SmellNoEvil
Harmless
Posts: 15
Registered: 23-2-2016
Member Is Offline
Mood: No Mood
|
|
Reaction Calculations.
I would like an explanation of how I would determine whether or not a reaction is possible. For instance: How would I determine whether or not the
reaction:
CO + H2O --> CO2 + H2
is possible?
I know for certain that the reaction above will not occur, but what about determining whether or not other reactions will happen? I've heard of
thermodynamic calculations to determine whether or not a reaction will proceed (Follow the link). Can somebody explain in more detail how to determine
whether or not a certain reaction will happen(Without actually trying to carry it out)?
Reference:
http://www.sciencemadness.org/talk/viewthread.php?tid=29710&...
[Edited on 4-3-2016 by SmellNoEvil]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by SmellNoEvil  | I would like an explanation of how I would determine whether or not a reaction is possible. For instance: How would I determine whether or not the
reaction:
CO + H2O --> CO2 + H2
is possible?
I know for certain that the reaction above will not occur, but what about determining whether or not other reactions will happen? I've heard of
thermodynamic calculations to determine whether or not a reaction will proceed (Follow the link). Can somebody explain in more detail how to determine
whether or not a certain reaction will happen(Without actually trying to carry it out)?
Reference:
http://www.sciencemadness.org/talk/viewthread.php?tid=29710&...
[Edited on 4-3-2016 by SmellNoEvil] |
This a complex, multi-faceted subject and I'm loathe to try and summarise it in a few lines but just for starters:
1. Thermodynamic considerations:
Take a simple reaction:
nA + mB ===> kC + lD
To determine whether such a reaction is thermodynamically favourable or not, we need to calculate the Change in Gibbs Free energy
ΔG<sup>0</sup> (where <sup>0</sup> stands for STP conditions) that it would cause (IF it proceeded).
This is calculated as:
ΔG<sup>0</sup> = k ΔG<sup>0</sup>C + l ΔG<sup>0</sup>D - [n
ΔG<sup>0</sup>A + l ΔG<sup>0</sup>B]
where the individual terms are the Gibbs Free Energy Changes of Formation of each reactant or reaction product. These values are usually obtained from
various databases like NIST webbook and others.
If ΔG<sup>0</sup> < 0, then the reaction is exo-energetic and thermodynamically favourable.
But if the reaction is carried out in conditions different from STP, corrections for temperature and/or pressure to ΔG<sup>0</sup> need
to be applied.
2. Kinetic considerations:
Even if ΔG < 0, this says precious little about reaction rate. For all we know, a reaction may be thermodynamically favourable in the
calculated conditions, yet only proceed at an imperceptibly slow rate in those conditions.
But maybe there are other conditions of temperature and/or pressure that favour higher reaction rates? That part of the story requires introduction
into various other concepts like Collision Theory and Activation Energy.
This resource is a quite a good low level entry one:
http://www.chemguide.co.uk/physmenu.html
So don't expect to learn all this overnight: most of us interested in it are perennial students of it! 
[Edited on 4-3-2016 by blogfast25]
|
|
|
Metacelsus
International Hazard
Posts: 2539
Registered: 26-12-2012
Location: Boston, MA
Member Is Offline
Mood: Double, double, toil and trouble
|
|
| Quote: Originally posted by SmellNoEvil | For instance: How would I determine whether or not the reaction:
CO + H2O --> CO2 + H2
is possible?
I know for certain that the reaction above will not occur |
What do you mean it won't occur? It definitely occurs (under proper conditions), and is well-known: https://en.wikipedia.org/wiki/Water-gas_shift_reaction
Basically, to determine whether a reaction occurs spontaneously, compute the Gibbs free energy change.
Edit: Ninja'd
[Edited on 3-4-2016 by Metacelsus]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Nope. Plenty of reactions have highly negative ΔG, yet don't proceed at RT. And they don't necessarily proceed quickly at higher T either (see
Activation Energy).
Most combustion reactions, e.g., don't proceed at RT. By your reasoning the world would be on fire all the time. 
Look at the ΔG < 0 as a 'necessary but not sufficient condition' for a reaction to proceed.
Please don't try and simplify a complex subject: you end up spreading misinformation.
[Edited on 4-3-2016 by blogfast25]
|
|
|
SmellNoEvil
Harmless
Posts: 15
Registered: 23-2-2016
Member Is Offline
Mood: No Mood
|
|
I came up with the reaction off the top of my head, I didn't actually think it was something that would occur.
Blogfast,can you please give an example calculation of the reaction between sodium and chlorine gas,or any other simple reaction, in a "For dummies"
type of explanation?
[Edited on 4-3-2016 by SmellNoEvil]
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by SmellNoEvil |
Blogfast,can you please give an example calculation of the reaction between sodium and chlorine gas,or any other simple reaction in a "For dummies"
type of explanation?
|
Tomorrow. The NaCl example is too stupid to kill because the values for Na and Cl2 are both zero. So the Free Energy Change for that
reaction is simply the tabulated value for the Free Energy Change of Formation of NaCl (about - 384.1 kJ/mol NaCl).
Nite.
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
Example: Classic Thermite:
Fe2O3 + 2 Al ====> 2 Fe + Al2O3
Data (all from Wolfram Alpha):
ΔGf<sup>0</sup> Fe2O3 = - 742.2 kJ/mol
ΔGf<sup>0</sup> Al2O3 = - 1582 kJ/mol
The elements Al and Fe have zero values.
So for the reaction:
ΔGR<sup>0</sup> = - 1582 - 0 - [- 742.2 - 0] = - 839.8 kJ/mol
So ΔGR<sup>0</sup> < 0: thermodynamically favourable. But note that a mixture of Fe2O3 + 2 Al does
not spontaneously ignite: it requires an ignition pill or mixture to light up and proceed at a practical rate.
[Edited on 5-3-2016 by blogfast25]
|
|
|
Metacelsus
International Hazard
Posts: 2539
Registered: 26-12-2012
Location: Boston, MA
Member Is Offline
Mood: Double, double, toil and trouble
|
|
Sorry for the confusion; I meant spontaneous in a (narrow) thermodynamic sense, with regards to the direction of a process. Free energy (Gibbs or Helmholtz, depending on whether it's constant pressure or
volume) can determine this kind of spontaneity. However, Blogfast is right in that just because a reaction is thermodynamically favorable under some
conditions does not mean that it will take place at a non-negligible rate under those conditions.
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
