aamc 11 PS #11--spontaneous

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Bumbl3b33

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This question is perplexing me. It asked

Based on the passage, Reaction 1 (decomposition of nitroglycerin) at 25C most likely has:

The answer is NEGATIVE (delta)G and positive (delta)S

I don't understand why delta-G is negative when in the passage it says that "it undergoes decomposition violently when heated or shocked". I thought, without calculating deltaH from the data, that since it needed an external source of energy, it'd be non-spontaneous. Is there something I'm misunderstanding about spontaneity?

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All bond-breaking requires an external source of energy. However, more energy is released when the bonds are formed. "violently" should have gave you the clue that a lot of energy was being released during the reaction.

With a -H and +S you can only have a -G
 
All bond-breaking requires an external source of energy. However, more energy is released when the bonds are formed. "violently" should have gave you the clue that a lot of energy was being released during the reaction.

With a -H and +S you can only have a -G
I didn't take the exam, but I'm sort of confused about this. I thought that when you decompose, entropy increases since there are more separate particles, but enthalpy is positive because when you decompose you break bonds. Breaking bonds requires energy, so energy was put into the sytem (endothermic reaction). Am I missing something in my reasoning? Therefore, Delta G = Delta H - TDelta S and Delta G would be negative only if Tdelta S > delta H
 
You should know: delta H = (delta H formation of bonds broken) - (delta H formation of bonds formed)

In decomposition, bonds are breaking in the reactants, but new ones are formed in the products.

For example in here:

2H2O(I) → 2H2 + O2

We are breaking 2OH bonds and forming O-O double bonds and H-H single bonds.
 
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This question is perplexing me. It asked

Based on the passage, Reaction 1 (decomposition of nitroglycerin) at 25C most likely has:

The answer is NEGATIVE (delta)G and positive (delta)S

I don't understand why delta-G is negative when in the passage it says that "it undergoes decomposition violently when heated or shocked". I thought, without calculating deltaH from the data, that since it needed an external source of energy, it'd be non-spontaneous. Is there something I'm misunderstanding about spontaneity?
It's not necessarily non-spontaneous if the rxn receives an external source of energy (usually heat). Spontaneity has to do with whether the rxn releases any free energy (Lowering its energy state afterward). When you heat/shock a rxn, you overcome its energy barrier (transition state) to proceed it forward. This is called endothermic process (added heat). In biological/physical process, we do this with catalysts. I don't have your references, so I can only tell you this much.

It's a pos. delta S hence the word "decomposition".
 
It's not necessarily non-spontaneous if the rxn receives an external source of energy (usually heat). Spontaneity has to do with whether the rxn releases any free energy (Lowering its energy state afterward). When you heat/shock a rxn, you overcome its energy barrier (transition state) to proceed it forward. This is called endothermic process (added heat). In biological/physical process, we do this with catalysts. I don't have your references, so I can only tell you this much.

It's a pos. delta S hence the word "decomposition".

this process was exothermic...can you maybe explain it in that context? i'm a little confused. it makes sense if it were endothermic
 
It's not necessarily non-spontaneous if the rxn receives an external source of energy (usually heat). Spontaneity has to do with whether the rxn releases any free energy (Lowering its energy state afterward). When you heat/shock a rxn, you overcome its energy barrier (transition state) to proceed it forward. This is called endothermic process (added heat). In biological/physical process, we do this with catalysts. I don't have your references, so I can only tell you this much.

It's a pos. delta S hence the word "decomposition".

yeah, endothermic/exothermic refer to the change in enthalpy, this is a state function and depends upon the products and reactants, not the activation energy.
 
this process was exothermic...can you maybe explain it in that context? i'm a little confused. it makes sense if it were endothermic
If it's an exothermic process, then heat energy was released. Under constant pressure, the change in heat (delta H) is negative. So it makes sense that the rxn is spontaneous. But you said it happens when being heated/shocked? Which one is it? You need to know the diff. in input/output energies to determine the process. There may be some relevant data/words that you overlook. A possibly misinterpretation?
 
You should know: delta H = (delta H formation of bonds broken) - (delta H formation of bonds formed)

In decomposition, bonds are breaking in the reactants, but new ones are formed in the products.

For example in here:

2H2O(I) → 2H2 + O2

We are breaking 2OH bonds and forming O-O double bonds and H-H single bonds.
Shouldn't it be molar enthalphy of products - molar enthalpy of reactants? So.... my question is..... I know for a fact that sometimes the bonds formed do not release more heat than the breaking of the bonds. How do you know in this case that the reaction was exothermic? Was there more information, like a table?
 
There was a table.

And your formula "molar enthalphy of products - molar enthalpy of reactants" is also correct.

bonds broken - bonds formed also works.
 
can anyone help me on #8 it seems like a super easy problem that i got totally wrong. i know one mole = 22.4 liters at STP. So what I thought to do was use a proportion. One mole = 22.4 Liters. so 22.4 * 6/29 = # moles of N2.

4 moles of reactant produces 29 moles of product, 6 of which come from N2. where did i go wrong?


8. At STP, the volume of N2(g) produced by the complete decomposition of 1 mole of nitroglycerin would be closest to which of the following?

A) 5 L B) 10 L C) 20 L D) 30 L


I believe the answer is supposed to be D.
 
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