2 SO2(g) + O2(g) 2 SO3(g) G° = -136.5 kJ and S° = -187.9 J/K at 326 K and 1

Chemistry
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SO2(g) O2(g)  SO3(g) 


G° = -136.5 kJ and S° = -187.9 J/K at 326 K and 1 atm.

This reaction is (reactant, product)  favored under standard conditions at 326 K.

The standard enthalpy change for the reaction of 2.15 moles of SO2(g) at this temperature would be  kJ.
Jun 2nd, 2015

Please let me know if you need any clarification. I'm always happy to answer your questions.

delta G is negative  so this is product favored reaction

G=H-TS

136.5=deltaH*-(-187.9)

deltaH=136.5-187.9

=-51.4 KJ/mol

so enthlphy change for 2.15mols=-51.4*2.15

=110.51 KJ



Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 3rd, 2015

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