Sciencemadness Discussion Board - Predicting Reactions and Outcomes

Predicting Reactions and Outcomes

aga - 17-6-2014 at 08:41

How do you tell whether a reaction will occur, under what conditions it may occur, and what the products will be ?

I have started with reaction types, charge states, reactivity series and delta H calculations.
e.g. What appears to be a Double Replacement reaction :-
NaCl + KOH -> NaOH + KCl = -16 kJ/mol
so that tells me it will be hardly a reaction at all.
in reverse, NaOH + KCl -> NaCl + KOH = 16 kJ/mol
so needs energy to happen, if indeed it can happen

All ions are either + or - 1 charges so no clues there.
K is more reactive than Na, however is Cl more reactive than OH ?

Probably a bad reaction to chose, but i did it anyway.
As no Gas is evolved, nothing appeared to happen.

Is there a logical sequence to follow to predict If a reaction will occur (under specified conditions), the outcome(s) of a reaction, and if so, what is it please ?

DraconicAcid - 17-6-2014 at 08:56

For a double replacement reaction, it will only occur if you're going to form a non-electrolyte. This could be water, a gas, or an insoluble solid, none of which can form here. The relative reactivities of Na and K, or Cl vs OH are irrelevant- these would only matter for a single replacement reaction.

blogfast25 - 17-6-2014 at 09:16

Whether reactions are thermodynamically favourable or not depends on whether ΔG is negative or not.

That calculation has to done using the correct data for the correct state of matter: reactions between pure reagents will lead to different ΔG than for reagents in solution for instance...

[Edited on 17-6-2014 by blogfast25]

aga - 17-6-2014 at 11:20

@draconicA :
I found a rule rule for double replacement reactions that says that 'both of the reactants have to be soluble, and only one of the products'

So is that where you'd stop, and say with confidence that the reaction will not occur ?
(i.e. a non-electrolyte is Not made ?)

@blogfast25: the ΔH is -ve (very slightly).
I guess that's not the ΔG then ?

[Edited on 17-6-2014 by aga]

DraconicAcid - 17-6-2014 at 11:30

Quote: Originally posted by aga

@draconicA :
I found a rule rule for double replacement reactions that says that 'both of the reactants have to be soluble, and only one of the products'

So is that where you'd stop, and say with confidence that the reaction will not occur ?
(i.e. a non-electrolyte is Not made ?)

That works for precipitation reactions. Soluble reactants giving an insoluble product will give a precipitation. Most acid-base reactions are also double replacements; HCl and NaOH -> NaCl + H₂O will go, not because either sodium chloride or water are insoluble, but because water is a non-electrolyte. HCl + NaCN -> NaCl + HCN will also go (HCN is a very weak acid, thus a very weak electrolyte. At higher than room temperatures, it's also a gas).

aga - 17-6-2014 at 11:42

Thanks for that.
I kinda get it, apart from 'weak electrolyte'.
So the rule is not non-electrolyte, but not-strng-enough electrolyte.
What would be the Limit of electrolyte-y-ness ?

The real question is how is it Known.
Look up in a table of 'what works' or is there a mathematical way to determine whether a reaction will happen or not ?

DraconicAcid - 17-6-2014 at 11:44

Quote: Originally posted by aga

Thanks for that.
I kinda get it, apart from 'weak electrolyte'.
So the rule is not non-electrolyte, but not-strng-enough electrolyte.
What would be the Limit of electrolyte-y-ness ?

Soluble ionic compounds are strong electrolytes. So are strong acids and bases.

Weak acids and bases are weak electrolytes.

aga - 17-6-2014 at 11:55

Quote:

For a double replacement reaction, it will only occur if you're going to form a non-electrolyte

Or, if one of the products isn;t enogh of an electrolyte to matter (as in the HCN case you cited)

So there must be a cut-off as to how much of an electrolyte the product has to be before it is too electrolytic to allow the reaction.

