Chapter 8 Thermodynamic Properties of

2012/3/29 1 Abstract

„ The thermodynamic description of mixtures, extended from pure fluids. „ The equations of change, i.e., energy and balance, for mixtures are developed. „ The criteria for phase and chemical equilibrium in mixtures

2012/3/29 2 8.1 THE THERMODYNAMIC DESCRIPTION OF MIXTURES

Thermodynamic property for pure fluids, θθ=()TPN , , where N is the number of moles. θθ=()TP , where the number of mole equals to 1.

Thermodynamic() property for mixtures, =TPN , , , N , , N where N is the number of moles of the ith component. θθ 12L c i θθ=()TPx , ,12 , x ,L , xci where x is the of the ith component.

For example()

UUTPNN= , ,12 , ,LL , Ncc or UUTPxx =() , , 12 , , , x () VVTPNN= , ,12 , ,L , Nc or VVTPxx =() , ,12 , ,L , xc

2012/3/29 3 Summation of the properties of pure fluids (before mixing at TP and ) C UTPxx(), ,12 , ,L , xci− 1= ∑ xUTPi () , (8.1-1) i=1 where UU is the molar , i is the internal energy of the pure i-th component at TP and . C ˆˆ UTPww()(), ,12 , ,L , wcii− 1= ∑ wUTP , (8.1-2) i=1 where wi is the mass fraction of component i.

2012/3/29 4 At the same T and P

50 cc 25 cc + 25 cc

H2O H2O

or 52 cc 25 cc + 25 cc 48 cc

+ 2 cc A B -2 cc Attractive Repulsive

2012/3/29 5 Property change upon mixing (at constantTP and ) C Δ=mixθθ θ()TPx,,ii −∑ xi () TP , i=1 Volume change upon mixing C Δ=mixVTP(),,,, VTPx()ii −∑ xVTPi () i=1 change upon mixing C Δ=mix HTP(),,,, HTPx()ii −∑ xHTPi () i=1

2012/3/29 6 Experimental data : properties changes upon mixing (H and V) Figure 8.1-1 Enthalpy- diagram for aqueous sulfuric acid at 0.1 MPa. The surfuric acid percentage Is by weight. Reference states: The of pure liquids at 0oC And their vapor are zero.

2012/3/29 8 Figure 8.1-2 (a) Volume change on mixing at 298.15 K: o methyl formate + methanol, zmethy formate + ethanol. (b) Enthalpy change on mixing at

298.15 K for mixtures of benzene (C6H6) and aromatic fluorocarbons (C6F5Y), with Y = H, F, Cl, Br, and I.

2012/3/29 9 Equations relating molar and partial molar properties

NTPNNNθφ= (),,12 , ,L , c At constant TP and ,

NNNN=θφ()12,,,L c Total differential of N is: ⎛⎞∂∂()NN ⎛⎞() ⎛⎞ ∂() N d() N=+⎜⎟ dN12 ⎜⎟ dN ++L ⎜⎟ dNc ∂∂NN12θ ∂ Nc ⎝⎠θ TPN,,jj12 ⎝⎠ TPN ,, ⎝⎠ TPN ,, j≠ c θθ≠≠ θ cc⎛⎞∂()N ==∑∑⎜⎟dNi i () T,, P x dNi ii==11⎝⎠∂Ni TPN,, jiθ≠ Definitionθ of the partial molar property ⎛⎞∂()Nθ i ()TPx, ,= ⎜⎟ θ (8.1-12) ∂N ⎝⎠i TPN,, θθ ji≠ c dN()= ∑ i dNi i=1

2012/3/29 10 At a constant T and P θ

50 cc + 1 mol of A θ + Δθ

A + B

Δθ is a partial molar property (at

Constant T, P, and NB)

2012/3/29 11 Total property and partial molar property

c dN()θθ= ∑ i dNi i=1 dN()θθθ=+ Nd dN (a)

dNiiii==+ d() Nx Ndx x dN (b) cc (a) (b) Nd+= dN ←⎯⎯ d Nii dN⎯⎯→ Ndx + x dN θθ θ θ θ ()∑∑iii() ii==11 cc Nddxx+=dN N∑∑iiii +dN θθ θ θii==11 ⎡⎤c ⎡⎤c ⎢d −+−∑ idxi ⎥ N ⎢⎥∑θ i xdNi = 0 ⎣ i=1 θθ⎦ ⎣⎦ θi=1 The NdN and are the number of moles in system and changing in number of moles, respectively. Thus, NdN and are arbitrary and not equal to zero.

2012/3/29 12 cc ddxθθ−==∑ ii θθii0; ddx∑ resulted from 1st term of LHS (c) ii==1 1 c c −θθ∑ i xi = 0; = ∑ i ()TPx,, xi resulted from 2nd term of LHS (8.1-13) i=1 i=1

Euler Theorem defined by (8.1-13) means can be calculated from i TPx,, . θθ () c Since: θθ= ∑ i xi i=1 θθcc θ θ ddxxd=+∑∑iiii ii==11 ccc θθ (c) θ ∑ iiidxiii ←+⎯⎯ddxd = ∑∑x i===11i 1 i c θθ ∑ xid i = 0 θ (8.2-9b) i=1 ⇒ Gibbs Duhem equation at constant TP and .

2012/3/29 13 2012/3/29 14 Property change upon mixing at constant T and P

50 cc 25 cc + 25 cc

H2O H2O

or 52 cc 25 cc + 25 cc 48 cc

+ 2 cc A B -2 cc Attractive Repulsive

2012/3/29 15 N1

θ1 Constant T and P Before After mixing mixing

N2

θ2

θi θi

θ = xθ θ = ∑ xiiθ ∑ ii

2012/3/29 16 Properties change upon mixing (i)

The total volume and enthalpy of the unmixed pure components are C VNVTP= ∑ i i (), i C HNHTP= ∑ i i (), i () whereas the volume and enthalpyTPx , ,i of the at the same and are, from Eq. 8.1-1 3, C VT(),,PN12 , N ,L == NVTPx() ,,∑ NVi i () TPx ,, i () C HTPNN,,12 , ,L == NHTPx() ,,∑ NHi i () TPx ,, i

2012/3/29 17 Properties change upon mixing (ii)

The isothermal volume change on mixing, C Δ=−mixVNNVNNNV()()TP,,,,, 1 2LL TPT ,,, 1 2 ∑ i i () ,P i C =−NVTPx⎡⎤i , , VTP , (8.1-14) ∑ i ⎣⎦()i () i ()()C Δ=−mixHTPNN,,,, 1 2LL HTPNN ,,,, 1 2 ∑ NHi i () T , P i C = NH⎡⎤i TPx, ,− H TP , (8.1-15) ∑ i ⎣⎦()i () i

2012/3/29 18 Thermodynamic relations for partial molar properties

AUTS=− ⎛⎞∂∂∂NA ⎛⎞ NU ⎛ NTS ⎞ ⎜⎟=− ⎜⎟ ⎜ ⎟ ∂∂NNii ∂ N i ⎝⎠TPN,,ji≠≠ ⎝⎠ TPN ,, ji ⎝ ⎠ TPN ,, ji ≠

AUTSi =−ii Also,

GHTSii=−i For each relationship among the thermodynamic variables in a pure fluid, there exists an identical relationship for the partial molar thermodynamic properties in a mixture.

