and work must be supplied by surroundings to make it occur, however the reverse reaction is spontaneous
Standard free energy of formation can be tabulated, similar to how it is done for standard enthalpy of formation
∆G° = ∑n∆G°f(products) - ∑m∆G°f(reactants)
The above equation gives the standard free energy change of a reaction

12. Gibb’s Free Energy and Equilibrium
∆G° = -RT lnQ can be used to find the reaction quotient
To find equilibrium constant:
∆G = ∆G° + RT lnK
Gibb’s free energy is zero when reaction is in equilibrium
0 = ∆G° + RT lnK
RT lnK = -∆G°
lnK = -∆G°/RT
Raise both sides as powers of e
K = e-∆G°/RT
A positive value for lnK means K>1, therefore the more negative the standard Gibb’s free energy, the larger the equilibrium constant
If Gibb’s free energy is positive, then lnK is negative (note lnK is negative, not K) which means that K<1 (K cannot equal zero or have a negative value)

13. Hess’s Law
Hess’s Law states that if a reaction is carried out in a series of steps, the enthalpy for the overall reaction will equal the sum of the enthalpy changes for the individual steps
C(s) + O2(g)  CO2(g) ∆H1 = kJ
CO(g) + 1/2 O2(g)  CO2(g) ∆H2 = kJ
Calculate the enthalpy for the combustion of C to CO.
C(s) + 1/2 O2(g)  CO(g) ∆H3 = ?
In order to cancel some of the molecules, one reaction must be reversed. The second reaction should be reversed because the enthalpy will remain negative if this is done
C(s) + O2(g)  CO2(g) ∆H1 = kJ
CO2(g)  CO(g) + 1/2 O2(g) ∆H2 = kJ
C(s) + 1/2 O2(g)  CO(g) ∆H3 = kJ