GChem - Phases/Calorimetry question

[yestomeds]

PHASES/PHASE CHANGES/CALORIMETRY
These 2 questions are from a free standing question from TPR 2014, whose answers I do not have.

If you are able to help, would you be able to do so in a very rudimentary, simple fashion please? I am slow.
Thank you!



A pot containing 0.5 L of water at sea level is brought to 100°C.
It is insulated around its sides to minimize heat loss to the environment, and heat is applied at the bottom of the container at a rate of 6 kJ/min over 3 minutes.
What is the resulting temperature of the water?
(ΔHvap = 40.7 kJ/mol; c(g) = 1.9 kJ/kg∙°C; c(l)=4.2 kJ/kg∙°C)

A. 115°C

B. 108°C

C. 104°C

D. 100°C






Which of the following salts would have the greatest effect on the melting point of water?

A. 1 g of KBr

B. 1 g of NaCl

C. 1 g of Na2CO3

D. 1 g of KNO3

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[jd989898]

.5L water at density 1kg/1L tells us the water weighs .5kg. Water has a molecular weight of ~18g/mol, which means the number of mols in .5kg of water is 500g÷18g=27.8mol. Since the water is already at 100 degrees C (its boiling point), it is in the process of vaporizing before the temperature rises any further. The Hvap is 40.7kJ/mol which means 40.7kJ must be added to the water to vaporize 1 mol. Since 6kJ/min over 3 minutes only amounts to 18kJ, only about 0.4mols of water will be vaporized. Thus, the temperature should stay at 100.........? Unless I'm just missing something

 

[jd989898]

.

 

[chemist16]

The second answer is C. Melting point displacement is directly proportional to the Vant Hoff factor (i) which is the number of ions that result from the dissociation of the compound. A, B, and D all have i=2, while C has i=3.

This is mostly incorrect. Your statement about the number of ions is correct. What's not correct is that you implicitly equate number of ions to mass. Note that you have the same mass of each substance in this question. The key deciding factor on melting point depression is the number of ions. So you have to convert the 1 g each to moles and then figure out how many moles of ions you have in solution. By my calculation, the answer is B. Feel free to verify, as I did the calculation very rapidly.

 

[jd989898]

This is mostly incorrect. Your statement about the number of ions is correct. What's not correct is that you implicitly equate number of ions to mass. Note that you have the same mass of each substance in this question. The key deciding factor on melting point depression is the number of ions. So you have to convert the 1 g each to moles and then figure out how many moles of ions you have in solution. By my calculation, the answer is B. Feel free to verify, as I did the calculation very rapidly.

My mistake. For some reason I read the question as 1 mole of each compound. 9 hours of studying really fries your brain lol. What did you think about the first problem in the OP?

 

[yestomeds]

This is mostly incorrect. Your statement about the number of ions is correct. What's not correct is that you implicitly equate number of ions to mass. Note that you have the same mass of each substance in this question. The key deciding factor on melting point depression is the number of ions. So you have to convert the 1 g each to moles and then figure out how many moles of ions you have in solution. By my calculation, the answer is B. Feel free to verify, as I did the calculation very rapidly.

@chemist16 @jd989898
Thanks guys. So this is specifically in reference to the 2nd question.

Would you mind it if we went about it step by step? So for number of ions (i), would it be:
A. KBr (i=2?)
B. 1 g of NaCl (i=2)
C. 1 g of Na2CO3 (i here = 2?! how to assess the C and O,O,O?)
D. 1 g of KNO3 (i here = 1 or do N and O count as well?)

Secondly,
What I learned so far is the changes to boiling point and freezing point. The question addresses MELTING point. Are we to use the delta T equation for... freezing point then?
So: deltaT = (constant "k subscript m")(i)(molality)

Third,
to get the molality for the first substance, KBr for instance, it would be:
(1g) / (120g/mol) = roughly 0.00833 (and then you just plug and chug).​

 

[yestomeds]

.5L water at density 1kg/1L tells us the water weighs .5kg. Water has a molecular weight of ~18g/mol, which means the number of mols in .5kg of water is 500g÷18g=27.8mol. Since the water is already at 100 degrees C (its boiling point), it is in the process of vaporizing before the temperature rises any further. The Hvap is 40.7kJ/mol which means 40.7kJ must be added to the water to vaporize 1 mol. Since 6kJ/min over 3 minutes only amounts to 18kJ, only about 0.4mols of water will be vaporized. Thus, the temperature should stay at 100.........? Unless I'm just missing something

Thank you! (This one is quicker).
So you're saying that since the heat added over that period of time, amounting to 18kJ, does not equal to the 40.7kJ required to vaporize a mol of water. Therefore temperature does not change.