2012/3/29 19 ILLUSTRATION 8.1-l Calculating the Energy Release of an Exothernuc Mixing Process Three moles of water and one mole of sulfuric acid are mixed isothermally at 0 oC. How much heat must be absorbed or released to keep the mixture at 0 o C? Water has a molecular weight of 18.015, and that of sulfuric acid is 98.078. Therefore, the mixture will contain 3 ×× 18.0154 + 1 98.078 = 152.12 g, and will have a (mass ) composition of 98.078 g ×=100% 64.5 wt% of sulfuric acid 152.12 g From Fig. 8.1-1 the enthalpy of the mixture is about − 315 kJ/kg. Therefore, when 3 mol water and 1 mol sulfuric acid are mix ed isothermally, kJ Δ=HwHˆˆˆHˆ − −wH =−315 mixmix 1 1 2 2 kg ˆˆoo since HT1 ()== 0 C = 0 and H2 T 0 C , so ()that a total of −× 315 kJ/kg 0.152 kg=− 47.9 kJ of energy must be removed to keep the mixture at a constant temperature of 0o C.

2012/3/29 20 Figure 8.1-1 Enthalpy-concentration diagram for aqueous sulfuric acid at 0.1 MPa. The sulfuric acid percentage Is by weight. Reference states: The Enthalpies of pure liquids at 0oC And their vapor pressures are zero.

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2012/3/29 64.5 21 COMMENT

Sulfuric acid and water are said to mix exothermically since energy must be released to the environment to mix these two components at constant temperature. The temperature rise that occurs when these two components are mixed adiabalically is considered in Illustration 8.4-1. Note also that to solve this problem we have, in effect, used an energy balance without explicitly writing a detailed balance equation.

2012/3/29 22 Partial molar property

θ + 1 mol of A θ + Δθ

A + B

Δθ is a partial molar property (at

Constant T, P, and NB)

2012/3/29 23 Physical meaning of partial molar property

For a binary mixture,

θθθ()TPN, ,12 , N=+ N 112() TPx , , N 2() TPx , , (8.1-17a)

Now suppose we add a small amount of species 1, ΔN1 , which is so small compared

with the total number of moles of species 1 and 2 that xx12 and are essentially unchanged θθθ , In this case we have

θθθ()()TPN, ,121+=+ΔΔNN11 , N N1 () TPx , , + N 22 () TPx , , (8.1-17b) Subtracting Eq. 8.1-17a from Eq. 8.1-17b, we find that the change in the property is

Δ+−=Δ=,,()()TPN112ΔNTP , N TPN ,,, 1 N21N θ 1 (), , x

Therefore, the amount by which a small addition of a species to a mixture changes the mixture property is equal to the product of the amount added and its partial molar property, that is, how the species behaves in a mixture , and not its pure component property.

2012/3/29 24 8.2 THE PARTIAL MOLAR GIBBS ENERGY AND THE GENERALIZED GIBBS-DUHEM EQUATION

Since the Gibbs energy of a multicomponent mixture is a function of temperature, pres- sure, and each species mole number, the total differential of the Gibbs energy function can be written as ⎛ ∂G ⎞⎛⎞∂∂GGC ⎛⎞ dG = ⎜ ⎟⎜⎟dT++ dP∑⎜⎟ dNi ⎝ ∂T ⎠⎝⎠PN,,∂∂PN TN i=1 i ii⎝⎠TPN,, j C =−SdT + VdP +∑ Gi dNi (8.2-1) i=1

The partial molar Gibbs energy, Gi = μi , the . ⎛⎞∂∂HH ⎛⎞ C ⎛⎞ ∂ H dH=++⎜⎟ dP ⎜⎟ dS∑⎜⎟ dNi ⎝⎠∂∂PSSN,, ⎝⎠ PN i=1 ∂ Ni ii⎝⎠PSN,, j C ⎛⎞∂H =++VdP TdS∑⎜⎟ dNi (8.2-2) i=1 ∂Ni ⎝⎠PSN,, j ⎛⎞∂∂HH ⎛⎞ ⎜⎟≠= ⎜⎟ H i ∂∂NNii ⎝⎠PSN,,j ⎝⎠ TPN ,,

2012/3/29 25 Chemical Potential HGTS= + C dH=+ dG TdS + SdT =−+SdT VdP +∑GidNi + TdS + SdT i=1 C =++VdPTdSN∑Gdi i (8.2-3) i=1 Comparing Eqs. 8.2-2 and 8.2-3 establishes that ⎛⎞∂∂HG ⎛⎞ ⎜⎟==Gi ⎜⎟ ∂∂NNii ⎝⎠PSN,,ji≠≠ ⎝⎠ TPN ,, ji Using the procedure established here, it is also easily shown that C dH=− TdS PdV +∑ Gi dNi (8.2-4) i=1 C dA=− SdT − PdV +∑ Gi dNi (8.2-5) i=1 ⎛⎞∂∂∂∂HGUA ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟==Gi ⎜⎟ = ⎜⎟ = ⎜⎟ =μ ∂∂∂∂NNNNi ⎝⎠i PSN,, ⎝⎠i TPN ,, ⎝⎠ii SV,, N ⎝⎠ T ,,V N ji≠ ji≠ ji≠≠ ji

2012/3/29 26 Generalized Gibbs-Duhem equation

Any thermodynamic function θ can be written as

NNθθ= ∑ i i

dN()=+ Ndii dN (8.2-6) θθθ∑∑ii NNTPN,, =θθ()i ⎛⎞∂∂NN ⎛⎞ C ⎛⎞ ∂ N dN()= ⎜⎟ dT++ ⎜⎟ dP ∑⎜⎟ dNi ⎝⎠∂∂T PN,, ⎝⎠PN TN i ∂ i θ ii⎝⎠TP,,N ji≠ θθθθ θ ⎛⎞∂∂θθ ⎛⎞ C d() N=++N ⎜⎟ dTN ⎜⎟ dPi dNi (8.2-7) ∂∂TP∑ ⎝⎠PN,,ii ⎝⎠ TN i Subtracting Eq. 8.2-7 from Eq. 8.2-6 gives the relation θθ ⎛⎞∂∂ ⎛⎞ −−+NdTNdPNd⎜⎟ ⎜⎟ i i =0 (8.2-8a) ∂∂T P ∑ ⎝⎠PN,,ii ⎝⎠ TN and dividing by the total number of moles N yields ⎛⎞∂∂θθ ⎛⎞ θ −−+⎜⎟dT ⎜⎟ dP xi d i =0 (8.2- 8b ) ∂∂TP∑ ⎝⎠PN,,ii ⎝⎠ TN The Generalized Gibbs-Duhem equation. θ

2012/3/29 27 ⎛⎞∂∂θθ ⎛⎞ −−+⎜⎟dT ⎜⎟ dP xi d i =0 (8.2-8b) ∂∂TP∑ ⎝⎠PN,,ii ⎝⎠ TN Gibbs-Duhem equation at constant TP and

∑ Ndi θ i =0 θ (8.2-9a)

∑ xdi θi =0 (8.2-9b) Finally, for a change in any property Y at constant temperature and pressure, Eq. 8.2-9a can be rewritten as

⎛⎞dθ i Ni ⎜⎟=0 (8.2-10) ∑ ∂Y ⎝⎠TP, whereas for a change in the number of moles of species j at constant temperature, pressure, and all other mole numbers, we have

C ⎛⎞dθ i N =0 (8.2- 11) ∑ i ⎜⎟ i ∂N j ⎝⎠TN,,P k≠ j

2012/3/29 28 The Gibbs-Duhem equations for the Gibbs energy , obtained by setting θ = G in Eqs. 8.2-8, are C 0=−+SdT VdP∑ Ni dGi (8.2-12a) i=1 C 0=−+SdT VdP∑ xi dGi (8.2-12b) i=1 At constant T and P , C ∑ NdGi i = 0 (8.2-13a) i=1 C ∑ xdi Gi = 0 (8.2-13b) i=1 C ⎛⎞∂Gi Ni ⎜⎟ = 0 (8.2-14) ∑ ∂Y i=1 ⎝⎠TP,

