**Unformatted text preview: **Chemistry 260 Today in Chemistry 260 • spontaneity • entropy: a measure of disorder • statistical and thermodynamic de5initions of entropy This Week in Chemistry 260 • reading: (Fri.) 13.3, 13.4, 13.6 • 2nd and 3rd laws of thermodynamics, the Gibbs energy • …more on spontaneity • OWL Homework Due on Thursday (3 pm) • MS 8 due Friday, Nov. 11 Lecture 24 November 2, 2011 Demonstration ConcepTest 2NH 4SCN(s) + Ba(OH)2 • 8H 2O(s) ⎯⎯
→ 2NH 4SCN(s) + Ba(OH)2 • 8H 2O(s) ⎯⎯
→ 2 NH 3 (g ) + Ba(SCN)2 (s) + 10H 2O() 2 NH 3 (g ) + Ba(SCN)2 (s) + 10H 2O()
Based on your observations, we can infer that for this reaction: A. q < 0 B. q = 0 C. q > 0 Heat 5lows IN http://www.youtube.com/watch?v=yTzcoyzPQE0 1 Demonstration Spontaneity Δ H° = standard molar enthalpies 2NH 4 SCN( s ) + Ba(OH) 2 • 8H 2 O( s ) ⎯⎯→ 2 NH 3 ( g ) + Ba(SCN) 2 ( s ) + 10H 2 O() F T1 > T2 Non- spontaneous: can only be brought about by doing work 2 NH4SCN(s) + Ba(OH)2• 8H2O(s) Non- spontaneous changes can be MADE to occur by doing work m y B A T1 = T2
How can we account for the distinction between these two types of change? x Spontaneity is not determined by the First Law - 411.15 kJ/mol Common observation leads to two classes of chemical and physical processes: Spontaneous: has the tendency to occur without work being done 2NH3 (g) + Ba(SCN)2 (s) + 10 H2O(l) m . . . . . . Some things happen… some things don’t. - 240.12 What Spontaneity is NOT - 167.16 NaCl(s) - - - > Na+(aq) + Cl- (aq) Thermodynamics deals with the tendency to change; without more information it is silent on the rate at which that tendency is realized. Calculate ΔH in kJ/mol ΔH = - 167.16 - 240.12 - (- 411.15) = 3.87 kJ/mol A reaction happening fast (rate is due to barriers) “Spontaneous combustion” (all combustion is spontaneous) What is NOT an Increase in Entropy This is an endothermic reaction - but clearly spontaneous 2NH 4SCN(s) + Ba(OH)2 • 8H 2O(s) ⎯⎯ 2 NH 3 (g ) + Ba(SCN)2 (s) + 10H 2O()
→
The reverse reactions are not spontaneous - clearly there is more to this than conservation of energy. Qualitatively: “Nature prefers disorder” Better: "A situation with more available states is more commonly observed" NaCl(s) “ordered” ——> Na+(aq) + Cl- (aq) “disordered” “con;ined” “less con;ined” A bedroom/desk getting messy (“clean” and “messy” are just two different con5igurations) On the other hand – if there is one way to be clean and many ways to be messy … 2NH 4SCN(s) + Ba(OH)2 • 8H 2O(s) ⎯⎯ 2 NH 3 (g ) + Ba(SCN)2 (s) + 10H 2O()
→ NaCl(s) “ordered” ——> Na+(aq) + Cl- (aq) “disordered” Few States Many States 2 Entropy is the quantitative thermodynamic measure of disorder Consider a system with 3 balls A macrostate of the system is 1 ball in each bowl. and 3 bowls: 3 Possibilities # = N! N1! N2! N3! Disordered State A Another macrostate is 2 balls in bowl 1, 1 ball in bowl 2. 6 Possibilities # = 3! 1! 1! 1! # = o rdered More 3! 2! 1 State! 0! A macrostate of the system is 1 ball in each bowl. 6 Possibilities # = N! N1! N2! N3! # = 3! 1! 1! 1! B Two gas bulbs of equal size Gas Molecules Coin Flips The probability of any one molecule being in A is W = 1 2 1 1 Any two molecules in A is W = 1 • = 4 2 2 1 1 1 1 Any three molecules in A is W = • • = 2 2 2 8 1 ( 1 ) 2 2 ( 1 ) 2 3 ( 1 ) 2 Or 3 balls in bowl 1. 1 Possibility # = 3! Ordered State 3! 0! 0! Entropy is the quantitative thermodynamic measure of disorder Consider a system with 3 balls Disorder in terms of probability and 3 bowls: Probability of N molecules in A 1 N
