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14 Chemical Kinetics Solutions to Exercises (b) A homogeneous catalyst is in the same phase as the reactants; a heterogeneous catalyst is in a different phase and is usually a solid. (c) A catalyst has no effect on the overall enthalpy change for a reaction. A catalyst does affect activation energy, Ea, which is one way that it changes reaction rate. It can also affect the frequency factor, A. 14.82 (a) The smaller the particle size of a solid catalyst, the greater the surface area. The greater the surface area, the more active sites and the greater the increase in reac- tion rate. (b) Adsorption is the binding of reactants onto the surface of the heterogeneous cata- lyst. It is usually the first step in the catalyzed reaction. 14.83 Analyze/Plan. Use the structure and unit cell edge of Pt, along with the formulas for volume and surface area of a sphere, to calculate the number of Pt atoms in a 2-nm sphere and on the surface of a 2-nm sphere. (a) For a Pt sphere with a 2.0 nm diameter, radius = 1.0 nm. V = =4/3πr³ = 3 1³ = 4.188879 10³ = 4.2 10³ ų In a face-centered cubic metal structure, there are 4 metal atoms per unit cell. The volume of the unit cell is (3.924 Å)³ = 60.42 ų 4 Pt atoms 60.42 ų X 4.1889 10³ = 277.3 = 2.8 Pt atoms in a 2.0 - nm sphere (b) Assume that the "footprint" of an atom is its cross-sectional area, the area of a circle with the radius of the atom. The area of this circle is The diameter, d, of a Pt atom is 2.8 Å, so r = d/2 = 1.4 Å. The footprint of the Pt atom is then The surface area of the 2.0-nm sphere is = = 12.56637 10² = 1.3 10³ 1 Pt atoms 6.1575 Å 1.2566 X = 204.1 = surface Pt atoms on a 2.0 nm sphere. (c) 204 surface Pt atoms 100 = 74% Pt atoms on the surface 277 total Pt atoms (d) For a 5.0-nm Pt sphere, radius = 2.5 nm V = 3 = 3 = 65.4498 10³ = 6.5 10⁴ 4 Pt atoms 60.42 ų 65.4498 10³ ų = 4,333 = Pt atoms in a 5.0 nm sphere The surface area of the 5.0-nm sphere is 418