onnect Spring 2018 Chemistry 182 Burdge 4e: Spring 2018 CHEM 182-02 CHEMISTRY me
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onnect Spring 2018 Chemistry 182 Burdge 4e: Spring 2018 CHEM 182-02 CHEMISTRY mework Question 11 (of 15) 00 points 3 attempts left C Check my work Select all that apply. In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.0 cm. If the sphere were broken down into eight spheres each having a volume of 1.25 cm2, 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. References eBook & Resources Multipart Answer 0 Type here to searchExplanation / Answer
The volume of the sphere in run 1 is 10.0 cm3. Let the radius of the sphere be r cm; therefore,
4/3**r3 = 10.0 cm3
=====> r3 = (10.0*3)/(4) cm3 = (10.0*3)/(4*22/7) cm3 = (10.0*3*7)/(4*22) cm3 = 2.386 cm3
=====> r = 1.336 cm.
The surface area of the sphere in run 1 is 4**r2 = 4*(22/7)*(1.336 cm)2 = 22.439 cm2.
The volume of each of the eight spheres in run 2 is 1.25 cm3. Let r’ cm be the radius of each of the spheres in run 2. Therefore,
4/3**r’3 = 1.25 cm3
====> r’3 = (1.25*3*7)/(4*22) cm3 = 0.298 cm3
====> r’ = 0.668 cm.
The surface area of each of the spheres in run 2 is 4**r’2 = 4*(22/7)*(0.668 cm)2 = 5.610 cm2.
Since we have eight spheres in run 2, the total surface area of the all the spheres combined is (8*5.610 cm2) = 44.88 cm2.
The ratio of the surface areas of the spheres in the two runs is (Surface area of run 2)/(Surface area of run 1) = (44.88 cm2)/(22.439 cm2) = 2.0/1.0 2.0; therefore, the second run has twice the surface area. Hence, the 4th. statement is correct.
In case of heterogenous catalysis, the rate of the reaction depends on the available surface area of the catalyst. Since the catalyst has a higher surface area in run 2, hence, the rate of the reaction in run 2 is expected to be higher than that in run 1. Therefore, the 1st. statement is expected to be correct.
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