physical chemistry i for biochemists lecture 41...
TRANSCRIPT
4/25/2011
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Physical Chemistry I for BiochemistsChem340
Lecture 41 (4/25/11)
Yoshitaka Ishii
Ch 9.11‐12, 10.1‐2, 10.4Review for Ch1‐2
Announcement
• We will perform review classes in this week.
• Average of Exam 3 ~81 points
• Review Quiz on 4/29 (Ch1‐6 & Ch9, Ch10)
• Study previous Exams & Quiz for the Final Exams
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9.8 Redox Reaction in Electrochemical Cells and the Nernst Equation
• Left(anode):
• Right (cathode) :
Zn(s) Zn2+ (aq) + 2e‐
Cu2+ (aq) + 2e‐ Cu(s) Daniel Cell
• Overall:
• Anode: Red1Ox1+1e‐
• Cathode: Ox2 +2e‐Red2• Overall:
Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s)
Zn Zn2+ + 2e‐ Cu2+ + 2e‐ Cu
• Overall:
2Red1 + 1Ox2 2Ox1 +1Red2Q. What is overall reaxn for Cu/Zn?
Q2. How do you get Greaction
using chemical potential?
• P9.40) By finding appropriate half-cell reactions, calculate the equilibrium constant at 298.15 K for the following reactions:
• a. 2Cd(OH)2 2Cd + O2 + 2H2O( )2 2 2
• a) From
• Cd(OH)2 + 2e– → Cd + 2 OH– Eº = –0.809V
• O2 + 2H2O + 4e–→ 4OH– Eº = 0.401V
The half cell reactions areThe half cell reactions are
4OH–→ O2 + 2H2O + 4e– EOXº= -0.401V
2Cd(OH)2 + 4e– → 2Cd + 4 OH– ERed = –0.809VEcell = EOX
0 + ERed0 K = exp(-ΔG/RT)= exp(-nFE/RT)
[Q1]
[Q2]
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• P9.32) Calculate and the equilibrium constant at 298.15K for the reaction
U i h h lf ll i
Gr
Cr2O7
2 aq 3H2 g 8H aq 2Cr3 aq 7H2O l .• Using the half cell reactions:
Ox (Anode):
Red (Cathode:) )7(Cr 2 )14(e 6OCr 23
22
72 OHH
e 6H 6H 3 2[Q1]
ΔFΔG
• What is n? n = 6
ΔφFn ΔG reaction
[Q2]
9.11 The Determination of E0 and Activity Coefficients Using an Electrochemical Cell
• Cell: Ox: Ag Ag+ + e‐ Red: H+ +e‐1/2H2 (SHE)
• For Ag+/Ag Ag+ arises from the dissociation of AgCl• For Ag+/Ag, Ag+ arises from the dissociation of AgCl. Assume that aAg+ = aCl‐. How can we get the activity? Recall activities of individual ions cannot be measured directly.
a = a+
+ a‐‐a
2 = aAg+ aCl‐ a = aAg+ = aCl‐ (if aAg+= aCl‐)
• Similarly and m m m Then the
[Q1]
• Similarly, = Ag+ = Cl‐ and m= mAg+ = mCl‐ . Then the cell potential is
•
• For low mAg+,
E = E0Ag+/Ag+(RT/F)ln(aAg+) = E0
Ag+/Ag +(RT/F){ln(m)+ln }
ln =-1.173|z+z-|(I)1/2 at 298K (9.33)
(I)1/2 = ({m+zi+2+ m-zi-
2}/2)1/2=m1/2
[Q2] [Q3]
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• P9.18) (a) Calculate I, ±, and a± for a 0.0250 m solution of K2SO4 at 298 K. Assume complete dissociation.
(b) H fid h l l d(b) How confident are you that your calculated results will agree with experimental results?
