IChO 2009 · Theory problem 2

Interstellar production of H2

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Interstellar production of \(\ce{H2}\)

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 \(k_a = 1.4 \cdot 10^{-5}\ \mathrm{cm^3\,s^{-1}}\). The typical number density of \(\ce{H}\) atoms (number of \(\ce{H}\) atoms per unit volume) in interstellar space is \([\ce{H}] = 10\ \mathrm{cm^{-3}}\).

[Note: In the following, 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. As a result, the units of the rate constants may be unfamiliar to you. Reaction rates have units of numbers of atoms or molecules per unit time.]

2.1 Calculate the rate at which \(\ce{H}\) atoms adsorb onto a dust particle. You may assume that this rate is constant throughout.

Desorption of \(\ce{H}\) atoms is first order with respect to the number of adsorbed atoms. The rate constant for the desorption step is \(k_d = 1.9 \cdot 10^{-3}\ \mathrm{s^{-1}}\).

2.2 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.

The \(\ce{H}\) atoms are mobile on the surface. When they meet they react to form \(\ce{H2}\), which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants \(k_a\), \(k_d\), and \(k_r\), for adsorption, desorption, and bimolecular reaction, as given below.

\[ \begin{aligned} k_a &= 1.4 \cdot 10^{-5}\ \mathrm{cm^3\,s^{-1}}\\ k_d &= 1.9 \cdot 10^{-3}\ \mathrm{s^{-1}}\\ k_r &= 5.1 \cdot 10^4\ \mathrm{s^{-1}} \end{aligned} \]

Model A

Reaction to form \(\ce{H2}\) is assumed to be second order. On a dust particle the rate of removal of \(\ce{H}\) atoms by reaction is \(k_rN^2\).

2.3 Write down an equation for the rate of change of \(N\), including adsorption, desorption and reaction. Assuming steady state conditions, determine the value of \(N\).

2.4 Calculate the rate of production of \(\ce{H2}\) per dust particle in this model.

Model B

Model B attempts to analyse the probability that the dust particles carry 0, 1 or 2 \(\ce{H}\) atoms. The three states are linked by the following reaction scheme. The assumption is made that no more than 2 atoms may be adsorbed simultaneously.

Figure 1

\(x_0\), \(x_1\) and \(x_2\) are the fractions of dust particles existing in state 0, 1 or 2, respectively. These fractions may be treated in the same way as concentrations in the following kinetic analysis. For a system in state \(m\) with fraction \(x_m\), the rates of the three possible processes are

Adsorption \((m\rightarrow m+1)\) : rate \(=k_a[\ce{H}]x_m\)

Desorption \((m\rightarrow m-1)\) : rate \(=k_dmx_m\)

Reaction \((m\rightarrow m-2)\) : rate \(=\frac12 k_rm(m-1)x_m\)

2.5 Write down equations for the rates of change, \(\mathrm{d}x_m/\mathrm{d}t\), of the fractions \(x_0\), \(x_1\) and \(x_2\).

2.6 Assuming steady-state conditions, use the above rate equations to find expressions for the ratios \(x_2/x_1\) and \(x_1/x_0\), and evaluate these ratios.

2.7 Evaluate the steady state fractions \(x_0\), \(x_1\) and \(x_2\).

[If you were unable to determine the ratios in 2.6, use \(x_2/x_1=a\) and \(x_1/x_0=b\) and give the result algebraically.]

2.8 Evaluate the rate of production of \(\ce{H2}\) per dust particle in this model.

2.9 It is currently not possible to measure the rate of this reaction experimentally, but the most recent computer simulations of the rate give a value of \(9.4 \cdot 10^{-6}\ \mathrm{s^{-1}}\). Which of the following statements apply to each model under these conditions? Mark any box you consider to be appropriate.

