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+clear;
+//clc();
+
+// Example 4.3
+// Page: 80
+printf("Example-4.3 Page no.-80\n\n");
+
+//***Data***//
+// We have the system which consists of isobutane and normal butane and isomerisaation is taking place between them
+// The equilibrium constant for this reaction is given by
+// K = (mole fraction of isobutane)/(mole fraction of n-butane) = x_iso/x_normal
+
+// For this reaction, at 25C,
+K = 4.52;
+
+// and
+// x_iso + x_normal = 1
+// so
+// K = x_iso/(1-x_iso)
+
+// solving for x_iso
+deff('[y]=f(x_iso)','y = x_iso/(1-x_iso)-K');
+x_iso = fsolve(0,f);
+
+printf(" Mole fraction of isobutane isomer in equilibrium is %0.2f",x_iso);