thermodynamics 2

Question 8.14

The temperature achieved when two fluid streams of differing temperature and/or composition are adiabatically mixed is termed the adiabatic mixing temperature. Compute the adiabatic mixing temperature for the following two cases: 1. Equal weights of aqueous solutions containing 10 wt.-% sulphuric acid at 20oC and 90 wt.-% sulphuric acid at 70oC are mixed. Material balance around mixing point  1,in  m  2,in  m  out m Overall mass balance:  out m

H2SO4-balance

 1,in  m  2,in m  1,in  2m

(mixing of equals weights)

 1,in  x H SO ,2,in  m  2,in  x H SO ,out  m  out x H2SO4 ,1,in  m 2 4 2 4  1,in  0.9  m  1,in  x H SO ,out  2  m  1,in 0.1 m 2

4

x H2SO 4 ,out  0.5

Energy balance around mixing point:    H 1,in  H 2,in  H out

ˆ ˆ ˆ    H 1,in  m1,in  H 2,in  m 2,in  H out  m out

ˆ ˆ H 1,in  H 2,in ˆ H  out 2

The energy balance can be solved graphically (using graph on CD-ROM) or by substituting the values read off from Figure 8.1-1. Alternatively, the data in Felder and Rousseau can be used after transformation of the weight percentages into molar ratio of sulphuric acid to water ( r) Enthalpy of sulphuric acid solution relative to the pure substances at 0 oC: kJ ˆ H 1,in  40  kg

kJ ˆ H 2,in  0  kg

kJ Thus, Hˆout  20  kg

The final temperature can be found by looking at the curve going through the point of 50 wt.-% sulphuric acid and a specific enthalpy of 20 kJ/kg. This results in a temperature of ca. 100 oC (coming rather close to the vaporliquid region  boiling  danger!)

2. Equal weights of aqueous solutions containing 10 wt.-% sulphuric acid at 20oC and 60 wt.-% sulphuric acid at 0oC are mixed.

Following the same approach as above: x H2SO4 ,out  0.35

kJ ˆ H 1,in  40  kg

kJ ˆ H 2,in  325  kg kJ

Thus, Hˆout  142.5  kg The temperature will be close to 20oC

Explain why the adiabatic mixing temperature is greater than that of either of the initial solutions in one of the cases and intermediate between those of the initial solution in the other case. The adiabatic mixing temperature is a result of the heat of mixing. The heat of mixing can be seen as the difference in the A-B interaction and the A-A and B-B interactions of the pure fluid. Hence, mixing relatively pure substances (as in part a) will result in more A-B interactions than A-A and AB interactions. This will lead to an increase in temperature, if A-B interactions are energetically favored over either A-A and B-B interactions. In part (b) the relative pure mixture is mixed with a already dilute mixture. Furthermore, we can consider the shape of the enthalpy curves as a function of the weight percentage of sulphuric acid. It shows a minimum at ca. 60 wt.-% sulphuric acid (this corresponds to a molar water to sulphuric acid ratio of ca. 3.8). The resulting enthalpy upon mixing of two mixtures containing less than 60 wt.-% sulphuric acid will be approximately on the curve, whereas mixing solutions on both sides of the minimum will result in a much higher enthalpy than predicted by the isotherm(s).

Question 8.15 The molar integral heat of solution is defined as the change in enthalpy that results when 1 mole of solute is isothermally mixed with N 2 moles of solvent. The molar integral heat of solution is easily measured in an isothermal calorimeter by monitoring the heat evolved or adsorbed on successive addition of solvent to a given amount of solute.

a. Calculate the heat evolved when 100 g of pure sulphuric acid is added isothermally to 100g of water. The molar mass of H2SO4  98 g/mole The molar mass of H2O  18 g/mole Thus, we are mixing 1.02 mole of sulphuric acid with 5.55 mole of water (or a molar water to sulphuric acid ratio of 5.44). The integral heat of solution for the addition of 5.44 mole of water to 1 mole of sulphuric acid is given as 58.370 kJ. Hence the heat evolved is 1.02*58.370 = 59,561 kJ b. Calculate the heat evolved when the solution prepared in part a) is diluted with an additional 100 g of water The process can be thought to occur via two sub-processes: 1 Splitting the solution in 1.02 mole sulphuric acid and 5.55 mole of water: heat adsorbed = 59.561 kJ 2 Mixing 1.02 mole with 11.11 mole of water (molar ratio water to sulphuric acid 10.89). By interpolation the integral heat of solution of 1 mole of sulphuric acid with 10.89 mole of water is 65.37 kJ. Thus, the heat evolved is 1.02*65.37 kJ = 66.703 kJ For the overall process the heat evolved is thus7.142 kJ