• += 2 ‐ = 1 z+ = 1 z‐ = ‐1
• I =(m{+z+2+ ‐z‐2}/2)1/2[Q1]
log = -0.50926|z+z-|(I)1/2
a± = m± ±
(b) See Figures 9.8 to find the deviation from DH limiting law using Davis equation (9.34)
[Q2]
m± = (m+2 m-)1/3[Q3]
More Na+ (coutNa+ > cin
Na+) +
More K+ (cinK+ > cout
K+) +
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Ch10.4 Biological Membranes and the Energy of Ion Transport
• The chemical potential for
a neutral species A outside thea neutral species A outside the
membrane is
• out = 0out +RT ln(cout/c
0),
where cout is the concentration of A . Similarly, for A inside the cell is
in = 0in +RT ln(cin/c
0),
The work involved to move dn moles of A from inside to out is
dG ( 0 0 )dn +RT ln(c /c ) dG = (0out ‐ 0
in)dn +RT ln(cout/cin).
Assuming 0out ~ 0
in,
dG = out ‐ in = RT ln(cout/cin)
Q. Which chemical potential is higher when cout > cin?
Q. When does the system reach equilibrium?
Chemical potential across membrane when the electronic potential is present
zFccRT )()/ln(~~~ zF~
So the system reaches equilibrium when
= ‐(RT/zF) ln(cout/cin).
zFccRT
zFccRT
inout
inoutinoutinout
)/ln(
)()/ln(
0~ [Q1] = 0.
More Na+ (coutNa+ > cin
Na+) +
Q. Is “cout = cin” correct ?
More K+ (cinK+ > cout
K+) +
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• P10.6) Calculate the reversible work required to transport 1mol of K+ from a region where
to a region where if the C
K 5.25 mM CK 35.5 mM
potential change accompanying this movement is = 0.055 V. Assume T = 298 K.
• The reversible work is given by:
zFccRT inifininiinifin fin )()/ln(~~~
zFccRT inifin
inifininiinifin f
)/ln(
)()(
G for Sodium‐Potasium Pump
• 3Na+(in) + 2K+(out)+ ATP
3N +( t) 2K+(i ) ADP3Na+(out) + 2K+(in)+ ADP
The work required to transport
3Na+ to inside and 2K+ to outside is
outK
inoutNa
inG
~2~3
FccccRT
FccRT
FccRTG
Kout
Kin
Nain
Naout
outinKout
Kin
inoutNain
Naout
)}/ln(2)/ln(3{
)}()/ln({2
)}()/ln({3
Try Example 10.4
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9.12 Biochemical Standard State• pH = 7 (cH+
0 = 10‐7 M NOT 1M)
• cH2O0 is that for pure water (Not 1M)
• Standard state of molecule having multiple Sta da d state o o ecu e a g u t p eionic states is defined by a total concentration (unless otherwise defined)
(ex. cATP = [ATP4‐] + [H‐ATP3‐] .. See Ch10.5 )
● aH2O0 = 1 for dilute aquatic solution (or pure
water)
● aH+0 = 1 at pH =7
Neglect aH2O and aH+ from Q for dilute aquatic solution at pH =7. (We set this as Q’)
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10.1 Thermodynamics and Living Systems
• 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)
Autotrophic Cells
• G0’reaction = G0’f(C6H12O6) – 6G
0’f(CO2) – 6G’
0f(H2O)
G0’reaction = 2868 kJ mol‐1 > 0 endergonic
H0’reaction = 2813 kJ mol‐1
S0’reaction = (H0’reaction ‐ G0’
reaction )/T = ‐182.3 JK‐1
1mol‐1
Q1. Is the reaction spontaneous?
Q2. What is needed to drive the reaction?
Q3. Guess the reason why S0’reaction is negative here
Sm for O2 (p100)Q. What is the reason why Sm changes discontinuously between liquid and gas?
T
DqS reversible
onvaporizati
onvaporizati
onvaporizati
reversiblereversibleonvavorizati T
H
T
q
T
DqS
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• C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)
• G0’reaction= ‐ 2868 kJ mol‐1 < 0 exergonic
H0’reaction = ‐ 2813 kJ mol‐1
Heterotrophic Cells
S0’reaction = (H0’reaction ‐ G0’
reaction )/T = 182.3 JK‐1 mol‐1
Adenosine triphosphate (ATP) and adenosine diphosphate (ADP)
Read Ch. 10.3 & 10.5 by yourself
10.2 The Principle of Common Intermediates
ATP(aq) + H2O(l) ADP(aq) + Pi(aq)
GØ ‘ = ‐30.5 kJmol‐1
Ex. 10.1 (p249) This standard condition is not applied in a cell. In the cell, typical concentration of ATP, ADP, and inorganic phospate arecATP = 1850 uM, cADP = 138 uM, cP = 1.00 mMATP , ADP , P
How much is G at the concentrations? G = GØ’ + RTlnQ’ = GØ ‘ + RTln{(138 uM/1 M)(1 mM/1M)/(1850uM/1M) }
= GØ’ +RT ln(7.46 x 10‐5) = ‐55.0 kJ
[Q1]
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1.2 Basic DefinitionsClosed system: A system that does not exchange matters with the “surroundings”.
