Standard enthalpy of formation for liquid water

[the prodogy]

I'm having a hard time understanding standard enthalpy of formation. On the Gen chem EK book, problem 63 for Lecture 3, they ask:

The standard enthalpy of formation for liquid water is:

H2 (g) + fiO2 (g) --> H2O(l) delta H = -285.8 kJ/mol

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

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

I know that it has to be less negative, so the choices are between C and D, so I chose D as a guess. But the answer says, "the standard enthalpy of formation of water vapor will not be an endothermic process". I don't understand why it can't be an endothermic process. Don't you need to add heat to water for it to turn into water vapor?

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[chemtopper]

0

 

[chemtopper]

The standard enthalpy of formation for liquid water is:

H2 (g) + fiO2 (g) --> H2O(l) delta H = -285.8 kJ/mol

, "the standard enthalpy of formation of water vapor will not be an endothermic process". I don't understand why it can't be an endothermic process. Don't you need to add heat to water for it to turn into water vapor?[/quote]

Here in the question they are asking for enthalpy of formation of water vapors and not enthalpy of vaporization of water.
Enthalpy of vaporization is positive because you need to supply heat to change the state.

Enthalpy of formation is formation of water vapors from its elements
H2 (g) + O2 (g) --> H2O(g) + heat
H2O(l) Heat --> H2O(g)
So it is negative but less heat is liberated .
delta H < -285.8 kJ/mol

 

[JohnnytheNomad]

I'm having a hard time understanding standard enthalpy of formation. On the Gen chem EK book, problem 63 for Lecture 3, they ask:

The standard enthalpy of formation for liquid water is:

H2 (g) + fiO2 (g) --> H2O(l) delta H = -285.8 kJ/mol

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

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

I know that it has to be less negative, so the choices are between C and D, so I chose D as a guess. But the answer says, "the standard enthalpy of formation of water vapor will not be an endothermic process". I don't understand why it can't be an endothermic process. Don't you need to add heat to water for it to turn into water vapor?

Yes, but it is still exothermic due to entropy reasons. This of how much more positive entropy is verses a liquid.

 

[JasoniusUW]

Isn't enthalpy separate from entropy? The two are additive to form Gibbs Free Energy?

 

[VanBrown]

the original reaction is a condensation reaction and is exothermic meaning that the liquid water is in a more stable state. Therefore, to go from the liquid to the gas phase (less energetically favorable) you must add heat (+delta H). Therefore the overall reaction would be less negative than the condensation.

 

[rocketbooster]

just know this:

solid--->liquid---->gas = always endothermic because it has to take in energy to break strong bonds in order to form less stable states (thus positive enthalpy)

gas---->liquid---->solid = always exothermic because energy is released as you are forming new, stronger bonds to form the more stable state (thus negative enthalpy)

I actually remember reading in TBR that water vapor is considered H2O(l). H2O(g) is technically called water gas. If you followed this, then B would be the answer. However, I only saw that in BR and don't think this is universally accepted. Go with water vapor as being a gas.

 

[0Complications]

Isn't enthalpy separate from entropy? The two are additive to form Gibbs Free Energy?

Yes, they are both different.

Delta H is enthalpy, which for the MCAT will most likely just deal with changes in heat within a system. (-) delta H is indicative of exothermic rxns meaning heat is released from the bonds causing the temperature to increase in the area around the molecule where the bonds are broken (Think a hot pack, that you shake and it becomes hot. This system is releasing heat energy). (+) delta H is endothermic because the bonds require energy from the surround to form, this cools the environment (Think about a chemical ice pack that you break and it becomes cold, that is drawing in heat energy from the surround).

Delta S is entropy, which is essentially the disorder of a system. The larger the value the greater the disorder.

just know this:

solid--->liquid---->gas = always endothermic because it has to take in energy to break strong bonds in order to form less stable states (thus positive enthalpy)

gas---->liquid---->solid = always exothermic because energy is released as you are forming new, stronger bonds to form the more stable state (thus negative enthalpy)

I actually remember reading in TBR that water vapor is considered H2O(l). H2O(g) is technically called water gas. If you followed this, then B would be the answer. However, I only saw that in BR and don't think this is universally accepted. Go with water vapor as being a gas.

I would not go with B on that one. I can see where you're going with that, but I don't agree with it. If it is the case I think it would be minutia the MCAT wouldn't try to test.

I would go for something just a little more positive than the enthalpy of H2O (l), the only choice that would support that option without being endothermic is C.

 

[rocketbooster]

Yes, they are both different.

