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+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