c. Calculate the heat evolved when 100 g of a 60 wt.-% solution of sulphuric acid is mixed with 75 g of a 25 wt.-% sulphuric acid solution 60 wt.-% sulphuric acid solution 60 g sulphuric acid  0.612 mole H2SO4 40 g water  2.22 mole H2O Molar ratio water/sulphuric acid 3.63 25 wt.-% sulphuric acid solution

18.75 g sulphuric acid  0.191 mole H2SO4 56.25 g water  3.125 mole H2O Molar ratio water/sulphuric acid 16.333 Resulting solution will contain

0.804 mole H2SO4 5.347 mole H2O Molar ratio water/sulphuric acid 6.65

The process can be thought to occur via three sub-processes: 1 Splitting the 60 wt.-% sulphuric acid solution in 0.612 mole sulphuric acid and 3.125 mole of water. The integral heat of solution can be found by interpolation to be 52.278 kJ/mole H2SO4. Thus, the heat adsorbed upon splitting the solution = 32.006 kJ 2 Splitting the 25 wt.-% sulphuric acid solution in 0.191 mole sulphuric acid and 4.167 mole of water. The integral heat of solution can be found by interpolation to be 68.954 kJ/mole H2SO4. Thus, the heat adsorbed upon splitting the solution = 13.192 kJ 3 Mixing 0.804 mole H2SO4 with 5.347 mole of water (molar ratio water to sulphuric acid 6.65). By interpolation the integral heat of solution of 1 mole of sulphuric acid with 6.65 mole of water is 59.881 kJ/mole H 2SO4. Thus, the heat evolved is 0.804*59.881 kJ = 48.118 kJ For the overall process the heat evolved is thus 2.9 kJ

d. Relate H1  H1  and H2  H2  to only N1,, N2, Hs, and the derivatives of HS with respect to the ratio N2/N1 The integral heat of solution is defined as: Hs  1  N2   Hmix  H1  N2  H2







H s  H1  H1  N 2  H 2  H 2



(for a solution containing 1 mole of sulphuric acid and N 2 mole of water; for a solution containing N1 mole of sulphuric acid and N2 mole of water, i.e. the molar ratio of water to sulphuric acid to water in the solution is N 2/N: N Hs  H1  H1  2  H 2  H2 ) N1









This gives us one equation for the variables H1  H1  and H2  H2  The other equation for the variables can be obtained by differentiating the integral heat of solution with respect to the ratio of N2 to N1 is:

Hs  N2   N 1  

H1





 N2   N 1  

H1



 N2   N 1  







 H2  H2 



 N2    N1 



 N2   N 1  



 N2  H2  N  H2     2    N1   N2   N1   N2   N   N   1  1

 

The partial molar property of the pure substances H1 and H2 are independent of N1 and N2 (and thus independent of their ratio as well) H1 H2 0 0  N2   N2       N1   N1   N   2  N1   N 2  H2 Hs H1     H 2  H2     N2   N2   N2   N1   N2          N N N N  1  1  1  1





 N2   N  1



The differential

H s  N   2   N1 





 N   2   N1 

1



 N  H 2 H1  H2  H2   2    N   N1   N 2   2   N   N1   1

According to the Gibbs-Duhem equation: H1 H 2 N1   N2  0  N2   N2       N1   N1   N  H 2 H1   2   0  N 2   N1   N 2       N1   N1  H s  H2  H 2  N2     N1  This thus yields the 2nd equation for our variables. Solving for the variables:





H H

2

1



 H2 



Hs  N2    N1 



 H1  H s 

N2 Hs  N1  N 2     N1 

e. Compute the numerical values of H1  H1  and H2  H2  in a 50 wt.-% sulphuric acid solution. A solution containing 50 wt.-% sulphuric acid has a molar ratio of water to H s

sulphuric acid of 5.44. The easiest way to determine the differential

 N2  N1







 is graphically. Plot the integral heat of solution as a function of the ratio N 2 to N1 and draw the tangent:

The slope can be determined to be 4 kJ/mol water.





kJ

Thus, H 2  H 2  4  mol H O 2





kJ

And H1  H1  58.370  5.44  4  36.61 mol H SO . 2 4

It can thus be deduced that the interaction of sulphuric acid in the solution is quite different from the interaction in pure sulphuric acid and the high heat of mixing is mainly due to sulphuric acid.