Delta H is enthalpy, which for the MCAT will most likely just deal with changes in heat within a system. (-) delta H is indicative of exothermic rxns meaning heat is released from the bonds causing the temperature to increase in the area around the molecule where the bonds are broken (Think a hot pack, that you shake and it becomes hot. This system is releasing heat energy). (+) delta H is endothermic because the bonds require energy from the surround to form, this cools the environment (Think about a chemical ice pack that you break and it becomes cold, that is drawing in heat energy from the surround).

Delta S is entropy, which is essentially the disorder of a system. The larger the value the greater the disorder.



I would not go with B on that one. I can see where you're going with that, but I don't agree with it. If it is the case I think it would be minutia the MCAT wouldn't try to test.

I would go for something just a little more positive than the enthalpy of H2O (l), the only choice that would support that option without being endothermic is C.

haha I see what you're saying. doesn't really matter tho bc that's all based on an assumption that shouldn't be made. just go with water vapor being H2O(g).

 

[rhinoblitz]

I'm having a hard time understanding standard enthalpy of formation. On the Gen chem EK book, problem 63 for Lecture 3, they ask:

The standard enthalpy of formation for liquid water is:

H2 (g) + fiO2 (g) --> H2O(l) delta H = -285.8 kJ/mol

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

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

I know that it has to be less negative, so the choices are between C and D, so I chose D as a guess. But the answer says, "the standard enthalpy of formation of water vapor will not be an endothermic process". I don't understand why it can't be an endothermic process. Don't you need to add heat to water for it to turn into water vapor?

I realize that this question was probably answered quite a long time ago BUT, I'll go ahead and give it my best shot to answer it in a simpler way for future students. It's actually quite simple but also quite tricky. I'm learning that the MCAT examkrackers book (and possibly the MCAT itself.) puts answers there for you to choose that are just "too good to be true!" So number one rule, WATCH OUT FOR THOSE!! Okay, now on to the problem. Refer above for the original question if needed as there is no need to write it out again. Basically, from left to right or from solid to liquid to gas it is endothermic and it takes more positive or endothermic energy as we go from left to right. Now, the opposite holds true in the opposite direction, from right to left or from gas to liquid to solid, including the standard enthalpy numbers starting small and rising each step of the way, only it's exothermic or negative. So if we have gas to liquid (as the problem states), we're exothermic AND it's -285.8 kJ/mol as we're going from left to right. Now, many of you are wondering why the opposite direction is still negative or exothermic if we're in fact going the opposite direction but it's not and this is why it's a tricky question. It's endothermic in deed BUT the standard enthalpy number is only around +44 (I believe but please check it out.) So you essentially have -285.8 + 44 which gives us our magic number of -241.8. I don't know why everyone else needs to make it so complicated. I would go ahead and study intermolecular bonds etc b/c those might have come in handy but if we know that left to right the standard enthalpy rises and left to right it's the same thing, we can figure it out merely by thinking it through. Hope that helps and if I'm wrong, someone please correct me. Hope I'm right, that's how I came to that answer anyway....

 

[cama0106]

I realize that this question was probably answered quite a long time ago BUT, I'll go ahead and give it my best shot to answer it in a simpler way for future students. It's actually quite simple but also quite tricky. I'm learning that the MCAT examkrackers book (and possibly the MCAT itself.) puts answers there for you to choose that are just "too good to be true!" So number one rule, WATCH OUT FOR THOSE!! Okay, now on to the problem. Refer above for the original question if needed as there is no need to write it out again. Basically, from left to right or from solid to liquid to gas it is endothermic and it takes more positive or endothermic energy as we go from left to right. Now, the opposite holds true in the opposite direction, from right to left or from gas to liquid to solid, including the standard enthalpy numbers starting small and rising each step of the way, only it's exothermic or negative. So if we have gas to liquid (as the problem states), we're exothermic AND it's -285.8 kJ/mol as we're going from left to right. Now, many of you are wondering why the opposite direction is still negative or exothermic if we're in fact going the opposite direction but it's not and this is why it's a tricky question. It's endothermic in deed BUT the standard enthalpy number is only around +44 (I believe but please check it out.) So you essentially have -285.8 + 44 which gives us our magic number of -241.8. I don't know why everyone else needs to make it so complicated. I would go ahead and study intermolecular bonds etc b/c those might have come in handy but if we know that left to right the standard enthalpy rises and left to right it's the same thing, we can figure it out merely by thinking it through. Hope that helps and if I'm wrong, someone please correct me. Hope I'm right, that's how I came to that answer anyway....

standard enthalpy of what is +44 ?