Entropy generation in chemical reactions

AI Thread Summary
In the discussion on entropy generation in chemical reactions, the first and second laws of thermodynamics are applied to derive the relationship between internal energy change and entropy generation. It is clarified that constant entropy does not imply no heat flow into or out of the system. The conversation also addresses the confusion surrounding negative changes in internal energy (dU<0) and the implications for temperature, noting that positive entropy generation can occur alongside a decrease in temperature. An example involving the mixing of salt and ice illustrates how entropy and temperature interact in a practical scenario. Overall, the discussion emphasizes the complexities of thermodynamic principles in chemical reactions.
Urmi Roy
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So from the first law for a closed system,

dU=dQ-dW=dQ-PdV

From the second law,

dS=dQ/T + Sgenerated (i.e. the entropy generated)

Putting expression of dQ from second law into first law,
dU=T*dS-T*Sgen-PdV

If s and v are constant,
dU= -T*Sgen>0

Hence dU<0
This is a derivation that was given in class

My questions are as follows:
1. When we say s is constant, does it mean there is not heat flow into/out of the system?

2.If dU<0, does the temperature decrease? I find it hard to understand how there can be a positive Sgen and decrease in temperature/ internal energy simultaneously!
 
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Urmi Roy said:
My questions are as follows:
1. When we say s is constant, does it mean there is not heat flow into/out of the system?

2.If dU<0, does the temperature decrease? I find it hard to understand how there can be a positive Sgen and decrease in temperature/ internal energy simultaneously!

1. No, you can imagine that dQ=Sgen/T
2. What happens if you mix salt and ice in a thermo jar?
 
About your Q2...not too sure, but I know that having a salt in water lowers the freezing point of water...So the ice melts back?
 
Yes, and gets cooler!
 
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