Phase Change

[G1SG2]

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If I'm told in a question that I have ice at 0 degrees celsius or water at 100 degrees celsius, and that I need to calculate the heat change, can I just use q=mL to calculate the heat for the phase change, or do I need to do q=mL AND q=mcdeltat? If I'm told that I already have ice at 0 degrees celsius or water at 100 degrees celsius, can I ignore q=mcdeltat, since that equation just tells me the amount of heat that needs to be added to raise the temperature to that substance's MP or BP?

 

[loveoforganic]

Yes. If you use it, you'd be plugging in a dT of 0, and it would cancel out.

 

[G1SG2]

Yes. If you use it, you'd be plugging in a dT of 0, and it would cancel out.

Okay, thanks.

 

[G1SG2]

For positive deviations from Raoult's law-we would have unfavorable interactions with the solvent and solute, and the vapor pressure would be higher than normal, right? Since more molecules from the solution would be trying to escape into the gas phase, and so delta H would be positive?

 

[sv3]

For positive deviations from Raoult's law-we would have unfavorable interactions with the solvent and solute, and the vapor pressure would be higher than normal, right? Since more molecules from the solution would be trying to escape into the gas phase, and so delta H would be positive?

I know your right about Raoults law. I am lost as far as the delta H thing but everything you said about Raoult is correct if that helps

 

[loveoforganic]

For positive deviations from Raoult's law-we would have unfavorable interactions with the solvent and solute, and the vapor pressure would be higher than normal, right? Since more molecules from the solution would be trying to escape into the gas phase, and so delta H would be positive?

I think that's all correct. And the factor driving solvation would be the increase in entropy.

 

[G1SG2]

I think that's all correct. And the factor driving solvation would be the increase in entropy.

Okay, so the delta H would be positive, but what about delta G? I mean wouldn't delta G initially be negative since we have solvation (the particles are breaking apart and are dissolving into the solution)? I'm a bit confused with regards to the sign conventions.

 

[loveoforganic]

It depends on the temperature. If the particles are dissolving though, dG is negative.

 

[G1SG2]

It depends on the temperature. If the particles are dissolving though, dG is negative.

But delta H would be positive because the interactions are unfavorable? Or is the solvation unrelated to the interactions between the solvent and solute?

 

[loveoforganic]

Ok, here goes 😛

You have a flask of compound A and a flask of compound B. Compound A makes strong intermolecular bonds with itself, and compound B makes strong intermolecular bonds with itself. However, compound A makes weak bonds with compound B. Assuming they are miscible with one another, if you mix compound A and compound B, you are breaking the strong intermolecular bonds of the two compounds with themselves (A-A / B-B) and forming weak intermolecular bonds between the two compounds (A-B). When you break a strong bond and form a weaker bond, heat is overall absorbed (endothermic / positive dH). However, since these compounds are dissolving one another, something has to be driving the spontanaeity of the reaction (the negative dG). This factor is entropy. There is more randomness when the two compounds are mixed with one another than isolated in their individual components (ABABABAB vs AAAABBBB). By dG = dH - TdS, when dH is positive and dS is positive, whether the reaction (solvation) is spontaneous (negative dG) is determined by the temperature of the solution. As temperature increases, the TdS factor increases until it eventually outweighs the dH component.

 

[G1SG2]

Ok, here goes 😛

You have a flask of compound A and a flask of compound B. Compound A makes strong intermolecular bonds with itself, and compound B makes strong intermolecular bonds with itself. However, compound A makes weak bonds with compound B. Assuming they are miscible with one another, if you mix compound A and compound B, you are breaking the strong intermolecular bonds of the two compounds with themselves (A-A / B-B) and forming weak intermolecular bonds between the two compounds (A-B). When you break a strong bond and form a weaker bond, heat is overall absorbed (endothermic / positive dH). However, since these compounds are dissolving one another, something has to be driving the spontanaeity of the reaction (the negative dG). This factor is entropy. There is more randomness when the two compounds are mixed with one another than isolated in their individual components (ABABABAB vs AAAABBBB). By dG = dH - TdS, when dH is positive and dS is positive, whether the reaction (solvation) is spontaneous (negative dG) is determined by the temperature of the solution. As temperature increases, the TdS factor increases until it eventually outweighs the dH component.

Thanks a lot! I thought they would've been related from the delta G equation but wasn't sure 👍