initial commit / add all books

22 files changed, 711 insertions, 0 deletions

diff --git a/647/CH12/EX12.1/Example12_1.sce b/647/CH12/EX12.1/Example12_1.sce
new file mode 100755
index 000000000..13a8797f5
--- /dev/null
+++ b/647/CH12/EX12.1/Example12_1.sce
@@ -0,0 +1,13 @@
+clear;

+clc;

+

+// Example: 12.1

+// Page: 471

+

+printf("Example: 12.1 - Page: 471\n\n");

+

+// This problem involves proving a relation in which no mathematics and no calculations are involved.

+// For prove refer to this example 12.1 on page number 471 of the book.

+

+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");

+printf(" For prove refer to this example 12.1 on page 471 of the book.");
\ No newline at end of file
diff --git a/647/CH12/EX12.10/Example12_10.sce b/647/CH12/EX12.10/Example12_10.sce
new file mode 100755
index 000000000..3c1d4c9fa
--- /dev/null
+++ b/647/CH12/EX12.10/Example12_10.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+

+// Example: 12.10

+// Page: 489

+

+printf("Example: 12.10 - Page: 489\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: CO(g) + H2O(g) ------> CO2(g) + H2(g)

+P = 10;// [bar]

+T = 1000;// [K]

+K_1000 = 1.5;

+//***********//

+

+// Moles in feed:

+nCO_feed = 1;

+nH20_feed = 1;

+// Let e be the degree of completion at equilibrium.

+// Moles at Equilibrium:

+// nCO_eqb = 1 - e;

+// nH20_eqb = 1 - e;

+// nCO2_eqb = e;

+// nH2_eqb = e;

+// Total moles at equilibrium = 1 - e + 1 - e + e + e = 2

+// Mole Fractions at Equilibrium:

+// yCO_eqb = (1 - e)/2;

+// yH20_eqb = (1 - e)/2;

+// yCO2_eqb = e/2;

+// yH2_eqb = e/2;

+// Sum of stoichometric coeffecient:

+v = 1 + 1 - 1 - 1;

+K = K_1000*P^v;

+deff('[y] = f(e)','y = K - (e/2)*(e/2)/(((1 - e)/2)*(1 - e)/2)');

+e = fsolve(0.5,f);

+printf("Composition of the gas leaving the mixture\n");

+printf("yCO = yH20 = %.4f\n",(1 - e)/2);

+printf("yCO2 = yH2 = %.4f\n",e/2);
\ No newline at end of file
diff --git a/647/CH12/EX12.11/Example12_11.sce b/647/CH12/EX12.11/Example12_11.sce
new file mode 100755
index 000000000..e7424d731
--- /dev/null
+++ b/647/CH12/EX12.11/Example12_11.sce
@@ -0,0 +1,38 @@
+clear;

+clc;

+

+// Example: 12.11

+// Page: 489

+

+printf("Example: 12.11 - Page: 489\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: N2(g) + 3H2(g) --------> 2NH3

+nN2_feed = 1;

+nH2_feed = 5;

+T = 800;// [K]

+P = 250;// [bar]

+K = 1.106*10^(-5);

+//**************//

+

+// Let e be the degree of completion at equilibrium.

+// Moles at Equilibrium:

+// nN2_eqb = 1 - e;

+// nH2_eqb = 5 - 3*e;

+// nNH3_eqb = 2*e;

+// Total moles at equilibrium = 1 - e + 5 - 3*e + 2*e = 2*(3 - e)

+// Mole Fractions at Equilibrium:

+// yN2_eqb = (1 - e)/(2*(3 - e));

+// yH2_eqb = (5 - 3*e)/(2*(3 - e));

+// yNH3_eqb = 2*e/(2*(3 - e));

+// Sum of stoichometric coeffecient:

+v = 2 - 3 - 1;

+Ky = K*P^(-v);

+deff('[y] = f(e)','y = Ky - ((2*e/(2*(3 - e)))^2)/(((1-e)/((2*(3 - e))))*((5 - 3*e)/(2*(3 - e)))^3)');

+e = fsolve(0.5,f);

+printf("Molar Composition of the gases\n");

+printf("yN2 = %.4f\n",(1 - e)/(2*(3 - e)));

+printf("yH2 = %.4f\n",(5 - 3*e)/(2*(3 - e)));

