Arkansas State University Department of Chemistry and Physics

Using the tables on the web, calculate DS^o and DG^o  for the following reactions using:
DS^o  = SS^o_products - SS^o_reactants  and

DG^o_rxn =SDG^o_products - SDG^o_reactants
 

1.   CH_{4 (g)}   +    2 O_{2 (g)}   --------->     CO_{2 (g)}    +   2 H₂O _(l)

a)  DS^o  = SS^o_products - SS^o_reactants
 

Soproducts	Soreactants
(1 mol CO2)(213.7 J/Kmol) = 213.7 J/K	(1 mol CH4)(186.1 J/Kmol) = 186.1 J/K
(2 mol H2O)(69.940 J/Kmol) = 139.88 J/K	(2 mol O2)(205 J/Kmol) = 410 J/K
SSoproducts = 353.58 J/K	SSoreactants = 596.1 J/K
DSo = 353.58 J/K - 596.1 J/K = -242.52 J/K

b)   DG^o_rxn=SDG^o_products - SDG^o_reactants
 

DGoproducts	DGoreactants
(1 mol CO2)(-394.4 kJ/mol) = -394.4 kJ	(1 mol CH4)(-50.81 kJ/mol) = -50.81 kJ
(2 mol H2O)(-237.192 J/Kmol) = -474.384 kJ	(2 mol O2)(0 kJ/mol) = 0 kJ
SDGoproducts = -868.784 kJ	SDGoreactants = -50.81 kJ
DGorxn= -868.784 kJ - (-50.81 kJ) = -817.974 kJ	spontaneous reaction

2.   2 MgO _(s)  -------->      2 Mg _(s)   +  O_{2 (g)}

a)  DS^o  = SS^o_products - SS^o_reactants
 

Soproducts	Soreactants
(2 mol Mg)(32.69 J/Kmol)  = 65.38J/K	(2 mol MgO)(26.9 J/Kmol) = 53.8 J/K
(1 mol O2)(205 J/Kmol)  = 205 J/K
SSoproducts = 270.38 J/K	SSoreactants = 53.8 J/K
DSo = 270.38 J/K - 53.8 J/K = 216.58 J/K

b)   DG^o_rxn=SDG^o_products - SDG^o_reactants
 

DGoproducts	DGoreactants
(2 mol Mg)( 0 kJ/mol)  = 0 kJ	(2 mol MgO)(-569.0 kJ/mol) = -1138.0 kJ
(1 mol O2)( 0 kJ/mol)  = 0 kJ
SDGoproducts = 0 kJ	SDGoreactants = -1138.0 kJ
DGorxn= 0 kJ - (-1138.0 kJ) = +1138.0 kJ	nonspontaneous reaction