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Given the reactions:

$ H + H \rightarrow H_2$

$ H_2 + \gamma \rightarrow H + H$

Reaction rate 1 is given by:

$$\frac{d[H_2]}{dt} = c . n_H . [H]$$

and the Reaction 2 rate is given by:

$$\frac{d[H]}{dt} = c . [H_2]$$

where the square brackets represent the concentrations of the those species and c is just some number.

$n_H = [H] + 2[H_2]$ The number density of all hydrogen in the cloud

How can I find the concentrations of both [H] or $[H_2]$ from these equations, I keep trying but always end up stuck since I have two unknowns in everything I end up with, being [H] and $[H_2]$? Or an assumption that would simplify things.

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  • $\begingroup$ Note we do have MathJax enabled, so you can use equations rather than code blocks (see here for some details) $\endgroup$ – Kyle Kanos Mar 19 '17 at 23:00
  • $\begingroup$ [H] and [H2] are the concentrations of H and H2 $\endgroup$ – Pulchritude Mar 19 '17 at 23:14
  • $\begingroup$ No they're definitely the abundancies $\endgroup$ – Oh_canada Mar 19 '17 at 23:15
  • $\begingroup$ "where the concentration [A]" en.wikipedia.org/wiki/Reaction_rate#Rate_equation $\endgroup$ – Pulchritude Mar 19 '17 at 23:21
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    $\begingroup$ I guess so, I've always had it as abundancy. The more you know $\endgroup$ – Oh_canada Mar 19 '17 at 23:24
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Using your notation we can obviously use

$$[H_2] = \frac 1 c \frac {d[H]} {dt}$$

Unfortunately this leads (by substitution) to a decidedly non-linear second order differential equation for $[H_2]$ :

To simplify I'm going to use $\tau := ct$ and $D := \frac d {d\tau}$ and working it out we get :

$$D^2 [H] = [H]^2 + D( [H]^2 )$$

I don't know a solution to that offhand, but it's obviously something you can use numerically. As it's conveniently dimensionless that's a bonus. Maybe someone will recognize this equation.

Our friends on Mathematics SE might be able to offer advice on solving it, numerically, approximately or even exactly. I'd suggest replacing the $[H]$ with a simple $h$ to avoid confusion on Mathematics SE - they do strange things with brackets over there.

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  • $\begingroup$ I don't believe $c$ is the speed of light here, so your substitution, while valid, could confuse those more familiar with SR than astrophysical chemistry. $\endgroup$ – Kyle Kanos Mar 20 '17 at 12:21
  • $\begingroup$ It's not my symbol choice, so I'm just following the OP here. I personally don't find any confusion, and it's probably not a good practice for people to assume particular meanings for a symbol out of context anyway. My own normal practice would be to use a subscript like $c_0$ for a symbol like this, but YMMV as they say. $\endgroup$ – StephenG Mar 20 '17 at 12:42

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