In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.0 cm3. If the sphere were broken down into eight spheres each having a volume of 1.25 cm3, and the reaction is run a second time, which of the following accurately characterizes the second run? Choose all that apply.

a. The second run will be faster.
b. The second run will be slower.
c. The second run will have the same rate as the first.
d. The second run has twice the surface area.
e. The second run has eight times the surface area.
f. The second run has 10 times the surface area.

The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

[tex]A=\pi ^{1/3} (6V)^{2/3}[/tex]

The area of the 10.0 cm³-sphere is:

[tex]A=\pi ^{1/3} (6.10.0)^{2/3}=22.4cm^{2}[/tex]

The area of each 1.25 cm³-sphere is:

[tex]A=\pi ^{1/3} (6. 1.25)^{2/3}=5.61cm^{2}[/tex]

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²

The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00

Since the surface area is doubled, the second run will be faster.