I've been trying to computationally achieve the following (I have no lab, I'm 14 years old, and I want to make a Minecraft mod to teach real Chemistry):
I want to mathematically predict the range of temperature and pressure (if the reaction is between gases) or temperature and concentration (if the reaction is in a solution) in which a given chemical reaction can happen, and the rate at which it happens, with the variation of those two arguments.
EDIT 1:
Regarding the first part of @sixtytrees answer I suppose I must calculate $K$.
I assume $\Delta G$ would be:
$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt] &= -162.5 - 137.3 - 0\\[6pt] &= -299.8\ \mathrm{\frac{kJ}{mol}}\end{align}$$
Now to solve for $K$:
(I believe $R$ must be $0.0083144598\ \mathrm{\frac{kJ}{mol\ K}}$, because $\Delta G$ is in $\mathrm{\frac{kJ}{mol}}$
Also, I think we have to take $25\ \mathrm{^\circ C}$ as the temperature which is $298\ \mathrm K$)
$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt] &= \frac{-299.8}{-0.0083144598 \cdot 298}\\[6pt] &= \frac{-299.8}{-2.4777090204}\\[6pt] &= 120.99887336713996006405304847878\\[6pt] K &= \mathrm e^{120.99887336713996006405304847878}\end{align}$$
And now I realize I must have made some mistake because this number will be too big.
☹
EDIT 2:
I just woke up and I noticed I miscalculated $\Delta G$, because I added them together when I should be subtracting. Let's see how it goes now.
$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt] &= -162.5 - (-137.3 - 0)\\[6pt] &= -25.2\ \mathrm{\frac{kJ}{mol}}\end{align}$$
Now $K$ will be:
$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt] &= \frac{-25.2}{-0.0083144598 \cdot 298}\\[6pt] &= \frac{-25.2}{-2.4777090204}\\[6pt] &= 10.170685820053125395644218884808\\[6pt] K &= \mathrm e^{10.170685820053125395644218884808}\\[6pt] &= 26126\end{align}$$
Still not as small as 10 that @sixtytrees predicted.
$$26126 = \frac{[\ce{CH3OH}]^1}{[\ce{H}]^2[\ce{CO}]^1}$$
What now?
EDIT 3:
I've been thinking, and decided Chemical Equilibrium is not something I need to bother with, because as the reaction progresses, it is assumed that the final products are drained out as they are formed, and additional pressure is added to account for the reduction of total molarity. This way I don't need to distinguish between reversible reactions, irreversible reactions, and virtually irreversible reactions (when there are trace amounts of the reactants in the final products).
This only works if the reactants are gases, I don't know how to sustain the rate of the reaction if I take away the products from a solution.
The original question remains, now that equilibrium is irrelevant, is it possible to predict the temperature and pressure/concentration range of the reactants in which the reaction can occur?
I downloaded Orca, by the way. No idea how to use it.
EDIT 4:
Ok, so let´s say this is all happening in a sealed container with 1L of volume, so
$\begin{align} 26126 &= \frac{[CH_3OH]}{2^2 \cdot 1^1}\\ [CH_3OH] &= 26126 \cdot 4 = 104,504 \end{align}$
What does this mean? It can't mean I end up with more than one hundred moles in 1 liter. It makes no sense, I don't have enough reactants to get so much product. So, I am clearly doing it all wrong, can someone explain it to me?