MCAT Science Workbook 2010 PR Question (Ionization Energy)

[SeeEll]

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In passage 61 of the GChem section, we are given the following question:

"The reaction between hydrogen and oxygen occurs at n explosive rate, while that of hydrogen and nitrogen is sluggish at all temperatures. Which of the following correctly accounts for this?"

A) At any given termperature, nitrogen molecules will have a lower velocity than oxygen moelcules and thus collide with hyrogen less frequently
B) The ionization energy for atomic oxygen is less than that of atomic nitrogen
C) Molecular nitrogen lacks non-bonding electrons to interact with hydrogen molecule directly
D) The bond dissociation energy of molecular nitrogen is greater than that of molecular oxygen

I was stuck between choosing B or D. I was not sure of D, but with B, I thought that Nitrogen's half filled p-shell would give it a higher ionization energy when compared to Oxygen, making it the correct answer. I understand the triple bond between molecular nitrogen gives it greater bond dissociation energy, but can anyone explain why B is wrong?

Thanks.

 

[Thirteeen]

In passage 61 of the GChem section, we are given the following question:

"The reaction between hydrogen and oxygen occurs at n explosive rate, while that of hydrogen and nitrogen is sluggish at all temperatures. Which of the following correctly accounts for this?"

A) At any given termperature, nitrogen molecules will have a lower velocity than oxygen moelcules and thus collide with hyrogen less frequently
B) The ionization energy for atomic oxygen is less than that of atomic nitrogen
C) Molecular nitrogen lacks non-bonding electrons to interact with hydrogen molecule directly
D) The bond dissociation energy of molecular nitrogen is greater than that of molecular oxygen

I was stuck between choosing B or D. I was not sure of D, but with B, I thought that Nitrogen's half filled p-shell would give it a higher ionization energy when compared to Oxygen, making it the correct answer. I understand the triple bond between molecular nitrogen gives it greater bond dissociation energy, but can anyone explain why B is wrong?

Thanks.

Ionization energy increases as we move from left to right of the same row. This rules out B.

My answer is D, since BDE of nitrogen molecule is indeed higher than O2.

http://en.wikipedia.org/wiki/Bond_dissociation_energy

 

[SeeEll]

Ionization energy increases as we move from left to right of the same row. This rules out B.

My answer is D, since BDE of nitrogen molecule is indeed higher than O2.

http://en.wikipedia.org/wiki/Bond_dissociation_energy

So we ignore the stability of the half filled p-shell? Doesn't that play into ionization energy? Something like this?

For ionization energy questions on the MCAT, do I just accept left to right = increasing ionization energy? Serious question.

 

[Thirteeen]

For ionization energy questions on the MCAT, do I just accept left to right = increasing ionization energy? Serious question.

The general trend is that, within a row, the IE increases as the atomic number increases (ie, moving L-->R). It decreases as we go down the column. However, there are exceptions when there is half-filled stability of the energy level and when there is an S2-shell. I think as long as you are familiar with the general trend and those few exceptions (He to Li, Ne to Na and Ar to K) , you are set.

 

[SeeEll]

The general trend is that, within a row, the IE increases as the atomic number increases (ie, moving L-->R). It decreases as we go down the column. However, there are exceptions when there is half-filled stability of the energy level and when there is an S2-shell. I think as long as you are familiar with the general trend and those few exceptions, you are set.

I think I might be missing something because doesn't answer choice B read: "the ionization energy for atomic OXYGEN is LESS than that of atomic NITROGEN"?

I just want to determine if this statement is true or false. I thought it was true because of the trend we've been discussing (half filled p orbitals with greater ionization energies). Sorry for pushing this thread up to the top over and over again.

 

[Thirteeen]

I think I might be missing something because doesn't answer choice B read: "the ionization energy for atomic OXYGEN is LESS than that of atomic NITROGEN"?

I just want to determine if this statement is true or false. I thought it was true because of the trend we've been discussing (half filled p orbitals with greater ionization energies). Sorry for pushing this thread up to the top over and over again.

Actually, the statement is false since the IE increases as we go from left to right in a row. Oxygen comes after Nitrogen so oxygen has a higher IE. Generally speaking, the IE increases as the atomic number increases with a few exceptions (He to Li, Ne to Na and Ar to K)

 

[ilovemcat]

Nitrogen actually does have a higher IE than Oxygen. It's one of those strange exceptions you mentioned earlier (half-filled stablility).

I think the reason why choice B is wrong is due to this reaction. O-H and O-N are both molecular bonds (covalent bonds). Electrons are shared equally. In contrast, an Ionic Bond is hogging an electron for itself (like the chlorine atom in NaCl). Therefore, although choice B is a true statement, it doesn't explain why the reaction doesn't proceed. Remember, Ionization Energy is the energy needed to literally remove 1 electron. No electrons are lost for the reaction they're referring to.

Choice D on the other hand, does provide a reasonable explanation. Because molecular Nitrogen (triple bond) is more stable than Oxygen (double bond), more energy is required to break it's bond than say breaking the bond of an Oxygen molecule. In other words, the reaction between molecular Nitrogen and Hydrogen is a more endothermic reaction than the reaction between molecular Oxygen and Hydrogen (which is actually exothermic overall).

I actually calculated the values just to be sure:

O=O + H-H --> H-O-O-H

deltaH = (119 + 104.2) - (2(35) + 222)
deltaH = -68.8 (EXOTHERMIC overall)

NN + H-H --> H-N=N-H

deltaH = (226 + 104.2) - (2(93) + 109)
deltaH = +35.2 (ENDOTHERMIC overall)

 

[Thirteeen]

Nitrogen actually does have a higher IE than Oxygen. It's one of those strange exceptions you mentioned earlier (half-filled stablility).

