ProblemSet9 Solutions

8/12/2019 ProblemSet9 Solutions

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MSE 3001: Applied Thermodynamics

Homework Set #9 Solutions

1) Gaskell #12.6(i) Define a solution:

Solid iron and ironoxide, FeO, are in equilibrium with a mixture of CO and CO2gases at

1273 K. Upon decreasing the temperature we have to find out which of the two solid phases

disappears first.

(ii) Plan a solution:Since we start at equilibrium and change one of the independent variables the equilibrium

will shift. If we know the direction in which the equilibrium will shift we can predict, which

phase will disappear first. Say, the reaction shifts toward the side of the reaction with pureiron then FeO will disappear first and vice versa if the composition of the gas is kept

constant. We can determine the change in the reaction equilibrium with LeChateliers

principle. If the reaction is endothermic then a decrease in temperature will shift the reactionin the direction that releases energy and vice versa. The statement that the composition of the

gas mixture remains constant at the equilibrium pressures at 1273 K is important because if

the composition is allowed to equilibrate at any temperature, specifically upon lowering the

temperature then there will always be some Fe and FeO. Only if the composition remainsfixed does one of the two solid phases disappear since the system then tried to reach the new

equilibrium by either oxidizing or reducing the solids. We have to assume, though, that upon

reducing the temperature the amount of Fe and FeO is finite and equilibrium can not bereached before one of the two solid phases has disappeared.

(iii) Execute the solution:

I. Fe(s) + O2(g) = FeO(s); G0I= -263,700 + 64.35T JII. CO2(g) = CO(g) + O2(g); G0II= 282,400 - 86.81T JI. + II.: Fe(s) + CO2(g) = FeO(s) + CO(g); G

0= 18,700 22.46T J

The reaction is endothermic since H0= 18,700 J > 0 J. Upon lowering the temperature

at constant composition the reaction will shift from the FeO + CO side to the Fe + CO 2side to release heat that could counter the lowering of the temperature. Therefore FeO

will disappear first.

2) Gaskell #12.11(i) Define a solution:

A mixture of water vapor and argon gas streams over CaF and reacts to form CaO and

HF gas. Based on the gas flow rate at 298 K and the weight changes of the system at two

different temperatures we calculate the temperature dependence of the standard Gibbs freeenergy for the reaction between CaF, water vapor, CaO and Hf gas.


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(ii) Plan a solution:

From the gas flow rate at 298 and the partial pressure of H2O we can determine thenumber of mol of H2O that participate in the reaction. From the weight loss at the two

different temperatures we can determine the number of mol of HF gas formed and H2O gas

consumed. From that information we can determine the partial pressures of HF and H2O at

the two temperatures and thus determine the value of

G

0

at two temperatures. Assuming alinear temperature dependence for G0, i.e., G

0= H

0 TS

0we can determine the

standard enthalpy and entropy.

(iii) Execute the solution:

- 60 l of gas in one minute at a pressure of 0.9 atm and 298K:mol2.2mol

298latm082.0

l60atm9.0n O2H =

=

- According to the reaction:CaF + H2O = CaO + 2HF and the molar weight of CaF and CaO thetransformation of one mol of CaF to one mol of CaO reduces the weight by 22g. Per one mol of CaF transforming, or a weight loss of 22 g, two mol of HF

form. For 1 g of CaF transforming, then, 2/22 or 1/11 mol of HF forms and

1/22 mol of H2O is consumed. Therefore, for a weight loss of 2.69*10-4

g,1/11*2.69*10

-4mol of HF form and 2/22*2.69*10

-4mol of H2O are

consumed.

- 1/11*2.69*10-4mol of HF in a volume of 60 l and a temperature of 900 Krelate to a partial pressure of HF of

atm10*94.9atm

60

9000.0824-10*2.69*1/11p 6HF

=

=

- The consumption of H2O of the order of 10-5mol is negligible compared tothe 2.2 mol, therefore

atm9. 0p O2H =

- K(900K) = p2HF/pH2O= 1.1*10-10. Thus G0(900K) = -R*900K*ln(1.1*10-10)= 171,596 J.

- The same approach at 1100 K leads toG

0(1100K) = -R*1100K*ln(1.0*10

-7) = 146,926 J.

- From G0= A + BT and the two data points we obtain:G

0= 282,000 123T J.

(iv) Evaluate the solution:The standard Gibbs free energy of the reaction CaF2+ H2O = CaO + HF can be obtained

from four separate reactions:

1. CaF2(s) = Ca(s) = F2(g)2. Ca(s) + O2= CaO(s)3. H2O(g) = H2(g) + O2(g)


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4. F2(g) + H2(g) = 2HF(g)Using the data from Kubaschewskis book (Metallurgical Thermochemistry) a fair

agreement is obtained.

3) For the oxidation reaction of pure Ta, 2Ta + 2.5O2= Ta2O5determine the standardGibbs free energy of the reaction at a temperature T according to the following steps:

(i) Determine H0(T) and S

0(T) according to the loop calculations

(ii) From (i) determine G0(T)

(iii) Use the result of (ii) to obtain G0(T) in the form G

0(T) = A + BTlnT + CT

(iv) Use the result of (iii) to obtain G0(T) in the form G

0(T) = D + ET, where

A,B,C,D,E are constants.

