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QUIZZES

CHE 303 W17

QAQQCG

QUIZ #1 OAK

• Find the pressure of a mixture of

• CH4 (0.25 kmol) + n-C4H10 (0.25 kmol)

• at T=250°C and V= 0.241 m3.

• Use the van der Waals EOS.

• (R=0.08314 Bar.m3/kmol.K)

CH4 nC4H10

a 2.293 13.86

b 0.0428 0.1162Answer:

78.547 bar

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QUIZ #2 OAK

• Find the pressure of a mixture of

• CH4 (0.18 kmol) + n-C4H10 (0.274 kmol)

• at T=238°C and V= 0.241 m3.

• Use the Redlich-Kwong EOS.

• (R=0.08314 Bar.m3/kmol.K)

CH4 nC4H10

a 32.11 289.5513

b 0.02965 0.0806

Answer:

68.257

bar

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Quiz 3: H-x: Water+H2SO4 Mx

• Find the partial molar enthalpy!

• water and acid at mass frac. wacid=0.60.

• Also find HTotal at wAcid=0.60 from the relation

between PMH and total HT.

H

H H

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Quiz 3: Answer

QUIZ 4: Chemical potential of steam QAR

• Calculate the chemical potential of steam at

P=60 bar, and T= 500o C using the steam

tables.

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Q4 ANSWER

303W17f/p= 60.44295

QAW MW 18

T 773.15

H 3422.2

S 6.8803

H* 3488.1

S* 8.8342

H-H* -65.9

S-S* -1.9539 μ=G/n=h-Ts -1897.304RTln(f/P*) 1444.758

R 8.314

RT 6427.969

* ln(f/P*) 4.045701

f/P* 57.15123

P*=1 bar

f= 57.15123 bar

QUIZ 5: Fugacity of steam PAM

• Calculate the fugacity of steam at P=60 bar,

and T= 500o C using the steam tables.

h s

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Quiz 5: Answer

Quiz 6: Use the vdW EOS for Mx fi

• A tank of mixture of (mole%)

• 75% argon (1) and 25% ethylene (2) is

• at T= 298.15 K and P= 91.2 bar.

• Find the fugacities of 1 & 2 (use vdW), also their activity coefficients.

• (Use R = 0.08314 bar.m3/[K.kmol] ).

• For the vdw parameters:

• a1= 1.3681, b1 = 0.032289;

• a2 = 4.56149, b2 = 0.057443.

• [Units of a = bar(m3/kmol)^2, b = m3/kmol. ]

• Molar volume: v = 0.23187 m3/kmol

• Also find the activity coefficients γ1 and γ2

• (Note: f10= 75.75 bar, f20= 46.28 bar)

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QUIZ 6: ANSWER: Mx Fugacities ( EOS): Argon(1)+Ethylene(2)-- Use vdw for mixturesAr (39.94) + C2H4 (28.05)

Find: Pur+Mx

DATA CALC.1 Molar volume (Mx)T, K 298.15 Mixture v

P, bar 91.2 Use goalseek P=RT/(v-b)-a/v^2

m, lb 157 P= 91.3073

MW1 39.94 Mx v= 0.23187

MW2 28.05

y1 0.75

y2 0.25

MW, Avg 36.9675

nT, lbmol 4.246973693

R, gas con 0.08314 bar.L/(K.gmol)

a11=a1 1.3618

b1 0.032289

a22=a2 4.56149

b2 0.057443

a12= 2.492355729

bm 0.0385775

am 1.985739024

Tc,1 271.8 oR

Pc,1 47.976308 atm

Tc,2 509.4 oR

Pc,2 50.54294176 atm

f10,bar 75.75

f20,bar 46.28

): Argon(1)+Ethylene(2)-- Use vdw for mixtures

vdw-mx Find:

CALC.2 Fugacities in MxvdW Formula: ln(f1/y1P)= b1/(v-b)-2(x1*a11+x2*a12)/vRT-ln[P(v-b)/RT]

Fugacity, f1

b1/(v-b) 0.167047351

L/gmol 2(...)/vRT 0.57221383

ln[P(v-b)/RT] -0.34086317

ln(f1/y1P) sum= -0.06430331

f1/y1P 0.937720534

f1= 64.1401 bar, cf. 68.4 idgm

Fugacity, f2

b2/(v-b) 0.297181732

2(...)/vRT 1.047261284

ln[P(v-b)/RT] -0.34086317

sum= -0.40921639

f2/y1P 0.6641705

f2= 15.143 bar, cf. 22.8 idgm

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CALC.2 Activity coeff. .γ1, γ2b1/(v-b)-2(x1*a11+x2*a12)/vRT-ln[P(v-b)/RT] Pure f0 eq: ln(f/P)= b/(v-b)-2a/(vRT)-ln[P(v-b)/RT]

