IChO 2006 · Theory problem 3

Interstellar chemistry

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Early interstellar chemistry is thought to have been a prelude to life on Earth. Molecules can be formed in space via heterogeneous reactions at the surface of dust particles, often called the interstellar ice grains (IIGs). Imagine the reaction between \(\ce{H}\) and \(\ce{C}\) atoms on the IIG surface that forms \(\ce{CH}\). The \(\ce{CH}\) product can either be desorbed from the surface or further react through surface migration with adsorbed \(\ce{H}\) atoms to form \(\ce{CH2}\), \(\ce{CH3}\), etc.

Depending on how energetically a molecule “jumps” from its anchored site, it either leaves the surface permanently (desorption) or returns to a new position at the surface (migration). The rates of desorption and migratory jump follow the Arrhenius formula, \(k = A\exp(-E/RT)\), where \(k\) is the rate constant for desorption or migratory jump, \(A\) the jumping frequency, and \(E\) the activation energy for the respective event.

3.1 Desorption of \(\ce{CH}\) from the IIG surface follows first-order kinetics. Calculate the average residence time of \(\ce{CH}\) on the surface at \(20\ \mathrm{K}\). Assume that \(A = 1\times10^{12}\ \mathrm{s^{-1}}\) and \(E_{\mathrm{des}} = 12\ \mathrm{kJ\,mol^{-1}}\).

3.2 Consider the shortest time it would take for one \(\ce{CH}\) unit to move from its initial position to the opposite side of an IIG by successive migratory jumps. Assume that the activation energy for migration (\(E_{\mathrm{mig}}\)) is \(6\ \mathrm{kJ\,mol^{-1}}\), and the IIG is a sphere with a \(0.1\ \mathrm{\mu m}\) radius. Each migratory jump laterally advances the molecule by \(0.3\ \mathrm{nm}\). Show work and choose your answer from (a) – (e) below.

(a) \(t \leq 1\ \text{day}\)

(b) \(10\ \text{days} \leq t \leq 10^2\ \mathrm{yr}\)

c) \(10^3\ \mathrm{yr} \leq t \leq 10^6\ \mathrm{yr}\)

(d) \(10^7\ \mathrm{yrs} \leq t \leq 10^{10}\ \mathrm{yr}\)

(e) \(t \geq 10^{11}\ \mathrm{yr}\)

3.3 Consider the reaction of \(\ce{CO}\) with \(\ce{H2}\) to form \(\ce{H2CO}\). The activation energy on a metal catalyst is \(20\ \mathrm{kJ\,mol^{-1}}\) that is produced by formaldehyde at a rate of 1 molecule/s per site at \(300\ \mathrm{K}\). Estimate the rate of formaldehyde formation per site if the reaction takes place at \(20\ \mathrm{K}\).

3.4 Which is a set of all true statements? Circle one.

(a) Most \(\ce{CH}\) species are desorbed from the IIG surface before encountering other reactants by surface migration.

(b) IIGs can assist transformation of simple molecules to more complex ones in interstellar space.

(c) For a reaction on the IIG to occur at an appreciable speed during the age of the Universe (\(1\times10^{10}\ \mathrm{yr}\)), the reaction energy barrier must be absent or negligible.

\(\square\) (a)  \(\square\) (b)  \(\square\) (c)  \(\square\) (a, b)  \(\square\) (a, c)  \(\square\) (b, c)  \(\square\) (a, b, c)

Official English text of the problem.

Answers

  • 3.1: \(\boxed{\tau \approx 2.2\times10^{19}\ \mathrm{s} \approx 7.0\times10^{11}\ \mathrm{yr}}\).
  • 3.2: \(\boxed{t \approx 4.9\times10^6\ \mathrm{s} \approx 57\ \text{days}}\); choose (b).
  • 3.3: Formaldehyde (C=O) formation rate: \(\boxed{r_{20}\approx1.8\times10^{-49}\ \text{molecules}\,\mathrm{s^{-1}\,site^{-1}}}\).
  • 3.4: Circle (b, c), within the problem’s thermal Arrhenius model.

Solution

Use \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\) and \(1\ \mathrm{yr}\approx3.156\times10^7\ \mathrm{s}\).

3.1 — Average residence time

The adsorbed species is methylidyne, \(\ce{CH}\), SMILES [CH]. Its attachment to the grain is not represented by this SMILES.

For first-order desorption, the probability that a molecule remains adsorbed after time \(t\) is \[ P(t)=e^{-k_{\mathrm{des}}t}. \] Therefore its mean residence time is \[ \tau=\int_0^\infty P(t)\,dt=\frac{1}{k_{\mathrm{des}}}. \]

The desorption rate constant is \[ \begin{aligned} k_{\mathrm{des}} &=A\exp\left(-\frac{E_{\mathrm{des}}}{RT}\right)\\ &=(1.0\times10^{12}\ \mathrm{s^{-1}}) \exp\left[-\frac{12000\ \mathrm{J\,mol^{-1}}} {(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})(20\ \mathrm{K})}\right]\\ &=4.55\times10^{-20}\ \mathrm{s^{-1}}. \end{aligned} \]

Thus \[ \boxed{\tau=\frac{1}{k_{\mathrm{des}}} \approx2.2\times10^{19}\ \mathrm{s} \approx7.0\times10^{11}\ \mathrm{yr}}. \]

This is the mean lifetime; the half-life would instead be \((\ln 2)/k_{\mathrm{des}}\).

