double displacment

[jeff einstein]

recently I have been studying methods of separation in chemistry in school. When I browse up some past papers it involves some hard double displacement reactions I am not aware of and the state it results in. I have searched the internet for identifying products of double displacement reaction but I have not gotten any useful information. Please give me an easy method to find the state matter of a compound by looking at its chemical formula (I mostly need it for ionic compounds). Also, it helped me find the products of double displacement reactions.

[chenbeier]

Give an example, what fo you mean.

[Sensei]

8 hours ago, jeff einstein said:

recently I have been studying methods of separation in chemistry in school. When I browse up some past papers it involves some hard double displacement reactions I am not aware of and the state it results in. I have searched the internet for identifying products of double displacement reaction but I have not gotten any useful information. Please give me an easy method to find the state matter of a compound by looking at its chemical formula (I mostly need it for ionic compounds). Also, it helped me find the products of double displacement reactions.

It is very simple: from the state in which compounds have more energy (E=mc^2) it goes to the state in which they have less energy.. e.g. CO₂ and H₂O have less energy/mass per particle than sugar, fuel or so. The difference is extremely small, counted in eV/c^2 per molecule (and in chemistry expressed as J/mols or so), so by looking at g/mol is impossible to determine in advance, as the difference is billions times less visible.

Such reactions are exothermic/exoenergetic i.e goes without any additional energy from external source.

In the reverse direction, energy is needed from an external source. e.g. no energy source is needed to convert a metal to a metal oxide. But to convert a metal oxide into a metal, an external energy source is needed. This is why so few metals in nature occur in the metallic state (for example, gold, which does not readily form compounds with other elements).

The energy is need to break bounds between elements, and the energy is released in breaking bounds between elements. It depends on the element and compound. Look at reactivity chart of elements:

https://www.google.com/search?q=activity+chart+chemistry+wikipedia

https://en.wikipedia.org/wiki/Reactivity_series

 

 

Some (most?) elements/compounds need a certain "activation energy" to start the whole reaction:

https://en.wikipedia.org/wiki/Activation_energy

That's why fuels or explosives do not spontaneously combust or explode in most cases.

 

[exchemist]

7 hours ago, Sensei said:

It is very simple: from the state in which compounds have more energy (E=mc^2) it goes to the state in which they have less energy.. e.g. CO₂ and H₂O have less energy/mass per particle than sugar, fuel or so. The difference is extremely small, counted in eV/c^2 per molecule (and in chemistry expressed as J/mols or so), so by looking at g/mol is impossible to determine in advance, as the difference is billions times less visible.

Such reactions are exothermic/exoenergetic i.e goes without any additional energy from external source.

In the reverse direction, energy is needed from an external source. e.g. no energy source is needed to convert a metal to a metal oxide. But to convert a metal oxide into a metal, an external energy source is needed. This is why so few metals in nature occur in the metallic state (for example, gold, which does not readily form compounds with other elements).

The energy is need to break bounds between elements, and the energy is released in breaking bounds between elements. It depends on the element and compound. Look at reactivity chart of elements:

https://www.google.com/search?q=activity+chart+chemistry+wikipedia

https://en.wikipedia.org/wiki/Reactivity_series

 

 

Some (most?) elements/compounds need a certain "activation energy" to start the whole reaction:

https://en.wikipedia.org/wiki/Activation_energy

That's why fuels or explosives do not spontaneously combust or explode in most cases.

 

I don't think this is very helpful. It's obvious the reaction won't proceed unless ΔG for the process is -ve, but the poster will not have tables of Gibbs free energy to hand and should not need them. E=mc² is neither here nor there, and nor is activation energy in this context.  

Double displacement reactions generally involve ionic compounds in solution, exchanging partners, i.e. AB + CD -> AD + CB, A and C being cations and B and D being anions. What drives the reaction to the right is the removal of one of the two product species from solution, usually because it is insoluble and precipitates.

All our poster needs to do is apply this idea to the examples he or she has been given. But of course it does require considering  the charge on the various cations and anions involved, in order to come up with the right formulae for the reaction products, and knowing - or guessing - which of these salts are insoluble.

An example would be BaCl₂(aq) + Na₂SO₄(aq) -> BaSO₄↓ + 2NaCl(aq). Both ionic compounds on the left are water soluble but barium sulphate is insoluble and precipitates, leaving only sodium chloride in solution. The driving force is the affinity for Ba²⁺ for SO₄²⁻, which prefer to bind to each other rather than to water molecules.

(If you like to express it thermodynamically,  ΔG is -ve for the process  Ba²⁺(aq) + SO₄²⁻(aq) -> BaSO₄(s), or, conversely that barium sulphate has a +ve solvation energy in water, so it does not dissolve. However this is not really germane to the poster's question.)

 

Edited November 6, 2023 by exchemist

[Sensei]

1 hour ago, exchemist said:

I don't think this is very helpful. It's obvious the reaction won't proceed unless ΔG for the process is -ve, but the poster will not have tables of Gibbs free energy to hand and should not need them. E=mc² is neither here nor there, and nor is activation energy in this context.  