+printf("yNH3 = %.4f\n",(2*e)/(2*(3 - e)));
\ No newline at end of file
diff --git a/647/CH12/EX12.12/Example12_12.sce b/647/CH12/EX12.12/Example12_12.sce
new file mode 100755
index 000000000..d04ad310d
--- /dev/null
+++ b/647/CH12/EX12.12/Example12_12.sce
@@ -0,0 +1,39 @@
+clear;

+clc;

+

+// Example: 12.12

+// Page: 490

+

+printf("Example: 12.12 - Page: 490\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: SO2(g) + (1/2)O2 ------> SO3(g)

+P = 1;// [bar]

+T = 750;// [K]

+K = 74;

+//************//

+

+// Moles in Feed:

+nSO2 = 1;

+nO2 = 0.5;

+// Let e be the degree of completion at equilibrium.

+// Moles at Equilibrium:

+// nSO2_eqb = 10 - e;

+// nO2_eqb = 8 - 0.5*e;

+// nSO3_eqb = e;

+// Total no. of moles = 10 - e + 8 - 0.5*e + e = 18 -0.5*e;

+// Mole fraction at Equilibrium:

+// ySO2_eqb = (10 - e)/(18 - 0.5*e);

+// yO2_eqb = (8 - 0.5*e)/(18 - 0.5*e);

+// ySO3_eqb = e/(18 - 0.8*e);

+// Sum of stoichometric coeffecient:

+v = 1 - 1 -0.5;

+Ky = K*P^(-v);

+deff('[y] = f(e)','y = Ky - (e*(18 - (0.5*e)))/((10 - e)*(8 - 0.5*e))');

+e = fsolve(7,f);

+printf("Molar Composition of the gases\n");

+printf("nSO2 = %.2f\n",(10 - e));

+printf("nO2 = %.2f\n",(8 - 0.5*e));

+printf("nSO3 = %.2f\n",e);
\ No newline at end of file
diff --git a/647/CH12/EX12.13/Example12_13.sce b/647/CH12/EX12.13/Example12_13.sce
new file mode 100755
index 000000000..bebafb163
--- /dev/null
+++ b/647/CH12/EX12.13/Example12_13.sce
@@ -0,0 +1,30 @@
+clear;

+clc;

+

+// Example: 12.13

+// Page: 492

+

+printf("Example: 12.13 - Page: 492\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: PCl5 = PCl3 + Cl2

+T = 250;// [OC]

+Kp = 1.8;

+e = 0.5;

+//**************//

+

+// Basis: 1 mol of PCl5

+// At Equilibrium:

+n_PCl5 = 1 - e;

+n_PCl3 = e;

+n_Cl2 = e;

+n_total = n_PCl5 + n_PCl3 + n_Cl2;

+// Patrial Pressures:

+// P_PCl5 = (n_PCl5/n_total)*P

+// P_PCl3 = (n_PCl3/n_total)*P

+// P_Cl2 = (n_Cl2/n_total)*P

+deff('[y] = f(P)','y = Kp - ((n_PCl3/n_total)*P)*((n_Cl2/n_total)*P)/((n_PCl5/n_total)*P)');

+P = fsolve(7,f);// [atm]

+printf("Total Pressure Required for 50 %% conversion of PCl5 is %.1f atm",P);
\ No newline at end of file
diff --git a/647/CH12/EX12.14/Example12_14.sce b/647/CH12/EX12.14/Example12_14.sce
new file mode 100755
index 000000000..ee66f0525
--- /dev/null
+++ b/647/CH12/EX12.14/Example12_14.sce
@@ -0,0 +1,53 @@
+clear;

+clc;

+

+// Example: 12.14

+// Page: 494

+

+printf("Example: 12.14 - Page: 494\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: SO2(g) + (1/2)O2(g) -------> SO3(g)

+// Moles in Feed:

+nSO2_feed = 1;

+nO2_feed = 0.5;

+nAr_feed = 2;

+P = 30;// [bar]

+T = 900;// [K]

+K = 6;

+//*************//

+

+// Let e be the degree of completion at equilibrium.

