Blaughable's MCAT study Questions

[Blaughable]

Instead of making a new thread here everytime I come up with a question, I thought that I'd just make a thread to put all my questions in. (Hope it's okay )

1.) Chapter 1 of TBR - Organic Chemistry. Passage I, question 8.

It has a table of enthalpy for a set of hydrogenations of various alkenes. I apparently have a very incorrect view of deltaH, enthalpy. I always assumed that delta H was the amount of heat released by a reaction. (at least as far as chemists are concerned) So wouldn't a more negative enthalpy be a greater amount of heat released meaning there was more energy in the bond and thus the bond was stronger?

The answer states that a lower(less negative) deltaH implies a stronger bond. Help please.

New question!

2.) I'm a little confused on how to determine if a molecule is paramagnetic. Here's the question from TBR

TBR - G-Chem - Ch2 - P4 - #21
Which of the following [is paramagnetic]?

A.) CdCl2
B.) F2
C.) CoCl2.6H2O
D.) H2

I understand that for a molecule to be paramagnetic it has to have a lone unpaired electron. I'm just confused in general how to think of the electrons in molecule A and C. I understand the lone electrons of H and F pair making H2 and F2 diamagnetic.

I'm probably just being anal, but I want to make sure I understand everything thoroughly. In CdCl2, what is going on with the electrons. Why is the bond even being formed if Cd has a full outer valence "d" orbital? Does it become some strange sp3d5 hybridized orbital? If you take two electrons from Cd in order to bond to Cl2 wouldn't you be causing 2 other electrons to be unpaired?

In CoCl2.6H2o first of all what does the dot in between CoCl2 and 6H20 mean? Here I see that there is an odd number of electrons so there must be an unpaired electron. However, should I be worried with (not necessarily for this question) how the electrons are shared? I guess Cl2 being more electronegative would take 2 electrons in order to fill their octets. Does this leave Co with 5 unpaired electrons in it's 3d orbital? Or do the electrons come from the 4s orbital leaving Co with 3 unpaired electrons?

argh, i'm confused

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[blackandgold1]

Instead of making a new thread here everytime I come up with a question, I thought that I'd just make a thread to put all my questions in. (Hope it's okay )

1.) Chapter 1 of TBR - Organic Chemistry. Passage I, question 8.

It has a table of enthalpy for a set of hydrogenations of various alkenes. I apparently have a very incorrect view of deltaH, enthalpy. I always assumed that delta H was the amount of heat released by a reaction. (at least as far as chemists are concerned) So wouldn't a more negative enthalpy be a greater amount of heat released meaning there was more energy in the bond and thus the bond was stronger?

The answer states that a lower(less negative) deltaH implies a stronger bond. Help please.

Heat of hydrogenation is different from heat of enthalpy. The lower the heat of hydrogenation (less negative) the stronger the bond; the higher the heat of enthalpy (exothermic; more negative) the stronger the bond.

 

[Blaughable]

Heat of hydrogenation is different from heat of enthalpy. The lower the heat of hydrogenation (less negative) the stronger the bond; the higher the heat of enthalpy (exothermic; more negative) the stronger the bond.

Thanks for the quick reply. So what does heat of hydrogenation actually physically correspond to? Is it hydrogenating the alkene to an alkane?

 

[blackandgold1]

Yup, its hydrogenating double bonds to single bonds. So like if comparing between cyclohexane and cyclopropane, cyclohexane (which is more stable due to no ring strain), would have a lower heat of hydrogenation then cyclopropane.

Also, as substitution around a double bond increases, its heat of hydrogenation decreases, becuase it is more stable then an unsubstituted double bond

 

[Rabolisk]

There are lots of things here, and I'm not sure where to begin...

Heat of hydrogenation is different from heat of enthalpy. The lower the heat of hydrogenation (less negative) the stronger the bond; the higher the heat of enthalpy (exothermic; more negative) the stronger the bond.

This should forever stay on the list of the greatest quotes in the MCAT Study forum.

Heat of hydrogenation is the synonymous with enthalpy of hydrogenation. Technically speaking, change in enthalpy (ΔH) is equal to the heat transfer at constant pressure. Regardless, heat of X (where X = some process) is the same as enthalpy of X for all intended purposes. So that leaves the question, what in the world is heat of enthalpy? Should it be called enthalpy of enthalpy? Enthalpy of heat? Heat of heat? I don't know, and I don't think anybody does.

It is true that the lower (less negative or more positive) the heat of hydrogenation, the stronger the pi bond, or the more stable the starting compound. More on that later.

Yup, its hydrogenating double bonds to single bonds. So like if comparing between cyclohexane and cyclopropane, cyclohexane (which is more stable due to no ring strain), would have a lower heat of hydrogenation then cyclopropane.

Also, as substitution around a double bond increases, its heat of hydrogenation decreases, becuase it is more stable then an unsubstituted double bond

I don't quite think you are applying this correctly. Heat of hydrogenation measures the heat/enthalpy released/gained as you go from an alkene to an alkane. It makes no sense to compare the heat of hydrogenation between cyclohexane and cyclopropane, since both are already fully hydrogenated without breaking the carbon backbone. If you meant to say cyclohexene and cyclopropene, you still can't claim that the heat of hydrogenation of cyclopropene is higher than that of cyclohexene. Heat of hydrogenation measures the relative stability of an alkene compared to that of its analogous alkane. It is true that cyclopropane is rather unstable, but I assume cyclopropene is as well. Cyclohexane is rather stable, and so is cyclohexene. So the difference in stability between cyclopropene and cyclopropane (i.e. ΔH) may or may not be greater than that between cyclohexene and cyclohexane.

