Physical Chemistry Exam 2

a thermodynamic quantity equivalent to the total heat content of a system.
H=U+PV

The amount of (heat) energy necessary to raise the temperature of one mole of a substance by 1K

a measure of how the internal energy of a system changes when it expands or contracts at constant temperature. For an ideal gas, this equals 0

Energy cannot be created or destroyed
U=w+q

(P2/P1)=(V1/V2)^γ-1
γ=Cp/Cv

(T2/T1)=(V1/V2)^γ-1
γ=Cp/Cv

Ideal Gas: Cp-Cv=R
Solids and Liquids: Cp = approximately Cv
Ideal Monatomic: U = 3/2 RT
Cp= 5/2R Cv= 3/2R

The extent to which a reaction proceeds
Positive if in forward reaction
Negative if in reverse reaction

1. Find ξLR for every reactant
2. Pick the smallest ξLR
3. ni= 0 (because LR will be 0) = ni,o + Vi ξLR

The compound that gets completely consumed in the reaction

The amount of heat released or absorbed when a mole of matter is transformed by a chemical reaction at Standard State Conditions

∆rH°=(dH/dξ)T,P

*Temperature isn’t part of SSC
Liquids: pure at 1 bar pressure
Solids: pure at 1 bar pressure
Gases: ideal at 1 bar pressure (No interactions between molecules)
Solutions: ideal solutions with 1 molality at 1 bar pressure (All interactions are the same)

Molarity = moles solute/liters of solution
Easier to do, temperature dependent

Molality= moles of solute/kg of solvent
Temperature independent

species is dissociated as ions in solution

species does not dissociate further into ions

Reactants –> Products (exo/endo)

Elements –> 1 mole of compound (exo/endo) (majority is exo)

Solid –> Liquid (endo)

Liquid –> Solid (exo)

Liquid –> Gas (endo)

Gas –> Liquid (exo)

Solid –> Gas (endo)

Gas –> Solid (exo)

Phase α –> Phase β (both exo and endo)

Pure Substances –> Mixture (both exo and endo)

Solute and Solvent –> Solution (both exo and endo)

1 Mole of Compound + O2 (exo)
CO2 + H2O + N2 + Others

Different molecular arrangements of a given element
ex) C (gr), C (dia)

Element in its most stable phase and allotropic form at the temperature of interest

One mole of the compound of interest is made from its constituent elements in their most stable allotropic form and phase at the temperature of interest

∆H > 0 energy entering the system (intensive quantity, multiply by moles of extent of reaction to make extensive)

∆H < 0 energy leaving the system (intensive quantity, multiply by moles of extent of reaction to make extensive)

An observation that enthalpy is a state function calculating enthalpy changes at non-tabulated temperatures

HOFBrINCl

MAMA BEAR: ∆rH°(T2) = ∆rH°(T1) + ∫(T1-T2) ∆rCp°dT
BABY BEAR: ∆rCp° = 0 then ∆rH°(T2) = ∆rH°(T1)
∆rCp° do the integral?

A chemical reaction is run under adiabatic conditions (no heat transfer). Interconversion of chemical and thermal energy

adiabatic + isochoric (q=0; qv=∆u=0)
Use internal energy (U) and Cv,tot

adiabatic + isobaric (q=0; qp=∆H)
Use enthalpy at standard state conditions (H°)
Cp°tot equals Cp°solution (mass x heat capacity) + Cp°instrument

Measure of the dispersal of energy in a system. As entropy increases there is more dispersal of energy. ∆S is a state function on any reversible path

∆vS°=SUM N – i=1 ViSi

Take Zero: It is impossible to produce useful work from a heat engine connected to a single heat reservoir

Take One: in a spontaneous process, energy moves from being more “concentrated” to being more spread out and dispersed
-physically dispersed: light, high KE molecules dispersed over more available energy states

Take Two:
In a spontaneous process ∆S universe > 0 (irreversible)
In an equilibrium process ∆S universe = 0 (reversible)
An impossible process is one where ∆S universe < 0 ∆S is greater than or equal to 0 ∆S universe = ∆S system + ∆S surroundings

Extensive: ∆S = q/T

Translations
Rotations
Intermolecular Forces
Vibrations
Electronic Energy

– Ideal reversible closed thermodynamic cycle in which the working substance goes through isothermal and adiabatic paths to initial state. A heat engine converts thermal energy (heat) into useful work.

– Overall, ∆S = 0 is state function, and ∆U = 0

– Most efficient cycle that can be made!

E = 1 – Tc/Th

Zero Kelvin is the temperature at which a reversible heat engine runs with perfect efficiency (can never achieve 0 K)

This sets the zero of our temperature scale! The triple point of water. Ttr = 273.16 K

Find a different reversible path between the same states

1. Think about enthalpy on an isobaric path. dH=CpdT
2. On an isobaric path dS=đq/T = Cpdt/T

Cpdt/T vs T graph gives us ∆S

used to model Cp at low temperatures. Find “a” using the lowest experimental Cp(T) value.
Cp=aT^3

∆vapS = 88 ± 5 J/molK

Example of an exception…
∆vapS of water is approximately 110 J/molK liquid H-bonding limits energy dispersal

In liquid water, hydrogen bonding limits the amount of translations and rotations the molecule can have. Thus, the jump to the gas phase is an even bigger jump in dispersal of energy.

For reactions and phase transitions lim∆rS=0 as T approaches 0K

The entropy of a pure substance in perfect crystalline form is 0 J/molK at zero Kelvin

Absolute Entropy at 0K = 0 J/molK

formation reactions where we compare relative elements in their most stable phase and allotropic form at the temperature of interest

gravitational potential energy (PE)
electrical PE
chemical PE
thermal energy
sound
kinetic energy
electromagnetic radiation (light)

-increase in the number of energy states
-increase in the amount of energy in the system
-increase in the amount of stuff in the system (extensive)

Tb < Ta Therefore 1/Tb > 1/Ta

During an expansion, the temperature decreases
During a compression, the temperature increases

Element in its most stable phase and allotropic form at the temperature of interest