(v) Plot the three G0(T) functions obtained from (ii)-(iv) as a function of temperature.

Data: Heat of formation of Ta2O5: H0298K= -488.5 kcal/mol

Entropy of formation of Ta2O5:

S

0

298K= 34.2 cal/mol*KEntropy of formation of Ta: S0

298K= 9.9 cal/mol*K

cp(T) Ta = 5.8 + 0.71*10-3

T + 0.05*10-6

T2cal/mol*K

cp(T) Ta2O5= 29.2 + 10-2

T cal/mol*K

cp(T) O2(gas) = 7.16 + 10-3

*T 0.4*T-2

cal/mol*K


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Specific heat data input

In[5]:= cpTaT_ : 5.8 0.71 103 T 0.05 106 T2

In[17]:= cpO2T_ : 7.16 103 T 0.4 105 T2

In[4]:= cpTa2O5T_ : 29.2 102 T

In[18]:= cpT_ : cpTa2O5T 2.5 cpO2T 2 cpTaT

In[20]:= H0 488 500

Out[20]= 488500

Calculation of standard enthalpy, entropy, Gibbs free energy

In[21]:= HT_ : H0298

T

cpTemp Temp

In[22]:= S0 34.2 2.5 49.1 2 9.9

Out[22]= 108.35

In[23]:= ST_ : S0298

T cpTemp Temp Temp

In[24]:= GT_ : HT T ST


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In[25]:= PlotGT,T, 298, 1000

Out[25]=

400 500 600 700 800 900 1000

440000

430000

420000

410000

400000

390000

In[27]:= $OutputSizeLimit 1000

Out[27]= 1000

Fit standard free energy with equation of the type a + bTlnT + cT

In[50]:= Temperature Tablei,i, 298, 1000

Out[50]=

A very large output was generated. Here is a sample of it:

298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314,

315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330,

331, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 609,

954, 955, 956, 957, 958, 959, 960, 961, 962, 963, 964, 965, 966, 967, 968, 969,

970, 971, 972, 973, 974, 975, 976, 977, 978, 979, 980, 981, 982, 983, 984, 985,

986, 987, 988, 989, 990, 991, 992, 993, 994, 995, 996, 997, 998, 999, 1000

Show Less Show More Show Full Output Set Size Limit...

In[51]:= Gibbs1 FunctionT, GT Temperature

Out[51]=


456212., 456103., 455 995., 455 887., 455 778., 455 670.,

455562., 455453., 455345., 685, 382 617., 382513., 382410.,

382306., 382202., 382098., 381994., 381890., 381 786.


2 Problem3.nb


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In[52]:= DimensionsTemperature

Out[52]= 703

In[53]:= RiffleTemperature, Gibbs1

Out[53]=


298, 456 212., 299, 456 103., 300, 455995., 301, 455887., 302,

455778., 303, 455670., 304, 455 562., 305, 455 453., 1374,

993, 382513., 994, 382 410., 995, 382 306., 996, 382202.,

997, 382098., 998, 381 994., 999, 381 890., 1000, 381786.


In[54]:= Data PartitionRiffleTemperature, Gibbs1, 2

Out[54]=


298, 456 212., 299, 456 103., 300, 455995.,

301, 455887., 302, 455 778., 303, 455 670., 304, 455562.,

689, 994, 382 410., 995, 382306., 996, 382 202.,

997, 382098., 998, 381 994., 999, 381 890., 1000, 381786.


In[55]:= FindFitData, a b T LogT c T,a, b, c, T

Out[55]=


a 489 795., b 3.78203, c 134.158


Fit a+bTlnT+c with an equation of the type A + BT

In[56]:= G2T_ : 489795 3.78 T LogT 134.158 T

Problem3.nb 3


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In[57]:= Gibbs2 FunctionT, G2T Temperature

Out[57]=


456233., 456125., 456 016., 455 907., 455 798., 455 689.,

455581., 455472., 455363., 685, 382 583., 382478., 382374.,

382270., 382165., 382061., 381957., 381853., 381 748.


In[58]:= Data3 PartitionRiffleTemperature, Gibbs2, 2

Out[58]=


298, 456 233., 299, 456 125., 300, 456016.,

301, 455907., 302, 455 798., 303, 455 689., 304, 455581.,

689, 994, 382 374., 995, 382270., 996, 382 165.,

997, 382061., 998, 381 957., 999, 381 853., 1000, 381748.


In[59]:= FindFitData3, d e T,d, e, T

Out[59]= d 487 543., e 106.02

Graphical representation of the results

In[60]:= G3T_ : 487543 106 T

PlotGT, G3T, G2T,T, 298, 1000, AxesLabel "TemperatureK", "G0 cal"

Out[62]=

400 500 600 700 800 900 1000TemperatureK

440000

430000

420000

410000

400000

390000

380000

G0

4 Problem3.nb


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The graph shows a perfect overlap between the three graphs. This means that the linear dependence of the standard

Gibbs free energy of reaction is a very good approximation of the actual temperature dependence.

Problem3.nb 5

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ProblemSet9 Solutions