Thermodynamics Flashcards
what is the first law of thermodynamics and what is something it does not indicate
First law of thermodynamics: energy of the universe is conserved
This law does not indicate the extent of the process or which direction is favoured
Physical and chemical changes tend to favour one direction over the other
what does spontaneous mean, and what is the other option
A spontaneous process is one that occurs without outside intervention.
Processes that are spontaneous in one direction are non-spontaneous in the other direction.
Spontaneity of a reaction is about direction, not speed.
Spontaneity can depend on conditions, for example temperature.
what makes a process spontaneous
What makes a process spontaneous?
Is it that system goes to lowest enthalpy? (exothermic: ∆H is negative)
Yet evaporation, melting ice at T = 20ºC, dissolving of NH4NO3 in water, are all spontaneous. But they are all endothermic processes (∆H is positive)
In each of these processes, the spontaneous reaction increases “randomness”. Increases the dispersal of energy and molecular disorder.
Energy becomes more dispersed / Spread out. Molecular disorder increases.
Enthalpy increases
what is the second law of thermodynamics
Second law of thermodynamics: in a isolated system (constant energy), energy will become more dispersed in a spontaneous process.
what is entropy
Entropy is a thermodynamic function that describes how spread out the energy is within a system: Symbol = S, Units = J/K
Since the dispersal of energy is accompanied by an increase in molecular disorder, we can quantify entropy using a statistical description of “molecular disorder”
what changes when a gas expands into a vacuum change
Expansion of a gas considered as a collection of particles
Consider a gas expansion into a vacuum at constant T: ∆u = w = 0
But there is a change in entropy since the kinetic energy is more “spread out” after expansion.
how can we quantify entropy by number of ways
We quantify entropy by counting the number of “ways” the system can be arranged before and after expansion.
For the expansion of four gas molecules there are 16 energetically equivalent arrangements. Each way is called a microstate.
So we can understand the spontaneous expansion of the gas from a statistical point of view. In the expanded state there are more microstates. Therefore the expansion of the gas is more probable than the reverse process.
Energy disperse through random molecular motion, which does not increase the randomness of macroscopic objects. Moving microscopic objects requires work.
what did Ludwig Boltzmann propose
Ludwig Boltzmann proposed that the entropy of a state is related to the number of equivalent ways the state can be achieved.
For molecules a microstate is a single set of positions and kinetic energies for all the particles in the system.
S=KBlnΩ
KB=R/NA=1.38*10^-23 We call this Kb the Boltzmann constant
Ω is the number of microstates
What kind of function is entropy
Entropy change in the expansion process.
Entropy is a state function
∆S = Sfinal - Sinitial
∆S = KblnΩfinal - KblnΩinitial = Kbln(Ωfinal/Ωinitial)
So, the change in entropy is positive during a gas expansion based on the above equation
what are the three types of motion
Translational, Rotational, and Vibrational motion
Rotational and vibrational motion increases as molecular complexity increases.
H2O has 3 types of vibrations (vibrational modes)
CH4 has 9 types of vibrations (vibrational modes)
how does T impact KE, and what affects number of micro states
Kinetic energy increases as T increases, broader distribution of energies at higher T.
Number of microstates (Ω) and entropy: increases with volume, Temperature, Number of molecules, and molecular complexity.
what is the third law of thermodynamics
Third law: The entropy of a pure crystalline substance at 0K is zero.
Reduction in temperature causes energy stored in motions to decrease. i.e. entropy decreases
at 0K, motion is reduced to a minimum (only 1 microstate)
S=KBln(1)=0
- established a zero point for entropy
the absolute value of entropy, S, can be determined
How do you calculate the change in entropy for a phase change
Entropy depends on measurable quantities: heat and temperature
e.g. For a phase change at the transition temperature:
For water at bp, ∆Hvap = 40.7 KJ/mol
∆Svap=∆Hvap/T=(40.7*10^3)/373K=109 J/K
In general:
∆Strs=∆Htrs/Ttrs
Equation does NOT apply for phase transitions not a transition temperature: e.g. melting ice at 20ºC
Where “trs” is the phase transition: Fusion, Vaporization, etc.
What increases Entropy
Entropy increases with greater degrees of freedom.
Increases with energy, dispersal, and molecular disorder.
Entropy increases for these processes:
Solid → liquids → gas
Solids → solutions (usually)
Increase in number of molecules of gas
When comparing molecules entropy generally increases with more atoms in the molecule.
what is standard molar entropy
Standard molar entropy (Sº)
Defined for substances in their standard states: Pure substance at 1 atm pressure. Usually tabulated at 25ºC (298K)
- Standard molar entropies of elements are not zero.
(Sº[O2(g)] = 205.0 J/mol K
what are the standard molar entropies related to gases and solid
- Standard molar entropies of gasses are greater that for the corresponding liquids or solids.