Statement Model A Model B Neither model
The rate determining step is adsorption of \(\ce{H}\) atoms.
The rate-determining step is desorption of \(\ce{H2}\) molecules.
The rate determining step is the bimolecular reaction of \(\ce{H}\) atoms on the surface.
The rate determining step is adsorption of the second \(\ce{H}\) atom.
The implicit assumption that reaction can take place regardless of the number of atoms adsorbed leads to substantial error (at least a factor of two).
Limiting the number of atoms adsorbed on the particle to 2 leads to substantial error (at least a factor of two).

Official English text of the problem.

Answers

Let \(F=k_a[\ce{H}]\). The molecular product is dihydrogen, \(\ce{H2}\), SMILES [H][H]. All production rates below are per dust particle.

  • 2.1: \(F=\boxed{1.4\times10^{-4}\ \text{atoms}\,\mathrm{s^{-1}}}\).
  • 2.2: \(\boxed{N=7.4\times10^{-2}}\) adsorbed atoms on average.
  • 2.3: \(\boxed{\mathrm{d}N/\mathrm{d}t=F-k_dN-k_rN^2}\), giving \(\boxed{N=5.2\times10^{-5}}\) adsorbed atoms on average.
  • 2.4: \(\boxed{R_A=\tfrac12k_rN^2=7.0\times10^{-5}\ \text{molecules}\,\mathrm{s^{-1}}}\).
  • 2.5: \[ \boxed{\begin{aligned} \frac{\mathrm{d}x_0}{\mathrm{d}t}&=-Fx_0+k_dx_1+k_rx_2,\\ \frac{\mathrm{d}x_1}{\mathrm{d}t}&=Fx_0-(F+k_d)x_1+2k_dx_2,\\ \frac{\mathrm{d}x_2}{\mathrm{d}t}&=Fx_1-(2k_d+k_r)x_2. \end{aligned}} \]
  • 2.6: \[ \boxed{\frac{x_2}{x_1}=\frac{F}{2k_d+k_r}=2.7\times10^{-9}}, \qquad \boxed{\frac{x_1}{x_0} =\frac{F(2k_d+k_r)}{k_d(2k_d+k_r)+k_rF} =6.9\times10^{-2}}. \]
  • 2.7: \(\boxed{x_0\simeq0.936,\quad x_1\simeq0.064,\quad x_2\simeq1.8\times10^{-10}}\).
  • 2.8: \(\boxed{R_B=k_rx_2=9.0\times10^{-6}\ \text{molecules}\,\mathrm{s^{-1}}}\).
  • 2.9: In the order printed: A; neither; neither; B; A; neither, distinguishing overall adsorption supply from second-atom adsorption. Under a broader reading of the first statement, B can also be marked there, since adsorption of the second atom is itself adsorption of an \(\ce{H}\) atom.

Solution

Atomic hydrogen, \(\ce{H}\), has SMILES [H]. The balanced surface processes are

\[ \ce{H(g) <=> H(ads)} \]

and, including the rapid release of the molecular product,

\[ \ce{2H(ads) -> H2(g)}. \]

Here “ads” denotes adsorption on dust; surface binding is not represented by the SMILES.

2.1

The adsorption rate per particle is

\[ F=k_a[\ce{H}] =\left(1.4\times10^{-5}\ \mathrm{cm^3\,s^{-1}}\right) \left(10\ \mathrm{cm^{-3}}\right). \]

Thus

\[ \boxed{F=1.4\times10^{-4}\ \text{H atoms per particle per second}.} \]

2.2

With adsorption and desorption alone,

\[ \frac{\mathrm{d}N}{\mathrm{d}t}=F-k_dN. \]

At steady state, adsorption balances desorption:

\[ N=\frac{F}{k_d} =\frac{1.4\times10^{-4}}{1.9\times10^{-3}} =0.073684\ldots \]

Therefore,

\[ \boxed{N=7.4\times10^{-2}\ \text{atoms per particle}.} \]

This is an average over particles or time; individual particles carry integer numbers of atoms.