+// nSO2_eqb = 1 - e;

+// nO2_eqb = 0.5*(1 - e);

+// nSO3_eqb = e;

+// nAr_eqb = 2;

+// Total moles at equilibrium = 1 - e + 0.5*(1 - e) + e + 2 = (7 - e)/2

+// Mole fractions:

+// ySO2_eqb = 2*(1 - e)/(7 - e)

+// yO2_eqb = (1 - e)/(7 - e)

+// ySO3_eqb = 2*e/(7 - e)

+// yAr_eqb = 4/(7 - e)

+// Sum of Stoichiometric Coeffecient:

+v = 1 - 1 - 1/2;

+Ky = K*P^(-v);

+e = 0.8;

+err = 1;

+while err > 0.2

+    Ky_new = (2*e/(7 - e))/(((2*(1 - e))/(7 - e))*(((1 - e)/(7 - e))^0.5));

+    err = abs(Ky - Ky_new);

+    Ky = Ky_new;

+    e = e + 0.001;

+end

+printf("Degree of Conversion is %.3f\n",e);

+ySO2_eqb = 2*(1 - e)/(7 - e);

+yO2_eqb = (1 - e)/(7 - e);

+ySO3_eqb = 2*e/(7 - e);

+yAr_eqb = 4/(7 - e);

+printf("Equilibrium Composition of the reaction Mixture\n");

+printf("ySO2 = %.4f\n",ySO2_eqb);

+printf("yO2 = %.4f\n",yO2_eqb);

+printf("ySO3 = %.4f\n",ySO3_eqb);

+printf("yAr = %.4f\n",yAr_eqb)
\ No newline at end of file
diff --git a/647/CH12/EX12.15/Example12_15.sce b/647/CH12/EX12.15/Example12_15.sce
new file mode 100755
index 000000000..903734527
--- /dev/null
+++ b/647/CH12/EX12.15/Example12_15.sce
@@ -0,0 +1,47 @@
+clear;

+clc;

+

+// Example: 12.15

+// Page: 498

+

+printf("Example: 12.15 - Page: 498\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: CH3COOH(l) + C2H5OH(l) --------> CH3COOC2H5(l) + H2O(l)

+T = 373.15;// [K]

+nCH3COOH_feed = 1;

+nC2H5OH_feed = 1;

+deltaHf_CH3COOH = -484.5;// [kJ]

+deltaHf_C2H5OH = -277.69;// [kJ]

+deltaHf_CH3COOC2H5 = -480;// [kJ]

+deltaHf_H2O = -285.83;// [kJ]

+deltaGf_CH3COOH = -389.9;// [kJ]

+deltaGf_C2H5OH = -174.78;// [kJ]

+deltaGf_CH3COOC2H5 = -332.2;// [kJ]

+deltaGf_H2O = -237.13;// [kJ]

+R = 8.314;// [J/mol K]

+//******************//

+

+deltaH_298 = deltaHf_CH3COOC2H5 + deltaHf_H2O - deltaHf_CH3COOH - deltaHf_C2H5OH;// [kJ]

+deltaG_298 = deltaGf_CH3COOC2H5 + deltaGf_H2O - deltaGf_CH3COOH - deltaGf_C2H5OH;// [kJ]

+T0 = 298;// [K]

+K_298 = exp(-(deltaG_298*1000/(R*T0)));

+K_373 = K_298*exp((deltaH_298*1000/R)*((1/T0) - (1/T)));

+// Let e be the degree of completion at equilibrium.

+// nCH3COOH_eqb = 1 - e;

+// nC2H5OH_eqb = 1 - e;

+// nCH3COOC2H5_eqb = e;

+// nH2O_eqb = e;

+// Total moles at equilibrium = 1 - e + 1 - e + e + e = 2

+// Mole fractions:

+// ySO2_eqb = (1 - e)/2

+// yO2_eqb = (1 - e)/2

+// ySO3_eqb = e/2

+// yAr_eqb = e/2

+// Sum of Stoichiometric Coeffecient:

+v = 1 + 1 - 1 - 1;

+deff('[y] = f(e)','y = K_373 - ((e/2)*(e/2))/(((1 - e)/2)*((1 - e)/2))');

+e = fsolve(0.5,f);

+printf("Mole fraction of ethyl acetate is %.3f",e/2);
\ No newline at end of file
diff --git a/647/CH12/EX12.16/Example12_16.sce b/647/CH12/EX12.16/Example12_16.sce
new file mode 100755
index 000000000..3bb7463ad
--- /dev/null
+++ b/647/CH12/EX12.16/Example12_16.sce
@@ -0,0 +1,22 @@
+clear;

