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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+
+// Example: 12.8
+// Page: 485
+
+printf("Example: 12.8 - Page: 485\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: N2(g) + 3H2(g) = 2NH3(g)
+T1 = 298;// [K]
+T2 = 700;// [K]
+H_NH3 = -46.1;// [kJ]
+G_NH3 = -16.747;// [kJ]
+R = 8.314;// [J/mol.K]
+//**************//
+
+H_N2 = 0;// [kJ]
+H_H2 = 0;// [kJ]
+G_N2 = 0;// [kJ]
+G_H2 = 0;// [kJ]
+H_rkn = 2*H_NH3 - (H_N2 + 3*H_H2);// [kJ]
+G_rkn = 2*G_NH3 - (G_N2 + 3*G_H2);// [kJ]
+H_rkn = H_rkn*1000;// [J]
+G_rkn = G_rkn*1000;// [J]
+K1 = exp(-(G_rkn/(R*T1)));
+K2 = K1*exp((H_rkn/R)*((1/T1) - (1/T2)));
+printf("Equilibrium constant at 700 K is %.4e",K2); \ No newline at end of file