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

Question Description

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.

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Final Answer

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.

sachintha (127)
Cornell University

Anonymous
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