IChO 2021 · Theory problem 1

Hydrogen at a Metal Surface

Download PDFIChO 2021
Figure 1

Hydrogen is expected to be a future energy source that does not depend on fossil fuels. Here, we will consider the hydrogen-storage process in a metal, which is related to hydrogen-transport and -storage technology.

PART A

As hydrogen is absorbed into the bulk of a metal via its surface, let us first consider the adsorption process of hydrogen at the metal surface, \(\ce{H2(g) -> 2H(ad)}\), where the gaseous and adsorbed states of hydrogen are represented as (g) and (ad), respectively. Hydrogen molecules (\(\ce{H2}\)) that reach the metal surface (M) dissociate at the surface and are adsorbed as \(\ce{H}\) atoms (Fig. 1). Here, the potential energy of \(\ce{H2}\) is represented by two variables: the interatomic distance, \(d\), and the height relative to the surface metal atom, \(z\). It is assumed that the axis along the two \(\ce{H}\) atoms is parallel to the surface and that the center of gravity is always on the vertical dotted line in Fig. 1. Fig. 2 shows the potential energy contour plot for the dissociation at the surface. The numerical values represent the potential energy in units of kJ per mole of \(\ce{H2}\). The solid line spacing is \(20\ \mathrm{kJ\,mol^{-1}}\), the dashed line spacing is \(100\ \mathrm{kJ\,mol^{-1}}\), and the spacing between solid and dashed lines is \(80\ \mathrm{kJ\,mol^{-1}}\). The zero-point vibration energy is ignored.

Figure 2

Fig.1 Definition of variables. Drawing is not in scale.

Figure 3

Fig.2

1.1 For each of the following items (i)–(iii), select the closest value from A–G.

(i) The interatomic distance for a gaseous \(\ce{H2}\) molecule

(ii) The interatomic distance between metal atoms (\(d_{\mathrm M}\) in Fig. 1)

(iii) The distance of adsorbed \(\ce{H}\) atoms from the surface (\(h_{\mathrm{AD}}\) in Fig. 1)

A. \(0.03\ \mathrm{nm}\) B. \(0.07\ \mathrm{nm}\) C. \(0.11\ \mathrm{nm}\) D. \(0.15\ \mathrm{nm}\)
E. \(0.19\ \mathrm{nm}\) F. \(0.23\ \mathrm{nm}\) G. \(0.27\ \mathrm{nm}\)

1.2 For each of the following items (i)–(ii), select the closest value from A–H.

(i) the energy required for the dissociation of gaseous \(\ce{H2}\) to gaseous \(\ce{H}\) [\(\ce{H2(g) -> 2H(g)}\)]

(ii) the energy released during the adsorption of a gaseous \(\ce{H2}\) [\(\ce{H2(g) -> 2H(ad)}\)]

A. \(20\ \mathrm{kJ\,mol^{-1}}\) B. \(40\ \mathrm{kJ\,mol^{-1}}\) C. \(60\ \mathrm{kJ\,mol^{-1}}\) D. \(100\ \mathrm{kJ\,mol^{-1}}\)
E. \(150\ \mathrm{kJ\,mol^{-1}}\) F. \(200\ \mathrm{kJ\,mol^{-1}}\) G. \(300\ \mathrm{kJ\,mol^{-1}}\) H. \(400\ \mathrm{kJ\,mol^{-1}}\)

PART B

The adsorbed hydrogen atoms are then either absorbed into the bulk, or recombine and desorb back into the gas phase, as shown in the reactions (1a) and (1b). \(\ce{H(ab)}\) represents a hydrogen atom absorbed in the bulk.

Figure 4
Figure 5

The reaction rates per surface site for adsorption, desorption, and absorption are \(r_1[\mathrm{s^{-1}}]\), \(r_2[\mathrm{s^{-1}}]\) and \(r_3[\mathrm{s^{-1}}]\), respectively. They are expressed as:

\[ r_1=k_1P_{\ce{H2}}(1-\theta)^2 \tag{2} \]

\[ r_2=k_2\theta^2 \tag{3} \]

\[ r_3=k_3\theta \tag{4} \]

where \(k_1\ [\mathrm{s^{-1}\,Pa^{-1}}]\), \(k_2\ [\mathrm{s^{-1}}]\) and \(k_3\ [\mathrm{s^{-1}}]\) are the reaction rate constants and \(P_{\ce{H2}}\) is the pressure of \(\ce{H2}\). Among the sites available on the surface, \(\theta\) (\(0\leq\theta\leq1\)) is the fraction occupied by \(\ce{H}\) atoms. It is assumed that adsorption and desorption are fast compared to absorption (\(r_1,r_2\gg r_3\)) and that \(\theta\) remains constant.

1.3 \(r_3\) can be expressed as:

\[ r_3=\frac{k_3}{1+\sqrt{\dfrac{1}{P_{\ce{H2}}C}}} \tag{5} \]

Express \(C\) using \(k_1\) and \(k_2\).

