Question

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.
The second run will be faster.
The second run will be slower.
The second run will have the same rate as the first.
The second run has twice the surface area.
The second run has eight times the surface area.
The second run has 10 times the surface area.

Answer 1

Answer: The correct options are:

The second run will be faster.

The second run has twice the surface area.

Explanation:

It is known that,

  Reaction catalyzing power of catalyst $\propto$ its exposed surface area

  $S_{1}$ = ?,   $V_{1} = 10 cm^{3}$,    

As,   area = $\frac{4}{3}\pi r^{3}$

                r = 1.545 cm

   $S_{1} = 4\pi r^{2}$

Also,

            $S_{2} = 8 \times (4 \pi \times r'^{2})$

                r' = ?

Hence, expression for the volume conservation will be as follows.

          $\frac{4}{3}\pi r^{3} = 8 \times \frac{4}{3} \pi \times (r')^{3}$

Since,   $V_{1} = V_{2}$

                       r' = $\frac{r}{2}$

     $S_{2} = 8 \times 4r \times (\frac{r}{2})^{2})$

      $S_{2} = 2 \times 4\pi r^{2}$

Therefore,    $S_{2} = 2S_{1}$

Thus, we can conclude that the second run will be faster and the second run has twice the surface area.