DraconicAcid - 17-6-2014 at 12:05

Quote: Originally posted by aga

Quote:

For a double replacement reaction, it will only occur if you're going to form a non-electrolyte

Depends on what you define as a reaction "going". Many such reactions will a measurable equilibrium constant, which tells you how far a reaction will go. The more soluble the precipitate being formed, or the stronger the electrolyte you're forming, the smaller the equilibrium constant will be.

aga - 17-6-2014 at 12:06

In other words, look up Equilibrium Constant.

OK

aga - 17-6-2014 at 12:25

Single Replacement reactions :-

Check the Reactivity Series table.
If the Element is higher on the list, the reaction will occur.

Double Replacement reactions :-

This is more concise than i can ever be :-
http://www.kentchemistry.com/links/Kinetics/PredictingDR.htm

Synthesis, Combustion and Decompostion next.

blogfast25 - 17-6-2014 at 12:28

Quote: Originally posted by aga

@blogfast25: the ΔH is -ve (very slightly).
I guess that's not the ΔG then ?

No, they are related but not the same.

ΔG = ΔH - TΔS

The 'entropic term', TΔS, is the product of temperature (at which the reaction proceeds) and the change in Entropy the system undergoes (when it goes from initial state 1 to end state 2, so that ΔG = G2 - G1

.

In chemistry, Entropy is a measure of orderliness or probability of a system and the Universe seeks to minimise orderliness and maximise chaos (or randomness, if you will). S is high for high probability, high state of disorder, high level of chaos.

A positive change ΔS means the system has gone from more orderly to less orderly and because of - TΔS, such a change makes ΔG smaller.

So reactions are favoured when:

1. ΔH < 0
2. ΔS > 0
3. So that ΔG = ΔH - TΔS < 0

aga - 17-6-2014 at 13:10

Cool. Thanks.

So look up Entropy, Gibbs' Free Energy and Reaction Rates then.

Ok.

Hegi - 17-6-2014 at 13:38

The real game starts when you have reactions which are theoretically (calculating entropy, gibbs energy) impossible to do but anyway they exist beacuse of equilibrium shifts while the reaction proceeds

blogfast25 - 17-6-2014 at 13:50

Hmmm... to make things more intriguing reaction, rates have nothing to do with Gibb's Free Energy but I'm going to let you find that out for yourself!

"In"soluble?

chornedsnorkack - 17-6-2014 at 22:49

Between NaCl, KOH, NaOH and KCl, all of them are pretty soluble. But remember that all of them are solids at room temperature and none of them is a liquid miscible with water. So none of their solubilities is infinite, nor can they possibly be equal.
At 25 Celsius, the solubilities are:
NaCl: 360 g/1000 g water
KCl: also 360 g/1000 g water
NaOH: over 1100 g/1000 g water
KOH: 1210 g/1000 g water.

When you mix saturated solutions of NaCl and KOH, and then evaporate the solution to dryness, or freeze it, what do you get? Do you get your original NaCl and KOH crystals back, or something else? And in which order?

blogfast25 - 18-6-2014 at 10:49

Quote: Originally posted by chornedsnorkack

At 25 Celsius, the solubilities are:
NaCl: 360 g/1000 g water
KCl: also 360 g/1000 g water
NaOH: over 1100 g/1000 g water
KOH: 1210 g/1000 g water.

When you mix saturated solutions of NaCl and KOH, and then evaporate the solution to dryness, or freeze it, what do you get? Do you get your original NaCl and KOH crystals back, or something else? And in which order?

As you remove the water, the least soluble components reach their solubility limit first, so they start crystallising first (NaCl and KCl). Only when the solubility limits of the more soluble components are reached do they start crystallizing.

This is often encountered in large salt deposits: they are stratified with the least soluble salts at the bottom and the more soluble ones higher up, near the surface. Of course quantity also plays a part: minority constituents may not reach their solubility limit if there isn't much of them and be found near the top.