C ⎛⎞∂Gi N = 0 (8.2-15) ∑ i ⎜⎟ i=1 ∂N j ⎝⎠TPN,,kj≠

2012/3/29 29 For a binary mixture, 2 ∑ xdi θ i = 0 at constant T and P i=1

xd1212θθ+= xd 0

⎛⎞∂∂12θθ ⎛⎞ xx12⎜⎟+= ⎜⎟ 0 (8.2-19b) ∂∂xx ⎝⎠11TP,, ⎝⎠ TP =G θ

⎛⎞∂∂GG12 ⎛⎞ xx12⎜⎟+= ⎜⎟ 0 (8.2-20) ∂∂x x ⎝⎠11TP,, ⎝⎠ TP

STOP HERE

2012/3/29 30 8.3 A NOTATION FOR CHEMICAL REACTIONS

α A + βρBR+→LL + ραβ RAB+−LL − −=0 ν ∑ iI= 0 (8.3-1) i

ν i is the stoichiometric coefficient of species I, so defined that ν i is positive for reaction products, negative for reactants, and equal to zero for inert species.

Using Nii to represent the number of moles of species in a closed system at any

time tN , and i,0 for the initial number of moles of species iNN , then i and i,0 are related through the reaction variable X , the molar extent of reaction , and the stoichiometric

coefficient ν i by

NNii=+i,0 ν X (8.3-2a) NN− X = i i,0 (8.3-2b) ν i

2012/3/29 31 NNii=+i,0 ν X (8.3-2a) NN− X = i i,0 (8.3-2b) ν i

⎛⎞dNi dX ⎜⎟==iiX& (8.3-3) ⎝⎠dtrxn νν dt The total number of moles in a closed system at any time is CC C C NN==∑∑ii()() NXi,0 +=νν ∑ NX i,0 + ∑() i (8.3-4) ii i i

2012/3/29 32 The concept of independent chemical reactions

⎧⎫the smallest collection of reactions that, on forming ⎪⎪ Indepenlent reactions=⎨⎬ various linear combinations, includes all possible ⎪⎪ ⎩⎭chemical reactions among th e species present

No reaction in the set = {} a linear combination of the others For example, three reactions between carbon and oxygen:

C + O22 = CO

2C + O2 = 2CO

2CO + O22 = 2CO If we add the second and third of these equations, we get twice the first, so these three reactions are not independent . In this case, any two of the three reactions form an independent set.

2012/3/29 33 Remarks on the chemical reacting system

(1) To describe a chemically reacting system it is not necessary to consider all the chemical reactions that can occur between the reactant species, only the independent reactions. (2) Furthermore, the molar extent of reaction for any chemical reaction among the species can be computed from an appropriate linear combination of the known extents of reaction for the set of independent chemical reactions.

2012/3/29 34 Denbigh’s method (1981)

In the Denbigh's method, the procedure to get the independent chemical reactions are as follows: (1) writes the stoichiometric equations for the formation , from its constituent atoms , of each of the molecular species present in the chemical reaction system. (2) One of the equations that contains an atomic species not actually present in the atomic state at the reaction conditions is then used, by addition and/or subtraction , to eliminate that atomic species from the remaining equations. (3) In this way the number of stoichiometric equations is reduced by one. The procedure is repeated until all atomic species not present have been eliminated. (4) The equations that remain form a set of independent chemical reactions; the molar extents of reaction for these reactions are the variables to be used for the description of the multiple reaction system and to follow the composition changes in the mixture.

2012/3/29 35 An example using Denbigh’s method

Considering the reaction: the oxidation of carbon

Molecular species: O22 , CO, CO , and atomic species C were found in the system, write the reaction of formation for each molecular species 1 2O = O O = O 2 2 2 C + O = CO

C + 2O = CO2 Since a free oxygen atom does not occur, the first equation is used to eliminate O from the other two equations to obtain

2C + O2 = 2CO

C + O22 = CO All the remaining atomic species in these equations (here only carbon) are present in the reaction system, so no further reduction of the equations is possible, and these two equations form a set of independent reactions.

2012/3/29 36 Pop-quiz

„ A system contains the following substances:

N2(g), O2(g), NO(g) , NO2(g), N2O(g), N2O4(g), and N2O5(g), find an set of independent reactions for the system. For multiple-reactions (j reactions)

M NNi =+i,0∑ν ij X j (8.3-5) j=1 MM ⎛⎞dNi dX j ⎜⎟==∑∑ijνν ijX& j (8.3-6) ⎝⎠dtrxn j=1 dt j=1 ˆ Finally, with XXVjj = / defined to be the molar extent of reaction per unit volume,

MMMˆ d ()X jV dXˆ ⎛⎞dNi d V ˆ j ⎜⎟==∑∑∑νννij ijXV j+ ij ⎝⎠dtrxn j=1 dt dt j=1 j=1 dt MM dV ˆ =∑∑ijXV j + ij r j (8.3-7) dt j=1νν j=1 dXˆ r = j , the rate of reaction per unit volume , is the reaction variable most j dt frequently used by chemists and chemical engineers in chemical reactor analysis.

2012/3/29 38 8.4 THE EQUATIONS OF CHANGE FOR A MULTICOMPONENT SYSTEM

Species mass balance for a reacting system dN K ⎛⎞Rate at which species i is being i =+()Nkkth& is entry port ∑ i k ⎜⎟ dt k =1 ⎝⎠produced by chemical reaction ()K ⎛⎞dN =+N& i ∑ i k ⎜⎟ k =1 ⎝⎠dt rxn K M dNi = ()N& + ν X& (8.4-1a) ∑ i k ∑ ij j dt k =1 j =1 We can obtain a balance equation on the total number of moles in the system by sum- C ming Eq. 8.4-la over all species i, recognizing that ∑ Ni = N is the total number i=1 of moles, and that

CCKK K NN&&==N& ∑∑()iikk ∑∑ ()∑ k ik==11 k == 1i1 k =1 C where ()N& = N& is the total molar flow rate at the kth entry port. ∑ i k k i1=

2012/3/29 39 Total mass balance for a reacting system

Molar basis dN KMC =+∑∑∑NX&&k ν ij j (8.4-1b) dt k =1i=1j=1 Since neither the number of moles of species nor the total number of moles is a conserved quantity, we need information on the rates at which all chemical reactions occur to use Eqs. 8.4-1.

2012/3/29 40 Total mass balance for a reacting system Mass Basis Although we will be interested in the equations of change mainly on a molar basis , for completeness and for several illustrations that follow, some of the equations of change will also be given on a mass basis . To obtain a balance equation for the mass of

species iim , we need only multiply Eq. 8.4-1a by the molecular weight of species . i ,

and use the notation MNMmNii = mi and ( && i)kii = ( )k to get KM dM i =+∑∑(MmX&&ik ) iijjν (8.4-2a) dt k =1j=1 dM K = ∑ M& k (8.4-2b) dt k =1 C where MM&&kik= ∑( ) . i=1 In Eq. 8.4-2b the chemical reaction term vanishes since total mass is a conserved quantity. C M ∑∑mXiijjν & = 0 i=1 j=1

2012/3/29 41 Energy balance for a reacting system

dUK dV =++−()NQWP&&H & (molar basis) (8.4-3) ∑ k S dtk =1 dt K dUˆ dV =++−()MQWP&&H & S (mass basis) ∑ k dtk =1 dt C ()NH&&= () NH (8.4-4) kk∑ ii i=1 C UNU= ∑ ii i=1

2012/3/29 42 Continuous-flow stirred-tank reactor

Figure 8.4-1 A simple stirred-tank reactor.

