1

Physical Chemistry I for Biochemists

Chem340

Lecture 16 (2/18/11)

Yoshitaka Ishii

Ch4.6, Ch5.1-5.5 & HW5

4.6 Differential Scanning Calorimetry(Differential Thermal Analysis)

• Dqsample = Cp, sdTs + dHfusion = (Ts –T)Kdt, [1]

where K =(1/) is a heat conductance

• Dqref= Cp, refdTref = (Tref –T)Kdt [2]

By calculating [1]/dt – [2]/dt,

(Ts-Tref)K = Dqsample/dt –Dqref/dt

T (1/K){D /dt D /dt}T =(1/K){Dqsample/dt –Dqref/dt}

= (1/K)(Dqp/dt)T

By integrating Dqp/dt over time

qp = (Cp, s Ts–Cp, ref Tref) +Hfusion

2

Chemical and Structural Change Detected by DSC

Hfusion

S lid Liquid Dq Solid Liquid

Dq/

dtdt

dt

Dqp

Solid Liquid

Time

qp = (Cp, s Ts–Cp, ref Tref) +Hfusion

HW5 P4.31) The figure below shows a DSC scan of a solution of a T4 lysozyme mutant. From the DSC data, determine Tm, the excess heat capacity CP and the intrinsic and transition excess heat capacities CP

int and CPtrs at T = 308 K. In your calculations, use the

extrapolated curves, shown as dashed lines in the DSC scan.e po ed cu ves, s ow s d s ed es e SC sc .

3

DSC Curve Upon Protein DenaturationDenatured(Unfolded) Intrinsic excess

heat capacity:CP

int

(Dq/dT)p = Cp(T)(T

)

CP

Transient excess heat capacity:CP

trs

Cp(

Melting T

Native(Folded)

Heat Cap. for denaturationCp

den = CPD(T) - CP

N(T)Excess heat capacityCp = CP

int(T) + CPtrs(T)

V 1LV=1L P = 1 bar

A

Ch 5.1 The Universe has a natural direction of change

C500K 300K

V=1L P’ = 0 bar

V=1L P = 1 barT =300 K

2V P = 1/2 barT = 300K

Adiabatic expansionBD

500K

Q. Why cannot we expect B Aor D C in a spontaneous way?

400K

4

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?

5

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?

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( ifVvreversible TTnCT

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( ifPPreversible TTnCT

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( ifVif TTnCVVnRS

)/ln()/ln()/ln()/ln( ifPififPif TTnCPPnRTTnCVVnRS

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Sample Question (p90)

• For a reversible adiabatic expansion

(Ti, Vi) (Tf, Vf), show that S = 0 using (Tf/Ti) = (Vi/Vf)-1 and

1

)/ln()()/ln(

)/ln()/ln(

ifVif

ifVif

VVnCVVnR

TTnCVVnRS

)/ln()/ln( ifVif TTnCVVnRS

0

1

)/ln()(

)/ln())({

)()()(

ifvp

ifv

ifVif

VVnRnCnC

VVnRnC

[Q2]

[Q1]

Ex. 5.3 (p89)• One mole of a monatomic ideal gas initially at Ti =500K and

Pi =100 atm expands adiabatically against a constant external pressure of 1.00 atm until the gas pressure also equals 1.00 atm. Assume that CV m = 3R/2.q V,m

(a) Calculate Vi and Vf of the gas and Tf.

(b) Calculate S for the gas.

Q. How to calculate S once we obtain Vi, Vf, and Tf?

(b)

(a) Vi is obtained by Vi = nRTi/Pi.. [1]

)/ln()/ln( ifVif TTnCVVnRS [Q1]

Use Vf = nRTf/Pext. [2] & q = 0 nCV,m(Tf-Ti) = -Pext(Vf-Vi) [3].

By substituting Vf in [3] with [2],

nCV,m(Tf –Ti) = -Pext(nRTf/Pext – Vi) = -nRTf +nRTiPext/Pi

n(Cv,m+R)Tf = n(Cv,mTi +RTiPext/Pi)

Tf = (Cv,m+RPext/Pi)Ti/(Cv,m+R) =(3/2+1/100)Ti/(3/2+1) =302K

[Q2 ]

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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

How can we prove that S is a state function? HW6 1.

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

8

S for liquid & solid (p91)

TfVf C

dTT

CdVdTC

TdV

T

V

V

PdS v

VPT

1

)/ln()( ifvif

Ti

v

Vi

TTCVVdTT

CdVS

)/ln()( ifPif

TfP

Pf

TTCPPVdTC

dPVS

Correct for Solid & Liquid

[5.22]

)/ln()( ifPif

TiPi

TTCPPVdTT

dPVS

Q Is the equation valid for an ideal gas?