So are we essentially looking at the flat portion of the phase diagram, during which you add and add heat (q), and temperature doesn't change? IF the heat added over time or w/e is GREATER than 40.7kJ, THEN more water will be vaporized... but I'm confused then, because 'vaporizing' talks about phase changes (which occurs when temperature doesn't change), whereas the question talks about temperature changes. Sorry. Can you give it to me slow.

 

[chemist16]

Would you mind it if we went about it step by step? So for number of ions (i), would it be:
A. KBr (i=2?)
B. 1 g of NaCl (i=2)
C. 1 g of Na2CO3 (i here = 2?! how to assess the C and O,O,O?)
D. 1 g of KNO3 (i here = 1 or do N and O count as well?)

Secondly,
What I learned so far is the changes to boiling point and freezing point. The question addresses MELTING point. Are we to use the delta T equation for... freezing point then?
So: deltaT = (constant "k subscript m")(i)(molality)

Third,
to get the molality for the first substance, KBr for instance, it would be:
(1g) / (120g/mol) = roughly 0.00833 (and then you just plug and chug).

1) Your ion count is correct for A and B but incorrect for C and D. For C, you get two sodium ions and a single carbonate ion so i=3. For D, you get one potassium ion and one nitrate ion so i=2.

2) Melting point and freezing point are the same thing. It wouldn't make sense to say that a particular system freezes (liquid to solid) at some temperature and then the same system melts (solid to liquid) at a different temperature. Take water for example. At what temperature does water melt? At what temperature does it freeze?

3) You really just need the moles of the substance. When you do an ion count (i=whatever), that means that for one mole of the original substance, you get i moles of ions out of it. So you can figure out how many moles of ions you get in each case.

 

[chemist16]

So are we essentially looking at the flat portion of the phase diagram, during which you add and add heat (q), and temperature doesn't change? IF the heat added over time or w/e is GREATER than 40.7kJ, THEN more water will be vaporized... but I'm confused then, because 'vaporizing' talks about phase changes (which occurs when temperature doesn't change), whereas the question talks about temperature changes. Sorry. Can you give it to me slow.

You have to add more heat than 40.7 kJ. Specifically, you have to add 40.7 kJ/mol and you have like 30 moles of water. The question just asks you what the temperature is, not what the change in temperature is. That is, there is no change in temperature here because the extra heat you're adding just goes toward breaking the H-bonds holding the water molecules together.

 

[yestomeds]

You have to add more heat than 40.7 kJ. Specifically, you have to add 40.7 kJ/mol and you have like 30 moles of water. The question just asks you what the temperature is, not what the change in temperature is. That is, there is no change in temperature here because the extra heat you're adding just goes toward breaking the H-bonds holding the water molecules together.

In other words, you are saying that because we are adding 40.7 kJ or less (we are adding less), we are simply putting in heat to break H-bonds, which is one way of saying we are helping it undergo a PHASE CHANGE only. Only when we were to add MORE heat, for it to
a) overcome H-bonds and change phases, and THEN
b) to use the additional heat toward increasing its kinetic energy of its particles, does it go up in temperature.

Is this correct?

 

[chemist16]

In other words, you are saying that because we are adding 40.7 kJ or less (we are adding less), we are simply putting in heat to break H-bonds, which is one way of saying we are helping it undergo a PHASE CHANGE only. Only when we were to add MORE heat, for it to
a) overcome H-bonds and change phases, and THEN
b) to use the additional heat toward increasing its kinetic energy of its particles, does it go up in temperature.

Is this correct?

Conceptually, yes. But again, you need more than 40.7 kJ. You need 1129 kJ. Anything less than that and all you're doing is breaking H-bonds.

 

[yestomeds]

Conceptually, yes. But again, you need more than 40.7 kJ. You need 1129 kJ. Anything less than that and all you're doing is breaking H-bonds.

I'm sorry but if the H of vaporization is 40.7 kJ, where did the 1129 kJ come from?
Although I do get how we'll need more than 40.7 because 40.7 just barely changed the phase (thus no temp changes yet).

 

[chemist16]

I'm sorry but if the H of vaporization is 40.7 kJ, where did the 1129 kJ come from?

Any heat of vaporation, fusion, sublimation, etc. is an intensive property. Thus, it must be given in units of kJ/mol or its equivalent. In other words., the heat of vaporation of water is not 40.7 kJ but 40.7 kJ/mol. How many moles of water do you have?