Entropically or Enthelpically favorable?

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SUMMARY

The discussion centers on the thermodynamic evaluation of the reaction 6CO2 + 6 H2O → C6H12O6 + 6 O2. Participants conclude that the reaction is entropically and enthalpically unfavorable due to the increase in order from the synthesis of glucose and the absence of heat generation. Additionally, the presence of more moles of gas on the reactant side and the classification of the reaction as a reverse combustion reaction, which is typically exothermic, further supports this conclusion.

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Homework Statement


6CO2 + 6 H2O → C6H12O6 + 6 O2



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The Attempt at a Solution


I believe that this is entropically and enthalpically unfavorable because the reaction is going to more order (synthesis of glucose) and heat is not being generated.

Is my logic correct? Thanks anyone for any help you can give me.
 
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Somewhat ... you've got more molar gas on the left , in addition this is a reverse combustion reaction , you should know that it is combustion reactions that are exothermic.
 

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