PP

V

VTT

V

V TP

1111

&

[5.23]

V

nR

T

P

• 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

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

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S=0 for Carnot Cycle?• S = qab/Thot + qcd/Tcold

= nRThotln(Vb/Va)/Thot + nRTcoldln(Vd/Vc)/Tcold

= nRln(Vb/Va) + nRln(Vd/Vc)= nRln(VbVd/VaVc)

d a & bc are adiabatic processes:• TcoldVd

-1 = ThotVa-1 & ThotVb

-1 = TcoldVc-1

(V V ) 1 (V V ) 1 V V /V V 1 (VdVb )-1 = (VaVc)-1 VbVd/VaVc=1

S = qab/Thot + qcd/Tcold = 0wcycle = -(qab+qcd) = nRThotln(Vb/Va) + nRTcoldln(Vd/Vc)

= nRThotln(Vb/Va) - nRTcoldln(Vb/Va)= nR(Thot-Tcold)ln(Vb/Va)

• Q1. Explain the meaning of (H/P) T, n

d (H/T) ?and (H/T) P, n?

• (δH/δP)T,n = -CpμJ-T and (δH/δT)p,n = Cp

• (H/P) T, n 1st Derivative of H with respect to P for fixed T and n.

(H/P) is the pressure dependence of (H/P) T, n is the pressure dependence of enthalpy at constant temperature and constant number of particles.

11

• P4.6) From the following data at 25°C, calculate the standard enthalpy of formation of FeO(s) and of Fe2O3(s):

Hrection0

• Fe2O3(s) + 3C(graphite) 2Fe(s) + 3CO(g) 492.6A 2 3

• FeO(s) + C(graphite) Fe(s) + CO(g) 155.8

• C(graphite) + O2(g) CO2(g) –393.51

• CO(g) + 1/2 O2(g) CO2(g) –282.98

Hf(FeO(s)) is calculated for the following reaction:

B

C

D

Hf(FeO(s)) is calculated for the following reaction:(1) Fe(s) + 1/2O2(g) FeO(s)(-2A –B) (C+1/2D) -B (Fe2O3: A = 0; CO2: C+D =0)

=1 =1/2 =1-B = 1 B = -1 & A =0 C+1/2D =1/2 & C+D = 0 C + ½(-C) = ½ C = 1 D= -1

• P4.6) From the following data at 25°C, calculate the standard enthalpy of formation of FeO(s) and of Fe2O3(s)

Hrection0

• Fe2O3(s) + 3C(graphite) 2Fe(s) + 3CO(g) 492.6A

• FeO(s) + C(graphite) Fe(s) + CO(g) 155.8

• C(graphite) + O2(g) CO2(g) –393.51

• CO(g) + 1/2 O2(g) CO2(g) –282.98

(1) 2Fe(s) + 3/2O2(g) Fe2O3(s)

B

C

D

(-2A –B) (C+1/2D) -A (FeO: B = 0; CO2: C+D =0)

B =0 & -A = 1C+1/2D = 3/2

[Q1 ] [Q2 ] [Q3 ]

12

• 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 ')'(

TreactionH ,0

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

115.298

)2(6)(6)3(4 00015.298,

0 OHHNOHNHHH fffKreaction

CP = B(1) + B(2)T/K + B(3)(T/K)2 + B(4)(T/K)3

Tf

NH3 NO N2 H2O Tf Ti-4 -6 5 6 Coefficient 650 298.15

A1 29.29 33.58 30.81 33.8 38.21 650 298.15 T 1.34E+04A2 0.01103 -0.02593 -0.01187 -0.00795 0.00441 2.11E+05 4.44E+04 T2/2 7.36E+02A3 4.2446E-05 5.3326E-05 2.3968E-05 2.8228E-05 -0.0002 9.15E+07 8.83E+06 T3/3 -1.66E+04A4 -2 7706E-08 -2 7744E-08 -1 0176E-08 -1 3115E-08 1 48E-07 4 46E+10 1 98E+09 T4/4 6 30E+03

Ti

P dTTC ')'( B(1)(Tf –Ti) + B(2)(Tf2 -Tf

2 )/2K +C(3)(Tf

3 - …..

A4 -2.7706E-08 -2.7744E-08 -1.0176E-08 -1.3115E-08 1.48E-07 4.46E+10 1.98E+09 T4/4 6.30E+033.89E+03 J

Reaction enthalpy at 298.15 K Reaction enthalpy at 650 K-1.815E+03 kJ -1.811E+03 kJ

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P4.18• Calculate the single bond enthalpies and energies

for Si–F, Si–Cl, C–F, N–F, O–F, H–F:Substance SiF4(g) SiCl4(g) CF4(g) NF3(g) OF2(g) HF(g)

a) SiF4

• The average Si-F single bond enthalpy is calculated from the formation enthalpies and the following associated reactions: Hf

0

H f

kJ mol1 –1614.9 –657.0 –925 –125 –22 –271

f

• -1614.9 kJmol-1

• 79.4 kJmol-1

• 450.0 kJmol-1

• Hreaction0

g SiF g F s Si 42 2

g F g F 1/2 2

g Si s Si g F 4g Si g SiF4

[Q2]

[Q3]

[Q1]

• 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). reaction( ) g J-T, p,A, reaction( )Use dHA = CpAdT – Cp, AJ-T dP + (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 + Y.

• Step 1: P 1 bar • Step 2: 3B(1 bar) X(1 bar) + Y(1bar)• Step 3: 1 bar P