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+//ques-18.30
+//Calculating equilibrium constant for the given reaction
+clc
+T=278;//temperature (in K)
+S=-333.3;//entropy change (in J/K/mol)
+R=8.31;// J/K/mol
+h1=-110.5;//heat of formation for CO (in kJ/mol)
+h2=-74.8;//heat of formation for methane (in kJ/mol)
+h3=-286;//heat of formation for water (in kJ/mol)
+H=h2+h3-h1;//enthalpy change (in kJ/mol)
+G=(H*1000)-T*S;//free energy change (in kJ/mol)
+//On solving, log10(K) = -G/(2.303*R*T)
+K=2.97;//(*10^26)
+printf("The equilibrium constant required is %.2f*10^26.",K);