- #1
MathewsMD
- 433
- 7
If ##ΔG_{sys} = ΔH_{sys} - TΔS_{sys}; ΔS_{sys} = \frac {-ΔH_{surr}}{T} ##
Then, doesn't this expression just simplify to:
##ΔG_{sys} = ΔH_{sys} + ΔH_{surr}## and isn't ##ΔH_{sys} = -ΔH_{surr}##?
So then ##ΔG = 0##...this does not seem correct...could anyone please clarify my mistake and the formal proof of free energy if it differs from the one I have shown?
Also, what is the explanation for why there is no free energy change during phase changes? I was merely told this without any explanation and one would be very helpful.
Finally, just a related question:
If you have 1 gram of ammonia and it dissolves in 50 grams of water to release 1000 J (just hypothetical). When calculating the temperature change of the water, using q = mcΔT, is m = 50g or 51g since the ammonia is now aqueous and inseparable from the water. Yet, it has slightly different properties now. I am unsure if it is treated just like a normal sample of water or not.
Then, doesn't this expression just simplify to:
##ΔG_{sys} = ΔH_{sys} + ΔH_{surr}## and isn't ##ΔH_{sys} = -ΔH_{surr}##?
So then ##ΔG = 0##...this does not seem correct...could anyone please clarify my mistake and the formal proof of free energy if it differs from the one I have shown?
Also, what is the explanation for why there is no free energy change during phase changes? I was merely told this without any explanation and one would be very helpful.
Finally, just a related question:
If you have 1 gram of ammonia and it dissolves in 50 grams of water to release 1000 J (just hypothetical). When calculating the temperature change of the water, using q = mcΔT, is m = 50g or 51g since the ammonia is now aqueous and inseparable from the water. Yet, it has slightly different properties now. I am unsure if it is treated just like a normal sample of water or not.
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