[Edited on 18-6-2014 by blogfast25]

AJKOER - 19-6-2014 at 02:57

Some finer points, for one, the relative volatility of the reactants can be important. In instances of heating solids, the more volatile compound may exit first and drive the reaction in that direction. Similarly, for example, if boiling a reaction mixture involving acids, the reaction can be moved in the direction where the more volatile acid is boiled off, even if it is the 'stronger' acid.

On a more advance note, the reaction products can, at times, be determined by physical (or material) chemistry considerations. For example, adding Na to KClO3 in H2SO4 results in H2 gas and aqueous sulfate and chlorate. But adding Zn to KClO3 in H2SO4 forms less than 'expected' H2 gas, aqueous sulfate, chlorate and the chlorite!. The hydrogen/Zn in the latter case is capable of reducing KClO3 (see http://books.google.com/books?id=alVKAAAAYAAJ&pg=PA245&a... ).

For those seeking a more currrent reference, see for example an interesting 2012 published article at http://www.researchgate.net/publication/221934434_Chemical_r... , titled "Chemical reduction of an aqueous suspension of graphene oxide by nascent hydrogen" by Viet Hung Pham, Hai Dinh Pham, ... in Journal of Materials Chemistry (Impact Factor: 5.97). 05/2012; DOI:10.1039/C2JM30562C to quote from the abstract:

"ABSTRACT One of the major challenges in the chemical reduction of graphene oxide is increasing the C/O atomic ratio of the chemically-converted graphene. In this paper, we report a simple and effective method to reduce aqueous suspensions of graphene oxide using nascent hydrogen generated in situ by the reaction between Al foil and HCl, Al foil and NaOH and Zn powder and NaOH. The nascent hydrogen-reduced graphene oxides (nHRGOs) were characterized by elemental analysis, UV-vis spectra, Raman spectra, X-ray photoelectron spectroscopy, thermogravimetric analysis and electrical conductive measurements. The reduction efficiency of graphene oxide strongly depended on the reaction medium and the rate of nascent hydrogen generation. The best nHRGO achieved a C/O atomic ratio greater than 21 and a bulk electrical conductivity as high as 12,500 S/m, corresponding to the nascent hydrogen generated from the reaction between Al foil and HCl. Since nascent hydrogen could be produced on a metal surface upon oxidation in solution, other metals with low standard reduction potentials, such as Mg, Mn, and Fe, can be applied to reduce graphene oxide."

Note, the term 'nascent' is a historical misnomer as the likely nature of the reaction mechanics is due to H2 as created/residing on the surface of select low standard reduction potential metals per kinetics within the realm of materials chemistry.

[Edited on 19-6-2014 by AJKOER]

aga - 19-6-2014 at 13:21

On a very Basic note, does a System exist to Predict a reaction, given the Reactants, and the environment, or is it still try it-and-see, then measure it, then work out some maths, then claim that it is Science ?

The original question : is there a Process to determine if a reaction will proceed or not, and if so, under what conditions ?

(or is it mostly still guesswork and blowing stuff up)

100 years or less of actual Science.

"Ths knowledge has been handed down from father to son for generations."
"But we only just discovered it."
"OK. So we're mostly handing it Around at the moment."

Texium - 19-6-2014 at 18:46

I've been wondering the same thing as you, aga.
Earlier today I went around the internet trying to teach myself to calculate ∆G, but as I've only had one year of basic chemistry class and whatever I've been able to learn on my own, I found it rather hard to understand, well, with the information I had available for the reaction I'm trying to figure out anyway. I understand the basic concept, but it's getting all of that information simply from the reaction itself that is puzzling me.