+clc;

+

+// Example: 12.16

+// Page: 501

+

+printf("Example: 12.16 - Page: 501\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: CaCO3(s) = CaO(s) + CO2(g)

+T = 1000;// [K]

+deltaH_1000 = 1.7533*10^5;// [J]

+deltaS_1000 = 150.3;// [J/mol K]

+R = 8.314;// [J/mol K]

+//****************//

+

+deltaG_1000 = deltaH_1000 - T*deltaS_1000;// [J]

+K_1000 = exp(-(deltaG_1000/(R*T)));// [bar]

+P = K_1000;

+printf("The decomposition pressure of limestone is %.4f bar at 1000 K",P);
\ No newline at end of file
diff --git a/647/CH12/EX12.17/Example12_17.sce b/647/CH12/EX12.17/Example12_17.sce
new file mode 100755
index 000000000..5988e86e3
--- /dev/null
+++ b/647/CH12/EX12.17/Example12_17.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+

+// Example: 12.17

+// Page: 501

+

+printf("Example: 12.17 - Page: 501\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: A(s) ---------> B(s) + C(g)

+deff('[deltaG] = f1(T)','deltaG = 85000 - 213.73*T + 6.71*T*log(T) - 0.00028*T^2');// [J]

+T1 = 400;// [K]

+T2 = 700;// [K]

+Pc = 1;// [bar]

+R = 8.314;// [J/mol K]

+//**************//

+

+deltaG_400 = f1(400);// [J]

+deltaG_700 = f1(700);// [J]

+K_400 = exp(-(deltaG_400/(R*T1)));// [bar]

+K_700 = exp(-(deltaG_700/(R*T2)));// [bar]

+printf("The decomposition pressure is %.4f bar at 400 K\n",K_400);

+printf("The decomposition pressure is %.2f bar at 700 K\n",K_700);

+

+// Equilibrium constant for solid - gas reaction is:

+// K = aB*aC/aA = aC = fC = Pc

+deff('[y] = f2(T)','y = Pc - exp(-f1(T)/(R*T))');

+T = fsolve(900,f2);// [K]

+printf("The decomposition temperature is %.3f K",T);
\ No newline at end of file
diff --git a/647/CH12/EX12.18/Example12_18.sce b/647/CH12/EX12.18/Example12_18.sce
new file mode 100755
index 000000000..4c1dea6fb
--- /dev/null
+++ b/647/CH12/EX12.18/Example12_18.sce
@@ -0,0 +1,42 @@
+clear;

+clc;

+

+// Example: 12.18

+// Page: 502

+

+printf("Example: 12.18 - Page: 502\n\n");

+

+// Solution

+

+//*****Data******//

+T = 875;// [K]

+K = 0.514;

+P = 1;// [bar]

+//*************//

+

+// Reaction: Fe(s) + H2O(g) --------->FeO(s) + H2(g)

+// K = a_FeO*a_H2/(a_Fe*a_H2O)

+// Since the activities of the solid components Fe & FeO at equilibrium may be taken as unity.

+// a_Fe = a_FeO = 1

+// Ka = a_H2/a_H2O;

+// Feed:

+nH2O_feed = 1;

+nH2feed = 0;

+// Let e be the degree of completion at equilibrium.

+// Moles at Equilibrium:

+// nH2O_eqb = 1 - e;

+// nH2_eqb = e;

+// Total moles at equilibrium = 1 - e + e = 1

+// Mole Fractions at Equilibrium:

+// yH20_eqb = 1 - e;

+// yH2_eqb = e;

+// Sum of stoichometric coeffecient:

+v = 1 - 1;

+Ky = K*P^(-v);

+// Ky = e/(1 - e)

+e = Ky/(Ky + 1);

+yH2_eqb = e;

+yH2O_eqb = 1 - e;

+printf("Equilibrium Composition\n");

+printf("yH2 = %.2f\n",yH2_eqb);

+printf("y_H2O = %.2f\n",yH2O_eqb);
\ No newline at end of file
diff --git a/647/CH12/EX12.19/Example12_19.sce b/647/CH12/EX12.19/Example12_19.sce
new file mode 100755
index 000000000..cc82d408d
--- /dev/null
+++ b/647/CH12/EX12.19/Example12_19.sce
@@ -0,0 +1,13 @@
+clear;

+clc;

+

+// Example: 12.19

+// Page: 503

+

+printf("Example: 12.19 - Page: 503\n\n");

+

+// Mathematics is involved in proving but just that no numerical computations are involved.

