Predict the sign for delta Ssys CH3CH2CH3(g) +5O2(g)-> 3CO2(g) + 4H2

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SUMMARY

The discussion focuses on predicting the sign of the change in entropy (ΔSsys) for two chemical reactions: the combustion of propane (CH3CH2CH3) with oxygen (5O2) producing carbon dioxide (3CO2) and water (4H2O), and the decomposition of dinitrogen pentoxide (N2O5) into nitrogen dioxide (NO2) and nitrate (NO3). The consensus is that the ΔSsys for the combustion reaction is negative due to the formation of more stable products, while the ΔSsys for the decomposition reaction is positive because it results in an increase in the number of gaseous molecules, thus increasing entropy.

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predict the sign for delta Ssys CH3CH2CH3(g) +5O2(g)---> 3CO2(g) + 4H2

Homework Statement



predict the sign for delta Ssys CH3CH2CH3(g) +5O2(g)---> 3CO2(g) + 4H20(l)


The same question for N205(g) --> NO2(g) + NO3(g)

Homework Equations





The Attempt at a Solution



I said the first was negative because it is combustion rxn and is more stable after the rxn.

The second one I said positive because now you have more molecules?

Good or no? Thanks
 
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You're looking for a change in entropy. In general, the system with a larger number of moles of gaseous molecules has the higher entropy.
 

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