2012/3/29 43 Mass/Energy balances for a steady state process

For a steady-state process dN i ==0()NN&& − () +ν X & (single reaction) (8.4-5a) dt iiiin out dU CC ==0()NH&& −() NH + Q& (8.4-5b) ∑∑iiin ii out dt i=1 i=1 For the very simple isothermal case in which the inlet and outlet streams and the reactor contents are all at temperature T , and with the assumption that the partial molar enthalpy of each species is just equal to its pure-component enthalpy, we obtain C ⎡⎤ QNNH& = ()&&− ()i ∑⎣⎦iiout in i=1 Now if there were no chemical reaction ()NN&&= () and the heat flow rate Q& iiout in should be equal to zero to maintain the constant temperature T . However, when a chemical reaction occurs, ()NN&& and are () not equal in magnitude, and the steady heat iiout in flow Q& required to keep the reactor at constant temperature is () () ν CCC ⎡⎤ QNNH& = &&−==iiiX& HX &ν H=ΔHX& ∑∑⎣⎦iiout in i ∑ i rxn i=1 i=1 i=1

ΔHQXrxn can be calculated by & and & .

2012/3/29 44 Entropy balances for a steady state process For a steady-state process

dSK Q& ==0()NS& + + S& (8.4-6) ∑ k gen dtk =1 T C where ()NS&&= () N S kk∑ ii i=1 SdV& = σ gen ∫ &s MC σφνλμ2 1 TGX2 (8.4-7) &sijij=Δ+−2 () ∑∑ & TTTji==11

2012/3/29 45 2012/3/29 46 Table 8.4-1 continued

2012/3/29 47 2012/3/29 48 Table 8.4-2 continued

2012/3/29 49 ILLUSTRATION 8.4-1 Temperature Change on Adiabatic Mixing of an Acid and Water Three moles of water and one mole of sulfuric acid, each at 0o C, are mixed adiabatically . Use the data in Fig. 8.1-1 and the information in Illustration 8.1-1 to estimate the final temperature of the mixture. SOLUTION From the closed-system mass balance, we have MMf =+ M =×+×=3 18.015 1 98.078 152.2 kg HO224 HSO and from the energy balance, we have UHHMHMHffif===ˆˆ = + M H ˆ =×+×=3100 0 kJ mix H2 O H 2 O H 24 SO H 24 SO Thus finally we have a mixture of 64.5 wt % sulfuric acid that has an enthalpy of 0 kJ/kg , (Note that we have used the fact that for liquids and solids at low pressure, the internal energy and enthaipy are essentially equal, UH= ). From Fig. 8.1-1 we see thal a mixture containing 64.5 wt % sulfuric acid has an enthalpy of 0 kJ/kg at a bout 150 oC. Therefore, if water and sulfuric acid are adiabalically mixed in the ratio of 3:1, the mixture will achieve a temperature of 150oC , which is just below the boiling point of the mixture. This large temperature rise, and the fact that the mixture is just below its boiling point, makes mixing sulfuric acid and water an operation that must be done with extreme care.

2012/3/29 50 ≈150 oC

Figure 8.1-1 Enthalpy-concentration diagram for aqueous sulfuric acid at 0.1 MPa. The sulfuric acid percentage Is by weight. Reference states: The Enthalpies of pure liquids at 0oC And their vapor pressures are zero.

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2012/3/29 51 COMMENT If instead of starting with refrigerated sulfuric acid and water (at 0o C), one started with these components at 21.2o C and mixed them adiabatically, the resulting 3:1 mixture would be in the liquid + vapor region; that is, the mixture would boil (and splatter). Also note that because of the shape of the curves on the enthalpy-concentration diagram, adding sulfuric acid to water adiabalically (i.e., moving to the right from the pure water edge of the diagram) results in a more gradual temperature rise than adding water to sulfuric acid (i.e., moving to the left from the pure-sulfuric acid edge). Therefore, whenever possible, sulfuric acid should be added to water, and not vice versa.

2012/3/29 52 ILLUSTRATION 8.4-2 Mass and Energy Balances on a Nonreacting System A continuous-flow steam-heated mixing kettle will be used to produce a 20 wt % sulfuric acid solution at 65.6o C from a solution of 90 wt % sulfuric acid at 0oo C and pure water at 21.1 C. Estimate

a. The kilograms of pure water needed per kilogram of initial sulfuric acid solution to produce a mixture of the desired concentration b. The amount of heat needed per kilogram of initial sulfuric acid solution to heat the mixture to the desired temperature c. The temperature of the kettle effluent if the mixing process is carried out adiabatically

2012/3/29 53 SOLUTION We choose the contents of the mixing kettle as the system . The difference form of the equations of change will be used for a time interval in which 1 kg of concentrated sulfuric acid enters the kettle. a. Since there is no chemical reaction, and the mixing tank operates continuously, the total and sulfuric acid mass balances reduce to 33

0 = MM& and 0 = ()HSO ∑∑k 24k kk==11

Denoting the 90 wt % acid stream by the subscript 1 and its mass flow by M& 1 , the water stream by the subscript 2, and the dilute acid stream by subscript 3, we have, from the total mass balance,

0 = M& 1 + MM&&&231+=+ MZMM && 13 +

MZM&&31=−() 1 + where Z is equal to the number of kilograms of water used per kilogram of the 90 wt % acid. Also from the mass balance on sulfuric acid, we have

0 = 0.900MM&&12++0. 2 0MM &31=−0.90& 0.20()1+ ZM& 1 Therefore, 0.90 1+Z = = 4.5 or Z = 3.5 0.20 so that 3.5 kg of water must be added to each 1 kg of 90 wt % acid solution to produce a 20 wt % solution .

2012/3/29 54 b. The steady-state energy balance is

0 =+∑()MH& ˆ Q& k k since WPdV = 0 and = 0. From the mass balance of part (a). s ∫

MM&&21==−3.5 and M & 3 4.5 M & 1 From the enthalpy-concentration chart, Fig. 8.1-1, we have ˆ ˆ o H1 ===−HT()90 wt % H24 SO , 0 C 183 kJ/kg () HHˆˆ===pure H O, T 21.1o C 91 kJ/kg 22() ˆˆ o H324===HT20 wt % H SO , 65.56 C 87 kJ/kg Q = ()4.5×− 87 3.5 × 91−×− 1() 183 =() 391.5 − 318.5 + 183 =256 kJ/kg of initial acid solution c. For adiabatic operation, the enerey balance is

0 = ∑()MH& ˆ k k ˆˆˆ 0=×−×−×−() 4.5HH33 3.5 91 1() 183 ; 4.5 ×= 135.5 and H3 = 30.1 kJ/kg Referring to the enthalpy-concentration diagram, we find that T ~ 50 o C.

2012/3/29 55 8.5 THE HEAT OF REACTION AND A CONVENTION FOR THE THERMODYNAMIC PROPERTIES OF REACTING MIXTURES

In the ideal gas-phase reaction 1 Hg+Og222()2 ()→ HOg () it is observed that 241.82 kJ are liberated for each mole of water vapor produced when this reaction is run in an isothermal , constant-pressure calorimtere at 25o C and 1 bar with all species in the vapor phase. Clearly, then, the enthalpies of the reacting molecules must be related as follows: Δ=H TP==25o C, = 1 barH TP 25o C, = 1 bar rxn ()HO2 () kJ −H vapor, T = 25oo C, PTP = 1 bar−=−1 H = 25 C, = 1 bar 241.82 H2 ()2 O2 () mol of H2 O produced so that we are nto free to choose the values of the enthalpy of hydrogen, oxygen, and water vapor all arbitrarily.