+// For prove refer to this example 12.19 on page number 503 of the book.

+

+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");

+printf(" For prove refer to this example 12.19 on page 503 of the book.");
\ No newline at end of file
diff --git a/647/CH12/EX12.2/Example12_2.sce b/647/CH12/EX12.2/Example12_2.sce
new file mode 100755
index 000000000..773766121
--- /dev/null
+++ b/647/CH12/EX12.2/Example12_2.sce
@@ -0,0 +1,13 @@
+clear;

+clc;

+

+// Example: 12.2

+// Page: 473

+

+printf("Example: 12.2 - Page: 473\n\n");

+

+// This problem involves proving a relation in which no mathematics and no calculations are involved.

+// For prove refer to this example 12.2 on page number 473 of the book.

+

+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");

+printf(" For prove refer to this example 12.2 on page 473 of the book.");
\ No newline at end of file
diff --git a/647/CH12/EX12.20/Example12_20.sce b/647/CH12/EX12.20/Example12_20.sce
new file mode 100755
index 000000000..ae651ae35
--- /dev/null
+++ b/647/CH12/EX12.20/Example12_20.sce
@@ -0,0 +1,37 @@
+clear;

+clc;

+

+// Example: 12.20

+// Page: 505

+

+printf("Example: 12.20 - Page: 505\n\n");

+

+// Solution

+

+//Reactions:

+// CO + (1/2)O2 ------------> CO2 ......................................(1)

+// C + O2 ------------------> CO2 ......................................(2)

+// H2 + (1/2)O2 ------------> H2O ......................................(3)

+// C + 2H2 -----------------> CH4 ......................................(4)

+

+// Elimination of C:

+// Combining Eqn. (2) with (1):

+// CO + (1/2)O2 ------------> CO2 ......................................(5)

+// Combining Eqn. (2) with (4):

+// CH4 + O2 ----------------> 2H2 + CO2 ................................(6)

+

+// Elimination of O2:

+// Combining Eqn. (3) with (4):

+// CO2 + H2 ----------------> CO + H2O .................................(7)

+// Combining Eqn. (3) with (6):

+// CH4 + 2H2O -------------> CO2 + 4H2 .................................(8)

+

+// Equations 7 & 8 are independent sets. Hence

+r = 2;// [No. of independent rkn.]

+C = 5;// [No. of component]

+P = 1;// [No. of phases]

+s = 0;// [No special constraint]

+// Applying Eqn. 12.81

+F = C - P + 2 - r - s;// [Degree of freedom]

+printf("No. of independent reaction that occur is %d\n",r);

+printf("No. of Degree of freedom is %d",F);
\ No newline at end of file
diff --git a/647/CH12/EX12.21/Example12_21.sce b/647/CH12/EX12.21/Example12_21.sce
new file mode 100755
index 000000000..ad564f631
--- /dev/null
+++ b/647/CH12/EX12.21/Example12_21.sce
@@ -0,0 +1,18 @@
+clear;

+clc;

+

+// Example: 12.21

+// Page: 506

+

+printf("Example: 12.21 - Page: 506\n\n");

+

+// Solution

+

+// Reaction: CaCO3 -----------> CaO + CO2

+r = 1;// [No. of independent rkn.]