A metal sample with a surface area of \(S=1.0\times10^{-3}\ \mathrm{m^2}\) was placed in a container (\(1\mathrm{L}=1.0\times10^{-3}\ \mathrm{m^3}\)) with \(\ce{H2}\) (\(P_{\ce{H2}}=1.0\times10^2\ \mathrm{Pa}\)). The density of hydrogen-atom adsorption sites on the surface was \(N=1.3\times10^{18}\ \mathrm{m^{-2}}\). The surface temperature was kept at \(T=400\ \mathrm K\). As the reaction (1) proceeded, \(P_{\ce{H2}}\) decreased at a constant rate of \(v=4.0\times10^{-4}\ \mathrm{Pa\,s^{-1}}\). Assume that \(\ce{H2}\) is an ideal gas and that the volume of the metal sample is negligible.

1.4 Calculate the amount of \(\ce{H}\) atoms in moles absorbed per unit area of the surface per unit time, \(A\ [\mathrm{mol\,s^{-1}\,m^{-2}}]\).

1.5 At \(T=400\ \mathrm K\), \(C\) equals \(1.0\times10^2\ \mathrm{Pa^{-1}}\). Calculate the value of \(k_3\) at \(400\ \mathrm K\). If you did not obtain the answer to 1.4, use \(A=3.6\times10^{-7}\ \mathrm{mol\,s^{-1}\,m^{-2}}\).

1.6 At a different \(T\), \(C=2.5\times10^3\ \mathrm{Pa^{-1}}\) and \(k_3=4.8\times10^{-2}\ \mathrm{s^{-1}}\) are given. For \(r_3\) as a function of \(P_{\ce{H2}}\) at this temperature, select the correct plot from (a)–(h).

Figure 6

Official English text of the problem.

Answers

  • 1.1(i): B — \(0.07\ \mathrm{nm}\).
  • 1.1(ii): F — \(0.23\ \mathrm{nm}\).
  • 1.1(iii): A — \(0.03\ \mathrm{nm}\).
  • 1.2(i): H — \(400\ \mathrm{kJ\,mol^{-1}}\) required.
  • 1.2(ii): D — \(100\ \mathrm{kJ\,mol^{-1}}\) released.
  • 1.3: \(\displaystyle C=\frac{k_1}{k_2}\).
  • 1.4: \(\displaystyle A=2.4\times10^{-7}\ \mathrm{mol\,s^{-1}\,m^{-2}}\).
  • 1.5: \(\displaystyle k_3=1.1\times10^{-1}\ \mathrm{s^{-1}}\).
  • 1.6: Plot (b).

Solution

Hydrogen gas is dihydrogen, \(\ce{H2}\), SMILES [H][H]. Atomic hydrogen, \(\ce{H}\), has SMILES [H]. The labels (ad) and (ab) specify hydrogen at the surface and within the bulk, respectively; SMILES does not encode these environments or the metal–hydrogen coordination.

1.1

(i) Far from the surface, at large \(z\), the minimum-energy molecular configuration occurs at approximately

\[ d=0.07\ \mathrm{nm}. \]

This is the equilibrium bond length of gaseous dihydrogen. The answer is B.

(ii) The minimum corresponding to the adsorbed state is located at approximately

\[ (d,z)=(0.23,\ 0.03)\ \mathrm{nm}. \]

In Fig. 1, the two adsorbed hydrogen atoms occupy equivalent positions on opposite sides of the central metal atom. Their horizontal coordinates relative to that atom are \(-d_{\mathrm M}/2\) and \(+d_{\mathrm M}/2\), so their separation equals \(d_{\mathrm M}\). Thus,

\[ d_{\mathrm M}=d_{\text{adsorbed}}\approx0.23\ \mathrm{nm}, \]

giving F.

(iii) The vertical coordinate of the same minimum gives the adsorption height:

\[ h_{\mathrm{AD}}=z_{\min}\approx0.03\ \mathrm{nm}. \]

The answer is A.

1.2

The energy reference is the minimum for the gaseous molecule, approximately \(0\ \mathrm{kJ\,mol^{-1}}\). All energies in Fig. 2 are per mole of \(\ce{H2}\).

(i) For dissociation,

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

both atoms must be far from the surface and far from each other: large \(z\) and large \(d\). The high-energy contours indicate a dissociation energy closest to

\[ \boxed{400\ \mathrm{kJ\,mol^{-1}}}. \]

The answer is H.

(ii) For dissociative adsorption,

\[ \ce{H2(g) -> 2H(ad)}, \]

the final state is the minimum near \((0.23,0.03)\ \mathrm{nm}\). Following the solid contours in \(20\ \mathrm{kJ\,mol^{-1}}\) increments places this minimum at approximately \(-100\ \mathrm{kJ\,mol^{-1}}\). Consequently,

\[ \Delta E_{\text{ads}}\approx-100\ \mathrm{kJ\,mol^{-1}}, \]

and the positive amount of energy released is

\[ \boxed{100\ \mathrm{kJ\,mol^{-1}}}. \]

The answer is D.

