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I was going to ask whether there are software that could be used to predict the products of any given chemical reaction. However, I then noticed these two earlier questions

where it is said such predictions are too difficult to make. What makes the prediction difficult?

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  • $\begingroup$ Welcome to chemistry.SE! If you had any questions about the policies of our community, you can ‎visit the help center or take a ‎‎tour of the website. $\endgroup$ – M.A.R. Jan 26 '15 at 15:33
  • $\begingroup$ I can give you a short answer: Mainly because there isn't only one type of chemical reaction and there aren't only a few factors that have to be taken into account when predicting. But I think "good answers are going to be too long for this format". $\endgroup$ – M.A.R. Jan 26 '15 at 15:50
  • $\begingroup$ I don't get why this was put on hold. The answer does not necessarily need to be overly long. DavePhD already provided some good basic information. $\endgroup$ – char Jan 26 '15 at 18:25
  • $\begingroup$ Dave provided a simple (no, not really!) example of how predicting the result would be hard. He didn't give all the reasons, but one. And I'm not the only guy with this idea, I suppose. $\endgroup$ – M.A.R. Jan 26 '15 at 18:58
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Because you need would need to solve the Dirac equation (relativistic Schrodinger equation) for all the particles in the system to get a completely accurate result.

Even solving the time independent non-relativistic Schrodinger equation for $\ce{H2+}$ in isolation, where there is only one electron and two protons involves approximating the protons as fixed relative to each other.

Now imagine trying solving a system of $10^{23}$ molecules each having multiple nuclei, and dozens or even hundreds of electrons, and including time dependence!

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  • $\begingroup$ While technically it is true, there are easier ways to have good guesses than this. Just as in Physics macroscopic Newtonian equations are good approximation for the behaviour of a pendulum, and you don't need to solve the Dirac equation for the atoms in the pendulum neither. $\endgroup$ – Greg Jan 26 '15 at 17:52
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    $\begingroup$ Yes, but we could argue the other way too, that the Dirac equation doesn't even completely explain the hydrogen atom (Lamb shift), or para versus orthohydrogen, or the London dispersion interaction (the switch from 1/r^6 to 1/r^7 dependence at long distance that Cashmir explained). $\endgroup$ – DavePhD Jan 26 '15 at 18:05

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