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diff --git a/647/CH12/EX12.14/Example12_14.sce b/647/CH12/EX12.14/Example12_14.sce
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+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