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+//Chemical Engineering Thermodynamics

+//Chapter 14

+//Thermodynamics of Chemical Reactions

+

+//Example 14.7

+clear;

+clc;

+

+//Given

+//SO2 + (1/2)O2 - SO3

+//Basis: 1 Kgmole of SO2

+n_SO2 = 1;// SO2 fed in Kgmole

+n_O2 = n_SO2;//O2 fed in kgmole

+

+//To Calculate the conversion of SO2 to SO3 at 1atm and at various temperature

+//(1)Calculate the conversion of SO2 to SO3 

+P = 1;//Pressure in atm

+T = 850;//Temperature in K

+m = 1-1-(1/2);

+//From example 14.6

+Ta = [700 800 825 850 900 1000 1100 1300 1500];

+Ka = [395.40 52.51 34.60 23.44 11.59 3.527 1.48 0.398 0.0016];

+clf;

+xset('window',2);

+plot2d(Ta,Ka,style=3);

+xtitle("Equilibrium constant vs Temperature","Temperature in K","Ka");

+Ka1 = interpln([Ta;Ka],850);

+//Let Nc be the moles of SO3 at equilibrium

+Nc = [0.1 0.2 0.3 0.4 0.5 0.7 0.8 0.9 0.930 0.95 0.98 0.988 0.989 0.9895 0.9897 0.9899 0.9900];

+//From equation 14.49 (page no 320) and using the given data ,we got equation (b) (page no 323)

+for i = 1:17

+    Ka(i) = (((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));

+end

+xset('window',1);

+plot2d(Nc,Ka,style=2);

+xtitle("Equilibrium constant vs Kgmoles of SO3","Kg moles of SO3","Ka");

+Nc1 = interpln([Ka;Nc],Ka1);

+C = Nc1*100/n_SO2;

+mprintf('(1)The conversion of SO2 to SO3 at 1atm and 850K is %f percent',C);

+

+//(2)Calculation of conversion at 1 atm and 850 K under the following conditions

+//(i) Given:

+n_N2 = 3.75;//Kgmoles of N2 fed

+//Let Nc be the moles of SO3 at equilibrium

+Nc = [0.85 0.87 0.90];

+//From equation 14.49 (page no 320) and using the given data ,we got equation (c) (page no 324)

+for i = 1:3

+    Ka2(i) = (((+n_N2+n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));

+end

+xset('window',1);

+plot2d(Nc,Ka2,style=5);

+Nc2 = interpln([Ka2';Nc],Ka1);

+C2 = Nc2*100/n_SO2;

+mprintf('\n\n (2)(i)The conversion of SO2 to SO3 at 1 atm and 850 K when inert gas is also added is %f percent',C2);

+

+//(ii)SO3 is also sent along the original feed

+n_SO3 = 1;//Kgmoles of SO3 fed

+//Let Nc be the moles of SO3 at equilibrium

+Nc = [0.80 0.86 0.92];

+//From equation 14.49 (page no 320) and using the given data ,we got equation (d) (page no 326)

+for i = 1:3

+    Ka3(i) = (((+n_SO3+n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*((n_SO3+Nc(i))/(n_SO2-Nc(i)));

+end

+xset('window',1);

+plot2d(Nc,Ka3,style=6);

+Nc3 = interpln([Ka3';Nc],Ka1);

+C3 = Nc3*100/n_SO2;

+mprintf('\n   (ii)The conversion of SO2 to SO3 at 1 atm and 850 K when SO3 is also added along the original feed is %f percent',C3);

+

+//(iii)Variation of SO2 to O2 ratio:

+//(a)SO2:O2 = 1:1 ; This has been worked out in part 1

+mprintf('\n   (iii)(a)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:1 is %f percent',C);

+Xc = Nc1/(n_SO2+n_O2-0.5*Nc1);

+

+//(b)SO2:O2 = 1.1:0.5,Now

+n_SO2 = 1.1;//Kgmoles of SO2 fed

+n_O2 = 0.5;//Kgmoles of O2 fed

+//Let Nc be the moles of SO3 at equilibrium

+Nc = [0.9 0.91 0.92];

+//From equation 14.49 (page no 320) and using the given data ,we got equation (e) (page no 327)

+for i = 1:3

+        Ka4(i) = (((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));

+end

+xset('window',1);

+plot2d(Nc,Ka4,style=1);

+Nc4 = interpln([Ka4';Nc],Ka1);

+C4 = Nc4*100/n_SO2;

+mprintf('\n   (iii)(b)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1.1:0.5 is %f percent',C4);

+Xc1 = Nc4/(n_SO2+n_O2-0.5*Nc4);

+

+//(c)SO2:O2 = 1:0.5

+n_SO2 = 1;//Kgmoles of SO2 fed

+n_O2 = 0.5;//Kgmoles of O2 fed

+//Let Nc be the moles of SO3 at equilibrium

+Nc = [0.8 0.85 0.86 0.87];

+//From equation (a) 

+for i = 1:4

+    Ka5(i) = (((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));

+end

+xset('window',1);

+plot2d(Nc,Ka5,style=4);

+Nc5 = interpln([Ka5';Nc],Ka1);

+C5 = Nc5*100/n_SO2;

+mprintf('\n   (iii)(c)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:0.5 is %f percent',C5);

+Xc2 = Nc5/(n_SO2+n_O2-0.5*Nc5);

+

+if(Xc2>Xc) and (Xc2>Xc1)

+    mprintf('\n        SO2:O2 = 1:0.5 gives the maximum concentration of SO3 at equilibrium.');

+else

+    if(Xc1>Xc) and (Xc1>Xc2)

+        mprintf('\n        SO2:O2 = 1.1:0.5 gives the maximum concentration of SO3 at equilibrium');

+    else

+        if(Xc>Xc1) and (Xc>Xc2)

+            mprintf('\n        SO2:O2 = 1:1 gives the maximum concentration of SO3 at equilibrium');

+        end

+   end

+end

+

+//(3)Conversion of SO2 to SO3 at 50 atm and 850 K when SO2:O2 = 1:1

+n_SO2 = 1;//Kgmole of SO2 fed

+n_O2 = 1;//Kgmoles of O2 fed

+P = 50;//Pressure in atm

+//From figure A.2.9

+phi_SO2 = 0.99;

+phi_SO3 = 0.972;

+phi_O2 = 1;

+//From equation 14.48 (page no320), Ka = Ky*(P^m)*K_phi

+K_phi = phi_SO3/(phi_SO2*(phi_O2^2));

+//Let Nc be the moles of SO3 at equilibrium

+Nc = [0.99 0.985 0.97 0.96];

+for i = 1:4

+    Ka6(i) = K_phi*(P^m)*((((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i))));

+end

+xset('window',1);

+plot2d(Nc,Ka6,style=3);

+Nc6 = interpln([Ka6';Nc],Ka1);

+C = Nc6*100/n_SO2;

+mprintf('\n\n (3)The conversion of SO2 to SO3 at 50atm and 850K  when SO2:O2 = 1:1 is %f percent',C);

+legend("1 part","2.(i) part","2.(ii)part","2.(iii).(b)part","2.(iii).(c)part","3 part");

+//end
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