Dario Bressanini Entropy: Qualitative More disordered = higher
S For given substance: S gas > > S liquid > S solid For
given substance: S gas > > S liquid > S solid Same
substance: higher T higher S Temperature Entropy (S) Melting Point
Boiling Point Solid Liquid Gas Entropy of fusion Entropy of
vapourisation Fig 20.5
Slide 3
Dario Bressanini Relative Entropy Entropy is a measure of
disorder: Entropy is a measure of disorder: In a phase change:
Solid Liquid Gas Highly orderedLess orderedVery disordered
Similarly, a gas is more probable than a solid: Similarly, a gas is
more probable than a solid:
Slide 4
Dario Bressanini Entropy, S S (gases) > S (liquids) > S
(solids) S o (J/Kmol) H 2 O(liq)69.91 H 2 O(gas)188.8 S o (J/Kmol)
H 2 O(liq)69.91 H 2 O(gas)188.8
Slide 5
Dario Bressanini 1.000 mole of an ideal gas initially at 8.00
atm and 100.0 E C expands adiabatically against a constant external
pressure of 1.50 atm until its pressure is 2.00 atm and its
temperature is 21.3 E C. What is S for the gas? Is the process
reversible or irreversible? Can we evaluate S for the gas by
calculating: S gas = I dq irr / T = I 0 / T = 0 Is S for the gas
greater than, equal to, or less than zero? To calculate S for the
gas we have to construct a hypothetical reversible path between the
inital and final states: reversible isothermal expansion of an
ideal gas dq = 0 (8.00 atm, 100.0 E C) (2.00 atm, 21.3 E C)
irreversible (2.00 atm, 100.0 E C) reversible constant pressure
cooling of an ideal gas S irreversible = S reversible expansion + S
reversible cooling = n R ln (P 1 / P 2 ) + n Cp ln (T 2 / T 1 ) =
(1.000 mole) (8.314 J/mole K) [ln (8.00 atm / 2.00 atm) + (5/2) ln
(251.9 K / 373.2 K)] = + 11.53 J / K - 3.268 J / K = + 8.258 J /
K
Slide 6
Dario Bressanini S E vap, 298 K H 2 O (l)
-----------------------------> H 2 O (g) 298 K, 1.00 bar 1.000
mole of liquid water is vaporized at 25.0 E C and 1.00 atm: Is this
phase change reversible and, if not, why not? An alternate
reversible path via which we can calculate the S E 298 K is
sketched below: SE1SE1 SE3SE3 S E vap, 298.2 K = ? H 2 O (l)
-----------------------------> H 2 O (g) 298.2 K, 1.00 atm S E
vap, 373.2 K = ? H 2 O (l) -----------------------------> H 2 O
(g) 373.2 K, 1.00 atm Step 1: In this step the liquid water is
reversibly heated from 298.2 K to 373.2 K at a constant pressure of
1.00 atm: S E 1 = 298.2 K I 373.2 K n C p, H 2 O (l) dT / T =
(1.000 mole) (18.07 cal / mole K) ln (373.2 K / 298.2 K) = + 4.05
cal / K
Slide 7
Dario Bressanini S E 3 = 373.2 K I 298.2 K n C p, H 2 O (g) dT
= (1.000 mole) 373.2 K I 298.2 K [7.17 - 2.56x10 -3 T + 8x10 +3 T
-2 ] dT / T = (1.000 mole) [ + 7.17 ln (298.2 K / 373.2 K) +
2.56x10 -3 (298.2 K - 373.2 K) - (8x10 +3 / 2) [1 / (298.2 K) +2 -
1 / (373.2 K) +2 ] = - 1.82 cal / K The entropy change for the
irreversible vaporization at 298.2 K and 1.00 atm is therefore: S E
vap, 298.2 = S E 1 + S E vap, 373.2 + S E 3 = (+ 4.05 cal / K) + (+
26.04 cal / K) + (- 1.82 cal / K) = + 28.27 cal /K Step 2: In this
step the liquid water is reversibly vaporized at 373.2 K and 1.00
atm: S E vap, 373.2 K = H E vap, 373.2 K / 373.2 K = + 9,720 cal /
373.2 K = + 26.04 cal / K Why did we choose to vaporize the water
at 373.2 K? Step 3: In this step the water vapor is reversibly
cooled from 373.2 K to the initial temperature of 298.2 K at 1.00
atm:
Slide 8
Dario Bressanini Otto Cycle Approximation of gasoline engine.
