Asked by K
So, i did this lab, a solubility and thermodynamics lab of KNO3, and in it we had to determine the delta h delta g and delta s, the question i have is what is the relationship of the signs delta h, delta s, and delta g, to the dissociation of the salt and the temperature changes? The temperature decreased, and the delta H was +, and the delta s was +, however like in trial 1 my delta s was 117.2, then in trial 2 my delta s went to 115.6, then in trial 3 it went back up to 16.4 ( i am a little confused on why it did this), and the delta g was -, i am having the most trouble on establishing the relationship between the signs of delta h, s, ang g to the dissociation of the salt.
Thanks for your help explaining
Thanks for your help explaining
Answers
Answered by
K
it went back up to 116.4*
Answered by
DrBob222
I don't think I will answer your question completely but perhaps somewhat anyway.
When KNO3 dissolves in water the water gets colder. That means that the sum of the dissociation of KNO3 (dH is + since it takes energy to pull that crystal apart) and the solvation energy of the ions (dH is - because energy is released) is +. That just means that the dissociation of the crystal takes more energy that is gained by the solvation of the ions. dS is + and it follows the usual scheme that dS is positive when salts are dissolved and the disorder is more in the aqueous state than in the solid state. dG really is the sum of dH and dS from dG = dH - TdS. I hope this helps.
When KNO3 dissolves in water the water gets colder. That means that the sum of the dissociation of KNO3 (dH is + since it takes energy to pull that crystal apart) and the solvation energy of the ions (dH is - because energy is released) is +. That just means that the dissociation of the crystal takes more energy that is gained by the solvation of the ions. dS is + and it follows the usual scheme that dS is positive when salts are dissolved and the disorder is more in the aqueous state than in the solid state. dG really is the sum of dH and dS from dG = dH - TdS. I hope this helps.
Answered by
K
yes I understand, and this helps a lot, but what I don't understand, is why in one trial the delta s was higher than in another trial, could you explain why that happened?
Answered by
DrBob222
I don't know enough about what you did to answer; however, my first educated guesses are (1) the procedure is no more precise than that or (2) you're still learning the procedure.
Answered by
K
the procedure is no more precise really, I understand the lab, if i can ask a different question then, when delta g is negative and becoming less negative (or more positive), why is this happening over a decrease in temperature (increase in volume)
Answered by
DrBob222
dG = dH - TdS
So if dG is more + it must mean the cost of breaking the crystal structure is more than the gain in the entropy by solid to solution; i.e. dH + and dS + and dH>-TdS
So if dG is more + it must mean the cost of breaking the crystal structure is more than the gain in the entropy by solid to solution; i.e. dH + and dS + and dH>-TdS
Answered by
K
thanks for your help :)
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