+C = 3;// [No. of component]

+P = 3;// [No. of phases, solid CaO, solid CaCO3, gaseous CO2]

+s = 0;// [No special constraint]

+// Applying Eqn. 12.81

+F = C - P + 2 - r - s;// [Degree of freedom]

+printf("No. of Degree of freedom is %d",F);
\ No newline at end of file
diff --git a/647/CH12/EX12.22/Example12_22.sce b/647/CH12/EX12.22/Example12_22.sce
new file mode 100755
index 000000000..2a14fa441
--- /dev/null
+++ b/647/CH12/EX12.22/Example12_22.sce
@@ -0,0 +1,76 @@
+clear;

+clc;

+

+// Example: 12.22

+// Page: 508

+

+printf("Example: 12.22 - Page: 508\n\n");

+

+// Solution

+

+//*********Data*********//

+// Reaction: 

+// C4H10 -----------> C2H4 + C2H6 ....................................(A)

+// C4H10 -----------> C3H6 + CH4 .................................... (B)

+T = 750;// [K]

+P = 1.2;// [bar]

+Ka = 3.856;

+Kb = 268.4;

+//************************//

+

+// Let

+// ea = Degree of conversion of C4H10 in reaction (A)

+// eb = Degree of conversion of C4H10 in reaction (B)

+

+// Moles in Feed:

+nC4H10_feed = 1;

+nC2H4_feed = 0;

+nC2H6_feed = 0;

+nC3H6_feed = 0;

+nCH4_feed = 0;

+

+// Moles at Equilibrium:

+// nC4H10_eqb = 1 - ea - eb

+// nC2H4_eqb = ea

+// nC2H6_eqb = ea

+// nC3H6_eqb = eb

+// nCH4_eqb = eb

+

+// Total moles at equilibrium = 1 - ea - eb + ea + eb + eb = 1 + ea + eb

+

+// Mole Fraction:

+// yC4H10_eqb = (1 - ea - eb)/(1 + ea + eb)

+// yC2H4_eqb = ea/(1 + ea + eb)

+// yC2H6_eqb = ea/(1 + ea + eb)

+// yC3H6_eqb = eb/(1 + ea + eb)

+// yCH4_eqb = eb/(1 + ea + eb)

+

+// Sum of the stoichometric coeffecient:

+va = 1 + 1 - 1;

+vb = 1 + 1 - 1;

+

+// e = [ea eb]

+// Solution of simultaneous equation

+function[f]=F(e)

+    f(1) = (((e(1)/(1 + e(1) + e(2)))*(e(1)/(1 + e(1) + e(2))))/((1 - e(1) - e(2))/(1 + e(1) + e(2))))*P^va - Ka;

+    f(2) = (((e(2)/(1 + e(1) + e(2)))*(e(2)/(1 + e(1) + e(2))))/((1 - e(1) - e(2))/(1 + e(1) + e(2))))*P^vb - Kb;

+    funcprot(0);

+endfunction

+

+// Initial guess:

+e = [0.1 0.8];

+y = fsolve(e,F);

+ea = y(1);

+eb = y(2);

+yC4H10_eqb = (1 - ea - eb)/(1 + ea + eb);

+yC2H4_eqb = ea/(1 + ea + eb);

+yC2H6_eqb = ea/(1 + ea + eb);

+yC3H6_eqb = eb/(1 + ea + eb);

+yCH4_eqb = eb/(1 + ea + eb);

+

+printf("At Equilibrium\n");

+printf("yC4H10 = %.4f\n",yC4H10_eqb);

+printf("yC2H4 = %.4f\n",yC2H4_eqb);

+printf("yC2H6 = %.4f\n",yC2H6_eqb);

+printf("yC3H6 = %.4f\n",yC3H6_eqb);

+printf("yCH4 = %.4f\n",yCH4_eqb);
\ No newline at end of file
diff --git a/647/CH12/EX12.3/Example12_3.sce b/647/CH12/EX12.3/Example12_3.sce
new file mode 100755
index 000000000..435338b00
--- /dev/null
+++ b/647/CH12/EX12.3/Example12_3.sce
@@ -0,0 +1,29 @@
+clear;

+clc;

+

+// Example: 12.3

+// Page: 479

+

+printf("Example: 12.3 - Page: 479\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: C2H5OH(g) + (1/2)O2(g) = CH3CHO(g) + H2O(g)

+Temp = 298;// [K]

+G_CH3CHO = -133.978;// [kJ]

+G_H2O = -228.60;// [kJ]

+G_C2H5OH = -174.883;// [kJ]

+R = 8.314;// [J/mol K]

+//***************//

+

+G_O2 = 0;// [kJ]

+G_rkn = G_CH3CHO + G_H2O -(G_C2H5OH + G_O2);// [kJ]

+G_rkn = G_rkn*1000;// [J]

+if G_rkn < 0 

+    printf("Reaction is feasible\n");