Of course, it is everywhere. It's like learning a language from the alphabet (characters). First the alphabet (characters), then words, then sentences. Your description (though it's accurate) is like starting from the opposite direction. Starting to read the book from the middle. That is, the person will have no idea where this Gibbs energy comes from and why all this happened (or not in the case of reaction which does not occur)..

I was trying to say why the reaction occurs without bothering with what these A, B, C, D are.

In my country, physics is/was taught from the same direction as discoveries and laws in historical order. Bizarre. Hook's law is/was taught first. I would say, it is stupefying people.. People learn about the energy released by a reaction and have no idea where that energy comes from.

 

Edited November 6, 2023 by Sensei

[John Cuthber]

1 hour ago, Sensei said:

Hook's law is/was taught first. I would say, it is stupefying people..

Is that why you answered a post on double displacement reactions without there being anything related to double displacement reactions in your reply?

[exchemist]

2 hours ago, Sensei said:

Of course, it is everywhere. It's like learning a language from the alphabet (characters). First the alphabet (characters), then words, then sentences. Your description (though it's accurate) is like starting from the opposite direction. Starting to read the book from the middle. That is, the person will have no idea where this Gibbs energy comes from and why all this happened (or not in the case of reaction which does not occur)..

I was trying to say why the reaction occurs without bothering with what these A, B, C, D are.

In my country, physics is/was taught from the same direction as discoveries and laws in historical order. Bizarre. Hook's law is/was taught first. I would say, it is stupefying people.. People learn about the energy released by a reaction and have no idea where that energy comes from.

 

This is chemistry. If you insist on starting from first principles of physics every time when addressing a chemistry question,  you will never reach an answer for anything but the most trivially simple systems. Physics can't accurately model anything in chemistry more complex than the hydrogen molecule ion. In chemistry the systems are far too complex for that type of approach.

You don't seem to have understood the question that was asked. Nobody asked about energy release. You introduced that yourself and it's not relevant. I only included some comments about it in my reply because that was the aspect that seemed to interest you.  

Anyway, let's see if @jeff einstein responds to this thread. Then we may see if any of this has been useful. 

 

Edited November 6, 2023 by exchemist

[jeff einstein]

Like @exchemist said I don't want anything about energy. an example of what I am asking is a reaction between calcium chloride and sodium carbonate or a reaction between potassium chloride solution. I want to know two things what is the result of the two examples (i already know the answer) and what is the stat of the results, so i know if there is any precipitate at the end? I know the answer to this but what if I got a double displacement reaction between two unknown compounds how do I find all of this Isn't there supposed to be a method for finding out or something?

 

[exchemist]

38 minutes ago, jeff einstein said:

Like @exchemist said I don't want anything about energy. an example of what I am asking is a reaction between calcium chloride and sodium carbonate or a reaction between potassium chloride solution. I want to know two things what is the result of the two examples (i already know the answer) and what is the stat of the results, so i know if there is any precipitate at the end? I know the answer to this but what if I got a double displacement reaction between two unknown compounds how do I find all of this Isn't there supposed to be a method for finding out or something?

 

OK thanks for coming back to clarify. If you dissolve 2 salts and all the combinations of ions are soluble compounds, then you won't get any displacement reaction as such, you will just get a mixed solution with all the ions dissolved. For instance, NaCl(aq) + CuSO₄(aq) will just give a solution with separate Na⁺, Cu²⁺, Cl⁻ and SO₄²⁻ ions, all happily solvated and swimming around. That's because there is no combination that has markedly lower solubility than the others. (If you were to concentrate the solution enough you you would eventually exceed the solubility limit of the least soluble combination and you would start to get that one precipitating out. I don't know without looking it up which one that would be.)  

In your example. CaCO₃ (chalk, limestone) is well known to have limited solubility so that would precipitate, leaving you with NaCL(aq). KCl has similar solubility to NaCl.  But if you are asking if there is a rule to predict which salts have high solubility and which ones have low solubility that gets involved and is quite hard.  There are slightly handwavy explanations e.g. that when cations and anions are of similar size they pack more efficiently in the crystal structure, with small gaps, and thus tend to have a high lattice energy. This makes them reluctant to trade that stability for the attraction of a cage of polar water molecules. This is one explanation for why BaSO₄ is insoluble, for instance. Both are big ions, Ba because it is in the 6th row of the Periodic table and sulphate because it is compound ion in the shape of a nice tetrahedron so it can pack more or less like a sphere. Both also have a double charge on them which increases electrostatic attraction in the crystal and also makes it a bit harder for water to stabilise them fully when solvated. Carbonate does not pack like sulphate as it is a planar triangle, and you need to know a bit of crystallography to understand how efficiently it can form a structure with Ca. But Ca also has a double +ve charge so is a bit more challenging for water to stabilise in solution. So, to he honest, this kind of argument strikes me as being better at rationalising after the fact than really predicting with confidence. 

If this is a school level question I think you will just need to know which of the common salts are insoluble and which are soluble. You won't get asked about obscure combos.

There is a rule of thumb in the link here which may help a bit: https://chem.libretexts.org/Courses/College_of_the_Canyons/Chem_201%3A_General_Chemistry_I_OER/04%3A_Introduction_to_Solutions/4.05%3A_Solubility_of_Ionic_Compounds

 

[jeff einstein]

Thanks for your help!!