Sº[H2O(l)] = 69.91 J/mol K.
Sº[H2O(g)] = 188.83 J/mol K
what increases standard molar entropies
- Standard molar entropies increase with increasing molar mass and increase with increasing number of atoms
Sº[Na(s)] = 51.45 J/mol K
Sº[K(s)] = 64.47 J/mol K
Sº[CH4(g)] = 186.3 J/mol K
Sº[C2H6(g)] = 229.5 J/mol K
how do you calculate change in entropy for a reaction
The change in entropy for a reaction (system) can be calculated from tabulated standard molar entropies because entropy is a state function.
∆Ssysº = ∑VpSº(products) - ∑VrSº(reactants)
how does the second law of thermodynamics related to entropy
In any spontaneous process, the total entropy of the universe always increases.
∆S universe = ∆S system + ∆S surroundings > 0
consistent with the first version of 2nd law since the universe is an isolated system.
To evaluate the entropy change for the universe, we have to consider both the system and the surroundings.
The second law allows spontaneous formation of ordered systems
But such processes must increase the entropy of the surroundings such that ∆S universe > 0
how do you calculate the entropy of the surroundings
Now, how about the entropy of the surroundings? Chemical reactions are constant Temperature and usually constant pressure.
For an isothermal process:
∆Ssurroundings=-Qsys/T
at constant P
Qsys=∆Hsys
∆Ssurr=-∆Hsys/T
Overall prediction of spontaneity requires one to know the ∆S universe = ∆S system + ∆S surroundings
∆S universe depends on ∆H sys, ∆S sys, and temperature.
what did Josiah Willard Gibbs propose
Josiah Willard Gibbs proposed a new state function, G, now called Gibbs Free energy or just free energy.
G system = H system - TS system ∆G = ∆H - T∆S (for constant temperature)
∆G system = -T∆S universe (for constant P and T)
When ∆S universe is positive ∆G must be negative
Therefore: ∆G < 0 for a spontaneous process (constant P and T processes only, including chemical reactions)
what determines the spontaneity of a reaction
For a spontaneous process, the second law states
∆S universe = ∆S system + ∆S surrounds > 0
∆S universe = ∆S system - ∆H sys/T
(since ∆S surroundings = -∆Hsys/T for constant P and T)
-T∆S universe = -T∆S system + ∆H system
The spontaneity of a reaction depends on both enthalpy and entropy changes of the system.
How does Gibbs free energy relate to different types of reactions
In any spontaneous process carried out at constant T and P,
the free energy always decreases: ∆G is negative
For a reaction Reactants ← → Products
if ∆G < 0, reaction is spontaneous in forward direction
if ∆G > 0, forward reaction not spontaneous, reverse reaction is spontaneous
If ∆G = 0m the reaction mixture is at equilibrium
how can you determine the spontaneity of a reaction based on ∆H and ∆S
If ∆H is - and ∆S is +, then ∆G is - and the rxn is spontaneous at any temp
If ∆H is - and ∆S is -, then ∆G is - and the rxn is spontaneous at low temp, and nonspontaneous at high temp
If ∆H is + and ∆S is +, then ∆G is - and the rxn is nonspontaneous at low temp, and spontaneous at high temp
If ∆H is + and ∆S is -, then ∆G is - and the rxn is nonspontaneous at any temp
What is the free energy of formation
Free energy involved in the formation of 1 mole of a compound in the standard state.
Thermodynamics standard state:
- most stable form of the substance at: 1 atm pressure (gas phase): 1M concentrations (solutions)
- Pure liquids for liquids and pure solids for solids
T for tabulated values usually 25ºC or 298K
For elements in their most stable form at standard conditions: ∆Gºf = 0
How do we determine the Gibbs free energy to make something from the free energy of formation under any nonstandard conditions
Most chemical reactions do not occur under standard conditions, hence need to obtain ∆G from ∆Gº
It can be show that: under any nonstandard conditions
∆G = ∆Gº + RTlnQ
where R is the gas constant (8.314 J/mol K) and Q is the reaction quotient
How do Q and K impact spontaneity
If Q«K → the forward reaction is spontaneous
If Q»K → the reverse reaction is spontaneous
if Q = K → equilibrium
The direction of spontaneity can be changed by altering the concentrations of reactants and products, which changed Q and therefor changes the value of ∆G.
what is the relationship between ∆Gº and K
At equilibrium, ∆G = 0 and Q = K
∆G = ∆g) + RTlnQ
∆Gº = -RTlnK
Rearranging K = exp(-∆Gº/RT)
For gases this gives Kp since 1 atm is the standard state of a gas
A k = 1, ∆Gº = 0
The lower the K value, the higher the ∆Gº value is
The higher the K value, the more negative the ∆Gº value is.