Surrounding: All the matters in universe otherthan the “system”
Open system: A system exchanging matterswith the “surroundings”.
Isolated system: A system that does not h i hexchange matters or energy with
the “surrounding”
Boundary: An Interface between the systems. Boundaries determine if the matter and/or energy can be transferred.
Equilibrium
(P1, V1, n1, ..) (P2, V2, n2, ..)
Two isolated systems characterized by system variables such as P, T, V ..
T1 T2
(P1, V1, n1, ..) (P2, V2, n2, ..)
The exchange of energy and/or matters through the boundary
A system variable reaches a constant
(P’1, V’1, n’1, ..) (P’2, V’2, n’2, ..)
over time in any part of the systems Equilibrium
Thermodynamics equilibrium: Equilibrium with respect to P, T, and concentration ( or n/V)
T’ T’
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Thermal Equilibrium
• Equilibrium with respect to Temperature[Q1 ]
• For ideal gas,
P = nRT/V = RT T = P/R
Thus, if the molar density are common b t t t P P h thbetween two systems, P1 = P2 when the systems reach thermal equilibrium.
Ideal Gas Eq. using molar density
PV = nRT
P, T, : Intensive variable (variables that does not depend on the size of the system)
n, V: Extensive variable (those that depend on the size)
P = RT This depends on only three intensive variable.
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Isotherms, isobars, isochores
• Isotherm: T = constant PV/n = const
• Isobar: P =const V/nT = const
• Isochore (or isovolumetric): V = const
• P/nT = const
60
80
a)
Q. Which type ofgraphs are they?
T increase
0
20
40
0 0.5 1 1.5 2
1/V (1/m3)
P (
Pa
Sample Question on Charles’s law
• Sam has a balloon filled with N2, which has a volume of 1 00 L at 25 C He heated up thevolume of 1.00 L at 25 C. He heated up the balloon to 50 C to float the balloon. Assuming the pressure inside the balloon is unchanged at 1 atm, how much is the volume of the heated balloon?
• V /V = T /T• V1/V0 = T1/T0 V1 = V0x T1/T0
= 1.00 L x (273 +50)/(273 + 25)
= 1.08 L
[Q1 ]
[Q2 ]
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1.4 Equation of State & Ideal Gas Law
• PV = nRT
= NkBT,
where R and kB are called ideal gas constant and Boltzman constant.
R = 8.315 JK‐1mol‐1
kB = R/NA, where NA is Avogadro constant
Choose a right unit for R (see Table 1.2 in p.8)
Q. For the above R, what kind of units should you use for V and P? V [m3] P [Pa]
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A Mixed Ideal Gas & Partial Pressure
• For a mixture of ideal gasses, the partial pressurefor molecule i is given by
P RT/V ( B l ’ Ch l ’ t)Pi = niRT/V, ( Boyle’s, Charles’s are correct)
where ni denotes # of moles for the gas molecule i.
• The overall pressure of the mixed gas is
P = PiP = Pixi = Pi/P = ni/n is called molar fraction,
where n = ni
• P1.2) Consider a gas mixture in a 2.00-dm3
flask at 27.0°C. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the g pcomposition of the mixture in mole percent:
• a. 1.00 g H2 and 1.00 g O2
Pa 10 6.18
mol kg10 2.02 m 10 2.00
K 300.15mol K J 8.314472 kg 0.001
VM
T R m
V
T R np 5
13-33-
11
2
22
2
H
HHH
TRTR 11 Pa 10 3.90
mol kg10 32.0 m 10 2.00
K300.15molKJ8.314472kg0.001
VM
T R m
V
T Rnp 4
13-33-
11
2
22
2
O
OOO
Pa 10 57.6ppp 5total 22
OH
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• P1.15) Consider a 20.0‐L sample of moist air at 60.°C and 1 atm in which the partial pressure of water vapor is 0.120 atm. Assume that dry air has the composition 78.0 mol % N2, 21.0 mol % O2, and 1.00 mol % Ar.