Question 8.26 Mattingley and Fernby (1975) have reported that the enthalpies of triethylamine-benzene solutions at 298.15K are given by



Hmix   x B  HB  1  x B   HEA   x B  1  x B   1418  482.4  1  2  x B   187.4  1  2  x B 

3



Where xB is the mole fraction of benzene and Hmix, HB and HEA are the molar enthalpies of the mixture, pure benzene and pure triethylamine. a. Develop expressions for HB  HB  and HEA  HEA  The change in the molar enthalpy upon mixing is defined as



 mix H  Hmix   x B  HB  1  x B   HEA   x B  1  x B   1418  482.4  1  2  x B   187.4  1  2  x B 







 mix H  x B  HB  HB  1  x B   HEA  HEA



3



H H  mix H x  1  x B  HB HEA  HB  HB  B  x B   x B  B  HEA  HEA   1  x B    1  x B   EA x B x B x B x B x B x B x B









The pure component molar enthalpy Hi with respect to the mol fraction is independent of the mole fraction of the mixture.  mix H x  1  x B  HB HEA  HB  HB  B  x B   HEA  HEA   1  x B   x B x B x B x B x B









According to Gibbs-Duhem equation: NB 

HB HEA  NEA  0 NB NB

xB 

HB HEA  x EA  0 NB NB

xB 

HB HEA  1  x B   0 NB NB

 mix H x  1  x B   HB  HB  B  HEA  HEA  x B x B x B









 

 mix H  HB  HB  HEA  HEA x B











 mix H  x B  HB  HB  1  x B   HEA  HEA

H H





 mix H  x B  HB  HB  x B  HEA  HEA x B

xB 

H





EA



 HEA   mix H  x B 



 mix H  x B

 HB   mix H  1  x B  

B

EA



 HEA   mix H  x B 





 mix H x B

 mix H x B

 mix H  1  x B   1418  482.4  1  2  x B   187.4  1  2  x B  3 x B



 x B  1418  482.4  1  2  x B   187.4  1  2  x B 



 x B  1  x B   964.8  1124 .4  1  2  x B 

H

EA



 HEA   mix H  x B 



2

3





 mix H  x B

x B  1  x B   1418  482.4  1  2  x B   187.4  1  2  x B  3





 x B  1  x B   1418  482.4  1  2  x B   187.4  1  2  x B  3



 x B 2  1418  482.4  1  2  x B   187.4  1  2  x B  3



 x B 2  1  x B   964.8  1124 .4  1  2  x B  2

H

EA

H

B











 HEA   x B  1418  482.4  1  2  x B   187.4  1  2  x B  2



 x B 2  1  x B   964.8  1124 .4  1  2  x B  2









3



 HB  1  x B  2  1418  482.4  1  2  x B   187.4  1  2  x B  3



 x B  1  x B   964.8  1124 .4  1  2  x B  2

2





b. Compute the values for at HB  HB  and HEA  HEA  at xB = 0.5 Substitute the values:

H

B



 HB  475.1

J mol

H

EA



 HEA  233.9 

J mol

c. One mole of a 25 mole-% benzene mixture is to be mixed with one mole of a 75 mol-% benzene mixture at 298.15 K. How much heat must be added or removed for the process to be isothermal? Hmix 

Hmix   x B  HB  1  x B   HEA  



x B  1  x B   1418  482.4  1  2  x B   187.4  1  2  x B  3

Three sub-processes to consider: 1. demixing of 25 mol-% mixture 2. demixing of 75 mol-% mixture 3. mixing 50 mol-% mixture Overall process: -1418 J/mol

 Hdemix = -1200 J/mol Hmix = 1418 J/mol Hdemix = -1636 J/mol