Pure Ar Fugacity, f10

b/(v-b)

2a/(vRT)

ln([P(v-b)/RT]

sum

f10/P

f10 75.75 bar

a1=f1/f10 0.846733789 cf. 0.75

γ1= a1/y1 1.12898Pure C2- Fugacity, f20

b/(v-b)

2a/((vRT)

ln[P(v-b)/(RT)]

sum

f20/P

f20 46.28

a2=f2/f20 0.327205864 cf.0.25

γ2= a2/y2 1.30882

Quiz 7: Use the RK for Mx fi• A tank of mixture of (mole%)

• 75% argon (1) and 25% ethylene (2) is

• at T= 298.15 K and P= 91.2 bar.

• Find the fugacities of the components (by Redlich-Kwong EOS)

also the acitivity coefficients.

• (Use R = 0.08314 bar.m3/[K.kmol],

• v = 0.2389 m3/kmol).

• For the RK parameters:

• a1= 17.035, b1 = 0.02238;

• a2 = 77.756, b2 = 0.039815.

• [Units of a = bar(m3/kmol)2*K(1/2), b = m3/kmol. ]

• f10 = 75.75 bar, f20 = 46.28 bar.

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QUIZ 8:OBFPBO

Read nomograph for Ki values

QUIZ

8 :

Rachford-

Rice for

Flash

DATA CALC. Rach-Rice CALC. liquid xi

To,F 64.3 sum (Ki-1)zi/(1+psi(Ki-1)) x3 0.100171

Po,psi 200 ψ= 0.4051 x4 0.131529

K3 3.46 x5 0.465867

K4 2.285 fi x6 0.302424

K5 0.651 C3 0.246421 sum= 0.999991

K6 0.164 C4 0.169015

z3 0.2 C5 -0.16259

z4 0.2 C6 -0.25283

z5 0.4 sum Σfi=0 2.12E-05By Goalseek

z6 0.2

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QUIZ 9:OBFPBOQBJ

Also find xi at ψ =0.4

Ans. (Quiz OBFQBJ)

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QUIZ 10:

Not to scale !

QBM

QBM

409K

377K

400K

390K

Not to scale !

Quiz 10: For example, the final T-x-y should look like this !

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QUIZ 11: Establish product distribution

• The ammonia synthesis reaction is

• N2 + 3 H2 ���� 2 NH3

• or A + 3 B ���� 2 C

• If nA0= 100 kmol, nB0= 250 kmol, find the expressions for yi (mole fractions) in terms of ε. (Note nc0=0 kmol).

• If ε = 45% of nitrogen (or 45 kmol), what are the values of yi ?

• (i = A, B, C)

•

Ans. Q11

yA = (100-ε)(350-2 ε) = 0.212

yB = (250-3ε)(350-2 ε) = 0.442

yC = 2ε/(350-2 ε) = 0.346

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QUIZ12 : Find ΔGrxn for NH3

• Use the standard Gibbs free energy table to

find the ΔGrxn for the ammonia systhesis

reaction at 25oC. Evaluate the equilibrium K

value!

• N2 + 3 H2 ���� 2 NH3

*2

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QUIZ 13: Ammonia ReactionPCBQBX

Find K,Find y i !

Hints:

• N2

+ 3H2

�������� 2 NH3

•• � t=0 (initial) ni0 : 1 3 0

• ����t’>0, extent -ɛ -3 ɛ 2 ɛ

• of reaction

• �t’>0, remaining 1- ɛ 3-3 ɛ 2 ɛ

• moles nit

• ����Total moles nT at t’>0: nT = 1- ɛ + 3-3 ɛ+ 2 ɛ = 4 – 2 ɛ

• �Mole frac:

