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 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

Answer 1

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

$V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}$

$10=\frac{4}{3}*pi*{r_{L}}^{3}$

${r_{L}}^{3}=\frac{7.5}{pi}$

$r_{L}=1.337cm$

Step 2: Determine surface area of large sphere

$A_{L}=4*pi*{r_{L}}^{2}$

$A_{L}=4*pi*1.337^{2}$

$A_{L}=22.463 cm^{2}$

Step 3: Determine radius of small sphere

$V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}$

$1.25=\frac{4}{3}*pi*{r_{s}}^{3}$

${r_{s}}^{3}=\frac{0.9375}{pi}$

$r_{s}=0.668cm$

Step 4: Determine surface area of small sphere

$A_{s}=4*pi*{r_{s}}^{2}$

$A_{s}=4*pi*0.668^{2}$

$A_{s}=5.607 cm^{2}$

Step 5: Determine total surface area of 8 small spheres

$A_{S}=8*A_{s}$

$A_{S}=8*5.607$

$A_{S}=44.856 cm^{2}$

$A_{L}=22.463 cm^{2}$ - Surface area of 1 large sphere

$A_{S}=44.856 cm^{2}$ - Surface area of 8 small spheres

Options:

1. The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
2. The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
3. The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
4. The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
5. The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
6. The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

Answer 2

Answer:

a. The second run will be faster

d. The second run has twice the surface area

Explanation:

The velocity of a chemical reaction can be changed by some factors, the principal is the temperature, the concentration, the presence of a catalyst and the contact surface.

As higher is the temperature, as higher will be the velocity, because the molecules will have higher energy, and because of that will collide more frequently.

As higher is the concentration of the reactants, as higher will be the velocity, because, with more molecules, the collisions will happen more frequently.

A catalyst reduces the activation energy, thus the reaction occurs faster.

As higher is the contact surface, as higher will be the velocity because the collision will happen more efficiently between the molecules. As more split is the substance, as higher is the surface area. So, by broken the catalyst, the reaction will be faster.

For a sphere, the volume can be calculated by:

V = (4/3)*πr³, where r is the radius

And the surface area by:

S = 4πr²

So, when the volume is 10.0 cm³, the radius is:

10 = (4/3)*π*r³

4*π*r³ = 30

r³ = 2.3873

r = ∛2.3873

r = 1.3365 cm

Thus the contact surface will be:

S = 4*π*(1.3365)²

S = 22.45 cm²

When the sphere is broken, each piece has a volume of 1.25 cm³, thus the radius will be:

1.25 = (4/3)*π*r³

4πr³ = 3.75

r³ = 0.2984

r = ∛0.2984

r = 0.6682 cm

The contact surface will be 8 times the surface of one piece, thus:

S = 8*4*π*(0.6682)²

S = 44.90 cm²

Which is twice the area of the first run.