The 1st IE of oxygen is smaller than nitrogen. 🙂

 

[SeeEll]

Nitrogen actually does have a higher IE than Oxygen. It's one of those strange exceptions you mentioned earlier (half-filled stablility).

I think the reason why choice B is wrong is due to this reaction. O-H and O-N are both molecular bonds (covalent bonds). Electrons are shared equally. In contrast, an Ionic Bond is hogging an electron for itself (like the chlorine atom in NaCl). Therefore, although choice B is a true statement, it doesn't explain why the reaction doesn't proceed. Remember, Ionization Energy is the energy needed to literally remove 1 electron.

Choice D on the other hand, does provide a reasonable explanation. Because molecular Nitrogen (triple bond) is more stable than Oxygen (double bond), more energy is required to break it's bond than say breaking the bond of an Oxygen molecule. In other words, the reaction between molecular Nitrogen and Hydrogen is a more endothermic reaction than the reaction between molecular Oxygen and Hydrogen (less endothermic).

Thank you, I just needed to know why B was wrong. This cleared things up for me.

 

[ilovemcat]

Thank you, I just needed to know why B was wrong. This cleared things up for me.

No problem. I made slight revision. Molecular Oxygen reacting with molecular Hydrogen is an overall exothermic reaction. Regardless, you should be able to see how stability of the bonds plays a role in BDE. A higher BDE means more energy is required to break that bond, or if you're forming a bond, BDE refers to the energy released (for that bond). The overall energy released (heat of enthalpy) is found by subtracting all the bonds formed from all the bonds broken. If this value is negative, it's an exothermic reaction, if positive - an endothermic reaction.

 

[Thirteeen]

Thank you, I just needed to know why B was wrong. This cleared things up for me.

I guess I should have clarified my reasoning. I marked B as incorrect since it didn't specifically mention the 1st IE.

 

[Rabolisk]

I can never turn down a chance to critique people.. even ilovemcat.

Choice D on the other hand, does provide a reasonable explanation. Because molecular Nitrogen (triple bond) is more stable than Oxygen (double bond), more energy is required to break it's bond than say breaking the bond of an Oxygen molecule. In other words, the reaction between molecular Nitrogen and Hydrogen is a more endothermic reaction than the reaction between molecular Oxygen and Hydrogen (which is actually exothermic overall).

I actually calculated the values just to be sure:

O=O + H-H --> H-O-O-H

deltaH = (119 + 104.2) - (2(35) + 222)
deltaH = -68.8 (EXOTHERMIC overall)

NN + H-H --> H-N=N-H

deltaH = (226 + 104.2) - (2(93) + 109)
deltaH = +35.2 (ENDOTHERMIC overall)

A couple things here. First, the reaction between H2 and N2 would go on to produce NH3, rather than N2H2, and that reaction is exothermic, although much less so than the formation of water. More importantly, however, thermodynamics does not determine rate. A reaction could be endothermic and be faster than an exothermic reaction. The reason ammonia forms at a slower rate is because the activation energy is higher due to the very high bond enthalpy of the nitrogen triple bond.

I guess I should have clarified my reasoning. I marked B as incorrect since it didn't specifically mention the 1st IE.

Really? It seems more like you forgot that nitrogen has a higher 1st IE than oxygen to me. After all, when they ionization energy without any qualifier, you can assume that they mean the first ionization energy. Also, everything you've written suggests that you simply forgot. Of course, that didn't really matter for this question, since the reason why B is wrong and D is correct has nothing to do with the validity of B in itself.

 

[ilovemcat]

I feel like such a ditz now, thanks to Mr. Rabo here. Sorry about that SeeEell, but he's correct. 😀 Next time I should read the question more closely. 🙂

 

[ridethecliche]

Rabolisk for forum police!

(Really though, the answer came down to double vs triple bond?)

 

[Rabolisk]

Rabolisk for forum police!

Personally, I prefer Commissioner Rabo.

(Really though, the answer came down to double vs triple bond?)

As a student of chemistry, it's easy to see that nitrogen triple bond, which is one of the strongest bonds in chemistry (if not the strongest?) is stronger than the oxygen double bond. However, you can deduce D just by realizing that A, B, and C are not true.

 

[ridethecliche]

Personally, I prefer Commissioner Rabo.



As a student of chemistry, it's easy to see that nitrogen triple bond, which is one of the strongest bonds in chemistry (if not the strongest?) is stronger than the oxygen double bond. However, you can deduce D just by realizing that A, B, and C are not true.

Aye sir.

And yeah, I know that triple bonds are much more stable than doubles, but I just thought that it was interesting how everything was overthought to becoming so complicating and then just came back to such a simple concept that most seemed to overlook.

I was just using the moment to show that the answer is often the simple thing that many of us overlook.

 

[SeeEll]

Thank you for the clarification of ...the clarification. Quick question though. I thought that products formed from multiple reactants require energy to form bonds, hence they are all endothermic. The opposite holding true for reactions where bonds are broken. Am I simplifying things or am I missing something huge?

edit: Wow, sorry about that. I'm just doing passage after passage here and just reversed the concept in my head. Bonds cleaving = endothermic. bond formation = exothermic.

 

[ModusProbandi]

I thought that products formed from multiple reactants require energy to form bonds, hence they are all endothermic.

Wut? Example please.