1
Physical Chemistry I for Biochemists
Chem340
Lecture 17 (2/21/11)
Yoshitaka Ishii
Ch5.2, 5.4, 5.5-5.10 & HW5, HW6
Announcement
• HW6 is uploaded in the web site
• Questions from HW6 to be covered in Quiz 2:
P5.2, 5.5, 5.6, 5.7, 5.8, 5.13, 5.14, 5.20, 5.22, 5.30 & Q1-Q2,
where questions in red will be studiedwhere questions in red will be studied today.
• Review class on Wednesday/Friday
2
Incorrect answers in HW5
• P4.1(a) -1815 kJmol-1; -1818 kJmol-1
• P4.8 -1811 kJ
• P4.18 H-F 566 kJmol-1
• P4.19(b) 331 kJmol-1 328 kJmol-1
• (c) 590 kJmol-1 586 kJmol-1
• P4.29 (a) 5629 kJmol-1
Correction:
• P4.19) Given the data in Tables 4.1 and 4.2 and the data tables, calculate the bond
h l d f h f ll ienthalpy and energy of the following:
a. The CH bond in CH4
b. The CC single bond in C2H6
In b,c, use the data from a for CH bond enthalpy.
c. The CC double bond in C2H4
3
Ch 5.3 Introducing Entropy
T
DqdS reversibleEntropy: S [5.12]
0 T
DqdS reversible
S is a new state function
For a “natural” process S 0 in an isolated systemSo our world’s entropy is always increasing!
Second Law of Thermodynamics:
5.4 Calculating Changes in Entropy for Ideal Gas
T
DqdS reversible V=1L P = 1 bar
T =300 KV=1L P’ = P bar
Case 1. Adiabatic reversible processdS = Dqreversible/T = 0
Case 2. Isothermal reversible process (Vi, Ti) (Vf, Ti)
qreversible = -wreversible =
S = 0
)/ln( if
V
V
V
V
VVnRTdVV
nRTPdV
f
i
f
i
)/ln( ifreversiblereversible VVnR
T
q
T
DqS
Q. What is the sign of S for isothermal reversible compression?
4
Alternative Definition of Entropy in Statical Thermodynamics
S = kB ln(W),
where W is the number of possible configuration (or the degree of freedom)
For a “natural” process S 0 in an isolated system
Nature prefers a state with more randomness (or higher degree of freedom)
p y
Enthalpy Change for Irreversible Process
Case 2’ For isothermal irreversible process (Vi, Ti) (Vf, Ti)S must be calculated for an equivalent reversible process.
Ex. Isothermal adiabatic irreversible expansion (Vi, Ti) (Vf, Ti) Isothermal reversible expansion (Vi, Ti) (Vf, Ti)
)/ln( ifreversiblereversible VVnR
T
q
T
DqS
Q. What is the sign of S for isothermal irreversible expansion?
Q2. Does isothermal irreversible compression occur naturally?
V=1L P = 1 barT =300 K
V=1L P’ = 0 bar
Irreversible adiabatic & isothermal expansion
(1L, 1 bar, 300K) (2L, 0.5 bar, 300K)
2V P = 1/2 barT = 300K
Q. What is the sign of S for isothermal irreversible compression?
5
• dSL = -Dq/TL=CvdTL/TL
• dSR = Dq/TR=CvdTR/TR500K 300K
5.5 Using Enthalpy to Calculate the Natural Direction of a Process in an Isolated System
Dq
• dS = Dq(1/TR-1/TL)
400
300
400
500
R
K
K R
VL
K
K L
V dTT
CdT
T
CS
Dq > 0 dS >0 A
dS>0 or dS<0?
400K
0116
16
100400
400
100400100400
400
300400500400
22
2
2
lnln
ln
)/ln()/ln(
vv
v
VV
CKK
KC
KKKK
KC
KKCKKC
Q.How much is S for BA? Is S positive?
B
Calculated Change in Enthalpy (continued)• Case3 (3’). Reversible change in T for a fixed V
Case4 (4’) Reversible change in T for a fixed P
)/ln(,,
ifmVmvreversible TTnC
T
dTnC
T
DqS
Case4 (4 ). Reversible change in T for a fixed P
Case 5 (reversible) & 5’(irreversible) process: (Ti, Vi) (Tf, Vf)
By calculating S for (T V ) (T V ) (T V )
)/ln(,,
ifmPmPreversible TTnC
T
dTnC
T
DqS
Note: We assume Cv & Cp constant
By calculating S for (Ti, Vi) (Ti, Vf) (Tf, Vf)
Case 6 & 6’ (Ti, Pi) (Tf, Pf)
By calculating S for (Ti, Pi) (Ti, Pf) (Tf, Pf) (Pi/Pf = Vf/Vi)
)/ln()/ln( , ifmVif TTnCVVnRS
)/ln()/ln()/ln()/ln( ,, ifmPififmPif TTnCPPnRTTnCVVnRS
6
HW6• P5.5) Calculate S if the temperature of 1 mol of an
ideal gas with CV = 3/2 R is increased from 150 K to 350 K under conditions of (a) constant pressure and (b) t t l Ch th t(b) constant volume. Choose the correct one.
