International Chemistry Olympiad · 2009

Problems

Problem 1The Avogadro constant $N_A$ can be determined by three independent methods. (Method A) X-ray diffraction shows that gold crystallises in a face-centred cubic unit cell of edge $a = 0.408$ nm; the density of Au is $1.93 \times 10^{4}\ \text{kg m}^{-3}$ and $M(\mathrm{Au}) = 196.97\ \text{g mol}^{-1}$ . (Method B, Rutherford 1911) A purified sample of 192 mg of $\ce{^{226}Ra}$ stands 40 days, after which its total $\alpha$ -decay rate is 27.7 GBq; sealing it for a further 163 days produces 10.4 $\text{mm}^3$ of He measured at 101325 Pa and 273 K ( $\alpha$ particles become He atoms). (Method C, Perrin 1909) Colloidal spheres of radius $2.12 \times 10^{-7}$ m and density $1.206 \times 10^{3}\ \text{kg m}^{-3}$ suspended in water ( $\rho = 999\ \text{kg m}^{-3}$ ) at 15 \°C show a Boltzmann height distribution: a plot of $\ln(n_h/n_{h_0})$ versus $(h - h_0)$ has slope $-0.0235\ \mu\text{m}^{-1}$ . Determine $N_A$ from each method.Solutions: 1
Problem 2In interstellar space, $\ce{H2}$ forms on the surface of dust grains because two colliding H atoms cannot dissipate the bond energy. The rate constant for adsorption is $k_a = 1.4 \times 10^{-5}\ \text{cm}^3\ \text{s}^{-1}$ , the H-atom number density is $[\ce{H}] = 10\ \text{cm}^{-3}$ , the first-order desorption constant is $k_d = 1.9 \times 10^{-3}\ \text{s}^{-1}$ , and the surface-reaction constant is $k_r = 5.1 \times 10^{4}\ \text{s}^{-1}$ (surface atom numbers may be used like concentrations). Model A treats reaction as second order: $dN/dt = k_a[\ce{H}] - k_d N - k_r N^2$ . Model B limits each grain to 0, 1 or 2 atoms with fractions $x_0, x_1, x_2$ and rates: adsorption $k_a[\ce{H}]x_m$ , desorption $k_d\,m\,x_m$ , reaction $\tfrac{1}{2}k_r\,m(m-1)x_m$ . (a) Find the steady-state $N$ and the rate of $\ce{H2}$ production per grain in each model. (b) Computer simulations give a true rate of $9.4 \times 10^{-6}\ \text{s}^{-1}$ ; which model is consistent, and which step is rate-determining?Solutions: 1