EK Chem Lecture 3 - Enthalpy problem

[SChiO]

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I know how to compute enthalpy if there are two reactions but how do I go about figuring this out?? The explanation in the book didn't make much sense to me 😕

The standard enthalpy of formation for liquid water is:
H2(g) + fiO2(g) --> H2O(l) ∆H°f = -285.8 kJ/mol

Which of the following could be the standard enthalpy of formation for water vapor?

A. -480.7 kJ/mol
B. -285.8 kJ
C. -241.8 kJ/mol
D. +224.6 kJ/mol

The answer is C. I understand why it should be negative ruling out D, and based off the units given I can rule out B but what is the logic behind choosing C over A??

 

[Postictal Raiden]

I know how to compute enthalpy if there are two reactions but how do I go about figuring this out?? The explanation in the book didn't make much sense to me 😕

The standard enthalpy of formation for liquid water is:
H2(g) + fiO2(g) --> H2O(l) ∆H°f = -285.8 kJ/mol

Which of the following could be the standard enthalpy of formation for water vapor?

A. -480.7 kJ/mol
B. -285.8 kJ
C. -241.8 kJ/mol
D. +224.6 kJ/mol

The answer is C. I understand why it should be negative ruling out D, and based off the units given I can rule out B but what is the logic behind choosing C over A??

Could it be the fact that liquid water molecules form stronger intermolecular forces than vapor water is what contributes to this difference in energy release?

Besides, option A is way too high to be a possible answer.

 

[brood910]

When you form water vapor, you break hydrogen bonds.
This increases enthalpy, and thus, less negative.

 

[ScienceNerdUofU]

We know this reaction is exothermic so it's negative. I would think about it like this. When you form water liquid from elements in their standard states, you get the formation of H bonds, which stores energy. However, in order to to form gaseous water you need to have energy to break those H bonds during the phase transition. This is because water is liquid in its standard state at a pressure of 1 bar so it prefers, in a way, to go to liquid. But some of the energy being released can be used to convert the liquid to gas.

 

[sleeper11]

Can someone explain why it has to be exothermic?

 

[brood910]

Can someone explain why it has to be exothermic?

It is already given in the Q.
It is negative.

 

[SChiO]

Ok thank you. I understand both explanations. So is it correct in saying that every reaction will increase entropy or only certain reactions?

 

[brood910]

Ok thank you. I understand both explanations. So is it correct in saying that every reaction will increase entropy or only certain reactions?

If you are increasing the # of moles, then you are increasing entropy since more free particles = more randomness = more disorderly

Ex: Nacl => na + cl

If you are decreasing, then you are decreasing entropy.

Ex: na + cl => Nacl

You should realize that this question is about enthalpy, not entropy.

 

[Meredith92]

It is already given in the Q.
It is negative.

Where does it say in the question? It only says formation of liquid water is exothermic. I thought evaporation is endothermic ( breaking strong intermecular interactions) from what I learned in tbr... So couldn't the formation of water vapor be endothermic and positive?

 

[brood910]

Where does it say in the question? It only says formation of liquid water is exothermic. I thought evaporation is endothermic ( breaking strong intermecular interactions) from what I learned in tbr... So couldn't the formation of water vapor be endothermic and positive?

You just said the answer yourself.
Formation of liquid water was given as exothermic, so it is negative.

For formation of water vapor, it must be less exothermic than the formation of liquid water since hydrogen bonds were broken, and thus, less negative. You are required to compare to the given info.

 

[sdm33]

A common misconception is that the process of boiling creates oxygen and hydrogen gases (O2 and H2), which is simply not true. The act of boiling water imparts energy to the liquid water molecules. This energy is enough to break the hydrogen bonds that hold the water molecules together in liquid form. Once these bonds are broken, the water molecules vaporize and go from being H2O (liquid) to H2O (gas). Therefore, the vapor produced as water boils is water vapor.The truth is evaporation of water is Endothermic processes, you see watch how the temperature of the surroundings changes. An exothermic process releases heat, and causes the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings. That is why it is becoming less negative.

 

[sdm33]

option A is not too high, it is too low if you really think about it

 

[Meredith92]

A common misconception is that the process of boiling creates oxygen and hydrogen gases (O2 and H2), which is simply not true. The act of boiling water imparts energy to the liquid water molecules. This energy is enough to break the hydrogen bonds that hold the water molecules together in liquid form. Once these bonds are broken, the water molecules vaporize and go from being H2O (liquid) to H2O (gas). Therefore, the vapor produced as water boils is water vapor.The truth is evaporation of water is Endothermic processes, you see watch how the temperature of the surroundings changes. An exothermic process releases heat, and causes the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings. That is why it is becoming less negative.

I understand what your saying now so it's the exothermic formation of water liquid + endothermic formation of water vapor. Sorry to ask one more question but how can we be sure that the endothermic part isn't greater than the exothermic part? Which would add together to be a positive number?

Thanks again for your help

 

[sdm33]

because we only breaking a H bonds, This energy is enough to break the hydrogen bonds that hold the water molecules together in liquid form, so it is not going to be that much, O−H…:O (21 kJ/mol or 5.0 kcal/mol)

 

[Meredith92]

riiiight!! Thanks!!

 

[LoLCareerGoals]

Can someone explain why it has to be exothermic?

This is a much better question. If anything it was between C and D. The answer I think is that H-bond breakage is much smaller in scale to polar O-H bond formation, so it couldn't possible require more energy than was release during formation of such strong polar bonds.

 

[SChiO]

If you are increasing the # of moles, then you are increasing entropy since more free particles = more randomness = more disorderly

Ex: Nacl => na + cl

If you are decreasing, then you are decreasing entropy.

Ex: na + cl => Nacl

You should realize that this question is about enthalpy, not entropy.

Oh yes thank you. Sorry about that. I struggle with mixing the two up subconsciously but I think I have it together now

 

[Halcyon32]

I know this problem was resolved for a while now but for my own clarification: since energy is used in the actual reaction from liquid H2O to gaseous H2O to break the H bonds, that energy that was used in the reaction is subtracted from the enthalpy of formation because it is not released into the surroundings, but rather used up in the reaction?
Also how do you just know that breaking an H-bond wouldn't require so much energy that it would be greater than 285kj/mol making the whole answer positive?