(1) (2)
(3) (4)
)/ln( KKR
3501502
3
)/ln( KKR
1503502
3
)/ln( KKR 3501502
)/ln( KKR 1503502
(5) (6) )/ln( KKR
3501502
5)/ln( KKR
1503502
5
P5.6) One mole of N2 at 20.5°C and 6.00 bar undergoes a transformation to the state described by 145°C and 2.75 bar. Calculate S if
C T T 2
When Cp or Cv is not constant
•
CP,m
J mol1 K1 30.8111.87 103 T
K 2.3968105 T 2
K2
1.0176108
T 3
K3
Tmp,f
bar6 00
bar 2.75lnndT
T
C n
p
plnnR ΔS
f
R[Q1]
1
Ti
KJ K
Tdn
bar6.00Tpi
K
K T
KT
dKT
cKT
ba15418
65293
32
.
.
7
Q1-2 How can we prove that S is a state function?HW6.
Dqreversible = dU -Dw
PdVdTCdVPP
T V
(U/V)T
PdVdTCdVPT
T VV
dTCT
dVT
P
T
DqdS V
V
ereveresibl 1
1
For ideal gas,
dTCT
dVVnRdS V
1 /
For liquid/solid,
dTT
CdVdTC
TdV
T
V
V
PdS v
VPT
1
5.7 The Change of Entropy in the Surroundings and Stotal = S + Ssurroundings
• The entropy of an isolated system increases in a spontaneous process.
Q Then is it tr e that a process is spontaneo s• Q. Then, is it true that a process is spontaneous
if S >0? True only for an isolated system.
• Consider surroundings for which P or V is constant.
• For a constant V, Dqsurroundings = dU
• For a constant P Dq di = dHState Functions!
For a constant P, Dqsurroundings = dH
dSsurroundings = dqsurroundings/Tsurroundings
dStotal = dS+ dSsurroundings = Dqrev/T + dqsurroundings/Tsurrounding
A spontaneous change occurs when dStotal > 0
8
HW6 P5.7, P5.8P5.7) One mole of an ideal gas with CV = 3R/2
undergoes the transformations described in the following list from an initial state described by Ti = 300. K and Pi = 1.00 bar. Calculate q, w, U, H,i q, , , ,and S for each process.
(c) The gas undergoes a reversible isothermal expansion at 300 K until P is half of its initial value.
P5.8) Calculate Ssurroundings and Stotal for part (c) of Problem P5.7. Is the process spontaneous?Problem P5.7. Is the process spontaneous?
(c) S = -nRln(Pf/Pi)+ nCp,mln(Tf/Ti) = –nRln(0.5/1)
Because T is const Ssurrounding= -q/Tsurrounding = -q/T
So Stotal = 0
)/ln()/ln( ifif PPnRTVVnRTPdVwq [Q1]
[Q2]
5.2 Heat Engines and the Second Law ofThermodyamics
What is Efficiency =|wcycle|/|qab|?
9
5.2 Heat Engines and the Second Law of Thermodyamics
Isotherm: -PdV = -(nRT/V)dVa b: qab = -wab =nRThotln(Vb/Va)>0
d RT l (V /V )<0
Carnot Cycle
c d: qcd = -wcd = nRTcoldln(Vd/Vc)<0Adiabatic: q= 0da & bc: qcd = qbc=0
U = 0 wcycle + qab+qcd =0 wcycle = -(qab+qcd) <0 (|qab|>|qcd|)
Efficiency =|w|/|qab| = |qab+qcd|/|qab| < 1
S=0 for Carnot Cycle?• S = qab/Thot + qcd/Tcold
= nRThotln(Vb/Va)/Thot + nRTcoldln(Vd/Vc)/Tcold
Rl (V /V ) Rl (V /V )= nRln(Vb/Va) + nRln(Vd/Vc)= nRln(VbVd/VaVc)
d a & bc are adiabatic processes:• TcoldVd
-1 = ThotVa-1 & ThotVb
-1 = TcoldVc-1
(VdVb )-1 = (VaVc)-1 VbVd/VaVc=1
S = qab/Thot + qcd/Tcold = 0
wcycle = -(qab+qcd) = nRThotln(Vb/Va) + nRTcoldln(Vd/Vc)= nRThotln(Vb/Va) - nRTcoldln(Vb/Va)
= nR(Thot-Tcold)ln(Vb/Va)
10
5.6 Clausius Inequality (p95)dU=Dq – PextdV (5.29)
• For a reversible change,
dU=Dqreversible – PdV = TdS – PdV (5.30)
When dU and dV for (5 29) and (5 30) are common• When dU and dV for (5.29) and (5.30) are common,
TdS – PdV = Dq-PextdV
TdS – Dq = (P – Pext )dV (5.31)
If P – Pext >0 The system will expand or dV > 0.