Instead of making a new thread here everytime I come up with a question, I thought that I'd just make a thread to put all my questions in. (Hope it's okay )

1.) Chapter 1 of TBR - Organic Chemistry. Passage I, question 8.

It has a table of enthalpy for a set of hydrogenations of various alkenes. I apparently have a very incorrect view of deltaH, enthalpy. I always assumed that delta H was the amount of heat released by a reaction. (at least as far as chemists are concerned) So wouldn't a more negative enthalpy be a greater amount of heat released meaning there was more energy in the bond and thus the bond was stronger?

The answer states that a lower(less negative) deltaH implies a stronger bond. Help please.

This is a huge misconception that apparently is ingrained in some students due to misleading and/or confusing terms like "high-energy bond", "bond energy", and "energy stored in the bond". First, there is no such thing as "energy stored in the bond", per se. Bonds are a reduction of potential energy, and as such, only and just only breaking bonds cannot "release" any energy. Energy will have to be put in to break bonds. Energy will have to be used to increase the potential energy of the chemical system. An increase in the potential energy of the system means that this energy has to come from somewhere. Exothermic reactions are exothermic because of the bond formation step that forms the products. You can say that the products are more stable than the reactants.

You have a bunch of alkenes which can be hydrogenated to the same alkane. For example, think of the different isomers of butene that can be hydrogenated to butane (not sure what TBR uses as an example). They are all negative, so butane (the product) is more stable than butene + H2 (the reactants). Since each reaction produces the same product (butane), this can be used to compare the relative stability of the different isomers of butene. The lower, less negative, or more positive ΔH corresponds to less difference in enthalpy between the butene isomer and butane. Higher ΔH means greater difference. Since ΔH is negative for all of them, higher ΔH corresponds to the least stable or highest enthalpy.

http://www.wou.edu/las/physci/ch334/lecture/lect16.htm

There's a picture about halfway down next to "heat of hydrogenation" that is very informative.

 

[Blaughable]

Thank you so much for the reply! Very Informative, I think I understand it now. I just have to make sure to nail these little misconceptions early on in my studying and not let them snowball.

 

[justhanging]

I don't quite think you are applying this correctly. Heat of hydrogenation measures the heat/enthalpy released/gained as you go from an alkene to an alkane. It makes no sense to compare the heat of hydrogenation between cyclohexane and cyclopropane, since both are already fully hydrogenated without breaking the carbon backbone. If you meant to say cyclohexene and cyclopropene, you still can't claim that the heat of hydrogenation of cyclopropene is higher than that of cyclohexene. Heat of hydrogenation measures the relative stability of an alkene compared to that of its analogous alkane. It is true that cyclopropane is rather unstable, but I assume cyclopropene is as well. Cyclohexane is rather stable, and so is cyclohexene. So the difference in stability between cyclopropene and cyclopropane (i.e. ΔH) may or may not be greater than that between cyclohexene and cyclohexane.


.

I think he meant to say heat of combustion, which I believe would be higher in in cyclopropane per CH2 group than cyclohexane because of ring strain.

 

[Swagster]

So that leaves the question, what in the world is heat of enthalpy? Should it be called enthalpy of enthalpy? Enthalpy of heat? Heat of heat? I don't know, and I don't think anybody does.

I could be wrong, but I believe flubber has a really high heat of enthalpic energy of thermal motion because it's green.

 

[Blaughable]

New question!

2.) I'm a little confused on how to determine if a molecule is paramagnetic. Here's the question from TBR

TBR - G-Chem - Ch2 - P4 - #21
Which of the following [is paramagnetic]?

A.) CdCl2
B.) F2
C.) CoCl2.6H2O
D.) H2

I understand that for a molecule to be paramagnetic it has to have a lone unpaired electron. I'm just confused in general how to think of the electrons in molecule A and C. I understand the lone electrons of H and F pair making H2 and F2 diamagnetic.

I'm probably just being anal, but I want to make sure I understand everything thoroughly. In CdCl2, what is going on with the electrons. Why is the bond even being formed if Cd has a full outer valence "d" orbital? Does it become some strange sp3d5 hybridized orbital? If you take two electrons from Cd in order to bond to Cl2 wouldn't you be causing 2 other electrons to be unpaired?

In CoCl2.6H2o first of all what does the dot in between CoCl2 and 6H20 mean? Here I see that there is an odd number of electrons so there must be an unpaired electron. However, should I be worried with (not necessarily for this question) how the electrons are shared? I guess Cl2 being more electronegative would take 2 electrons in order to fill their octets. Does this leave Co with 5 unpaired electrons in it's 3d orbital? Or do the electrons come from the 4s orbital leaving Co with 3 unpaired electrons?

argh, i'm confused

 

[Blaughable]

still confused...

 

[tkronen]

So for F2--> the electrons are shared, so there are no uneven electrons. The same goes for H2. For CdCl2--> the Cl has a -1, so because there are 2 of them, it has a -2 charge in total, which makes Cd 2+. The electrons would come from the 5s2, so there would be only 4p10 left over, which is an even number of electrons for that orbital. So all of those are diamagnetic. For CoCl2 x H2o, the period in between them just means a multiplication sign, so you can disregard the water. Again the Cl has a -1, and with 2 of them, it has a -2 charge, so the Co has a 2+ charge. This would come out of the 4s2, so you would be left with 3d7, which is an uneven number of electrons. A B field can align these electrons to create a dipole moment, which would then make this molecule weakly magnetic. Does that make sense?