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Ch.14 - Chemical Kinetics

Chapter 14, Problem 113c

Platinum nanoparticles of diameter 2 nm are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 Å. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0-nm nanoparticle.

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well everyone. So in this video we're trying to calculate for the percentage of, of gold atoms there on the surface of a 5.0 nanometer gold sphere were also given the equation to solve for the surface area of a sphere right over here and we're given the equation for the volume of sphere right over here and were also given the diameter of a gold atom right over here. So let's first calculate for the total atoms. But first things first, we need to go ahead and calculate the volume. The volume is equal to four thirds times pi times r. So radius cute. Of course we have those values. So we'll say four thirds. And then for pie, we're gonna simplify to 3.14. That's approximately how much equals to 3. dot right there. And then the radius is 5.0 nm. And then we're gonna go on to convert that into angstrom. So we'll go ahead and divide this by two and multiply by a strong For every one nano meter. Of course, like we said this are going to be cute. So putting that into my calculator, I'll get the value of 65416. angstrom cute. So now calculating the volume of the unit cell, that's going to equal to 4.07 Ekstrom cute. And that's just going to equal to 67. cute. Alright, now finally calculating for the atoms. Alright, so we have as in the problem says we have a face centered cubic arrangement. So then in that case this just means one second. Okay so just ignore this. So we're coupling for the items down again. We have a face centered cubic arrangement. So that means that we have four atoms per unit cell. Let's go ahead and scroll down for a more room. Alright so let's go ahead and get this dimension analysis started. Then the volume which we just solved for. We got the value of (654) 16. and some cute. We're gonna go ahead and calculate that. So again we have the volume of the unit cell being 67.42 and that's per unit cell. And of course that's just a conversion factor then. So we go ahead and put the one unit cell on top and then the angstrom on the bottom. So we can go ahead and cancel. So be 67.42 Angstrom cubed on the bottom. Then of course with the FCC that we have will put the four atoms on top and then the one unit still on the bottom. You see why parents ordered Because now our ancient cube unit will be canceled and the unit cell will be canceled. Leaving us with just the atoms like we want. So then putting all this into my calculator, I'll get the numerical value Of 3881. And of course my unit is going to be our gold atoms. All right, so one part is complete. Now let's go ahead and move on to calculating for our atoms on the surface. Alright, so let's first calculate the cross sectional area of a gold atom. So that area is going to equal two pi r squared. So then pie again is approximately equal to 3. and then This radius is 2.92, ancient, Of course divided by two and then squared. So my numerical value, once I put all these values into the calculator will be 6. and my units will be ancient squirt. So that's just going to be the cross sectional area of the atom. Now calculate the surface area of the sphere we had that the area is equal to four times pi times radius squared. So we have four and then on pi is approximately 3.2 and the radius is 5.0 nanometers. Of course divided by two, converting that nanometers into angstrom. We have the conversion factor of 10 angstrom per one nanometer. And all this is going to be squared. Again, putting all this into my calculator, my numerical value is going to be and my unit is going to be angstrom squared. And of course we see here that the nanometers will go ahead and cancel. Alright now calculating for the number of our gold atoms. Some number of gold Adams. Okay, so we'll start off by what we just saw for it right over here. So we have the 7850 Asian squared we have then the This right here, so 6.69 So that those can cancel and top will be one gold adam. All right. As you see then the units of angstrom squared can cancel. And we're left with the unit of gold atoms which we are solving for. Right. Alright, so putting that into my calculator, I'll get the value of 1000 173 units being cold atoms. This is on the surface. Alright. Now, finally, finally, we're calculating for the percentage. All right. So we have the Adams On the surface divided by the total number of atoms, which is 3,881 atoms. Of course you do percentage. We need to go ahead and multiply this by 100%. And that total is actually going to equal to 30.2%. So there's 30.2% is going to be my final answer for this problem. Hopefully just helped you out. Thank you all so much for watching
Related Practice
Textbook Question

In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism has been proposed for the decomposition of (CH3)3AuPH3:

Step 1: (CH3)3AuPH3 k1 k -1 (CH3)3Au + PH3 (fast)

Step 2: (CH3)3Au k2 C2H6 + (CH3)Au (slow)

Step 3: (CH3)Au + PH3 ¡k3 1(CH3)AuPH3 (fast)

(e) What is the rate law predicted by this mechanism?

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Textbook Question

In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism has been proposed for the decomposition of (CH3)3AuPH3:

Step 1: (CH3)3AuPH3 k1 k -1 (CH3)3Au + PH3 (fast)

Step 2: (CH3)3Au k2 C2H6 + (CH3)Au (slow)

Step 3: (CH3)Au + PH3 ¡k3 1(CH3)AuPH3 (fast)

(f) What would be the effect on the reaction rate of adding PH3 to the solution of (CH3)3AuPH3?

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Textbook Question

Platinum nanoparticles of diameter 2 nm are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 Å. (b) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere 14pr22 and assuming that the 'footprint' of one Pt atom can be estimated from its atomic diameter of 2.8 A .

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Textbook Question

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the CO2 accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to 107 CO2 molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

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Textbook Question

Enzymes are often described as following the two-step mechanism: E + S Δ ES 1fast2 ES ¡ E + P 1slow2 where E = enzyme, S = substrate, ES = enzyme9substrate complex, and P = product. (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor.

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Textbook Question

The reaction between ethyl iodide and hydroxide ion in ethanol 1C2H5OH2 solution, C2H5I1alc2 + OH- 1alc2 ¡ C2H5OH1l2 + I - 1alc2, has an activation energy of 86.8 kJ>mol and a frequency factor of 2.10 * 1011 M-1 s-1. (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?

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