Wh t i th l t f O ?• a What is the mole percentage of O2?
PO2/Ptotal = {(1.0 ‐0.12)atm x 0.21} /1.0 atm
• b. The percent relative humidity is defined as
where is the partial pressure of water in the sample and atm is the
%RH PH2O PH2O
*
PH2O* 0.197
PH2O
[Q1]
equilibrium vapor pressure of water at 60°C. The gas is compressed at 60 °C until the relative humidity is 100%. What volume does the mixture contain now? Q. How do you solve this? PH2O(20L ) = PH2O* (V)
Practice for c.
Parameters & Units (in International units of system or SI unit)
In SI, there are 7 base units: m, kg, s, A (ampere), K (kelvin), mol, cd (candela), g, , ( p ), ( ), , ( )
• Weight (kg)• Force (N; newton) 1 N = 1 kg m/s2
• Pressure: Force/Area‐ In SI units, Pa (pascal) is used.1 Pa = 1 N/m2 = 1 N (= 1 x 105 bar) / ( )
• Volume m3
1 (dm)3 = 1x10‐3 m3 ( = 1 L)
About the SI units, seehttp://physics.nist.gov/cuu/Units/units.html
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Non‐ideal Gas
• Two assumptions for the ideal gas
(1) No interactions between gas molecules
(2) The gas molecule can be treated as a point mass (no volume is considered)
The assumption is correct when gas density is low. (roughly speaking P < 10 bar and T > 100K)
Isothermal P‐V Plots for Real Gas (Experiment for CO2)
50 C Above Tc, G L
Tc: Critical temperatureQ. Why does the curve drop at 40 C?
40 C31.4 C (Tc)
20 C
0 CLiquid CO2
x
,transition is not possible
X denotes inflection point in isotherm(dP/dV 0 d2P/dV2 0)
Condensation from gas to liquid below Tc
Raff (p21‐41)
(dP/dV = 0, d2P/dV2=0)P & V at this point are named Pc and Vc.
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Non‐ideal Gas: two types of interactions
r > rtransition
otential Energy rV < r < rtransition
V(r) < 0
Po
V(r) < 0
V(r) > 0
r < rV
2
2
V
an
van der Waals Equation of State
nbV
nRTP
VnbV b: finite size of 1 mol of the moleculesa: attractive force constant
RTP
a
bVP
m
2mV
a
nVVm /
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van der Waals Equation for CO2
120
140 P (bar) (ideal gasat 300K)P (CO2) at 290K
Well reproduces the experimental curves above Tc!
60
80
100
120
P (
bar
)
(CO ) at 90
P (CO2) 304.3K
P (CO2) 315K
P (CO2) 350K
0
20
40
0 0.1 0.2 0.3 0.4 0.5 0.6
V (Lmol-1)
Critical temp obtained from dP/dVm = 0and d2P/dVm
2=0
Maxwell constructionreproduces G‐L transition
Calculate Tc from Van der Waals Eq.
2mm V
a
bV
RTP
At Tc the following relationships are expected
bVVmCmC
46
0
232
mm
c
m V
a
bV
RT
V
P
0
62432
2
c aRTP
At Tc, the following relationships are expected.
32
2
mCmC
C
V
a
bV
RT
26 RTa C 432 mmm VbVV 34 bVV mCmC
Vmc = 3b, Tc = 8a/(27bR), Pc = a/27b2
83 /cmcc RTVP
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HW21 (a) Calculate the molar volume, temperature,
pressure of H2O at the critical point using van der Waals parameters a and b (a = 5.537 bar mol‐1, b = 0.03049 Lmol‐1).0.03049 Lmol ).
(b) Obtain the density of H2O at the critical point, and compare it with that of liquid H2O (~1 g/mL). Assume that the molar mass of H2O is 18.0.
a) Easy as long as you remember equations.