2012/3/29 56 Enthalpy of formation

The enthalpy of formation is ΔΔHGoo and the Gibbs energy of formation is . By definition, ffHO2 Δ=HHo TP= 25oo C, = 1 bar TP= 25 C, = 1 bar f HO22()HO() −HHvapor, TP= 25oo C, = 1 bar− 1 TP= 25 C, = 1 bar HO2 ()2 2 () H =Δ HHo ++1 H H2O f HO22 H2 O 2 o o Appendix A.IV contains a listing of ΔΔf HG and f for a large collection of sub- stances in their normal states of aggregation at 25oC and 1 bar. Isothermal heats (enthalpies) and Gibbs energies of formation of species may be summed to compute the enthalpy change and change that would occur if the molecular species at 25o C, 1 bar , and the state of aggregation listed in oooo Appendix A.IV. We will denote these changes by ΔΔrxnHG(25 C, 1 bar) and rxn (25 C, 1 bar), respectivel. y

2012/3/29 57 Enthalpy of reaction and enthalpy of formation

For the gas-phase reaction 3NO22 +H O = 2HNO 3 +NO, we have Δ=HHoo25 C,1 bar 2 25 o C,1 bar + H 25 o C,1 bar rxn ()HNO3 () NO () −− 3HH 25oo C,1 bar 25 C,1 bar NO22() H O () ⎡⎤o 3 1 = 2 Δ++f HHHNO 2 H O 2 + H N ⎣⎦3H2 2 2 25o C,1 bar ⎡⎤o 1 1 +Δf HHNO +2 H O +2 N ⎣⎦2 2 25o C,1 bar ⎡⎤o 1 −Δ 3 f HHNO +H O +2 ⎣⎦2H2 2 25o C,1 bar ⎡⎤o 1 −Δf HHHO + +2 H O ⎣⎦2H2 2 25o C,1 bar oooo =Δ⎡⎤ν 2ffffHHHHNO +Δ NO −Δ 3 NO −ΔH H O ⎣⎦322 25o C,1 bar o o =Δ∑ i f H i () 25 C,1 bar ()ooo o Δ=ΔrxnH 25 C,1 bar∑ν i f H i () 25 C,1 bar (8.5-1) ()ooo o Δ=ΔrxnGG25 C,1 bar∑ iν f i () 25 C,1 bar (8.5-2)

2012/3/29 58 At T and 1 bar

ooo o Δ=ΔrxnHH()25 C,1 bar∑ν i f i ( 25 C,1 bar) (8.5-1) ()ooo o Δ=ΔrxnG 25 C,1 bar∑ iν f Gi () 25 C,1 bar (8.5-2) The standard heat of reaction at any temperature T o o Δ=ΔrxnHT(),1 bar∑ν i f HTi () ,1 bar (8.5-3) ()o o Δ=ΔrxnG TT,1 bar∑ iν f Gi () ,1 bar (8.54)- T oo o HTPii(), =1 bar= H()()T , P =1 bar+ C TP , =1 bar dT o ∫T P,i o T ()ooo o Δ=ΔHT,1 barννH i () 25 C,1 bar + CdT rxn∑ ii f ∑ ∫T =25Co P,i ()T = Δ H o 25o C,1 bar + CdTo (8.5) -5 rxn ∑ i ∫T =25o C P,i o CP,i is the heat capacity of species i in its standard state. ν

2012/3/29 59 Heat of reaction at T and 1 bar.

T + ν CdTo ∑ i ∫T =25o C P,i Heat of reaction at 25 oC and 1 bar.

o Molecular species at 25 C and 1 bar. H2O, HNO3, NO2

Enthalpy of formation

o Atomic species at 25 C and 1 bar. O2,H2, N2 H(25 oC, 1 bar)

2012/3/29 60 Heat of combustion

ILLUSTRATION 8.5-1 Calculation of the Standard Heat of Reaction at 25o C Compute the standard heat of reaction for the hydrogenation of benzene to cyclohexane,

C66 H + 3H 2→ C 612 H from the standard-heat-of-combustion data. SOLUTION The standard heat of reaction can, in principle, be computed from Eq. 8.5-3; however, for illustration, we will use the heat of combustion for cyclohexane. From the standard-heat-of- o combustion data in Appendix A.V, we haveΔ−c H = 3,919,906 J/mol of cyclohexane for the following reaction:

C612 H() l +→ 9O 2 6CO 2 + 6H 2 O l () Thus Δ=+−−=−HHHo 66H 9 H 39 19906 J/mol c CH612 CO 2 HO 2C612H O 2 or H =−ΔHHHHo − 9 + 6 + 6 CH612 c CH612 O 2 CO 2 HO 2 Similarly, HHHHH=−Δo − 71 + 6 + 3 CH66c CH 662 O 2 CO 2 HO 2 and 3HH=−Δo − 1 1 HH +3 HHOHO222c 2 2

2012/3/29 61 Therefore, Δ=HHo − H − 3 H rxn CH612 CH 66 H 2 =−ΔH o −9H +6H +6H c CH612 O2 CO2 HO2 −−ΔH o − 7 1 H +6H +3H ()c CH6 6 2 O2 CO2 H2O − −Δ H o −1 1 H + 3H ()c H2 2 O2 HO2

=−ΔHHHooo +Δ +Δ =− ν Δ H o cccCH612 CH 662 H ∑ i c i i J =−−×= 3,919,906 3,267,620 3 285,840− 205,234 mole of benzene kJ = − 205.23 mole of benzene

2012/3/29 62 COMMENT oo Note that in the final equation, Δrxn HH =− ∑ν iΔc i , the enthalpies of the reference-state atomic species cancel, as they must due to conservation of atomic species on chemical reaction.

The equation developed in this illustration,

oooo ΔΔrxn HH()25 C,1 bar =− ∑ν i c i () 25 C,1 bar (8.5-6) is always valid and provides a way of computing the standard heat of reaction from standard-heat-of-combustion data.

2012/3/29 63 ILLUSTRATION 8.5-2 Calculation of the Standard Heat of Reaction as a Function of Temperature

Compute the standard-state heat of reaction for the gas-phase reaction NO24 = 2NO 2 over the temperature range of 200 to 600 K. oo Data: See Appendices A.II(ideal Cp ) and A. IV(ΔΔffH and G ). SOLUTION The heat of reaction at a temperature T can be computed from

oo Δ=ΔrxnHT()∑ν i f HTi ()

o At T = 25 C we find, from the data in Appendix A.IV, that for each mole of N24 O reacted, o o Δ==×−rxn HT()25 C[] 2 33.18 9.16 kJ/mol=57.20 kJ/mol To compute the heat of reaction at any temperature T , we start from Eq. 8.5-5 and note that since o* the standard state for each species is a low-pressure gas, CCp,i = p,i . Therefore,

ν T ooo* Δ=Δ=+rxnHT() rxn HT()25 C ν i CdT p,i ∫T =298.2K ∑ i For the case here we have, from Appendix A.II,

CdT***=−2 C C ∑ i p,i p,NO224 p,N O i J =−12.804 7.239×+× 10−−25283TT 4.301 10 + 1.5732 × 10 − T mol K