W = 2 3 Possibilities # = 3! 2! 1! 0! Or 3 balls in bowl 1. 1 Possibility # = 3! 3! 0! 0! W = N ( 1 ) 2 Entropy is the quantitative thermodynamic measure of disorder Consider a system with 3 balls Another macrostate is 2 balls in bowl 1, 1 ball in bowl 2. N consecutive heads ( ) and 3 bowls: Another state is 2 balls in any bowl, one in another ball 3 Possibilities # = 3! 2! 1! 0! 6 ways to do this – with three possibilities for each 3 × 6 = 18 Possibilities 3 Entropy is the quantitative thermodynamic measure of disorder Consider a system with 3 balls Or 3 balls in bowl 1. and 3 bowls: Consider a system with 3 balls 1 Possibilities A state of the system is 1 ball in each bowl. # = 3! 3! 0! 0! Or 3 balls in bowl 2. 1 Possibilities # = 3! 1! 1! 1! 6 + 18 + 3 = 27 Total possible states 6 Statistical Debinition of Entropy W ∝ probability of a state, # W = 3 ! = 1 = 3 27 ! 2! 1! 09 Or 3 balls in bowl 1. 1 Possibility # =W 31 = ! 3! 27 0! 0! k = Boltzmann Constant ∴ΔS has units of J K- 1 ⎛ 1⎞
W =⎜ ⎟
⎝ 2⎠ 10 = 1
1024 S = - 6.93 k S is a state function. The value of S depends on the state, not on how the state was reached. W = probability Flip three coins “isoenergetic” degeneracy of a state, and/or number of microstates S is an extensive function. 5 3 Possibilities Let’s start using “microstates” 1.3805 × 10- 23 J K- 1 1
⎛ 1⎞
W =⎜ ⎟ =
⎝ 2⎠
32
S = - 3.47 k 2 W = 27 = 9 # = 3! 3! 0! 0! Boltzmann Equation Another state is 2 balls in bowl 1, 1 ball in bowl 2. # = N! N1! N2! N3! 3 Possibilities 1 Possibilities S = k ln W and 3 bowls: 6 Possibilities # = 3! 3! 0! 0! Or 3 balls in bowl 3. Entropy is the quantitative thermodynamic measure of disorder A B C D H H H H H T H T H T H H T T H T H T H T T T T T 4 possible “states” 8 possible outcomes S - 2 - 2.08 k - 0.98 k - 0.98 k - 2.08 k S = k ln W = k ln(1) = 0 0 - 1 1/8 3/8 3/8 1/8 S B C A D By this de5inition using probabilities entropy = 0 when probability =1. ⇒ All possible states are accessible. More ordered states (lower probability) have negative entropy. 4 … doing so allows us to de;ine a conceptual zero point for entropy 4 possible “states” 8 possible outcomes 1 3 3 1 0 k 1.10 k 1.10 k 0 k S 0 - 2 S - 2.08 k - 0.98 k - 0.98 k - 2.08 k 1/8 3/8 3/8 1/8 S - 1 W = # of states B A ⇒ C D 1 B C A D We’ve only shifted the scale. 1 3 3 1 0 k 1.10 k 1.10 k 0 k 4 possible “states” 8 possible outcomes If there is only one accessible microstate, the entropy of the system is given by: S = k ln W = k ln(1) = 0 Entropy is a state function; heat is not. Heat that would be transferred in a reversible process ΔS = SB - SA = q>0 S - 2.08 k - 0.98 k - 0.98 k - 2.08 k 1/8 3/8 3/8 1/8 Total number of accessible microstates: 8, so S = k ln(8) = 2.1 k Thermodynamic Debinition of Entropy q>0 W = # of states 2 0 dS = dqrev
T A B C D S W = probability Flip three coins H H H H H
T H T H T H H T T H T H T H T T T T T … doing so allows us to de;ine a conceptual zero point for entropy B q
∫A dT rev Adding heat increases the entropy of the system. The entropy of the system increases in an isothermal expansion. Irreversible expansion q = –w = 1.42 kJ w = –P2 ΔV = –1.42 kJ M P1 = 15 atm, V1 = 1 L P1 = 15 atm V1 = 1 L T = 298 K Pressure “isoenergetic” A B C D H H H H H T H T H T H H T T H T H T H T T T T T S W = probability Flip three coins Let’s start using “microstates” “isoenergetic” Let’s start using “microstates” P2 = 1 atm V2 = 15 L T = 298 K M Volume Reversible expansion ⎛ྎ V ⎞ྏ
w = −nRT ln⎜ྎ 2 ⎟ྏ = −4.12kJ
⎜ྎ V ⎟ྏ
⎝ྎ 1 ⎠ྏ
qrev = –wrev = 4.12 kJ P2 = 1 atm, V2 = 15 L ΔSsys = q rev 4.12 kJ
=
= 13.8 J K −1
T
298 K Entropy change in a system is calculated from reversible heat 5 Thermodynamics meets statistics… Thermodynamic de5inition: ΔS = SB − SA = ∫ B A dq rev 1 B
q
= ∫ dq rev = rev
A
T
T
T (V ) qrev = nRT ln isothermal ⇒ ΔU= q + w = 0 wrev = - nRT ln V2
1 (V ) ΔS = nR ln V2
1 (V )
V
2
1 entropy change in the isothermal expansion of an ideal gas. Statistical de5inition (counting states): S = k ln W V1 W1∝ (V1 )N ΔS = k ln (V2)N - k ln (V1)N ( V ) ( V ) ΔS = k ln V2 N = N k ln V2 1
1 N molecules of gas V2 W2∝ (V2 )N ( N ) ( V ) = N NA k ln V2 1
n A
R ⎛ྎ V ⎞ྏ
ΔS = nR ln⎜ྎ 2 ⎟ྏ
⎜ྎ V ⎟ྏ
⎝ྎ 1 ⎠ྏ 6 ...

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