Vmc = 3b, PC = a/27b2, PCVMC= (3/8)RTC[Q1] [Q2] [Q3] mc C C MC C
b) = M / Vmc
(Do not forget to convert the unit to g/mL !)
[ ] [ ] [ ]
[Q4]
Problem for HW2
Vmc = 3bTc = 8a/(27bR)P = a/27b2
[Q1 ]
[Q2]
b = Vc/3 a = 3 (Pc x 9b2) = 3Pc x Vc2
How can we get the radius from a Pc = a/27b or b?
Assume that there is no space between sphere. 4R3/3 = b/NA
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Kinetic Energy of Gas• v2 = vx
2 + vy2 + vz
2
• The probability of finding a molecule flying at vx is
p(v ) = exp( mv 2/2k T)/Z
V = (vx, vy, vz)
p(vx) = exp(‐mvx2/2kBT)/Z,
where Z = {2kBT/m}1/2 and a = m/2kBT.
The expected kinetic energy due to vx
2222
22
)()( dvvpvm
dvvpmvmv
xxxxxxx
24
22
22 /)/exp( TkaaZ
mdvTkmvv
Z
mBxBxx
2222 222 //// Tkmvmvmv BZYX
2322 // TkmvE Btrans Q. Is this energy for 1 mol?
Temperature Dependence of Distribution of |vx|
• P(vx) = exp{‐mvx2/(2kT)}/Z
0 001
0.0015
0.002
0.0025
ob
abil
ity
O2 150K
O2 300K
O2 600K
0
0.0005
0.001
-1200 -800 -400 0 400 800 1200Vx (m/s)
Pr
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Distribution of speed C, P(C) (Raff994)
Distribution of Speed for O2 in Gas
C = {vx2+vY
2+vZ2}1/22
223
224 C
kT
mC
kT
mCP
exp)(
/
2.00E-03
3.00E-03
P(C
)
150K O2
300 K O2
600K O2
0.00E+00
1.00E-03
0 500 1000 1500 2000
Speed (C) (m/s)
P
How can p(C) have a very different shape from p(vx)?
ZYx
zyxzyxzyxzYx
dvdvdvZvvvm
dvdvdvvpvpvpdvdvdvvvvP
3222
/exp
)()()(),,(
zyx dvdvdvZ
kT2/exp
Change Cartesian velocity (VX, VY, VZ) Polar Velocity (C, , )
ddCdCdvdvdv zyx sin2
)(),,( CPvvvPC zYx 24 )(),,( zYx
ddCdCZkT
CmddCdC
C
CPsin/expsin
)( 232
22 24
32
2
24 Z
kT
CmCCP /exp)(
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<C>, <C2>1/2, Most probable C(Raff 998‐999)
• Average speed, <C>:2123 //
(Home work 2)
• Root‐Mean‐Square speed Crms:
21
0
3223
8
224
//
exp
M
RTdCC
kT
mC
kT
mC
2121
4223
212 34
///
exp
RTdCC
mCmC /
• Most probable C: Cmp
0 224 exp
MdCC
kTkTC
212
/
M
RTCmp
dP(C)/dC = 0
Q. Which is greater?
Kinetic Theory of Gas An ideal gas of N molecules (mass m) is enclosed in a cubic box of length L
‐mvx
Force by one molecule F = mdv/dt = dp/dt~ p/t = 2mvx/(2L/vx)= mvx
2/L
By N molecules,
LvmF kx
N
k
/
2
1
mvx
L
LvmN
LvN
mN
x
N
kkx
/
/
2
1
21
P = F/L2 P =mN<vx2>/L3 = NkBT/L
3= nRT/V
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Internal Energy of Translation, Rotation, Vibration (Text Ch23 p600)
Translational Energy for N molecules33 nRTTNk
Rotational Energy
• UR = NkBT = nRT (linear)
• UR = 3NkBT/2 =3nRT/2 (non‐linear)
U 0 ( t i )
2
3
2
3 nRTTNkU B
T Internal Energy of Ideal Gas
[Q1] UR = 0 (atomic)
Vibrational Energy
1
)/~exp(
~
kThc
NhcUV
[Q1]
Ch. 2
• First law of Thermodynamics:
The internal energy, U, of an isolated
system is constant.
