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Chemical Thermodynamics. Taylor Walsh Jeff Shannon Phil Nirav. Spontaneous Processes(19.1). Problems: (#7-18) Sample Exercise 19.1. Vocabulary. Spontaneous Process—proceeds on its own without outside assistance
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Chemical Thermodynamics Taylor Walsh Jeff Shannon Phil Nirav
Spontaneous Processes(19.1) • Problems: (#7-18) • Sample Exercise 19.1
Vocabulary • Spontaneous Process—proceeds on its own without outside assistance • Reversible Process – a system is changed in such a way that the system and surroundings can be restored to the original state by reversing the change. • Irreversible Process – cannot be reversed to restore the system and its surroundings to their original states. • Isothermal -- a process that occurs at a constant temperature
Energy is conserved • It can be transformed or converted, but the total remains the same. • Example: • A nail rusting is spontaneous • Reversing this reaction is nonspontaneous • Temperature and pressure are important in determining whether the reaction is spontaneous
Equilibrium • When the reaction is at equilibrium, the products and reactants are changing at an equal rate.
Reversible and Irreversible • Reversible change produces the greatest amount of work (qrev, qmax) • Any spontaneous process is irreversible because if it is returned to the original condition, the surroundings will have changed (done work)
Can reverse direction whenever a very small change is made to some property of the system Reversible Processes
Entropy and the Second Law of Thermodynamics(19.2) • Problems: (#19-26) • Sample Exercise: 19.2
Vocabulary • Entropy: a thermodynamic state function related to the spontaneous nature of processes. • Second law of thermodynamics: the way entropy controls the spontaneity of processes, governs the change in the entropy of the universe.
The Molecular Interpretation of Entropy (19.3) • Problems: (#27-40) • Sample Exercise: 19.3, 19.4
Vocabulary • Translational motion: the entire molecule moves in space. • Vibrational motion: the atoms of the molecule move toward and away from one another in periodic fashion. • Microstate: a particular combination of motions and locations of the atoms and molecules of a system at a particular instant. • Third law of thermodynamics: the entropy of a pure crystalline solid at 0K = 0.
Molecular Motions and Energy • Molecules undergo 3 kinds of motion: Translational, vibrational, and rotational. • These different forms of motion are ways the molecule can store energy (motional energy) • Vibrations(1-3) • Rotation(4) http://wps.prenhall.com/wps/media/objects/3312/3392504/imag1903/AAAYXDM0.JPG 1 2 4 3
Boltzmann’s Equation and Microstates • Statistical thermodynamics: uses statistics + probability to provide link b/w microscopic + macroscopic worlds. • Total # of microstates of a thermodynamic system can be determined w/ statistics + probability. Boltzmann’s equation: S = k ln W • S(entropy), W(# of microstates), k( Boltzmann’s constant, 1.38 x 10-23 J/K) • Entropy is a measure of how many microstates are associated w/ a particular macroscopic state.
Entropy change accompanying any process: • ΔS = klnWfinal-klnWinitial=k(lnWfinal/Winitial) • Entropy increases with the number of microstates of the system, which increases w/ volume, temperature, or # of independently moving molecules in the system (increase in possible positions and energies of the molecules of the system)
Making Qualitative Predictions about ΔS • (s)(l) = more microstates+entropy • When an ionic(s)ionic(l) in water, ions have more entropy, but water(l) has less entropy b/c ion-dipole attractions. • The greater the ion’s charge, the greater the attractions holding the water+ion together. • While solution process generally is a net increase in entropy, dissolving of salts w/ highly charged ions could be a net decrease in entropy. • The entropy of the system increases for processes w/: 1. Gases are formed from either solids or liquids, 2. Liquids or solutions are formed from solids, 3. The number of gas molecules increases during a chemical reaction.
Example • H2(g)+ 2 O2(g) H2O(l) ΔS would be negative; decrease in # of molecules and volume. ΔS decreases with a decrease in the number of independently moving particles, and a decrease in volume (less freedom of motion).
