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I'm no chemist, but even though this should probably be obvious to me, I'm still unsure. If we were trying to figure out the time scale for a gas-phase reaction between two hydrogen atoms in a molecular cloud (which has density $~10^4/$cm$^3$), apparently the reaction would happen on a time scale proportional to the inverse of the density multiplied by $10^{15}$ years.

Aside from the cloud not being dense and the probability that a collision will surpass the activation energy is small, is the time elongated because you need to induce a dipole moment between two hydrogen atoms to actually bind them together?

Edit: I recently read somewhere that even if the reactants would go into the unbound state of $H_2$ they can't emit a photon that would let it go into the bound state of $H_2$. I guess they are talking about quantum mechanics, but it is not obvious to me what the bound and unbound states of molecular hydrogen are. (If someone could clarify that too)

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Is your timescale a rate, a half-life, the time it takes to reach a certain completion (i.e. 99%), the time it takes to reach equilibrium, etc.? If you can tell me what that timescale is, I will update my answer to be better. –  Ben Norris Nov 14 '12 at 16:51

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up vote 8 down vote accepted

Why is the time-scale so long?

The time-scale is so long because the density is not just low, it's virtually nonexistent.

The particle density of any gas can be determined at any temperature and pressure using the ideal gas law (PV=nRT). Let's use air as an example. At 25 oC (298 K), and 1 atmosphere, the particle density of air (n/V) is:

$$ \frac{n}V =\frac{RT}P $$ $$ \frac{n}V =\frac{(0.08026 \frac{\text{L atm}}{\text{K mol}})(298 \text{ K})}{1 \text{ atm}} = 23.9 \frac{\text{mol}}{\text{L}} $$

Converting to molecules per cm3:

$$23.9 \frac{\text{mol}}{\text{L}} \times 6.022\times10^{23}\frac{\text{molecules}}{\text{mol}}\times\frac{1 \text{ L}}{1000 \text{ cm}^3}=1.44\times10^{22}\frac{\text{molecules}}{\text{cm}^3}$$

The density of gasses we are used to dealing with is 18 orders of magnitude higher than the density you quoted. If we have a gas of hydrogen atoms at air density, then the timescale for formation of H2 is:

$$ 10^{15} \frac{\text{years}\times\text{cm}^3}{\text{particles}} \times \frac{1 \text{ cm}^3}{1.44\times10^{22}\text{ molecules}}=6.94\times 10^{-8}\text{ years} $$

$$6.94\times 10^{-8}\text{ years}= 2.54\times 10^{-5}\text{ days}=6.08\times 10^{-4} \text{ hours}=2.19 \text{ seconds}$$

At "normal" densities, the timescale is much shorter.

The second part of your question deals with the quantum mechanics of the activation barrier to this reaction, which I cannot do justice to at this time.

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Thanks for your help! –  Atreyu Nov 14 '12 at 20:54

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