Here's my situation: I was trying to prepare copper(II) chloride from copper(II) sulfate and calcium chloride, so I combined .01mol of each of them (I remembered to add in the 5H₂O on the CuSO₄ mass) both already in solution, and allowed the reaction to proceed to completion. I weighed the CaSO₄ that precipitated, but found that I had about half as much as there should have been, and I did take it into consideration that it is the dihydrate that was produced. My conclusion was that the reaction was at equilibrium, and the reactants and products were all in the system in equal ratios. Is this true, or did I make some simpler blunder that I carelessly missed? I tried to figure out the ∆G but was unsuccessful in finding it online or calculating it myself with information that I could find.

Also, one other thing that stumped me when trying to calculate the ∆G: when a hydrate is formed, do you have to consider water as a reactant when making the calculations and plug in its information too?

gdflp - 19-6-2014 at 19:33

Don't forget that calcium sulfate is NOT insoluble in water 2.4g/L at 25C. Especialy on a small scale like this it makes a difference. Try the same experiment with barium chloride, the results might be much closer. Delta G will not really be relevant in aqueous solution because ions will exchange freely with no or little change in energy (excludings acids and bases).

aga - 20-6-2014 at 00:03

Quote: Originally posted by zts16

trying to prepare copper(II) chloride from copper(II) sulfate and calcium chloride

It would be awesome, and greatly appreciated, if the workings-out of the delta G calculations for this reaction (or any other) were to be shown ...

chornedsnorkack - 20-6-2014 at 02:05

Quote: Originally posted by blogfast25

This is often encountered in large salt deposits: they are stratified with the least soluble salts at the bottom and the more soluble ones higher up, near the surface. Of course quantity also plays a part: minority constituents may not reach their solubility limit if there isn't much of them and be found near the top.

But if water is evaporated to dryness, ALL salts will reach their solubility limit in water, and precipitate.

On the other hand, salts may be miscible or have nonzero solubility limits in other solid salts.

If you have the metathesis... assuming you are using water in catalytic amounts then you do not need to know the exact delta-G of saturated solution mixture. You can find the delta-G for the 4 solids, and this will tell you the direction of the reaction.

But will you get your products (in the case, KCl and NaOH) as large crystals which you can identify by crystal habit and pick out, or as a mixture of very small crystals hard to segregate?

blogfast25 - 20-6-2014 at 05:32

Quote: Originally posted by zts16

My conclusion was that the reaction was at equilibrium, and the reactants and products were all in the system in equal ratios. Is this true, or did I make some simpler blunder that I carelessly missed?

I'm not sure about the 'equal ratios' but you are correct in saying that the reaction was at equilibrium, due to the smallish solubility of CaSO4:

Ca2+(aq) + SO42-(aq) < === > CaCO4(s)

Quote: Originally posted by gdflp

Don't forget that calcium sulfate is NOT insoluble in water 2.4g/L at 25C. Especialy on a small scale like this it makes a difference. Try the same experiment with barium chloride, the results might be much closer. Delta G will not really be relevant in aqueous solution because ions will exchange freely with no or little change in energy (excludings acids and bases).

Correct. Delta G isn't really necessary to predict reactions where an ionic precipitate forms. And better results would be obtained with barium.

But Delta G is nonetheless negative for such a reaction because of the lattice energy that is released when an ionic substance crystallised (precipitates) out. So these reactions are by no means some 'exception to a general rule': the rule is upheld.

Quote: Originally posted by chornedsnorkack

But will you get your products (in the case, KCl and NaOH) as large crystals which you can identify by crystal habit and pick out, or as a mixture of very small crystals hard to segregate?

The long and short of it is that this [evaporating to dryness] is not a great way of separating things, even though separation does occur. NaOH and KOH would have to be evaporated out under vacuum or inert blanket because they turn into carbonates in air.

[Edited on 20-6-2014 by blogfast25]

Texium - 20-6-2014 at 08:11

Alright, thanks for the help. I don't have any barium chloride, but I do have barium nitrate, so I'll try preparing copper(II) nitrate using that and copper(II) sulfate and see if that works.