The Third Law of Thermodynamics • At absolute zero all the units of a pure crystalline solid’s lattice have no thermal motion; 1 microstate. S = kln1 = 0. • As temperature increases, atoms/molecules gain energy in vibrational motion, entropy increases. • Entropy jumps up w/ solid liquidgas • Solidliquid = bonds holding atoms/molecules together are broken • Liquidboiling(gas), increased volume of molecules. • Ssolid < Sliquid < Sgas http://www.chem.umass.edu/~botch/Chem122S08/Chapters/Ch7/Entropy/Entropy.html
Entropy Changes in Chemical Reactions (19.4) • Problems: (#41-48) • Sample Exercise: 19.5
Vocabulary • Standard molar entropy (S°): the entropy of a mole of a substance under standard conditions
Absolute entropies are based on a reference point of zero entropy for perfect crystalline solids at 0K • Measured as molar quantities (J/mol*K)
Do NOT equal zero at the reference temperature 298K Generally increase with molar mass Standard Molar Entropies
Change in Entropy S=ΣnS °(products) - ΣmS °(reactants) Example • Calculate S for the synthesis of ammonia from O2 (g) and H2(g) at 298K. 2 H2(g) + O2(g) 2 H2O(g)
Answer • S° = 2S° (H2O) – [S° 2(H2) + S° (O2)] • = (2mol)(69.91J/molK) – [(2mol)(130.58J/molK) + (1mol)(205.0J/molK] • = -326.3 J/K
Entropy • Processes in which the disorder of the system increases tend to be spontaneous • Symbol -- (S) • The more random a system, the higher its entropy • Entropy is a state function, heat is not • S = q/T (q = heat, T = temperature) • At constant tempurature
Entropy Changes of the Surroundings • Isothermal process: Ssurr = (-qsys) / T Example: What is the change in entropy of the surroundings at 298K? CH4 + 2O2 CO2 + 2H2O
Hsys: [( -393.509 kJ/mol)+(2* -285.83 kJ/mol)]-[(2*0)+( -74.87 kJ/mol)] = -890.229 kJ/mol Ssurr : 890.229/298 = 2.9873 kJ/mol
Entropy changes of the Universe Suniv = Ssys + Ssurr CH4 + 2O2 CO2 + 2H2O Ssys = [(.21374 kJ/mol)+(2* .06995 kJ/mol)]-[(2*.20507 kJ/mol)+( .18626 kJ/mol)] = -.24276 kJ/mol Suniv = (-.24276 kJ/mol) + (2.9873 kJ/mol) = 2.745 kJ/mol
The change in entropy for the universe for a spontaneous reaction will always be positive. • The reaction of the system will always move towards the products.
Gibbs Free Energy(19.5) • Problems: (#49-70) • Sample Exercise: 19.6, 19.7
Vocabulary • Gibbs free energy(a.k.a. free energy) (G): a thermodynamic state function that combines the two state functions enthalpy and entropy: G = H-TS. • Standard free energies of formation, ΔG°f , the value for a prue element in its standard state is defined to be zero
The extent of randomness or spontaneity involves two concepts: enthalpy and entropy • Gibbs Free Energy: combines enthalpy and entropy in the form ∆G=∆H-T∆S • ∆G provides us with spontaneity of processes that occur at constant temperature and pressure • ∆G < 0: Reaction is spontaneous in forward direction • ∆G > 0: Reaction in forward direction is non-spontaneous • ∆G = 0: Reaction is at equilibrium • Free energy of a chemical system dereases until it reaches a minimum value
Gibbs Free Energy (cont’d) • In any spontaneous process at constant temp. and pressure, the free energy always decreases • When free energy reaches the minimum value, the reaction is at equilibrium • Ex. • When a boulder rolls down a hill, the potential energy in gravity drags the boulder down until it reaches the valley (equilibrium)
Standard Free-Energy Changes • Free energy is a state function b/c the standard free energies of formation represent a specific state of a substance • ∆G = Sum of G (products) – Sum of G (reactants) • The free energy values are set to zero for elements in their standard states
Free Energy and Temperature(19.6) • Unfortunately, No problems • Sample Exercise: 19.8
How is the change in free energy affected by the change in temperature? • ΔG = ΔH - T ΔS = ΔH + (-T ΔS) Enthalpy Term Entropy Term • T is a positive number at all temperatures other than absolute zero; only ΔS changes negative/positive of entropy term. • Value of T directly affects magnitude of -T ΔS; as temperature increases, magnitude + importance increases in determining ΔG. • Standard conditions: ΔG° = ΔH° - T ΔS°
Effect of Temperature on Spontaneity of Reactions ΔH ΔS -TΔS ΔG = ΔH – TΔS Reaction Characteristics Example - + - - Spontaneous at 2O3(g)->3O2(g) all temperatures + - + + Nonspontaneous at 3O2(g)->2O3(g) all temperatures - - + +/- Spontaneous at low T; H2O(l)->H2O(s) nonspontaneous at high T + + - +/- Spontaneous at high T; H2O(s)->H2O(l) nonspontaneous at low T
Free Energy and Equilibrium Constant (19.7) • Problems: (#71-80) • Sample Exercise: 19.9, 19.10, 19.11
Analyzing Chemical Reactions • There are two more ways of using free energy to analyze chemical reactions • First, you can relate ∆G° to the value of the equilibrium constant • ∆G = ∆G° + RT ln(Q) • You can use this equation to derive the relationship between ∆G° and Q, and we can calculate the value of K if we know the value of ∆G° • K = e^(-∆G°/RT)
Conventions for the Equilibrium Constant Expression (19.7) • Gas pressures are measured in atm • Solution concentrations are measured in molarity (mol/L) • Solids, liquids, and solvents don’t appear in the equilibrium constant expression
Free Energy and the Equilibrium Constant Continued (19.7) • If ∆G° is negative ln(K) > 0 - ln(K) > 0, then K is greater than zero • The more negative ∆G° is, the larger K is • If ∆G° is positive, then ln(K) is negative