2012/3/29 64 o T Δ=+HT()57200( 12.804 −×+×+× 7.239 10−−25283 T 4.301 10 T 1.5732 10 − TdT) rxn ∫298.2 7.239× 10−2 =+57200 12.804()TT − 298.15 −()22 − 298.15 2 4.301×× 10−−58 1.5732 10 +−+−()TT33298.15() 44 298.15 34 =+56189 121.804TT −×+×+ 3.619 10−−22 1.4337 10 53 T J +3.933× 10−94T mol N24 O T ()K 200 300 400 500 600

o ⎛⎞kJ Δrxn HT()⎜⎟57.423 57.192 56.538 55.580 54.448 ⎝⎠mol N24 O

2012/3/29 65 Third law reference state In some databases for example, in the very extensive NIST Chemistry Web- Book - the data reported for each substance are the the standard state heat of o o o formation Δf H and the absolute entropy S , both at 25 C. Here by absolute entropy is meant entropy based on the third law of as defined in Sec. 6.8. The reason for reporting these two quantities is that they ar e determined directly by thermal or calorimetric measurements, unlike the Gibbs energy of formation, which is obtained by measuring chemical equilibrium constants. If the standard state heat of formation and the absolute entropy of each substance are known, the Gibbs energy of reaction can be compute d as follows. First, the heat of reaction is computed using C oo Δ=ΔrxnHH∑ν i f i i=1 and then the entropy change for the reaction is computed from C o o Δ=rxn SS∑ν i i=1 The standard-State Gibbs energy change on reaction at T = 25o C can then be computed from C Δ=Δ−GHoooo298.15 ⋅Δ= S⎡⎤ Δ− H 298.15 ⋅Δ S o rxn rxn rxn∑ ii⎣⎦ fi rxn i=1

2012/3/29 66 ν Then using the heat capacities reported in the NIST Chemistry WebBook and Eq. 8.5-5 the Gibbs free energy of reaction at any other temperature can be obtained. As an example of the use of data in this form, we return to the gas-phase reaction of hydrogen and oxygen to form water considered at the beginning of this section. Using the NIST Chemistry WebBook, we obtain the following data. kJ J Species Δ HSoo f mol mol K

H2 0 130.680

O2 0 205.150

H2 O -241.826 188.835

This results in Δ−frxnrxnHS = 241.826 kJ/mol, Δ = 44.43 J/(mol K), and Δ G = − 228.579 kJ/mol. This last value agrees with the Gibbs energy of formation for water () − 228.6 kJ/mol as a vapor in Appendix A.IV.

2012/3/29 67 8.6 THE EXPERIMENTAL DETERMINATION OF THE PARTIAL AND ENTHALPY

Experimental data

(1) PVT measurement,

VVVmixture→Δ mix → i

(2) Heat capacity (enthalpy) measurement,

HHmixture→Δ mix→ H i

(3) VLE measurement

VLE data →→γγiiGRT = ln i

SGHTiii =()− / .

2012/3/29 68 2012/3/29 69 2012/3/29 70 Equation for the calculation of the partial molar volume

Δ=mixVxVxVxVxVxVVxVV() 1 1 + 2 2 − 112 − 2 = 1( 1 −+ 1) 2() 2 − 2 (8.6-1)

⎛⎞∂Δ()mixV ⎛⎞∂∂VV12 ⎛⎞ ⎜⎟=−()VV1112 + x⎜⎟ −−() VV 2 + x 2 ⎜⎟ (8.6-2) ∂∂∂xxx ⎝⎠111TP, ⎝⎠TP,, ⎝⎠ TP Due to Gibbs-Duhem equation,

⎛⎞∂∂VV12 ⎛⎞ xx12⎜⎟+= ⎜⎟ 0 ∂∂xx ⎝⎠11TP,, ⎝⎠ TP

⎛⎞∂Δ()mixV ⎜⎟=−()VV1212 −−() VV (8.6-3) ∂x ⎝⎠1 TP,

⎛⎞∂Δ()mixV Δ−mixVx 1⎜⎟ =−() VV 2 2 (8.6-4a) ∂x ⎝⎠1 TP,

⎛⎞∂Δ()mixV ⎛⎞∂Δ()mixV Δ−mixVx 2⎜⎟ =Δ+ mix Vx 2 ⎜⎟=−()VV1 1 (8.6-4b) ∂x ∂x ⎝⎠2 TP, ⎝⎠1 TP,

2012/3/29 71 2012/3/29 72 Redlich-Kister expansion (can be used for the change or mixing of any molar property ) n i Δ=mixV xx 1 2∑ ai () x 1 − x 2 (8.6-5) i=0 nn i i Δ=mixH xx 1 2∑∑ ai () x 1 − x 2 or Δ=mixG xx 1 2 ai () x 1 − x 2 ii=0 =0 n i Δ=mixVx 1 ()(12−−xx1 ∑ ai 1 1 ) i=0 nn ⎛ ∂ΔmixV ⎞ ii+−11 ⎜ ⎟ =− axaiii()2x1 −− 1+21 1xx11 ()() 2 − 1 (8.6-5) ∂x ∑∑ ⎝ 1 ⎠TP, ii=0 =0

⎛⎞∂ΔmixV VV2mix1−=Δ−2 Vx⎜⎟ ∂x ⎝⎠1 TP, n ii−1 =−−−x2 axx⎡⎤()2 ixxx () (8.6-6a) 112212∑ i ⎣⎦ i=0

⎛⎞∂ΔmixV VV1mix− 1 =Δ Vx + 2 ⎜⎟ x ⎝⎠∂ 1 TP, n =−−−xaxx2 ⎡⎤ii2 ixxx−1 (8.6-6b) 212112∑ i ⎣⎦()() i=0

2012/3/29 73 Redlich-Kister expansion n i Δ=mixVxxaxx 1 2∑ i () 1 − 2 (8.6-5) i=0

n ⎛⎞∂Δ V ii−1 VV−Δ+ Vxmix x2 axx⎡⎤()() −−2 ixxx − (8.6-6b) 1mix21 ⎜⎟ 212112∑ i ⎣⎦ ⎝⎠∂x1 i=0 TP, n ⎛⎞ ii−1 ∂ΔmixV 2 ⎡⎤ VV2mix1−Δ−2 Vx⎜⎟x112212 axxi ()−−2 ixxx − ()(8.6-6a) ∂x ∑ ⎣⎦ ⎝⎠1 TP, i=0

2012/3/29 74 An accurate representation of the water-methanol data has been obtained using Eq 8.6-5 with (in units of m3 /mol) n i Δ=mixV xx 1 2∑ ai () x 1 − x 2 (8.6-5) i=0 −6 a0 -4.0034× 10 −6 a1 −× 0.17756 10 −6 a2 0.54139× 10 −6 a3 0.60481× 10 and the partial molar volumes in Table 8.6-2 have been computed using these constants.

2012/3/29 75 The partial molar enthalpy of a species in a binary mixture A steady-state flow calorimeter

2012/3/29 76 Two streams, one of pure fluid 1 and the second of pure fluid 2, both at a temperature TP and a pressure , enter this steady-state mixing device, and a single mixed stream, also at TP and , leaves. Heat is added or removed to maintain the temperature of the outlet stream. Taking the contents of the calorimeter to be the system, the mass and energy balances are

0=NNN&&&123++ (8.6-7)

−NNN&&&312=+

0 = NH&&1212mix++ NH NH & 3 + Q& (8.6-8) ⎡⎤ ⎡⎤ QNNH& =⎣⎦&&12+−−=+Δmix NHNHNN & 1 1 & 2 2 ⎣⎦ && 12mix H ⎡⎤ Δ=mixHQN& / ⎣⎦&& 1 + N 2

Δmix HQ can be calculated from & .