• yi= (1- ɛ)/(4-2 ɛ), 3(1- ɛ)/(4-2 ɛ), 2 ɛ/(4-2 ɛ),

• ����Change in moles: Δv = 2-3-1 = -2

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*P-21

Excel:

p.525 Tsys, K 800Psys, bar 200GFERxn= 77324 J/mol

lnK=-DG/(RT)= -11.62557133 •∆v = 2-3-1 = -2K= 8.93467E-06 yRatio= 0.357387

N2+3H2===2NH3 yRatio= K* P^- ∆v= (8.93467E-6)*(200)^2

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x yRatio0.249749 0.3574 4x^2(4-2x)^2//27/(1-x)^4

Constraint: Ka 0.35738728

yN2 (1-x)/(4-2x) 0.21432669 yi= (1- ɛ)/(4-2 ɛ), yH2 3(1-x)/(4-2x) 0.642980069 3(1- ɛ)/(4-2 ɛ), yAmmo 2x/(4-2x) 0.142693242 2 ɛ/(4-2 ɛ),

1

By Goalseek

QUIZ 14: Find ε (extent) and yi (mole frac.) for the

reaction at 2500 K (P=10 atm).

Initial feed is: 10 kmols of CO and 9 kmols of O2

(coming from dry air) to the reactor.

In dry air, the nitrogen/oxygen ratio is 3.76.

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QUIZ 15: OCDQCC

• A closed vessel initially contains a gaseous mixture consisting of 3 lbmol of CO2, 6 lbmol of CO, and 1 lbmol of H2. An equilibrium mixture at 2600°C, 1 atm is formed containing CO2, CO, H2O, H2, and O2. (1)Formulate the expressions for the equilibrium constants KA and KB. (2) Make the 5-step stoichiometric balances. (3) Express the mole fractions yi in terms of the two extents of reaction, x and y. (No need to find the values of x and y)

•n.g.jhwSimultaneous Reactions:

(A)… CO2 + H2 ���� CO + H2O

(B)… CO2 ���� CO + (1/2) O2

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ANSWERSUsing Solver!!![DATA] [CALC.]T= 2600KP= 1 atm I. Mole balances

Rxn A: CO2+ H2 ? CO +

t=0 3 6Rxn B: CO2 ? CO ½ O2Feed: t=0 1 0Extent A .-x .-x .+x

Extent B: .-y .+y .+0.5y

Remain: A&B 3-x-y 1-x 6+x+y 0.5y

total nT nT = 10+0.5y

From Table A27 Thus Table Value Calc. Value

= 6.3387

= 0.060395

= 6.3387

= 0.060395

HINTS:

� �

� �

+ H2O

0

.+x

x

ANSWERSUsing Solver!!!

[DATA] [CALC.]T= 2600KP= 1 atm I. Mole balances

Rxn A: CO2+ H2 ? CO +

t=0 3 6Rxn B: CO2 ? CO ½ O2Feed: t=0 1 0Extent A .-x .-x .+x

Extent B: .-y .+y .+0.5y

Remain: A&B 3-x-y 1-x 6+x+y 0.5y

total nT nT = 10+0.5y

From Table A27 Thus Table Value Calc. Value

= 6.3387

= 0.060395

= 6.3387

= 0.060395

+ H2O

0

� �

� �

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From Table A27 Thus Table Value Calc. Value

log10 KA=0.802 0.802 KA= 6.338697113 6.213673887

log10 KB= -1.219 KB= 0.060394863 0.060414529

Use Solver:x= 0.682y= 0.0087nT= 10.00435

Calc. mol.frac yCO2 0.2308yH2 0.0318yO2 0.0004yH2O 0.0682yCO 0.6688

sum 1

QUIZ 16: Use van der Waals Eq. for Mixture Eq. for

fugacities: Find the fugacity coefficient,φ3, for NH3.

Mx molar volume, v = 0.2353 lit/mol

T, K= 773.2 R= 0.08314P, bar= 300

a12 0.584063

a13 2.412986

a23 1.028713

b

3 1 2

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Excel Calculations:

[Calc.1] Molar volume of mx

Goal seek v,L/gmol P, bar

0.2353 300

0.2353

am 0.810828

bm 0.030969

Mx. Fugacities

ln(f1/y1P) 0.097235

ln(f2/y2P) 0.137307

ln(f3/y3P) -0.01476

Φ1 1.102119

Φ2 1.147181

Φ3 0.98535

Φ-product Φ3^2/[Φ1*Φ2^3]

mx fugs ==> 0.583522

pur fugs ==> 0.48147

Compare with φ from pure vdW

φ3 φ1 φ2

Φo = 0.885 1.098 1.114