In other words, if the surrounding exchanges energy with the system,the total energy of the surrounding and the system should not change.
0= UTotal = Usystem + Usurrounding
Usystem = ‐Usurrounding
In a closed system: ‐Usurrounding = q(heat) + w(work)
U = q + w
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Work by pressurePressure: P = F/A, where A is area F = PexternalA
A
)( dzAPdzAP
dLFw
externalexternal
The work done by the system:A
dz
)(VdPexternal
Note: In general, Pexternal Psystem
VPw externalFor a constant Pext
W > 0
U = w > 0
The sign of w (Q. Which process gives us a positive work?)
Initial Final
W 0
PiVi PfVf
Initial Final
W < 0 U = w < 0
P’iV’iP’fV’f
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Sample Question (Ex. 2.1 p18)
• Calculate the work involved when a human adult exhales. Assume that an adult expels 0.5 L of air at 1 atmatm.Pexternal = 1 atmV = 0.5 L
W = ‐ Pexternal V = ‐1 atm x 0.5 L = ‐1x 1.01x105Pa x 0.5 x 10‐3 m3
= ‐50.0 J (~ ‐0.2 kcal) ( )
Coke ~ 97 Food cal = 97 kcal ~ 500 breathsDiet Coke 1 Food cal = 1 kcal ~ 5 breaths
Math in Expansion of Gas)(VdPw external Case [3] Isothermal Reversible
Expansion. Pexternal = Pint = nRT/V& T = const (see p27)We assume P // dV
finV
V
Vini
external dVPwfin
Case [1] If Pexternal = const (p24)
fin
Vini
externaldVPw
)/(
)/(
V
Vini
V
Vini
dVVnRT
dVVnRTw
fin
fin
1
)( inifinexternalVVexternal
Vini
VVPVP fin
ini
)}/{ln(
)}ln(){ln(
)][ln(
inifin
inifin
VV
VVnRT
VVnRT
VnRT fin
ini
Case [2] Reversible Expansion Pexternal = Pint
finV
Vini
dVTVPw ),(int
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Work by Electrostatics
The work required for transporting charge q across the electrical potential
• Welectrical = q
Using a current I and (q = It),
• Welectrical = I t [A V sec] = [W sec] = [J]
(Ch 9.1)
-Fdx = (kx)dx
2.3 Heat
• Heat is defined as the quality of energy that flows across the boundary between theflows across the boundary between the system and the surrounding because of a temperature difference
Heat
When w = 0U = qsys
System
Heat qsys < 0
System
Heat qsys > 0
Tsurrounding < Tsys
Surrounding Surrounding
Tsurrounding > Tsys
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2.4 Heat Capacity
• Assume that the work w of It is done by a heating coil in a system and the work is completely converted to heat q (ie q It)completely converted to heat q (ie. q = It)We expect temperature change from Tini to Tfin
The response of the system to a heat flow is described by heat capacity Cdescribed by heat capacity C
dT
Dq
TT
qC
inifinT
0
limD is denoted as d in the text. The meaning will be discussed in a later section.
C denotes heat needed to increase T by 1K
Cm = C/n: molar heat capacity
[J/K]
Heat Capacity Depends on the Conditions (V, P, T)
• CV: Heat capacity for a constant volume
T dependence of Cmp for Cl2constant volume
• CP: Heat capacity for a constant pressure
CV CP
C and C also depend on
mp 2
CP and CV also depend on temperature and phase in general.
Q. How much heat is needed to increase T by 2K for 1 mol of Cl2 at T = 100, 200, 300K?
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2.5 State Functions and Path Functions
• An alternate statement of first law:
U i i d d t f th th b t thU is independent of the path between the initial and final states.
U only depends only on the initial and final states as
UUUdUU
fin Any function that satisfies this condition is called a state function.
inifinUUUdUU
ini
dU is called an “exact differential”, the integration of which (U) is a state function. (Dq is not an exact differential).