2012/3/29 77 2012/3/29 78 Figure 8.6-3 Heat of mixing data for the water(1)-methanol(2) system At T = 19.69 oC

2012/3/29 79 Once the composition dependence of the heat of mixing is known,

HH12 and may be computed in a manner completely analogous to the procedure used for the partial molar volumes, inparticular, it is easily established that

⎛⎞∂Δ()mix H ()HH2mix1−=Δ−2 Hx⎜⎟ (8.6-9a) ∂x ⎝⎠1 TP,

⎛⎞∂Δ()mix H ()HH1mix2−=Δ−1 Hx⎜⎟ (8.6-9b) ∂x ⎝⎠1 TP,

2012/3/29 80 2012/3/29 81 General equation relating the partial molar property to theθθ pure component θ property and the property change on mixing θθ θ ⎛⎞∂Δ( mixθ ) 1mix2()TPx, ,−=Δ−1 () TP ,() TPx , , x⎜⎟ (8.6-10a) θ ∂x ⎝⎠2 TP, θ ⎛⎞∂Δ ()mix θ 2mix1()TPx, ,−=Δ−2 () TP ,() TPx , , x⎜⎟ (8.6-10b ) ∂x θ ⎝⎠1 TP, θ ⎛⎞∂()mix 12()TPx, ,=−mix () TPx , , x⎜⎟ (8.6-11a) θ ∂x ⎝⎠2 TP,

⎛⎞∂()θ mix 21()TPx, ,=−mix () TPx , , x⎜⎟ (8.6-11b) ∂x ⎝⎠1 TP,

2012/3/29 82 θθ

θθ θ θ

θθθ ⎛⎞∂Δ( mixθ ) 12()TPx,,−=1 () TP ,Δmix ()TPx,, − x⎜⎟ (8.6-10a) Δmix ()T,,P x ∂x ⎝⎠2 TP,

Δ=mix −x 1− x 2 mixθθ 1θθ 2 θθ θ θ ⎛⎞⎛⎞∂Δ()mix ∂()mix ⎜⎟⎜⎟ θ= + 12− ∂∂x x ⎝⎠⎝⎠22TP,, TP θθ θ⎧⎫ ⎛⎞⎛⎞∂Δ()mix ⎪⎪ ∂()mixθ ⎛⎞ ∂() mix xx22⎜⎟⎜⎟ = ⎨⎬+12−= x 2θ⎜⎟+−xx22θθ12 ∂∂x x θ ∂x ⎝⎠22TP, ⎩⎪⎪θθ⎝ ⎠ TP, ⎭ ⎝⎠ 2TP, θ θ ⎛⎞∂Δ()mix 1mix2()TPx,,−=Δ−1 () TP ,θ () TPx ,, x⎜⎟ ∂x ⎝⎠2 TP, ⎧⎫θ ⎪⎪⎛⎞∂()mix =−mixxx12 1− x2 22 −⎨⎬⎜⎟ + x 2 1 − x2 θθ∂x ⎩⎭⎪⎪⎝⎠2 TP,

⎛⎞∂()mix =−mix x2 ⎜⎟ −1 ∂x ⎝⎠2 TP, θ Thus,

⎛⎞∂()mix 12()TPx, ,=−mix () TPx , , x⎜⎟ (8.6-11a) θ mix ∂x ⎝ 2 ⎠TP,

2012/3/29 83 ILLUSTRATION 8.6-1 Calculations of Partial Molar Enthalpies from Experimental Data Using the data in Fig. 8.1-1 , determine the partial molar enthalpy of sulfuric acid and water at 50 mol % sulfuric acid and 65.C6o . SOLUTION First we must obtain values of enthalpy versus concentration at 65.6o C. The values read from this figure and converted to a molar basis are given below.

(MW(H24 SO ) = 98.078 ; MW(H2O) = 18.015)

2012/3/29 84 278

92

85 Figure 8.1-1 Enthalpy-concentration diagram for aqueous sulfuric acid at 0.1 MPa. The sulfuric acid percentage Is by weight. Reference states: -78 The Enthalpies of pure liquids at 0oC And their vapor pressures are zero. -175

-153

-60

2012/3/29 85 dHΔmix 2 =−82.795 + 278.965xxHSO − 170.049 HSO dx 24 24 HSO24

2

0

-2 Plot of Example 8.6-1

-4 Slope= Slope= -82.8 +26.11 -6 kJ/mol

, -8 Η mix

Δ -10 Slope= -12 +14.17 -14

-16 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 x 1 These data are fit reasonably well with the simple expression ⎛⎞kJ Δ=−+mixHxx⎜⎟ H SO H O()82.79 56.683 x H SO ⎝⎠mol 24 2 24 ()( ) =−−+xx 1 82.79 56.683 x HSO24 HSO 24 HSO 24 =− 82.795x + 139.478xx23− 56.683 HSO24 HSO24 HSO 24 dHΔ and mix =− 82.795 + 278.965xx − 170.049 2 dx HSO24 HSO 24 HSO24 dHΔΔΔkJ dH dH kJ For x = 0.5; mix = 14.17 ; mix =− mix =− 14.17 HSO24 dxmol dx dx mol HSO24 HO2 HSO24 kJ Δ==−Hx()0.5 13.61 mix H24 SO mol

2012/3/29 87 kJ kJ 98.708 g kJ H =×92 × =9.02 from Fig. 8.1-1 HSO24 kg 1000 g mol mol LLL kJ kJ 18.015 gkJ H =××278 =5.01 from Fig. 8.1-1 HO2 kg 1000 g mol mol LLL Finally, from Eq. 8.6-9b, we have H xT= 0.5,= 65.6Co HSO24()HO 24S ⎛⎞∂Δ H = HT()=+Δ=+65.6o CHx() 0.5 x ⎜⎟mix HS24O mix H24 SO H 2 O ⎜⎟∂x ⎝⎠HSO24 x =0.5 HSO24 kJ =−9.02 13.61 +−=− 0.5() 14.7 11.68 mol kJ H −=H −+−=13.61 0.5()14.7− 0.7 HSO24 HS 2O4 mol Hx = 0.5,T = 65.6o C HO22()H SO4 ⎛⎞∂Δ H ==+Δ=+HT()65.6o C H()x 0.5 x ⎜⎟mix HO22424mix H SO H SO ⎜⎟∂x ⎝⎠HS2 O4 x =0.5 HSO24 kJ =−5.01 13.61 + 0.5() 14.7 =− 1.52 mol kJ HH−=−+13.61 0.5() 14.7 =− 6.5 HO22 HO mlo

2012/3/29 88 At θθinfinite

θ θθ

⎛⎞∂Δmixθ 11()TPx,,−=1 Δmix() TP ,,x −x 2 ⎜⎟ ∂x θθ⎝⎠2 T ,P

⎛⎞⎛⎞∂Δmix ∂Δ mix 1 ()TP,,θθx1 → 0 −=−1 ⎜⎟⎜⎟ =+ (8.6-12a) ∂∂xx21 ⎝⎠⎝⎠xx21==10 and θθ ⎛⎞⎛⎞∂Δmix ∂Δ mix 2 ()TPx,0, 2 → −2 =−⎜⎟⎜⎟ =+ (8.6-12b) ∂∂xx12 ⎝⎠xx12==10⎝⎠

The largest difference be tween θθii− is at infinite dilution.

