Kinetics of Interstellar Chemistry — Applying steady state to formation of $\ce{H2}$
I am trying to solve problem 2 from the 2009 International Chemistry Olympiad (IChO), which is as follows:
If two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\ce{H2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\ce{H2}$ rapidly desorbs. This question examines two kinetic models for $\ce{H2}$ formation on the surface of a dust particle.
In both models, the rate constant for adsorption of $\ce{H}$ atoms onto the surface of dust particles is $k_a = 1.4\cdot10^{−5}/\mathrm{cm^3}$. The typical number density of $\ce{H}$ atoms (number of H atoms per unit volume) in interstellar space is $[\ce{H}] = 10/\mathrm{cm^3}$. (You may treat numbers of surface adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations).
Desorption of $\ce{H}$ atoms is first order with respect to the number of adsorbed atoms. 374
The rate constant for the desorption step is $k_d = 1.9 \cdot 10^{-3}/\mathrm{s}$. Assuming that only adsorption and desorption take place, calculate the steady-state number, $N$, of $\ce{H}$ atoms on the surface of a dust particle.
First off, what do they mean by steady-state number here? I tried searching online but I could not find anything. The solution they give has the equation:
$$k_dN − k_a[\ce{H}] = 0$$ from which they solve for $N$. I know this has something to do with steady state, but in this case why must the rate of absorption be equal to the rate of desorption. Is it because you are considering a single $\ce{H}$ atom which is considered to be an intermediate? Also, why is it $k_dN$?
Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?
In the above, how did they come up with those rates? Why does absorption have $\ce{[H]}x_m$ but desorption have $\ce{m}x_m$? Finally, for the reaction rate, where did they get $m(m-1)$? I think the $\frac{1}{2}$ comes from the fact that the rate of production of $\ce{H2}$ is half the rate of absorption/desorption, is that right?
From these, they calculate the rate of change of $x_0$, $x_1$, and $x_2$ as follows:
\begin{aligned} \frac{dx_{0}}{dt}&=-k_{a}\left[ \ce{H} \right]x_{0}\; +\; k_{d}x_{1}\; +\; k_{r}x_{2}\ \frac{dx_{1}}{dt}&=k_{a}\left[ \ce{H} \right]x_{0}\; -\; \left( k_{a}\left[ \ce{H} \right]\; +\; k_{d} \right)x_{1}\; +\; 2k_{d}x_{2}\ \frac{dx_{2}}{dt}&=k_{a}\left[ \ce{H} \right]x_{1}\; -\; \left( 2k_{d}\; +\; k_{r} \right)x_{2}\ \end{aligned}
I sort of understand how they got the value for rate of change of $x_0$ as it is the rate of production minus the rate of consumption, and desorption from state 1 and reaction produce state 0, while absorption consumes it. However, why does the $k_a$ term have [H] while the other terms do not? What happened to the $m x_m$ that was mentioned above?
Further, how did they get the rate of change for $x_1$ and $x_2$? 374