If P – Pext <0 The system will shrink or dV < 0. ext y
Namely, TdS – Dq > 0 when P Pext (when 5.29 is irreversible)
T
DqdS leirreversib
0T
DqdS leirreversib For an irreversible process in
an isolated system
T
DqdS leirreversib0
Sample Question
• Which relationship is correct for an i ibl i l t dirreversible process on an isolated system? Choose all.
(a) (b)
(c) (d)
T
DqdS leirreversib T
DqdS leirreversib
0dS 0dS(c) (d)
(e) (f)
0dS 0dS
T
Dq leirreversib0 T
Dq leirreversib0
11
Sample Question2• Which relationship is correct for an
irreversible process? Choose all. p
(a) (b)
(c) (d)
T
DqdS leirreversib
T
DqdS leirreversib
0dS 0dS
(e) (f)
(g)
T
Dq leirreversib0 T
Dq
T
dS leirreversib
T
DqdS leirreversib
S in vaporization and fusion (p90)
• During fusion and vaporization, T is const.Li idG d PLiquidGas under a constant P
Solid Liquid under a constant P
onvaporizati
onvaporizati
onvaporizati
reversiblereversibleonvavorizati T
H
T
q
T
DqS
fusion
fusion
fusion
reversiblereversiblefusion T
H
T
q
T
DqS
12
5.8 Absolute Entropies and The Third Law of Thermodynamics
• The entropy of an element or a compound is experimentally determined from – Dqreversible =CpdT
Cp,m for O2
T gasTb Liquid
f
mfusionTf Solid
pmmm
dTCHdTC
T
H
T
dTCKSTS
'""
'
')()( ,
0
0
Molar Entropy for Gas
T
Tb
gasmp
b
monvaporizatiTb
Tf
Liquidmp
T
dTC
T
H
T
dTC
'"
'"
"
" ,,,
The entropy of a pure, perfectly crystalline
substance (element or compound) is zero at 0K.
Third Law of thermodynamics - What is Sm(0K)?
T Dependence of SmCp,m /T > 0 and Svaporization, Sfusion > 0 Sm increases as T increases
13
The origin of Cp,m and Sm for Solids• Number of degrees of freedom for molecule made
of n atoms.
NT: Translational: 3
NR: Rotational: Atom 0, Linear 2, Non-Linear 3
NV: Vibrational: 3n – NT – NR
Cp,m & Sm higher for a molecule with more atoms.
Calculated Change in Enthalpy (continued)• Case3 (3’). Reversible change in T for a fixed V
Case4 (4’) Reversible change in T for a fixed P
)/ln(,,
ifmVmvreversible TTnC
T
dTnC
T
DqS
Case4 (4 ). Reversible change in T for a fixed P
Case 5 (reversible) & 5’(irreversible) process: (Ti, Vi) (Tf, Vf)
By calculating S for (T V ) (T V ) (T V )
)/ln(,,
ifmPmPreversible TTnC
T
dTnC
T
DqS
By calculating S for (Ti, Vi) (Ti, Vf) (Tf, Vf)
Case 6 & 6’ (Ti, Pi) (Tf, Pf)
By calculating S for (Ti, Pi) (Ti, Pf) (Tf, Pf) (Pi/Pf = Vf/Vi)
)/ln()/ln( , ifmVif TTnCVVnRS
)/ln()/ln()/ln()/ln( ,, ifmPififmPif TTnCPPnRTTnCVVnRS
14
5.9 Standard States in Entropy Calculation (P102
• For enthalpy, we defined Hf,A0 for the most stable
pure elements at 298.15K (T0) and 1 bar (P0) as 0.
• We define standard state of Entropy as S0
m = Sm(P0, T0) • What is the relationship between S0
m(P) and S0
m(1 bar)?