+    K = exp(-(G_rkn/(R*Temp)));

+    printf("Equilibium Constant is %.3e",K);

+else

+    printf("Reaction is not feasible\n");

+end
\ No newline at end of file
diff --git a/647/CH12/EX12.4/Example12_4.sce b/647/CH12/EX12.4/Example12_4.sce
new file mode 100755
index 000000000..bfaa2cfca
--- /dev/null
+++ b/647/CH12/EX12.4/Example12_4.sce
@@ -0,0 +1,24 @@
+clear;

+clc;

+

+// Example: 12.4

+// Page: 479

+

+printf("Example: 12.4 - Page: 479\n\n");

+

+// Solution

+

+//*****Data******//

+// H2O(l) = H2O(g)

+deltaH = 9710;// [cal]

+deltaS = 26;// [e.u.]

+Temp = 27 + 273;// [K]

+P = 1;// [atm]

+//**************//

+

+deltaG = deltaH - Temp*deltaS;// [cal]

+if deltaG > 0

+    printf("Vaporisation of liquid water is not spontaneous\n");

+else

+    printf("Vaporisation of liquid water is spontaneous")

+end
\ No newline at end of file
diff --git a/647/CH12/EX12.5/Example12_5.sce b/647/CH12/EX12.5/Example12_5.sce
new file mode 100755
index 000000000..e5435c7de
--- /dev/null
+++ b/647/CH12/EX12.5/Example12_5.sce
@@ -0,0 +1,24 @@
+clear;

+clc;

+

+// Example: 12.5

+// Page: 481

+

+printf("Example: 12.5 - Page: 481\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: N2(g) + 3H2(g) = 2NH3(g)

+Temp = 298;// [K]

+G_NH3 = -16.750;// [kJ/mol]

+R = 8.314;// [J/mol K]

+//***************//

+

+G_N2 = 0;// [kJ/mol]

+G_H2 = 0;// [kJ/mol]

+G_rkn = 2*G_NH3 - G_N2 - 3*G_H2;// [kJ/mol]

+printf("Standard Gibbs free energy change is %.1f kJ/mol\n",G_rkn);

+G_rkn = G_rkn*1000;// [J/mol];

+K = exp(-(G_rkn/(R*Temp)));

+printf("Equilibrium constant is %.2e",K);
\ No newline at end of file
diff --git a/647/CH12/EX12.6/Example12_6.sce b/647/CH12/EX12.6/Example12_6.sce
new file mode 100755
index 000000000..cb95afa2f
--- /dev/null
+++ b/647/CH12/EX12.6/Example12_6.sce
@@ -0,0 +1,24 @@
+clear;

+clc;

+

+// Example: 12.6

+// Page: 481

+

+printf("Example: 12.6 - Page: 481\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: CO(g) + H2O(g) = CO2(g) + H2(g)

+G_CO = -32.8;// [kcal]

+G_H2O = -54.64;// [kcal]

+G_CO2 = -94.26;// [kcal]

+Temp = 273 + 25;// [K]

+R = 1.987;// [cal/mol.K]

+//***************//

+

+G_H2 = 0;// [kcal]

+G_rkn = G_CO2 + G_H2 - (G_CO + G_H2O);// [kcal]

+G_rkn = G_rkn*1000;// [cal]

+K = exp(-(G_rkn/(R*Temp)));

+printf("Equilibrium Constant is %.3e",K);
\ No newline at end of file
diff --git a/647/CH12/EX12.7/Example12_7.sce b/647/CH12/EX12.7/Example12_7.sce
new file mode 100755
index 000000000..51916217e
--- /dev/null
+++ b/647/CH12/EX12.7/Example12_7.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+

+// Example: 12.7

+// Page: 485

+

+printf("Example: 12.7 - Page: 485\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: CO(g) + H2O(g) = CO2(g) + H2(g)

+T1 = 298;// [K]

+T2 = 1000;// [K]

+P = 1;// [bar]

+K1 = 1.1582*10^5;

+H_CO = -110.532;// [kJ]

+H_H2O = -241.997;// [kJ]

+H_CO2 = -393.978;// [kJ]

+R = 8.314;// [J/mol.K]

+//***************//

+

+H_H2 = 0;// [kJ]

+H_rkn = H_CO2 + H_H2 - (H_CO + H_H2O);// [kJ]