Work and Heat Are “Path Functions” (not “State Function”)
• W and q are not state functions, but path functionsfunctions.
inifin
fin
iniqqDqq
inifin
fin
iniwwDww
W + q = U are actually a state function
Q.Calculate the work performed along the path in the right. Assume (Pext =P)
W = -P2(V2-V1) - P1(V1-V2) = (P1 – P2)(V2-V1)
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Additional Questions
Q2. How much heat is needed for this cycle?
W + q = U: a state function
q = U –w = -w
Q3. How much work is needed for the reverse cycle?
W1 = -P2(V1-V2) - P1(V2-V1) = -(P1 – P2)(V2-V1)
W1 = -w
2.7 Work for Reversible Process• In reversible process, P = Pext
finfin VV
dVPdVPw
)/(V
Vini
dVVnRTwfin
ViniVini
ext dVPdVPw int
For isothermal reversible process,
)}/{ln()}ln(){ln(
)][ln(
)/(
inifininifin
VV
V
Vini
VVnRTVVnRT
VnRT
dVVnRT
fin
ini
fin
1Q.How much w is needed for the cycle Pi Pf Pi?
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2.9 Calculating q, w , U, H for Ideal Gas
(Note: all the equations in this section of the text are valid only for Ideal Gas)
)()( inifinV
T
T VV TTCdTTCqUfin
ini
TnRTCTnRUnRTU
PVUH
V
)(
)(
HTTCdTTCq inifinP
T
T PP
fin
ini
)()(
TCTnRC pV
V
)(
)(
HW3
• P2.3) 3.00 moles of an ideal gas are compressed isothermally from 60.0 to 20.0 L using a constant external pressure of 5 00using a constant external pressure of 5.00 atm. Calculate q, w, U, and H.
• How can we calculalte U and H?Which equation should we use?
)( inifinVV TTCqU )( inifinPP TTCqH
Which equation should we use?
VPUH extqwU
VPUH int
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HW 3
• P2.7 For 1.00 mol of an ideal gas, Pexternal = P = 200.0 103 Pa. The temperature is changed from 100 0°C to 25 0°C and C = 3/2Rfrom 100.0 C to 25.0 C, and CV,m = 3/2R.Calculate q, w, U, and H.
)( inifinVV TTCqU
)( TTCqH
Which equation should we use first?
VPUH extqwU )( inifinPP TTCqH qwU
VPUH intVP CCR
• P9.28) The principal ions of human blood l d th i l l t tiplasma and their molal concentrations are
• Calculate the ionic strength of blood plasma. m
Na 0.14 m, m
Cl 0.10 m, m
HCO3 0.025 m.
2222 1025.0110.0114.011
I mmmzm ii [Q1] 22i
ii [Q1]
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• P9.29) Estimate the degree of dissociation of a 0.100 m solution of acetic acid (Ka = 1.75 10–5) that is also 0.500 m in the strong electrolyte given in parts (a)–(c) Use the dataelectrolyte given in parts (a) (c). Use the data tables in Appendix B to obtain ±, because the electrolyte concentration is too high to use the Debye–Hückel limiting law.
HA H+ + A-
Ka =a+a-/(0.100 m –x) = m22 /(0.100 m –x)
• P9.50) Consider the couple with the oxidized and reduced species at unit activity. What must be the value of E° for this half-cell if the reductant R is to liberate hydrogencell if the reductant R is to liberate hydrogen at 1 atm from
• a. an acid solution with aH+ =1
• b. water at pH = 7? (aH+ ~10-7)
2 + 22H+ +2e–→ H2
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• P10.9) Suppose a membrane is passively permeable to water and to Cl– ion, but not to H+. The electrostatic potential and the equilibrium concentrations of Cl– and H+ inside and outside the membrane are given here. Assume that the ionic strength inside the membrane is sufficiently small so that the
i i ffi i f i l i b k b 1activity coefficients of univalent ions may be taken to be 1 inside the membrane. Assume T = 298 K.
a. What is the equilibrium concentration
of Cl– inside the membrane ?
Outside Membrane Inside Membrane
0 150V 0 000V
out
in
R T cΔφ ln
z F c
out
in
c zFΔφ
c R TExp
Assume cout is constant for Cout cl-
b. Calculate the difference between
the chemical potentials of [H+] inside
and outside the membrane.
0.150V 0.000V
C
H 5 106 M C
H 107 M
C
Cl 5 102 M