1) That helps a lot haha so in this case the couch represents the lower energy while the kitchen represents the higher energy. But what would your wife represent here? If i were to use the decomposition of MgCO3 in this case, MgCO3 would break into Mg2+ CO2 and O2- so in the intermediate stage. So this would prefer to go back to form MgCO3 as it's energetically favorable. However, when i cool down this solution the 'wife' would cause the particles to reform as MgO and CO2 rather than MgCO3. So what is the 'wife' that doing this to the intermediate stage particles?
My wife represents a second reaction that is taking me farther away from the couch. If my wife wasn't around, I'm not going to stand around in the kitchen. I'm going to go back to the couch becuse that's the lower energy state. But instead my wife tells me I need to run some errands. Having a peaceful home is an even lower energy state than watching TV on the couch, so then instead of going back to the couch I go to the store.
The point here is this:
Given a process A --> B: Even if the process is thermodynamically unfavorable, from simple probability
it does happen because there is latent energy in the environment. The thermodynamic favorability and unfavorability only impacts whether the equilibrium tends to favor A or B
, not whether the reaction happens. That is, at equilibrium the thermodynamical favorability impacts the average
relative concentrations of A and B (or, put another way, if you were to zoom in at a single molecule in the system at a given time, it impacts the relative probability that the molecule would be A or B). This is something students new to chemical thermodynamics don't typically grasp.
In addition, in a real, open system, one must consider the thermodynamical quantities of all possible reactions.
That is, a process like A --> B may be better described as A --> B --> C. Even if B is thermodynamically unfavorable (in which case the rate constant for A --> B is probably slow), C may be most favorable of all (in which B --> C rate contant is enormous). We know that A --> B is unfavorable, but it DOES happen. Even if it is so unfavorable that the ratio of A to B in solution at equilibrium is 1000:1, if the rate constant of the second conversion is very fast, every B that is formed in solution almost automatically gets transformed into C. Going from C back to B might be even MORE unfavorable than going from A to B (say - 10 billion to 1 for equilibrium concentrations). What this means is that over time, all A will be consumed and transformed to C, despite the fact that the first conversion is "thermodynamically unfavorable". The amount of time this takes will depend on the temperature, of course, the level of unfavorability of B wrt A, and other non-thermodynamical factors.
But, given infinite time, it will happen.
In my analogy, my couch is A and the kitchen is B. In an isolated system, my equilibrium will favor A. However in an open system, the world around me is summarized as C. As a representative example I've chosen my wife (a thermodynamical driving force if there ever was one) but it can be anything. While I'm sure we all love sitting on the couch and watching TV, we don't spend our lives there because we have other things that draw us away (jobs, the need to sleep, whatever). So while it's thermodynamically unfavorable to get up and do these other things, we do it because there are other thermodynamically favorable endpoints down the road.
2) Hmm I'm not very sure what you mean. Because i'm thinking that when i heat up the decomposing substance, bonds would be broken. So now that all the bonds required to be broken are broken, new bonds should form as in all decomposition reactions. However, if i were to leave the heating unit on, would those new bonds be able to form? For example in this reaction, MgCO3 MgO+CO2 the bonds between Mg2+ and CO3 2-, CO2 and O2- are being broken during the heating phase. So now the new bonds MgO has to form. But I'm thinking if i don't let the solution cool down the Mg2+ and the O2- cannot form. So i'd have a gaseous solution of Mg2+, O2- and CO2 (since all bonds are broken technically it has to be a gas rather than a molten liquid)?
Well yes, if you kept the heat on indefinitely you would not form stable molecules - at least, in the statistical average. (If you took a snapshot in time, you'd see some molecules anyway because there is a distribution of molecular energies.) Bond formation is a low energy state (a potential well or trough) and you need to remove energy to "trap" molecules there. In a closed fantasy system, you'd probably have some kind of really rapid interconversion between highly excited products and reactants. In reality, if you kept the heat high you'd access all kinds of downstream side products - in common parlance, you'd get charred goop. The trick to any reaction is heating long enough to favor your product state but not so long as to form too many unwanted downstream (and even lower energy) side products. In a kitchen, we call this "overcooking", and it's no good.
In the scheme above, we want to go from A --> B but cool down before getting too much of C. Of course, you're always going to have some of each because that's the nature of probability and kinetics, but if you use the right combination of temperatures and times, you can heat then cool and trap most of your molecules in a stable B intermediate state. Otherwise known as a burger cooked to medium!
Lastly, by 'then we remove the heat source and now maybe there isn't enough energy to surmount the activation energy for the reverse reaction.', do you mean now that we remove all the heat, there is not enough energy energy for the reverse reaction of the decomposition? Like MgO+CO2 MgCO3?
Yes. Though again there is ALWAYS some kind of equilibrium. In a binary system that's closed, the equilibrium may favor the products so much that the reverse reaction is virtually negligible at the temperature under consideration. This is especially true in real world systems where you might have a chain of fifty chemical conversions before you get to the end product. This is why, for example, cooking a chicken breast is considered "irreversible", because the probability of a single "cooked" molecule going through all the reverse reactions is small - then how low must be the probability of an entire 6 oz hunk of cooked meat (gazillions of molecules!) all going through the reverse process at the same time to yield its raw starting material? Probably so low that it wouldn't happen once in many, many lifetimes of the universe.
It's all probability, friend. That's the universe you live in.