Phase Change & Enthalpy of _____

[MedPR]

http://en.wikibooks.org/wiki/File:Phase_Heat_Diagram.png

So in this graph the horizontal lines are the phase changes because during a phase change there is no temperature change because all the heat being added is going into breaking bonds, correct?

Since this is a graph of water, the first horizontal line is for T=0C, and the second horizontal line is for T=100C assuming 1atm of pressure. Also, in order to calculate the amount of heat at T=0, we would need to know what the starting temperature is (at the origin) right? Then we could use q=m*c*dT since we know there is 1mol of water and c=4.187kJ/mol?

Also, this is completely idealized right? In other words, in a real situation the temperature during a phase change will increase slightly since the phase change happens at different times for different molecules, but in these graphs we are kind of assuming that all molecules change phase uniformly (where the average temp equals the BP/MP/FP/whatever point). So on a graph where the phase change is completely horizontal, it's easy to give a value to the enthalpy of fusion and enthalpy of vaporization, since they are just the temperature at which the horizontal line begins and/or ends.

If the line was slightly sloped, would the enthalpy of the reaction be the average temperature along that line? Would it be the start of the phase change (so the lowest temperature on the line)? Or would it be the highest temperature just before the phase change ended at the temperature of the system started to rapidly increase again (the endpoint of the line)?

Edit: I realize the second part (about the non-perfectly horizontal lines) might be confusing so I drew a graph.

Which arrow would represent the heat of fusion and which would represent the heat of vaporization?

---

Members don't see this ad.

 

[milski]

Yes on all of these. You may want to check c for solid/liquid/gaseous water - it's most likely different. To all mcΔt you'll have to add the heat of fusion and heat of evaporation - the two flat areas in the graph.

Ok, you aded more questions after I replied. The water can exist as both solid/liquid and liquid/gas at the transition temperatures. For infinitely small amount of water the graph will be horizontal. As you get more water, you run into problems with distributing the heat uniformly - some parts will get heated up and melted earlier and will start warming up before others are completely melted. At that point it really depends what you want to graph - the average temp of the whole system? Lowest temp? In other words, there is no need to worry about the case where the line is not horizontal.

 

[MedPR]

Yes on all of these. You may want to check c for solid/liquid/gaseous water - it's most likely different. To all mcΔt you'll have to add the heat of fusion and heat of evaporation - the two flat areas in the graph.

What do you mean by that last sentence?

 

[milski]

What do you mean by that last sentence?

The total heat to get from the leftmost part of the graph to the rightmost is the heat that you need to change the temp (the steep parts of the curve) + the heat that you need to change state (the two flat areas). The first is mcΔt, the second is a mX where X is some sort of book value.

 

[MedPR]

The total heat to get from the leftmost part of the graph to the rightmost is the heat that you need to change the temp (the steep parts of the curve) + the heat that you need to change state (the two flat areas). The first is mcΔt, the second is a mX where X is some sort of book value.

Oh ok, I think I know what you mean.

You would have to do two q= mcΔt calculations and just add them together since the c will be different for different phase changes.

More heat is required for fusion than vaporization though, right? Because of lattice interactions?

 

[milski]

Oh ok, I think I know what you mean.

You would have to do two q= mcΔt calculations and just add them together since the c will be different for different phase changes.

More heat is required for fusion than vaporization though, right? Because of lattice interactions?

You need three mcΔt - from -X to 0, from 0 to 100, and from 100 to Y. These are the sloped parts of the graph.

Heat of fusion/evaporation will be the level parts - these are given as book values per mole or per kg. So for them you'll have mHf and mHe.

The total heat for the whole process will be mc1Δt1 + mHf + mc2Δt2 + mHe + mc3Δt3. For what it's worth, Δt2=100-0=0.

Heat of fusion for water is about 10 times smaller than heat of evaporation (2260 J/g vs 334 J/g). You have to break a lot of H-O bonds to make water evaporate.

 

[MedPR]

You need three mcΔt - from -X to 0, from 0 to 100, and from 100 to Y. These are the sloped parts of the graph.

Heat of fusion/evaporation will be the level parts - these are given as book values per mole or per kg. So for them you'll have mHf and mHe.

The total heat for the whole process will be mc1Δt1 + mHf + mc2Δt2 + mHe + mc3Δt3. For what it's worth, Δt2=100-0=0.

Heat of fusion for water is about 10 times smaller than heat of evaporation (2260 J/g vs 334 J/g). You have to break a lot of H-O bonds to make water evaporate.

Ah that makes sense. I'm confusing myself because I made this thread with one graph while I'm watching the lecture with the graph flipped.

In this graph, the q value would just be the enthalpy value right? But since they give all the enthalpy values on the y axis, we wouldn't have to do any calculations to find the overall enthalpy change, correct? We might be asked to find the q for the periods between the phase changes (which are the somewhat horizontal lines on this graph) and for that we would use q=mcΔt, where c would be for solid, liquid, or gas depending on which part of the graph we're looking at.

 

[milski]

Ah that makes sense. I'm confusing myself because I made this thread with one graph while I'm watching the lecture with the graph flipped.

In this graph, the q value would just be the enthalpy value right? But since they give all the enthalpy values on the y axis, we wouldn't have to do any calculations to find the overall enthalpy change, correct? We might be asked to find the q for the periods between the phase changes (which are the somewhat horizontal lines on this graph) and for that we would use q=mcΔt, where c would be for solid, liquid, or gas depending on which part of the graph we're looking at.

That sounds right. Note that the y axis is in J/kg, so you may have to multiply by some factor if the question is talking about different amount of water.

 

[MedPR]

That sounds right. Note that the y axis is in J/kg, so you may have to multiply by some factor if the question is talking about different amount of water.

Ok thanks. Also, he says that the heat of vaporization has a much greater change in electrostatic potential energy than does the heat of fusion.

I understand that P=kq1q2/r and that when r increases, P will decrease (become less negative & approach 0) when q1 and q2 have opposite charges. So if P increases in going from liquid to gas, doesn't that mean that gas particles have opposite charges? How does that work out?

 

[milski]

Ok thanks. Also, he says that the heat of vaporization has a much greater change in electrostatic potential energy than does the heat of fusion.

I understand that P=kq1q2/r and that when r increases, P will decrease (become less negative & approach 0) when q1 and q2 have opposite charges. So if P increases in going from liquid to gas, doesn't that mean that gas particles have opposite charges? How does that work out?

I'm not sure what he means by that. My only is guess is that the transition from liquid to gas increases the distance between the particles much more than the transition from solid to liquid. That will make any electrostatic forces between them much smaller.

 

[MedPR]

I'm not sure what he means by that. My only is guess is that the transition from liquid to gas increases the distance between the particles much more than the transition from solid to liquid. That will make any electrostatic forces between them much smaller.

Maybe that's what he meant.. that the electrostatic forces decrease much more in vaporization than they do in fusion.