Raoult's Law Deviation and Temperature

[matth87]

If the vapor pressure of a solution is higher than predicted by Raoult's law, then the temperature of the solvents when put into solution will:

A. increase due to the energy absorbed by the breaking of bonds.
B. increase due to the energy released by the formation of bonds.
C. decrease due to the energy absorbed by the breaking of bonds.
D. decrease due to the energy released by the formation of bonds.

If its higher then predicted that means bonds were broken and we now have weaker bonds then before, and (delta)H would be positive because we are breaking bonds. This eliminates choice B and D since bonds are broken not formed.

Now I had to find out what happens to the temperate of the solvents. I reasoned that they would absorb more energy because (delta)H = positive and since energy is proportional to temperature I thought that their temp would increase. I choose A.

The correct answer listed in C, the temperate of the solvents will decrease.

This is the solution given.

Solution
Since Ptotal is higher than predicted by Raoult's Law, we must consider the second graph in Figure 1, which illustrates the case where (delta)Hs > 0. Since the mixture is an endothermic reaction, the result will be a decrease in the temperature of the solution.

The answer says the result will be a decrease in the temp of the solution... Is this the same as a decrease in the temp of the solvents? I still dont really understand why the temp would decrease and not increase if delta h is positive.

I could see how maybe the energy is coming from the solution to break the bonds but the question is asking about the temperature of the solvents. Would these be the same or diff then the solution?

Any help would be awesome! Thanks.

---

Members don't see this ad.

 

[wanderer]

If the vapor pressure of a solution is higher than predicted by Raoult’s law, then the temperature of the solvents when put into solution will:

A. increase due to the energy absorbed by the breaking of bonds.
B. increase due to the energy released by the formation of bonds.
C. decrease due to the energy absorbed by the breaking of bonds.
D. decrease due to the energy released by the formation of bonds.

If its higher then predicted that means bonds were broken and we now have weaker bonds then before, and (delta)H would be positive because we are breaking bonds. This eliminates choice B and D since bonds are broken not formed.

Now I had to find out what happens to the temperate of the solvents. I reasoned that they would absorb more energy because (delta)H = positive and since energy is proportional to temperature I thought that their temp would increase. I choose A.

The correct answer listed in C, the temperate of the solvents will decrease.

This is the solution given.

Solution
Since Ptotal is higher than predicted by Raoult’s Law, we must consider the second graph in Figure 1, which illustrates the case where (delta)Hs > 0. Since the mixture is an endothermic reaction, the result will be a decrease in the temperature of the solution.

The answer says the result will be a decrease in the temp of the solution... Is this the same as a decrease in the temp of the solvents? I still dont really understand why the temp would decrease and not increase if delta h is positive.

I could see how maybe the energy is coming from the solution to break the bonds but the question is asking about the temperature of the solvents. Would these be the same or diff then the solution?

Any help would be awesome! Thanks.

Okay if vapor pressure is higher than predicted that means that there isn't as much as an attraction between solvent and solute as predicted. So if the net attraction between solvent and solute is weak it means that the enthalpy of solvation must have been positive (H bonds broken - H bonds formed > 0) Since dH solvation is positive (endothermic) it must absorb heat from the surroundings (the solution) and therefore the temperature goes down. Basically, since solvation is endothermic, energy is absorbed to break bonds.

 

[matth87]

Okay if vapor pressure is higher than predicted that means that there isn't as much as an attraction between solvent and solute as predicted. So if the net attraction between solvent and solute is weak it means that the enthalpy of solvation must have been positive (H bonds broken - H bonds formed > 0) Since dH solvation is positive (endothermic) it must absorb heat from the surroundings (the solution) and therefore the temperature goes down. Basically, since solvation is endothermic, energy is absorbed to break bonds.

Alright, so the heat of the solution goes down, that makes perfect sense.

 

[Tokspor]

Okay if vapor pressure is higher than predicted that means that there isn't as much as an attraction between solvent and solute as predicted. So if the net attraction between solvent and solute is weak it means that the enthalpy of solvation must have been positive (H bonds broken - H bonds formed > 0) Since dH solvation is positive (endothermic) it must absorb heat from the surroundings (the solution) and therefore the temperature goes down. Basically, since solvation is endothermic, energy is absorbed to break bonds.

Are you saying the temperature of the unmixed solvents, separately, will decrease while the temperature of the solution will increase? That would make sense.

However, the "temperature of the solvents when put into the solution" reads to me like the "temperature of the solution," which would increase. The wording seems misleading to me.

 

[wanderer]

Are you saying the temperature of the unmixed solvents, separately, will decrease while the temperature of the solution will increase? That would make sense.

However, the "temperature of the solvents when put into the solution" reads to me like the "temperature of the solution," which would increase. The wording seems misleading to me.

It's one system, everything undergoes the same temperature change.

Endothermic processes lead to a decrease in temperature, just like exothermic processes lead to an increase in temperature.

 

[thebillsfan]

yeah, the temperature of the solutions will DECREASE not increase. if you separated the components out, would their temperatures increase?

 

[Tokspor]

It's one system, everything undergoes the same temperature change.

Endothermic processes lead to a decrease in temperature, just like exothermic processes lead to an increase in temperature.

If you look at the one of the forms of the Clausius-Clayperon equation here, the vapor pressure, which is written as P here, is directly proportional to the temperature. Wouldn't increasing the vapor pressure also increase the temperature?

 

[matth87]

yeah, the temperature of the solutions will DECREASE not increase. if you separated the components out, would their temperatures increase?

lol ummmmmm, the temperature of the air around them would increase????? since the opposite would be exothermic.

 

[matth87]

If you look at the one of the forms of the Clausius-Clayperon equation here, the vapor pressure, which is written as P here, is directly proportional to the temperature. Wouldn't increasing the vapor pressure also increase the temperature?

o snap...

 

[wanderer]

If you look at the one of the forms of the Clausius-Clayperon equation here, the vapor pressure, which is written as P here, is directly proportional to the temperature. Wouldn't increasing the vapor pressure also increase the temperature?

Is T in this referring to actual temperature or change in temperature due to a physical process?

 

[Tokspor]

Is T in this referring to actual temperature or change in temperature due to a physical process?

Well, I would think it is just a given temperature which results in a certain vapor pressure. I was just reminded of this equation from EK Chemistry. According to them, "vapor pressure is a function of temperature...so the equation indicates that vapor pressure increases with temperature."

Which text did the original question come from by the way?

 

[Tokspor]

Has anyone figured out why the answer is C? If more energy was absorbed by the system to break the bonds than was released from the formation of the new bonds, why wouldn't the answer be A, that the temperature for the system increases overall?