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diff --git a/389/CH2/EX2.1/Example2_1.sce b/389/CH2/EX2.1/Example2_1.sce
new file mode 100755
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+++ b/389/CH2/EX2.1/Example2_1.sce
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+clear;
+clc;
+
+// Illustration 2.1
+// Page: 30
+
+printf('Illustration 2.1 - Page 30\n\n');
+
+// solution
+
+//***Data***//
+// a = O2 & b = CO
+Dab = 1.87*10^(-5);//[square m/s]
+Pt = 10^5;//[N/square m]
+z = 0.002;//[m]
+R = 8314;//[Nm/kmol]
+T = 273;//[K]
+Pa1 = 13*10^(3);//[N/square m]
+Pb1 = 10^(5)-13*10^(3);//[N/square m]
+Pa2 = 6500;//[N/square m]
+Pb2 = 10^(5)-6500;//[N/square m]
+//********//
+
+// Calculation from Eqn. 2.30
+Pbm = (Pb1-Pb2)/log(Pb1/Pb2);//[N/square m]
+Na = Dab*Pt*(Pa1-Pa2)/(R*T*z*Pbm);//[kmol/square m.s]
+printf('Rate of diffusion of oxygen is %e kmol/square m.sec',Na);
+\ No newline at end of file