Can someone help me out with understanding inorganic chem?

In summary: NH3 + H2O + CO2 —> (NH4)2CO3The first one is a reduction because the two nitrogen atoms are reducing agents.
  • #1
Vriska
138
2
2Ca(NO3)2 -> 2CaO + 4NO2 + O2

Why is this happening? Are there any steps I can follow to get to the product? Something like with CaCO3 you remove CO2 because Ca polarizes the radical reducing bond strength..

Honestly I'm having a lot of difficulty with inorganic chemistry, id like to think there's a reason for why stuff happens (like reaction mechanism in organic chemistry, tendencies and general steps you could follow - like that CaCO3 thing above or like this one :

2NH3 + H2O + CO2 —> (NH4)2CO3

It looked random to me but i realized i could :

2NH3 + H2CO3 -> (NH3)2CO3 as an acid base reaction

Or perhaps something like this

2NaHCO3 -> Na2CO3 + CO2 + H2O

For this I'll juggle

NaHCO3 -> NaOH +CO2

, NaHCO3 +NaOH as an acid bass reaction because that H on the bicarbonate looks kinda acidic (like in organic chem)

Yields Na2CO3 + H2O

The above might not be correct bc everything is in solid state but you see, i want something like this so i don't have to memorize a buttload of reactions. Am I supposed to be doing this? Am I on the right track? If not what the hay should i do with these billion reactions, they seem random like the first one(how do i do that btw (if actually i can do anything at all))? Someone help, inorganic is hell for me : /
 
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  • #2
We want to help you. But some details are needed for us to get an idea of where you are, which at the moment seems to cornered by the prospect of rote memorization.

Undergrad inorganic chem class? Or Freshman Chemistry? Home study?

And there is sense behind all this, but you seem not to have encountered any set of "rules of thumb" or anything else more explanatory to help you see. Example - the old idea of redox potential.

@TeethWhitener may have some ways to get up to speed.
 
  • #3
Note: in most cases in inorganic chemistry there are no mechanisms to consider. For most of the time we deal with rather small, thermodynamically stable molecules (think water, carbon dioxide) and it is this stability that plays an important role. Plus, several simple rules about acid-base reactions, salt creation from different compounds and solubility. Things will get more complicated when you get to redox reactions, but then you should already "feel" the logic behind most simple problems.
 
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Likes jim mcnamara
  • #4
jim mcnamara said:
We want to help you. But some details are needed for us to get an idea of where you are, which at the moment seems to cornered by the prospect of rote memorization.

Undergrad inorganic chem class? Or Freshman Chemistry? Home study?

And there is sense behind all this, but you seem not to have encountered any set of "rules of thumb" or anything else more explanatory to help you see. Example - the old idea of redox potential.

@TeethWhitener may have some ways to get up to speed.

I'm home studying. I'm somewhat comfortable with using reduction potential tables , somewhat alright with periodic trends and molecular structure. Not quite clear with Gibbs yet or HSAB theory.

And no, I'm not sure if I've heard of many rules of thumb except 'remove stable molecule' which I'm not having a lot of success with. I get thrown off badly with stuff that looks random, like the first reaction i mentioned, borax stuff, acids of phosphorus etc..

Edit : and just now this - > KO2 -> KOH + H2O2 + O2?
I get if it where K2O2 how the H2O2 might come there but what's with this oxygen
 
Last edited:
  • #5
Borek said:
Note: in most cases in inorganic chemistry there are no mechanisms to consider. For most of the time we deal with rather small, thermodynamically stable molecules (think water, carbon dioxide) and it is this stability that plays an important role. Plus, several simple rules about acid-base reactions, salt creation from different compounds and solubility. Things will get more complicated when you get to redox reactions, but then you should already "feel" the logic behind most simple problems.

Can you give me an example using the stability of small molecules? I've read about removing the stable molecule from larger ones but am not quite clear on how to do it.>salt creation from different compounds and solubility

What is this exactly?
 
  • #6
Vriska said:
Can you give me an example using the stability of small molecules? I've read about removing the stable molecule from larger ones but am not quite clear on how to do it.

Here you go:

Vriska said:
2Ca(NO3)2 -> 2CaO + 4NO2 + O2
Vriska said:
2NaHCO3 -> Na2CO3 + CO2 + H2O

Vriska said:
>salt creation from different compounds and solubility

What is this exactly?

What methods of salt synthesis do you know?

And google "solubility rules".
 
  • #7
Borek said:
Here you go:

Heyyyyy, i get the second one as i rationalized it but what about the first? Why not make it Ca(NO3)2 -> N2 +CaO2 + 2O2, they're all small, stable molecules.

What are salt synthesis methods? Google isn't helping with that
 
  • #8
Vriska said:
Heyyyyy, i get the second one as i rationalized it but what about the first? Why not make it Ca(NO3)2 -> N2 +CaO2 + 2O2, they're all small, stable molecules.
This is a fantastic question. In short, you would get there eventually if you went to high enough temperatures. NO2 dissociates at moderately high temperatures to NO and O2, but this is an entropically driven transformation (NO2 is actually enthalpically more stable than NO). Eventually, at even higher temperatures, you would start to see a transformation of NO back to N2 and O2. As you indicated, these are the stable, small molecules that you would conclude would be the products if you only took thermodynamics into account. However, kinetics comes into play too, and at more moderate temperatures, the system gets trapped in local energetic minima: these might not be the most stable products overall, but the activation barrier in getting from these products to the most stable products might be prohibitively high at lower temperatures.
Vriska said:
Someone help, inorganic is hell for me : /
I sympathize, and these sorts of exceptions frequently pop up and are not guided by overarching principles but rather by considerations that are unique to particular systems (especially in your example of calcium nitrate). That's unfortunately the nature of the game at this level of chemistry.
 