3.2 — Shortest migration time

Use the same temperature and jumping frequency as in 3.1. The migration rate constant is \[ \begin{aligned} k_{\mathrm{mig}} &=(1.0\times10^{12}\ \mathrm{s^{-1}}) \exp\left[-\frac{6000}{8.314\times20}\right]\\ &=2.13\times10^{-4}\ \mathrm{s^{-1}}. \end{aligned} \] The mean waiting time per jump is consequently \[ \tau_{\mathrm{jump}}=\frac{1}{k_{\mathrm{mig}}} =4.69\times10^3\ \mathrm{s}. \]

The shortest surface path between opposite points of a sphere is half a great circle: \[ L=\pi r=\pi(0.1\times10^{-6}\ \mathrm{m}) =3.14\times10^{-7}\ \mathrm{m}. \]

If every jump advances along this path, the required number of jumps is \[ N\approx\frac{L}{\ell} =\frac{3.14\times10^{-7}\ \mathrm{m}} {0.3\times10^{-9}\ \mathrm{m}} =1.05\times10^3. \]

Hence the shortest-path travel-time estimate is \[ \begin{aligned} t&=N\tau_{\mathrm{jump}}=\frac{N}{k_{\mathrm{mig}}}\\ &\approx4.91\times10^6\ \mathrm{s}\\ &\approx56.8\ \text{days}\approx0.156\ \mathrm{yr}. \end{aligned} \]

Thus \[ \boxed{\text{Choose (b): }10\ \text{days}\leq t\leq10^2\ \mathrm{yr}.} \]

This estimate assumes that all jumps advance toward the opposite point. Random migration generally requires more jumps.

3.3 — Formaldehyde formation at \(20\ \mathrm{K}\)

The balanced reaction is \[ \ce{CO + H2 -> H2CO}. \]

The species are carbon monoxide, \(\ce{CO}\), [C-]#[O+]; molecular hydrogen, \(\ce{H2}\), [H][H]; and formaldehyde (methanal), \(\ce{H2CO}\), C=O.

Assuming the same activation energy and effective rate prefactor at both temperatures, the Arrhenius rate ratio is \[ \frac{r_{20}}{r_{300}} =\exp\left[-\frac{E_a}{R} \left(\frac{1}{20\ \mathrm{K}}-\frac{1}{300\ \mathrm{K}}\right)\right]. \]

Substituting \(E_a=20000\ \mathrm{J\,mol^{-1}}\), \[ \frac{r_{20}}{r_{300}} =\exp\left[-\frac{20000}{8.314} \left(\frac{1}{20}-\frac{1}{300}\right)\right] =e^{-112.26} =1.76\times10^{-49}. \]

Since \(r_{300}=1\ \text{molecule}\,\mathrm{s^{-1}\,site^{-1}}\), \[ \boxed{r_{20}\approx1.8\times10^{-49}\ \text{molecules}\,\mathrm{s^{-1}\,site^{-1}}}. \]

3.4 — Evaluation of the statements

(a) False. Migration is enormously faster than desorption: \[ \frac{k_{\mathrm{mig}}}{k_{\mathrm{des}}} =\exp\left(\frac{E_{\mathrm{des}}-E_{\mathrm{mig}}}{RT}\right) =\exp\left(\frac{6000}{8.314\times20}\right) \approx4.7\times10^{15}. \] Thus, in the absence of reaction, a methylidyne species can make approximately \(4.7\times10^{15}\) migratory jumps on average before desorbing. Migration provides extensive opportunity to encounter other adsorbed reactants.

(b) True. The grains retain reactants and permit encounters through surface migration. Successive hydrogen additions can build more complex species: \[ \ce{C + H -> CH}, \] \[ \ce{CH + H -> CH2}, \] \[ \ce{CH2 + H -> CH3}. \] Here carbon and hydrogen atoms have SMILES [C] and [H]; methylene, \(\ce{CH2}\), has SMILES [CH2]; and the methyl radical, \(\ce{CH3}\), has SMILES [CH3]. These equations describe species on the surface. The SMILES do not specify their surface bonding or spin multiplicities.

(c) True in the intended thermal-activation model. At \(20\ \mathrm{K}\), even the \(20\ \mathrm{kJ\,mol^{-1}}\) barrier in 3.3 makes the reaction negligible over the stated age of the Universe. The expected number of formaldehyde molecules formed per site in that time is \[ \begin{aligned} r_{20}t_{\mathrm{Universe}} &=(1.76\times10^{-49}\ \mathrm{s^{-1}\,site^{-1}}) (10^{10}\times3.156\times10^7\ \mathrm{s})\\ &\approx5.6\times10^{-32}\ \mathrm{site^{-1}}. \end{aligned} \] Appreciable thermally activated chemistry therefore requires an absent or sufficiently small reaction barrier.

Consequently, circle \[ \boxed{(b,\ c)}. \]

The solution was checked against the official solution.