S (1 bar P) = −Rln(P/P ) Ent
ropy
Sm(1 bar P) = −Rln(P/P0)
Mol
ar E
S0m
Sm(P) = S0m(P0) −Rln(P/P0)
5.10 Entropy Change in Chemical Reaction
At 298.15 K and 1 barA + 2B 2C + D • Sreaction
0 = 2SC,m0 + SD,m
0 – SA,m0 – 2SB,m
0
I lIn general
A(T) + 2B(T) 2C(T) + D(T)For a constant pressure
X
mXXR STS 00,)(
T
p
T
dTCKSTS
298
0
298'
')()(
00
XmpXp CC ,
Q. How much is Cp0 for the above reaction?
00000 22BmpAmpDmpCmpp CCCCC ,,,,
15
• P5.20) Consider the formation of glucose from carbon dioxide and water, that is, the reaction of the following photosynthetic process: 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g). The following table of information will be useful in working this problem:p
• Calculate the entropy and enthalpy changes for this chemical system at (a) T = 298 K and (b) T = 330. K. Calculate also the entropy of the surrounding and the universe at both temperatures.
(a) SR0 = XSX
0
(b)
T
p
T
dTCKSTS
298
0
298'
')()(
00 pXXp CC
HW5 P4.32) Using the protein DSC data from Problem P 4.31, calculate the enthalpy change ΔH between the T=288 K and T=318 K. Give your answer in units of kilojoules per mole. Assume the molecular weight of the protein is 14000. grams.
CPtrs
CP
ΔH /g
Area
(1) Obtain the orange area in mm2
ΔH/mol =ΔH /g Mlysozyme
Tm
CPint
(2) Obtain the area for 0.418 Jg-1x 10
in mm2
(3) Convert (1) into Jg-1
16
• P4.2) Calculate and for the oxidation of benzene. Also calculate
Hreaction Ureaction
• The chemical equation for the oxidation of benzene is:
lOH 3gCO 6gO 7lHC 22221
66 (*)
Hreaction
in P4.2
• P4.3) Use the tabulated values of the enthalpy of combustion of benzene and the enthalpies of formation of CO2(g) and H2O(l) to determine for benzene.H f
• Q3(Mod). Now, we would like to derive Hreaction(P) at a pressure slightly higher than 1 bar at a standard temperature. Derive Hreaction(P) using J-T, Cp,A, Hreaction(1 bar). Use dHA = CpAdT – Cp, AJ-T dP + (H/n)T PdnA (A = B X or Y) assuming that J T and(H/n)T, PdnA (A B, X, or Y) assuming that J-T and Cp,A are constant. Specify the path of (P, T, n) you would like to use. Assume a reaction 3B X + 2Y.
Step 1: 3B(P) 3B(1 bar) H1 = – 3Cp, BJ-T,B (1 bar-P)
• Step 2: 3B(1 bar) X(1 bar) + 2Y(1bar)
H H (1 b )H2 = Hreaction(1 bar)
• Step 3: X(1 bar) + 2Y(1bar) X(P) + 2Y(P)
H3 = – (Cp, XJ-T,X + 2Cp, YJ-T,Y) (P- 1 bar)
Hreaction(P) = H1 + H2 +H3
[Q1]
[Q2]
17
P4.19) Given the data in Tables 4.1 and 4.2 and the data tables, calculate the bond enthalpy and energy of the following:
a. The CH bond in CH4
b. The CC single bond in C2H6g 2 6
In b,c, use the data from a for CH bond enthalpy.
c. The CC double bond in C2H4
a) CH4(g) C(g) + 4H(g)
∆Hreaction0= ∆Hf
0(C(g))+ 4∆Hf0(H(g)) - ∆Hf
0(CH4(g)) [Q1]
∆Hbond0(C-H) = ∆Hreaction
0/4
b) C2H6(g) 2C(g) + 6H(g)
∆Hreaction0= 6 ∆Hbond
0(C-H) + ∆Hbond0(C-C)
[Q2]
[Q3]
• P4.8) Calculate at 650 K for the reaction 4NH3(g) +6NO(g) 5N2(g) + 6H2O(g) using the temperature dependence of the heat capacities from the data tables. (-1811 kJ)
T
dTTCHH 00 ')'( K
PKreactionTreaction dTTCHH15298
1529800
.
.,, ')'(
(See Table 2.4 in Appendix B for Hf0)
CP = {-4ANH3(1) - 6ANO(1) + 5 AN2(1) + 6AH2O(1) }+ {XAX(2) }(T/K) + {XAX(3) }(T/K)2 + {XAX(2) }(T/K)3
XXX
T
K
n
n
T
K
P nAnBn
TnBdTTC )}({)(;
')(')'(
.
298
4
115298