  • #9
Vriska said:
What are salt synthesis methods? Google isn't helping with that

There are several possible combinations of acid (and acid anhydride) and base (and base anhydride) that can produce salt (and sometimes water). Say, CaO+SO2, or NaOH+CO2, this kind of things. These are important not only during synthesis, but also quite often hint at what will happen during decomposition.
 
  • #10
TeethWhitener said:
This is a fantastic question. In short, you would get there eventually if you went to high enough temperatures. NO2 dissociates at moderately high temperatures to NO and O2, but this is an entropically driven transformation (NO2 is actually enthalpically more stable than NO). Eventually, at even higher temperatures, you would start to see a transformation of NO back to N2 and O2. As you indicated, these are the stable, small molecules that you would conclude would be the products if you only took thermodynamics into account. However, kinetics comes into play too, and at more moderate temperatures, the system gets trapped in local energetic minima: these might not be the most stable products overall, but the activation barrier in getting from these products to the most stable products might be prohibitively high at lower temperatures.

I sympathize, and these sorts of exceptions frequently pop up and are not guided by overarching principles but rather by considerations that are unique to particular systems (especially in your example of calcium nitrate). That's unfortunately the nature of the game at this level of chemistry.

Ah i see, but now how am I supposed to guess activation energies at that temperature?
 
  • #11
8
Borek said:
There are several possible combinations of acid (and acid anhydride) and base (and base anhydride) that can produce salt (and sometimes water). Say, CaO+SO2, or NaOH+CO2, this kind of things. These are important not only during synthesis, but also quite often hint at what will happen during decomposition.

where can I read up more on this? I'm really not finding anything googling "methods of salt synthesis"
 
  • #12
This is strange - it was part of my course eons ago, it is a real chemistry, yet apparently it is not taught this way any more (or at least it is not taught this way in English speaking countries).
 
  • #13
Vriska said:
Edit : and just now this - > KO2 -> KOH + H2O2 + O2?
I get if it where K2O2 how the H2O2 might come there but what's with this oxygen

KO2 contains the superoxide anion ##\mathrm{O}_2^-##. The product of hydrolysis ##\mathrm{HO_2}## will disproportionate immediately into H2O2 and O2.
This process is known as "dismutation" and is a special case of a disproportionation reaction. There exist also enzymes in the body, the superoxide dismutases, which catalyse this process .
 
  • #14
Vriska said:
Ah i see, but now how am I supposed to guess activation energies at that temperature?
Unfortunately, you really can't. That's why I said I sympathize. In the case of calcium nitrate, there's no really good system for reasoning it out. You just have to know it. (I will freely admit that if I had to work on these types of problems for my research, I'd simply look up the answers).
 
  • #15
Vriska said:
Heyyyyy, i get the second one as i rationalized it but what about the first? Why not make it Ca(NO3)2 -> N2 +CaO2 + 2O2, they're all small, stable molecules.
CaO2 is certainly not too stable upon heating and will decay into CaO +1/2 O2. But as a matter of fact, N2 and O2 are more stable at not too high temperatures than NO2, so yes, this decay is possible and these products are also thermodynamically the most stable, however, the decay of NO2 into O2 and N2 is kinetically unfavourable.
 
  • #16
DrDu said:
KO2 contains the superoxide anion ##\mathrm{O}_2^-##. The product of hydrolysis ##\mathrm{HO_2}## will disproportionate immediately into H2O2 and O2.
This process is known as "dismutation" and is a special case of a disproportionation reaction. There exist also enzymes in the body, the superoxide dismutases, which catalyse this process .

Ah thank you, this helps a lot. what about Na2O2 though? it looks like it degrades into H2O2 without oxygen?
 
  • #17
Vriska said:
Ah thank you, this helps a lot. what about Na2O2 though? it looks like it degrades into H2O2 without oxygen?
Which anion does Na2O2 contain?
 
  • #18
DrDu said:
Which anion does Na2O2 contain?

peroxide, gottit thanks
 

1. What is inorganic chemistry?

Inorganic chemistry is the branch of chemistry that deals with the study of inorganic compounds, which are substances that do not contain carbon-hydrogen bonds. It also includes the study of elements, their properties, and their reactions.

2. What are the main topics covered in inorganic chemistry?

The main topics covered in inorganic chemistry include atomic structure, periodic table trends, chemical bonding, chemical reactions, and the properties and uses of inorganic compounds.

3. What is the difference between inorganic and organic chemistry?

Inorganic chemistry deals with the study of inorganic compounds, while organic chemistry deals with the study of organic compounds, which are substances that contain carbon-hydrogen bonds. Inorganic compounds are generally simpler and have fewer variations compared to organic compounds.

4. Why is inorganic chemistry important?

Inorganic chemistry is important because it helps us understand the properties and behavior of elements and compounds, which are essential to many areas of science and technology. It also plays a crucial role in the development of new materials and technologies, such as pharmaceuticals, semiconductors, and catalysts.

5. How can I improve my understanding of inorganic chemistry?

To improve your understanding of inorganic chemistry, it is important to first have a strong foundation in basic chemistry principles. You can also practice by solving problems and doing experiments to gain hands-on experience. Additionally, seeking help from a tutor or joining a study group can also be beneficial.

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