2.3

The problem defines \(k_rN^2\) as the rate of removal of atoms by reaction. Hence

\[ \boxed{\frac{\mathrm{d}N}{\mathrm{d}t} =F-k_dN-k_rN^2.} \]

The steady-state condition gives

\[ k_rN^2+k_dN-F=0. \]

Only the nonnegative root is physically meaningful:

\[ N=\frac{\sqrt{k_d^2+4k_rF}-k_d}{2k_r}. \]

Substitution gives

\[ N= \frac{ \sqrt{(1.9\times10^{-3})^2+ 4(5.1\times10^4)(1.4\times10^{-4})} -1.9\times10^{-3}} {2(5.1\times10^4)} =5.2375\times10^{-5}. \]

Thus

\[ \boxed{N=5.2\times10^{-5}\ \text{atoms per particle}.} \]

2.4

Each molecule consumes two atoms, so the molecular production rate is half the atom-removal rate:

\[ R_A=\frac12k_rN^2. \]

Using the unrounded value of \(N\),

\[ R_A =\frac12(5.1\times10^4)(5.2375\times10^{-5})^2 =6.9950\times10^{-5}\ \mathrm{s^{-1}}. \]

Therefore,

\[ \boxed{R_A=7.0\times10^{-5}\ \text{molecules per particle per second}.} \]

The steady-state atom balance provides a check:

\[ 2R_A=F-k_dN. \]

2.5

The transition rates are

Transition Rate
\(0\rightarrow1\), adsorption \(Fx_0\)
\(1\rightarrow0\), desorption \(k_dx_1\)
\(1\rightarrow2\), adsorption \(Fx_1\)
\(2\rightarrow1\), desorption \(2k_dx_2\)
\(2\rightarrow0\), reaction \(k_rx_2\)

The factor of two for desorption from state 2 arises because either atom can desorb. For reaction in state 2,

\[ \frac12k_rm(m-1)x_m =\frac12k_r(2)(1)x_2=k_rx_2. \]

Adsorption from state 2 is excluded by the imposed two-atom limit. Taking incoming minus outgoing flux for each state gives

\[ \boxed{\begin{aligned} \frac{\mathrm{d}x_0}{\mathrm{d}t} &=-Fx_0+k_dx_1+k_rx_2,\\[2pt] \frac{\mathrm{d}x_1}{\mathrm{d}t} &=Fx_0-k_dx_1-Fx_1+2k_dx_2,\\[2pt] \frac{\mathrm{d}x_2}{\mathrm{d}t} &=Fx_1-2k_dx_2-k_rx_2. \end{aligned}} \]

Their sum is zero, as required for conservation of the total fraction:

\[ \frac{\mathrm{d}}{\mathrm{d}t}(x_0+x_1+x_2)=0. \]

2.6

At steady state, the equation for \(x_2\) gives

\[ Fx_1=(2k_d+k_r)x_2. \]

Consequently,

\[ a\equiv\frac{x_2}{x_1} =\frac{F}{2k_d+k_r} =\frac{1.4\times10^{-4}} {2(1.9\times10^{-3})+5.1\times10^4} =2.74510\times10^{-9}. \]

Thus

\[ \boxed{\frac{x_2}{x_1}=2.7\times10^{-9}.} \]

The steady-state equation for \(x_0\) gives

\[ Fx_0=k_dx_1+k_rx_2 =(k_d+k_ra)x_1. \]

Therefore,

\[ b\equiv\frac{x_1}{x_0} =\frac{F}{k_d+k_ra} =\frac{F(2k_d+k_r)} {k_d(2k_d+k_r)+k_rF}. \]

Numerically,

\[ b=0.06862745\ldots, \]

so

\[ \boxed{\frac{x_1}{x_0}=6.9\times10^{-2}.} \]

2.7

The fractions satisfy

\[ x_0+x_1+x_2=1, \qquad x_1=bx_0, \qquad x_2=abx_0. \]

It follows that

\[ \boxed{ x_0=\frac{1}{1+b+ab},\qquad x_1=\frac{b}{1+b+ab},\qquad x_2=\frac{ab}{1+b+ab}. } \]

Using the unrounded ratios,

\[ x_0=0.935779816\ldots,\qquad x_1=0.064220184\ldots,\qquad x_2=1.7629069\times10^{-10}. \]

Hence

\[ \boxed{x_0\simeq0.936,\qquad x_1\simeq0.064,\qquad x_2\simeq1.8\times10^{-10}.} \]

The extremely small \(x_2\) reflects the very short lifetime of a particle carrying two atoms.

