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+// Example 15_14
+clc;funcprot(0);
+// Given data
+p=0.100;// MPa
+T_a=298;// K
+T_b=2000;// K
+R=0.0083143;// MJ/kgmole.K 
+
+// Calculation
+// (a)
+gbar0_f_H2O=-228.583;// kJ/kgmole
+// since H2 and O2 are elements, their molar specific Gibbs function of formation is zero. Then, from Table 15.7, 
+gbar0_f_H2=0;// kJ/kgmole
+gbar0_f_O2=0;// kJ/kgmole
+K_e=exp(gbar0_f_H2O/(R*T_a));// The equilibrium constant
+printf("\n(a)The equilibrium constant,K_e=%1.2e",K_e);
+// (b)
+T_b_R=T_b*1.8;// R
+// Eq. (15.34) with Tables 15.7 and C.16c in Thermodynamic Tables to accompany Modern Engineering Thermodynamics give
+h_a_H2O=4258.3;// Btu/lbmole
+h_b_H2O=35540.1;// Btu/lbmole
+h_a_H2=3640.3;// Btu/lbmole
+h_b_H2=26398.5;// Btu/lbmole
+h_a_O2=3725.1;// Btu/lbmole
+h_b_O2=29173.5;// Btu/lbmole
+s_a_H2O=188.833;// kJ/(kgmole.K)
+s_b_H2O=63.221;// Btu/(lbmole.R)
+s_a_H2=130.684;// kJ/(kgmole.K)
+s_b_H2=44.978;// Btu/(lbmole.R)
+s_a_O2=205.138;// kJ/(kgmole.K)
+s_b_O2=64.168;// Btu/(lbmole.R)
+// Note: The multipliers 2.3258 and 4.1865 in these equations are necessary to convert the Btu/lbmole and Btu/(lbmole.R) values in Table C.16c into kJ/kgmole and kJ/(kgmole.K), respectively.
+gbar_f_H2O=(gbar0_f_H2O*10^3)+((h_b_H2O-h_a_H2O)*2.3258)-[((T_b*s_b_H2O)*4.1865)-(T_a*s_a_H2O)];// kJ/kgmole
+gbar_f_H2=gbar0_f_H2+((h_b_H2-h_a_H2)*2.3258)-[((T_b*s_b_H2)*4.1865)-(T_a*s_a_H2)];// kJ/kgmole
+gbar_f_O2=gbar0_f_O2+((h_b_O2-h_a_O2)*2.3258)-[((T_b*s_b_O2)*4.1865)-(T_a*s_a_O2)];// kJ/kgmole
+K_e=exp([gbar_f_H2O-gbar_f_H2-((1/2)*gbar_f_O2)]/(R*10^3*T_b));// The equilibrium constant
+printf("\n(b)The equilibrium constant,K_e=%1.2e",K_e);