a-b adiabatic compression *V > rV b-c constant volume heat
addition *Q H c-d adiabatic expansion d-a constant volume heat
release a-e, e-a exhaust, intake QHQH QCQC W e b c d a V rV p
Slide 9
Dario Bressanini Idealized Diesel Cycle Approximation of diesel
engine. a-b adiabatic compression *rV > V b-c constant pressure
heat addition *Q H c-d adiabatic expansion d-a constant volume heat
release a-e, e-a exhaust, intake QHQH QCQC W c e b d a V rV p
Slide 10
Dario Bressanini Spontaneous Reactions reaction happening or
arising without apparent external cause; self- generated reaction
happening or arising without apparent external cause; self-
generated
Slide 11
Dario Bressanini The Solution Process For the dissolution of
KCl (s) in water For the dissolution of KCl (s) in water KCl (s) K
+ (aq) + Cl - (aq) Low entropyHigh entropy The formation of a
solution is always accompanied by an increase in the entropy of the
system! Entropy is the reason why salts like NaCl (s), KCl (s), NH
4 NO 3 (s) spontaneously dissolve in water. Entropy is the reason
why salts like NaCl (s), KCl (s), NH 4 NO 3 (s) spontaneously
dissolve in water.
Slide 12
Dario Bressanini Entropy, S Entropy usually increases when a
pure liquid or solid dissolves in a solvent.
Slide 13
Dario Bressanini Entropy, S negative sign indicates system is
more ordered negative sign indicates system is more ordered reverse
the reaction and sign changes reverse the reaction and sign changes
S o 298 = +0.1758 kJ/K where the + sign indicates system is more
disordered S o 298 = +0.1758 kJ/K where the + sign indicates system
is more disordered
Slide 14
Dario Bressanini Entropy, S Example 15-15: Calculate S o 298
for the reaction below. Use appendix K. Example 15-15: Calculate S
o 298 for the reaction below. Use appendix K.
Slide 15
Dario Bressanini Entropy, S Changes in S are usually quite
small compared to E & H. Changes in S are usually quite small
compared to E & H.
Slide 16
Dario Bressanini Fig. 20.9
Slide 17
Dario Bressanini The Standard Entropy of Reaction, S o rxn For
many chemical reaction:S o = S o products - S o reactants > 0
Standard entropy of reaction, S o rxn : mS o products -nS o
reactants S o rxn = But for reactions in which the moles of product
substances decrease, particularly gases which have very high
entropy, we predict that the entropy of the products is less than
that of the reactants and the entropy decreases during the
reaction: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) S o = S o products - S o
reactants < 0 S o rxn = (2 mol NH 3 x S o of NH 3 ) - [(1 mol x
191.5 J/mol K) + (3 mol x 130.6 J/mol K)] S o rxn = -197 J/K.. As
we predicted, S o < 0
Slide 18
Dario Bressanini Calculating the Standard Entropy of Reaction,
S o rxn I Problem: Calculate the S o rxn for the oxidation of one
mole of S 8 to form either SO 2 (g), or SO 3 (g) at 25 o C: S 8 (s)
+ 8 O 2 (g) 8 SO 2 (g) or S 8 (s) + 12 O 2 (g) 8 SO 3 (g) Plan: To
determine S o rxn, we apply Equation 20.3. We predict the sign of S
o rxn from the change in the number of moles of gas: 8 = 8 or 12 =
8, so the entropy will decrease ( S o rxn < 0). Solution:
Calculating S o rxn. From Appendix B values, Rx #1 S o rxn = ( 8
mol SO 2 x S o of SO 2 ) - [(1 mol S 8 x S o of S 8 ) + ( 8 mol O 2
x S o of O 2 )] = ( 8 mol x 248.2 J/mol K) - [(1 mol x 430.211
J/mol K) + (8 mol x 205.0 J/mol K)] = (1,985.6 J/K) - [(430.211
J/K) + (1,640.0 J/K)] = 1985.6 J/K - 2,070.211 J/K = - 84.611
J/K...
Slide 19
Dario Bressanini Calculating the Standard Entropy of Reaction,
S o rxn - II Rx #2 S o rxn = ( 8 mol SO 3 x S o of SO 3 ) - [(1 mol
S 8 x S o of S 8 ) + ( 12 mol O 2 x S o of O 2 )] = ( 8 mol x
256.66 J/mol K) - [(1 mol x 430.211 J/mol K) + (12 mol x 205.0
J/mol K)] = (2,053.28 J/K) - [(430.211 J/K) + (2,460.0 J/K)] =
2,053.28 J/K - 2,890.211 J/K = - 836.931 J/K Summary: For Reaction
#1 (SO 2 is product) S o rxn = - 84.611 J/K For Reaction #2 (SO 3
is the product) S o rxn = - 836.931 J/K As we predicted they are
both negative, but Rx#1 is close to zero, and we would also predict
it would be close to zero, since the number of moles of gaseous
molecules did not change from reactant to product....