Chapter 19: Chemical Thermodynamics
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Transcript of Chapter 19: Chemical Thermodynamics
Joe DiLosaHannah Robinson
Justin WhetzelEmily Penn
Casey White
Spontaneous Process – a process that proceeds on its own without any outside assistance
Reversible Process – A process that can go back and forth between states along exactly the same path
Irreversible Process – A process that is not reversible
Isothermal Process – A process that occurs at a constant temperature
Occur in a definite direction Ex: When a brick is dropped and it falls
down (gravity is always doing work on the brick)
Opposite direction (reverse) would be nonspontaneous (takes work to go against gravity)
If the surroundings must do work on the system to return it to its original state, the process is irreversible
Any spontaneous process is irreversible
http://wps.prenhall.com/wps/media/objects/3312/3392504/blb1901.html
A spontaneous reaction does not mean the reaction occurs very fast
Iron rusting is spontaneous Experimental conditions are important in
determining if a reaction is spontaneous (temperature and pressure)
To determine spontaneity, distinguish between the reversible and irreversible paths between states
Which of the following processes are spontaneous and which are nonspontaneous: (a) the melting of ice cubes at -5°C and 1 atm pressure, (b) dissolution of sugar in a cup of hot coffee, (c) the reaction of nitrogen atoms to form N2 molecules at 25°C (d) alignment of iron filings in a magnetic field, (e) formation of CH4 and O2 molecules from CO2 and H2O at room temperature and 1 atm of pressure?
(a) is nonspontaneous because the ice will not change state
(b) is spontaneous because to get the sugar back, the temperature of the system must be altered.
(c) is spontaneous because work must be done to form N2 and it is an irreversible process
(d) is spontaneous because the system must do work against the magnetic field to go back to its original state
(e) is nonspontaneous because a change in temperature and work must be done to break the products up and form the original reactants
Practice Problems - #7, 8, and 9
Entropy – A thermodynamic function associated with the number of different equivalent energy states or spatial arrangements in which a system may be found
Second Law of Thermodynamics – in general, any irreversible process results in an overall increase in entropy whereas a reversible process results in no overall change in entropy
Infinitesimal difference - an extremely small difference
Entropy = q/T = S A reversible change produces the
maximum amount of work that can be achieved by the system on the surroundings (wrev = wmax)
Reversible processes are those that reverse direction whenever an infinitesimal change is made in some property of the system
Example of Reversible Process◦ Flow of heat between a system and its
surroundings Example of Irreversible Process
◦ Gas that has been limited to part of a container expands to fill the entire container when the barrier is removed
∆S = Sfinal – Sinitial
∆S = qrev/T (where T = constant, isothermal process)
qrev = ∆Hfusion when calculating ∆Sfusion for a phase change
∆Ssystem = qrev/T ∆Ssurroundings= -qrev/T ∆Stotal = ∆Suniv = ∆Ssystem + ∆Ssurroundings
For a reversible process ∆Stotal = 0 For an irreversible process ∆Stotal > 0
The normal boiling point of methanol (CH3OH) is 64.7˚C and its molar enthalpy of vaporization is ∆Hvap=71.8 kJ/mol. When CH3OH boils at its normal boiling point does its entropy increase or decrease? Calculate the value of ∆S when 1.00 mol CH3OH (l) is vaporized at 64.7 ˚C
Increases, the process is endothermic∆S = qrev/T = ∆Hvap/T
64.7 ˚C + 273 = 337.7 K(1 mol)(71.8 kJ/mol)/(337.7 K)= 0.213 kJ/K = 213 J/Kentropy increases
Practice Problems- # 20 and 21
http://ffden-2.phys.uaf.edu/212_fall2003.web.dir/Anca_Bertus/page_7.htm
Translational Motion – movement in which an entire molecule moves in a definite direction
Vibrational Motion – the atoms in the molecule move periodically toward and away from one another
Rotational Motion – the molecules are spinning like a top Microstate – the state of a system at a particular
instant, one of many possible states of the system Third Law of Thermodynamics – the entropy of a
pure crystalline substance at absolute zero is zero
Vibration Vibration
Vibration Rotation
Img src = wps.prenhall.com
S = k ln W (W = number of microstates) k = Boltzmann’s constant = 1.38 x 10-23
J/K Entropy is a measure of how many
microstates are associated with a particular macroscopic state
∆S= k ln Wfinal - k ln Winitial = k ln Wfinal/Wintial
Entropy increase with the number of microstates of the system
The # of microstates available to a system increases with1. Increase in volume2. Increase in temperature3. Increase in the number of molecules Because any of these changes increases the
