Question

Calculate ?H^o,?S⁰,?G⁰ ( 25.0^oC)for the formation of sulfur trioxide from oxygen gas and sulfur dioxide ;

Wtite balance equation and show all units, substances in your equation.

Substance	Enthalpy value (Kj/mole)	Entropy value(J/K-mole)
Sulfur dioxide gas	-300	250
Sulfur trioxide gas	-395	260
Oxygen gas	0	210

Answer 1

The balanced reaction for the formation of sulfur trioxide from oxygen gas and sulfur dioxide can be written as

2SO2(g) + O2(g) --------------> 2SO3(g)

DeltaH0(rxn) for the above reaction can be calculated as

DeltaH0(rxn) = H0(products) - H0(reactants)

=  2xH0(SO3, g) - [ 2xH0(SO2, g) + H0(O2, g)]

= 2x(-395KJ/mol) - [2x(-300KJ/mol) +0]

=> DeltaH0(rxn) = - 190KJ/mol

DeltaS0(rxn) for the balanced reaction for the formation of sulfur trioxide from oxygen gas and sulfur dioxide can be calculated as

2SO2(g) + O2(g) --------------> 2SO3(g)

DeltaS0(rxn) = S0(products) - S0(reactants)

=  2xS0(SO3, g) - [ 2xS0(SO2, g) + S0(O2, g)]

= 2x(260J/Kmol) - [2x(250J/Kmol) +210J/Kmol]

=> DeltaS0(rxn) = -190 J/Kmol = - 0.190KJ/Kmol

Now DeltaG0(rxn) for the balanced reaction

2SO2(g) + O2(g) --------------> 2SO3(g)

can be calculated as

DeltaG0(rxn) = DeltaH0(rxn) - TxDeltaS0(rxn)

=  - 190KJ/mol - 298Kx(- 0.190KJ/Kmol)

= - 133.4 KJ