2.8

Each \(2\rightarrow0\) reaction produces one molecule. Its transition rate already counts molecular formation events, so

\[ R_B=k_rx_2. \]

Therefore,

\[ R_B=(5.1\times10^4)(1.7629069\times10^{-10}) =8.990825\times10^{-6}\ \mathrm{s^{-1}}, \]

or

\[ \boxed{R_B=9.0\times10^{-6}\ \text{molecules per particle per second}.} \]

The corresponding atom balance is

\[ F(x_0+x_1) =k_d(x_1+2x_2)+2R_B, \]

which the calculated fractions satisfy.

2.9

Distinguishing overall adsorption supply from the specific second-atom arrival gives the following markings.

Statement Model A Model B Neither model
Adsorption of \(\ce{H}\) atoms is rate determining. ✓
Desorption of \(\ce{H2}\) molecules is rate determining. ✓
The bimolecular surface reaction is rate determining. ✓
Adsorption of the second \(\ce{H}\) atom is rate determining. ✓
Allowing reaction regardless of the number of adsorbed atoms causes an error of at least a factor of two. ✓
Limiting adsorption to two atoms causes an error of at least a factor of two. ✓

For Model A, desorption removes atoms at only

\[ k_dN=(1.9\times10^{-3})(5.2375\times10^{-5}) =9.95\times10^{-8}\ \mathrm{s^{-1}}, \]

which is negligible compared with \(F=1.4\times10^{-4}\ \mathrm{s^{-1}}\). Almost every arriving atom is therefore consumed in the model:

\[ R_A\simeq\frac{F}{2}. \]

The adsorption supply controls production.

For Model B, a singly occupied particle usually loses its atom before another arrives. The relevant mean times are

\[ \tau_{\mathrm{arrival}}=\frac1F\simeq7.1\times10^3\ \mathrm{s}, \qquad \tau_{\mathrm{desorption}}=\frac1{k_d}\simeq5.3\times10^2\ \mathrm{s}. \]

The probability that the second atom arrives before the first desorbs is

\[ \frac{F}{F+k_d}=0.0686\simeq6.9\%. \]

Once two atoms are present, their reaction is extremely fast:

\[ \tau_{\mathrm{reaction}}=\frac1{k_r} \simeq2.0\times10^{-5}\ \mathrm{s}. \]

Thus the crucial slow event is adsorption of a second atom while the first remains on the particle. Indeed, since \(k_r\) is very large,

\[ R_B\simeq Fx_1 \simeq\frac{F^2}{k_d+2F}. \]

The surface reaction is fast in both models, and the statement explicitly assumes rapid desorption of the newly formed molecule. Neither of those processes is rate determining.

Finally, comparison with the supplied simulation rate gives

\[ \frac{R_A}{R_{\mathrm{sim}}} =\frac{6.9950\times10^{-5}}{9.4\times10^{-6}} =7.44, \]

whereas

\[ \frac{R_B}{R_{\mathrm{sim}}} =\frac{8.9908\times10^{-6}}{9.4\times10^{-6}} =0.956. \]

Model A overestimates production by about 7.4-fold. A fractional mean occupancy is valid, but using \(N^2\) fails to represent the requirement that two atoms coexist on the same particle. Model B explicitly enforces that requirement.

Model B is only about 4.4% below the simulation value. Its two-atom limit therefore does not produce an error approaching a factor of two under these conditions.

The solution was checked against the official solution.