Chemistry, 12.12.2019 05:31 williamslyric

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

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Answer from: kyle65

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:

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

Answer from: Quest

one negative thing it that it can kill water animals. another negative thing is it can cause the destruction of ecosystem. one way water pollution can be reduced is to conserve energy.

Answer from: Quest

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )

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