1.3

The reactions are

\[ \ce{H2(g) <=> 2H(ad)} \]

and

\[ \ce{H(ad) -> H(ab)}. \]

Each adsorption event creates two adsorbed atoms, each desorption event removes two, and each absorption event removes one. Constant coverage therefore requires

\[ \frac{d\theta}{dt}=2r_1-2r_2-r_3=0. \]

Because \(r_1,r_2\gg r_3\), adsorption and desorption are in approximate equilibrium:

\[ k_1P_{\ce{H2}}(1-\theta)^2\simeq k_2\theta^2. \]

Taking the nonnegative square root gives

\[ \frac{\theta}{1-\theta} =\sqrt{\frac{k_1P_{\ce{H2}}}{k_2}}. \]

Within this approximation,

\[ \theta =\frac{1}{1+\sqrt{\dfrac{k_2}{k_1P_{\ce{H2}}}}}, \]

so

\[ r_3=k_3\theta =\frac{k_3}{1+\sqrt{\dfrac{k_2}{k_1P_{\ce{H2}}}}}. \]

Comparison with equation (5) yields

\[ \boxed{C=\frac{k_1}{k_2}}. \]

Its units are \(\mathrm{Pa^{-1}}\), making \(P_{\ce{H2}}C\) dimensionless.

1.4

Use the ideal-gas equation with \(R=8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}\):

\[ n_{\ce{H2}}=\frac{P_{\ce{H2}}V}{RT}. \]

At constant volume and temperature, the rate of disappearance of gaseous dihydrogen is

\[ -\frac{dn_{\ce{H2}}}{dt} =\frac{Vv}{RT} =\frac{ (1.0\times10^{-3}\ \mathrm{m^3}) (4.0\times10^{-4}\ \mathrm{Pa\,s^{-1}}) }{ (8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}) (400\ \mathrm K) } =1.20\times10^{-10}\ \mathrm{mol\,s^{-1}}. \]

Since the surface coverage remains constant, hydrogen does not accumulate on the surface. The net process is therefore

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

Two moles of hydrogen atoms enter the bulk for each mole of gas consumed. Dividing by the surface area,

\[ A=\frac{2Vv}{RTS} =\frac{2(1.20\times10^{-10}\ \mathrm{mol\,s^{-1}})} {1.0\times10^{-3}\ \mathrm{m^2}}. \]

Thus,

\[ \boxed{A=2.4\times10^{-7}\ \mathrm{mol\,s^{-1}\,m^{-2}}}. \]

1.5

At the stated pressure,

\[ P_{\ce{H2}}C =(1.0\times10^2\ \mathrm{Pa}) (1.0\times10^2\ \mathrm{Pa^{-1}}) =1.0\times10^4. \]

Hence,

\[ \theta=\frac{1}{1+(P_{\ce{H2}}C)^{-1/2}} =\frac{1}{1+0.010} =0.9901. \]

Each absorption event transfers one hydrogen atom. Using \(N_{\mathrm A}=6.022\times10^{23}\ \mathrm{mol^{-1}}\), the molar absorption rate and the rate per surface site satisfy

\[ A=\frac{Nr_3}{N_{\mathrm A}} =\frac{Nk_3\theta}{N_{\mathrm A}}. \]

Therefore, using the unrounded result from 1.4,

\[ k_3=\frac{AN_{\mathrm A}}{N\theta} =\frac{ (2.405\times10^{-7}\ \mathrm{mol\,s^{-1}\,m^{-2}}) (6.022\times10^{23}\ \mathrm{mol^{-1}}) }{ (1.3\times10^{18}\ \mathrm{m^{-2}})(0.9901) } =0.1125\ \mathrm{s^{-1}}. \]

To two significant figures,

\[ \boxed{k_3=1.1\times10^{-1}\ \mathrm{s^{-1}}}. \]

1.6

Rewrite equation (5) as

\[ r_3=k_3\frac{\sqrt{CP_{\ce{H2}}}} {1+\sqrt{CP_{\ce{H2}}}}. \]

Over the plotted pressure range,

\[ CP_{\ce{H2}} \leq (2.5\times10^3\ \mathrm{Pa^{-1}}) (2.5\times10^{-6}\ \mathrm{Pa}) =6.25\times10^{-3}\ll1. \]

Thus, approximately,

\[ r_3\simeq k_3\sqrt{CP_{\ce{H2}}}. \]

The curve starts at the origin, rises with increasing pressure, and bends downward. This selects the family (a)–(d).

To identify its magnitude, take \(P_{\ce{H2}}=1.0\times10^{-6}\ \mathrm{Pa}\). Then

\[ \sqrt{CP_{\ce{H2}}} =\sqrt{(2.5\times10^3)(1.0\times10^{-6})} =0.050, \]

and the full expression gives

\[ r_3=(4.8\times10^{-2}) \frac{0.050}{1.050} =2.29\times10^{-3}\ \mathrm{s^{-1}}. \]

The low-pressure approximation gives \(2.4\times10^{-3}\ \mathrm{s^{-1}}\), matching the scale of curve (b). Thus,

\[ \boxed{\text{Plot (b)}}. \]

The solution was checked against the official solution.