possible positions and energies of the molecules of the system
Generally expect the entropy of a system to increase for processes in which1. Gases are formed from either solids or
liquids2. Liquids or solutions are formed from
solids3. The # of gas molecules increases during
a chemical reaction
At absolute zero there is no thermal motion so there is only one microstate
As the temperature increases the molecules gain energy in the form of vibrational motion
When the molecules increase their motion they have a greater number of microstates
http://wpscms.pearsoncmg.com/au_hss_brown_chemistry_1/57/14649/3750314.cw/content/index.html
Practice Problems- # 28, 30, 38, 40
Standard Molar Entropy – The molar entropy value of substances in their standard states
The standard molar entropies of elements are not 0 at 298K
The standard molar entropies of gases are greater than those of liquids and solids
standard molar entropies generally increase with increasing molar mass
Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance
Pictures taken from p820 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay, and Bursten
∆S˚= ΣnS˚(products) - ΣmS˚(reactants) ∆Ssurr= -qsys/T
For a reaction at constant pressure qsys= ∆H
Calculate ∆S˚ values for the following reaction:
C2H4 (g) + H2 (g) C2H6 (g)
∆S˚= ΣnS˚(products) - ΣmS˚(reactants) = 229.5 J/molK – (219.4 J/molK + 130.58 J/molK)
= -120.5 J/molK
Practice Problems - # 48
Gibbs free energy- a thermodynamic state function that combines entropy and enthalpy
Standard free energies of formation- the change in free energy associated with the formation of a substance from its elements under standard conditions
Free energy (G) is defined by:G = H-TS
Where T is absolute temperature, S is entropy, and H is enthalpy
When temperature is constant the equation for the change in free energy for the system is
ΔG=ΔH-TΔS If T and P are constant then the sign of ΔG and
spontaneity of a reaction are related.1) If ΔG < 0, reaction proceeds forward2) If ΔG = 0, reaction is at equilibrium3) If ΔG > 0, the forward reaction is not spontaneous
because work must be supplied from the surroundings but the reverse reaction is be spontaneous.
In any spontaneous process at constant temperature and pressure the free energy always decreases
Conditions of Standard free energies of formation (implied)
1 atm pressure for gases pure solids pure liquids 1M solutions 25˚ C (normally)
The free energy of elements is 0 in their standard states.
ΔGf˚ = standard free energies of formation Calculated the same way as ΔH
ΔG˚ = ΣnΔGf˚ (products) - ΣmΔGf˚ (reactants)
ΔG = -wma
x
http://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Gibbs_free_energy_graph
Practice Problems - #49, 51, 53, 54, 56, 58, 63, 66
For a certain reaction ΔH = -35.4 kJ and ΔS = -85.5 J/K. Calculate ΔGo for the reaction at 298 K. Is the reaction spontaneous at 298 K?
ΔG = ΔH – ΔS*T ΔG = -35.4KJ – (-85.5 J/K * 298K) ΔG = -9920 J
No new vocabulary
When ΔH and –TΔS are negative, ΔG is negative and the process is spontaneous
When ΔH and –TΔS are positive, ΔG is positive and the process is non-spontaneous
When ΔH and –TΔS are opposite signs, ΔG will depend on the magnitudes
http://www.chem1.com/acad/webtext/thermeq/TE4.html
Picture taken from p828 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay, and Bursten
For a certain reaction ΔH = -35.4 kJ and ΔS = -85.5 J/K. is the reaction exothermic or endothermic? Does the reaction lead to an increase or decrease of disorder in the system? Calculate ΔGo for the reaction at 298 K. Is the reaction spontaneous at 298 K?
A) The reaction is exothermic because ΔH is negative
B) It leads to a decrease in the disorder of the system (increase in the order of the system) because ΔS is negative
C) ΔG = ΔH – ΔS*T ΔG = -35.4KJ – (-85.5 J/K * 298K) ΔG = -9920 J D) Yes, if all reactants are in their
standards states because ΔG is negative
Practice problems (same as 19.5) - #49, 51, 53, 54, 56, 58, 63, 66
Nonstandard conditions – conditions other than the standard conditions
ΔG = ΔGo + RT ln Q Where R = 8.314 J/mol-K, T = absolute temperature, Q = reaction quotient
ΔGo = ΔG under standard conditionsB
ecause Q = 1 and lnQ = 0
ΔGo = -RT ln KA
t equilibrium, ΔG = 0
K = e –(delta G standard)/RT
Smaller (more negative ΔG0), the larger the K value
Direction of Reaction Keq ΔG
Toward forming more products >1 negative
Toward forming more reactants <1 positive Products and reactants equal 1 0
http://www.wiley.com/legacy/college/boyer/0470003790/reviews/thermo/thermo_keq.htm
#71 – Explain Quantitatively how ΔG changes for each of the following reactions as the partial pressure of O2 is increased:
A) 2 CO(g) + O2(g) → 2 CO2 (g)
2 CO(g) + O2(g) → 2 CO2 (g)
-137.2 kJ/mol 0 kJ/mol -394.4 kJ/mol
ΔG = products – reactantsΔG = -394.4 kJ/mol + 137.2 kJ/molΔG = -257.2 kJ/mol
Practice Problems: #71,76,79
#82, 87, 88, 89, 90, 92