2012/3/29 89 ILLUSTRATION 8.6-2 Calcutation of Infinite Dilution (x = 0) Partial Molar Enthalpies from Experimental Data Compute the difference between the infinite dilution partial molar enthalpy and the pure component molar enthalpy for sulfuric acid and water at 65.6o C using the information in the previous illustration. SOLUTION ⎛⎞dHΔ From the previous illustration ⎜⎟mix =− 82.795 + 278.965x − 170.049x2 ⎜⎟dx HSO24 HSO24 ⎝⎠HSO24 TP, ⎛⎞dHΔΔkJ ⎛⎞ dH kJ Therefore, ⎜⎟mix=+26.11 and ⎜⎟ mix =−82.80 ⎜⎟dxmol ⎜⎟ dx mol ⎝⎠HSO24xx==1 ⎝⎠ HS 24O 0 H2SO4 H24SO kJ so that Hx()==0 , T 65.6oo C − HT() = 65.6 C =− 82. 8 0 HSO24 HSO 24 HSO2 4 mlo

2012/3/29 90 ILLUSTRATION 8.6-2 in which case kJ Hx()==0, T 65.6o C = 9.02 − 82.8 =− 73.8 HSO24 HSO 24 mol kJ Hx()==0, T 65.6oo C − HT() = 65.6 C =− 26.11 HO22 HO HO 2 mol kJ and Hx()==0 , T 65.6o C = 5.01 − 26.11 =− 21.1 HO22 HO mol Note that for the sulfuric acid + water system at T = 65.6o C the differences between the pure component and partial molar properties at infinite dilution are considerably greater than at the mole fraction of 0.5 in the previous illustr ation. (-82.8, -26.11 ; -20.7, -6.5 kJ ). mol

2012/3/29 91 θθ There is a simple physical interpretation for partial molar properties at infinite dilution .

In general, we have from Eq. 8.1-13 for any total property θ ()TPx, , in a binary mixture that θθ ()TPx,, =+ N11() TPx, , N 2θ 2 ()TPx, , (8.6-13)

Now consider the case when NN2121 = 1 and >> N so that x ~ 1 , in which case

11()TP, ,xθ ~ 1≅ 1 ()TP , , since species 1 is essentially at the pur e component

limit. Also, 2 ()TPx , , ~ 1 is the partial molar property of species 2 at infinite θθ1 dilution, so that at in this limit (NN12 >> and N 2 = 1 ) () θ TPx, ,11 ~ 1=+NTPθ 1 ()T, P 2 () , ,x 2 ~ 0 (8.6-14)

From this equation we see that the infinite dilution partial molar property θ 2 ()TPx, ,2 ~ 0 is the amount by which the total property changes as a result of the addition of one

mole of species 2 to an infinitely large amount of species 1 (so that x2 remains about zero). Note that if the solution were ideal, the total property θ would change by an amount equal to the pure component molar property θ : however, since most θ 2 are nonideal, the change is instead equal to 2 .

2012/3/29 92 ILLUSTRATION 8.6-3 Calculation of the Isothermal Enthalpy Change of Mixing One mole of sulfuric acid at 65.6o C is added to 1000 moles of water at the same temperature. If the mixing is done isothennally, estimate the change in enthalpy of the mixture . SOLUTION From Eq. 8.6-14, HT======65.6oo C, N1000, N1 − HT 65.6 C, N1000, N 0 ()()HO224 HSO HO 224 HSO kJ ==HT()65.6o C, x =≈=0.001 0 −73.8 HSO24 HSO 24 mol

= ΔmixH ()1000 mol H 2 O + 1 mol HSO24 [The numerical value for HT ( = 65.6o C, x≈ 0) was obtained from the previous HSO24 HSO 24 illustration.]

2012/3/29 93 8.7 CRITERIA FOR PHASE EQUILIBRIUM IN MULTICOMPONENT SYSTEMS

For a closed system, we have for both the pure component and multicomponent cases dU dV =+QW& & − P (8.7-1) dtS dt dS Q& =+S& (8.7-2) dt T gen For the pure component system (molar basis) UNUTP= (), SNSTP= (), and for a multicomponent system (molar basis) C UNUTPx= ∑ i i (),, i=1 C SNSTPx= ∑ i i (),, i=1 The equilibrium criteria for a closed multicomponent mixture are SMUV = maximum for equilibrium at coinstant , , and AMTV = minimum for equilibrium at constant , , and (8.7-4) G = minimum for equilibrium at constant MT , , and P

2012/3/29 94 The equilibrium criterion The equilibrium criterion for a closed, adiabatic, constant-volume system is S = maximum subject to the constraints of constant UV , , and total number of moles of each species

Ni. For the two-phase system, each extensive property (e.g., NSUVi , , , is the sum of the properties for the individual phases, for example, III NNNiii=+ dS=+ dSIII dS (8.7-5) 11P C dS=+− dU dV∑ Gii dN TTTi=1 dUIII=− dU dVIII=− dV III dNii=− dN IIIC I II ⎛⎞11 II⎛⎞⎛PP GGii ⎞ I dS=−⎜⎟III dU +−⎜⎟⎜ III dV −∑ I − II ⎟ dNi =0 ⎝⎠TT⎝⎠⎝ TTi1= T T ⎠

2012/3/29 95 First criteriaon for phase equilibrium TTIII= (8.7-7a) Second criteriaon for phase equilibrium PPIII= (8.7-7b) Third criteriaon for phase equilibrium IIIIII GGii== or μμ ii (8.7-7c)

2012/3/29 96 8.8 CRITERIA FOR CHEMICAL EQUILIBRIUM, AND COMBINED CHEMICAL AND PHASE EQUILIBRIUM

A single chemical reaction occurring in a single phase in a closed system at constant temperature and pressure, CC GNGN==+∑∑ii,0iii()ν XG i=1 i=1 Since the only variation possible in a one-phase, closed system at constant temperature and pressure is in the extent of reaction X , the equilibrium criterion is ⎛⎞∂G ⎜⎟= 0 ⎝⎠∂X TP, Criterion for chemical equilibrium of a single reaction C 0=∑ν i Gi (8.8-1) i=1

2012/3/29 97 For the multiple chemical reactions in a closed, single-phase, constanl-temperalure and constant-pressure system. M NNii,0ijj=+∑ν X (8.3-5) j=1 CCM⎛⎞ GN= ∑∑∑iiGN=+⎜⎟ i,0ν ijj XGi i=1 i=1⎝⎠ j=1 CCMν =+∑∑∑NGi,0i ij XG j i (8.8-2) i=1 i=1 j=1 The condition for chemical equilibrium in this multireaction system is G = minimum or dG = 0 for all variations consistent with the stoichiometry at constant temperature, pressure, and total mass. For the present case this implies ⎛⎞∂G = 0 j = 1, 2, 3, , M (8.8-3) ⎜⎟ L ∂X j ⎝⎠TPX,,i

CC ⎛⎞∂∂GG ⎛⎞i ==0 ν GNi + ⎜⎟∑∑ij i ⎜⎟ ∂∂XXjji=1 i=1 ⎝⎠TPX,,i ⎝⎠ TPX ,,

C ⎛⎞∂Gi for all independent reactions j=1, 2, 3, , MN . = 0 due to the Gibbs-Duhem equation L ∑ i ⎜⎟ i=1 ∂X j ⎝⎠TPX,,i Chemical equilibrium criteria for multiple reactions C 0=∑ν ijGMi j=1, 2, 3, L , (8.8-4) i=1

2012/3/29 98 8.9 SPECIFICATION OF THE EQUILIBRIUM THERMO- DYNAMIC STATE OF A MULTICOMPONENT, MULTIPHASE SYSTEM; THE GIBBS PHASE RULE

IIIIII P TT== T ==L T (8.9-1) IIIIII P PP== P ==L P (8.9-2) I II III P GGii== G i ==L G i (8.9-4) For M independent chemical reactions, C ∑ν ijGi = 0 (8.9-5) i=1

2012/3/29 99 Gibbs phase rule

⎛⎞Number of unknows ⎜⎟⎛⎞Number of independent relations F =−⎜⎟thermodynamic ⎜⎟ ⎜⎟⎝⎠among the unknows parameters ⎝⎠parameters =+−−+−+PC()()()12⎣⎦⎡⎤ P 1 CP 1 M =−CMP −+2 (8.9-6) where P = number of phases C = number of components M = number of independent reactions F = degrees of freedom

2012/3/29 100 Problems

„ 8.7, 8.9, 8.17, 8.26, 8.28, 8.31

2012/3/29 101