+H_rkn = H_rkn*1000;// [J]

+// From Van't Hoff Equation,

+K2 = K1*exp((H_rkn/R)*((1/T1) - (1/T2)));

+printf("Equilibrium constant at 1000 K is %.4f",K2);
\ No newline at end of file
diff --git a/647/CH12/EX12.8/Example12_8.sce b/647/CH12/EX12.8/Example12_8.sce
new file mode 100755
index 000000000..614127b0c
--- /dev/null
+++ b/647/CH12/EX12.8/Example12_8.sce
@@ -0,0 +1,30 @@
+clear;

+clc;

+

+// Example: 12.8

+// Page: 485

+

+printf("Example: 12.8 - Page: 485\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: N2(g) + 3H2(g) = 2NH3(g)

+T1 = 298;// [K]

+T2 = 700;// [K]

+H_NH3 = -46.1;// [kJ]

+G_NH3 = -16.747;// [kJ]

+R = 8.314;// [J/mol.K]

+//**************//

+

+H_N2 = 0;// [kJ]

+H_H2 = 0;// [kJ]

+G_N2 = 0;// [kJ]

+G_H2 = 0;// [kJ]

+H_rkn = 2*H_NH3 - (H_N2 + 3*H_H2);// [kJ]

+G_rkn = 2*G_NH3 - (G_N2 + 3*G_H2);// [kJ]

+H_rkn = H_rkn*1000;// [J]

+G_rkn = G_rkn*1000;// [J]

+K1 = exp(-(G_rkn/(R*T1)));

+K2 = K1*exp((H_rkn/R)*((1/T1) - (1/T2)));

+printf("Equilibrium constant at 700 K is %.4e",K2);
\ No newline at end of file
diff --git a/647/CH12/EX12.9/Example12_9.sce b/647/CH12/EX12.9/Example12_9.sce
new file mode 100755
index 000000000..0cdd2d1e7
--- /dev/null
+++ b/647/CH12/EX12.9/Example12_9.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+

+// Example: 12.9

+// Page: 486

+

+printf("Example: 12.9 - Page: 486\n\n");

+

+// Solution

+

+//*****Data******//

+// Reaction: CO(g) + H2O(g) ----> CO2(g) + H2(g)

+T1 = 298;// [K]

+T2 = 1000;// [K]

+deltaH_298 = -41450;// [J/mol]

+deltaGf_298 = -28888;// [J/mol]

+alpha_CO2 = 45.369;// [kJ/kmol K]

+alpha_H2 = 27.012;// [kJ/kmol K]

+alpha_CO = 28.068;// [kJ/kmol K]

+alpha_H2O = 28.850;// [kJ/kmol K]

+beeta_CO2 = 8.688*10^(-3);// [kJ/kmol square K]

+beeta_H2 = 3.509*10^(-3);// [kJ/kmol square K]

+beeta_CO = 4.631*10^(-3);// [kJ/kmol square K]

+beeta_H2O = 12.055*10^(-3);// [kJ/kmol square K]

+R = 8.314;// [J/mol K]

+//*************//

+

+delta_alpha = alpha_CO2 + alpha_H2 - (alpha_CO + alpha_H2O);

+delta_beeta = beeta_CO2 + beeta_H2 - (beeta_CO + beeta_H2O);

+// To obtain the standard heat of reaction:

+deltaH_0 = deltaH_298 - (delta_alpha*T1 + (delta_beeta*T1^2)/2);// [kJ/mol]

+// From Eqn. 12.52:

+IR = (deltaH_0 - delta_alpha*T1*log(T1) - (delta_beeta*T1^2)/2 - deltaGf_298)/T1;// [kJ/mol K]

+// Substituting T = T2 and IR in Eqn. 12.51:

+deltaG_1000 = deltaH_0 - delta_alpha*T2*log(T2) - (delta_beeta*T2^2)/2 - IR*T2;// [kJ/mol]

+printf("Standard Gibbs free energy at 1000 K %.3f kJ\n",deltaG_1000/1000);

+

+// Standard Equilibrium Constant at 1000 K

+K_1000 = exp(-(deltaG_1000)/(R*T2));

+printf("Standard Equilibrium Constant is %.1f",K_1000);
\ No newline at end of file