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diff --git a/3718/CH5/EX5.3/Ex5_3.sce b/3718/CH5/EX5.3/Ex5_3.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 3
+clc;
+
+//Declaration of Variables
+t0 = 37.0                   //in cm cube of KMnO4
+t5 = 29.8                   //in cm cube of KMnO4
+t15 = 19.6                  //in cm cube of KMnO4
+t25 = 12.3                  //in cm cube of KMnO4
+t45 = 5.00                  //in cm cube of KMnO4
+
+// Solution
+K5 = 2.303 / 5 * log10(t0 / t5)
+K15 = 2.303 / 15 * log10(t0 / t15)
+K25 = 2.303 / 25 * log10(t0 / t25)
+K45 = 2.303 / 45 * log10(t0 / t45)
+
+mprintf("At t = 5 min, K = %.3e /min\n",K5)
+mprintf(" At t = 15 min, K = %.3e /min\n",K15)
+mprintf(" At t = 25 min, K = %.3e /min\n",K25)
+mprintf(" At t = 45 min, K = %.3e /min\n",K45)
+mprintf(" As the different values of K are nearly same, the reaction is of first-order\n")
+K = (K45 + K25 + K5 + K15) / 4
+mprintf(" The average value of K = %.3e /min",K)