BP (boron phosphide) is a valuable wear-resistant hard coating that is produced by the reaction of b — Analytical Chemistry Chemistry Question
A Ceramic Hard Coating
BP (boron phosphide) is a valuable wear-resistant hard coating that is produced by the reaction of boron tribromide and phosphorus tribromide under a hydrogen atmosphere at high temperature (>750 °C). This ceramic material is used as a protecting thin film on metal surfaces. BP crystallizes in a cubic-close-packed structure with tetrahedral surrounding.
Give the equation for the formation of BP.
Model Answer
BBr3 + PBr3 + 3 H2 → BP + 6 HBr
Draw the Lewis structures of boron tribromide and phosphorus tribromide.
Model Answer
[VISUAL] Boron tribromide: planar and trigonal.
Phosphorus tribromide: trigonal pyramidal.
Draw the structure of BP in the crystalline state.
Model Answer
[VISUAL] BP crystallizes in a zinc blende (sphalerite) type structure: a cubic close-packed (FCC) lattice of boron atoms with phosphorus atoms occupying half of the tetrahedral interstitial sites (or vice versa), each tetrahedrally coordinated.
Give the overall composition of the unit cell corresponding with the formula BP.
Model Answer
A FCC-structure of the B atoms gives:
Angular points (corners): 8 × 1/8 = 1
Planes (face centers): 6 × 1/2 = 3
Total B atoms = 4
In each unit cell, there are also 4 phosphorus atoms present which are tetrahedrally surrounded by boron atoms.
Thus, the overall composition of the unit cell is B4P4, corresponding to 4 formula units of BP.
Calculate the density of BP in kg m–3 when the lattice parameter of the unit cell is 4.78 Å.
Model Answer
The relative atomic masses of boron and phosphorus are 11 and 31, respectively (M = 42 g/mol).
Density (d) = (n × M) / (V × Na)
d = (4 × 42 × 10^-3 kg/mol) / ((4.78 × 10^-10 m)^3 × 6.022 × 10^23 mol^-1) = 2554 kg m^-3
Calculate the distance between a boron and a phosphorus atom in BP.
Model Answer
The distance between a corner atom and the nearest tetrahedral site is 1/4 of the body diagonal of the unit cell.
Distance B–P = 1/4 × √3 × a = 1/2 × √3 × 1/3 a = 2.069 Å (where a = 4.78 Å).
Calculate the lattice energy of BP using the Born-Landé formula:
U_lattice = - (Z^+ * Z^- * A * e^2) / r_0 * (1 - 1/n)
The factor f * e^2 amounts to 1390 when the ionic radii r+ and r– are given in Å. The Madelung constant is 1.638. The Born exponent n is 7. The charges of the ions Z+ and Z– are integer numbers.
Model Answer
Assuming BP is a trivalent ionic compound (B^3+ P^3-), Z+ = 3 and Z- = 3.
U_lattice = - (3 × 3 × 1.638 × 1390) / 2.069 × (1 - 1/7) = - 8489 kJ mol^-1.
Determine the order of the reaction leading to BP and give the equation, based on the following rate data:
[VISUAL] Table:
Temperature, °C | [BBr3], mol dm–3 | [PBr3], mol dm–3 | [H2], mol dm–3 | r, mol s–1
800 | 2.25×10–6 | 9.00×10–6 | 0.070 | 4.60×10–8
800 | 4.50×10–6 | 9.00×10–6 | 0.070 | 9.20×10–8
800 | 9.00×10–6 | 9.00×10–6 | 0.070 | 18.4×10–8
800 | 2.25×10–6 | 2.25×10–6 | 0.070 | 1.15×10–8
800 | 2.25×10–6 | 4.50×10–6 | 0.070 | 2.30×10–8
800 | 2.25×10–6 | 9.00×10–6 | 0.070 | 4.60×10–8
880 | 2.25×10–6 | 9.00×10–6 | 0.070 | 19.6×10–8
Note: The last row's temperature is used as 880 °C in calculations.
Model Answer
The reaction is first order in [BBr3] and first order in [PBr3]. Thus, the overall order of the reaction is 2.
Rate equation: r = k [BBr3] [PBr3]
Calculate the rate constants at 800 and 880 °C.
Model Answer
k800 = r800 / ([BBr3] [PBr3]) = 4.60×10^–8 mol s^–1 / (2.25×10^–6 mol dm^–3 × 9.00×10^–6 mol dm^–3) = 2272 dm^6 mol^–1 s^–1
k880 = r880 / ([BBr3] [PBr3]) = 19.60×10^–8 mol s^–1 / (2.25×10^–6 mol dm^–3 × 9.00×10^–6 mol dm^–3) = 9679 dm^6 mol^–1 s^–1
Calculate the activation energy for the formation of BP.
Model Answer
Using Arrhenius Equation:
ln(k2 / k1) = - Ea/R × (1/T2 - 1/T1)
T1 = 800 + 273 = 1073 K
T2 = 880 + 273 = 1153 K
ln(9679 / 2272) = - Ea / 8.314 × (1 / 1153 - 1 / 1073)
1.4497 = Ea / 8.314 × 6.463 × 10^–5
Ea = 186 kJ mol^–1.