Assume that 1.588 g of an alkali metal undergoes complete
reacti...

Question

Assume that 1.588 g of an alkali metal undergoes complete 
reaction with the amount of gaseous halogen contained 
in a 0.500 L flask at 298 K and 755 mm Hg pressure. In 
the reaction, 22.83 kJ is released 1ΔH = -22.83 kJ2. The 
product, a binary ionic compound, crystallizes in a unit cell 
with anions in a face-centered cubic arrangement and with 
cations centered along each edge between anions. In addition, 
there is a cation in the center of the cube. 
(c) Sketch a space-filling, head-on view of the unit cell, 
labeling the ions. Are the anions in contact with one 
another?

Accepted Answer

Hi everyone for this problem. It reads assume that at 298 kelvin and 789 millimeters of mercury pressure, 8.4634 g of an alkali metal completely reacts with gaseous halogen in a 0.750 liter container, 12.9 kg bulls are released during the reaction with a entropy change equal to negative 12.9 kg jewels. The end product, a binary ionic compound crystallizes in a unit cell with cad ions aligned along edge, each edge between an ions and ann ions and a face centered cubic arrangement. The edge length of the unit cell is 696 PICO meters. Additionally, the cube center contains a cat eye on dry, detailed illustration of the unit cell from above. Labeling the ions are the ann ions and contact with each other. Okay, so what we want to do here is draw detailed illustration of the unit cell and we're going to label the ions and answer the question of whether or not they're in contact with each other for us. Before we do that, let's go ahead and first write out our reaction. Okay, so we know we have a medal and this metal is reacting with a gashes halogen. Alright, so we have we'll say our gaseous halogen is X. Okay, and this is going to create the product M. X. Solid. So this is our reaction from this. We're going to calculate the moles of gas using the ideal gas law. So ideal gas law is pressure times volume is equal to the number of moles times gas constant R times T, which is temperature and kelvin. So let's go ahead and write out what we know. We know that the pressure is equal to well before we do that. We want to calculate the moles. Alright, so let's go ahead and right here calculate moles of gas. All right, And we're going to do that using the ideal gas law. So we need to rearrange this ideal gas law so that we're solving for moles. So we'll do that by dividing both sides of the equation by R. T. And so what we're going to get is our moles is going to equal pressure times volume over R times T. Alright, So as I was saying, we'll write out what we know. All right. So we know that the pressure Given in the problem is equal to 789 millimeters of mercury. We're going to want this in atmospheres because our gas constant R is in units of atmospheres. So to convert this from millimeters of mercury, two atmospheres, we're going to use the conversion and one atmosphere there is 760 millimeters of mercury. Alright, so with that are units of mm of mercury cancel and we're left with units of atmospheres. So when we do this calculation, what we're going to get is 1.04 atmospheres. So let's go ahead and plug that in for our pressure. Our pressure is 1.4 atmospheres. Okay, the volume given in the problem is 0.750 liters r gas constant r is a value we should know and that value is 0. leaders atmosphere over mole Kelvin. And the temperature given in the problem is 298 Kelvin. So we have everything that we need to solve for the moles and when we do this, what we're going to get is 0. 184 moles of our gaseous halogen which were writing as X. Okay, So now that we know the moles of gas using the ideal gas law, we're going to calculate the molds of metal using the multi mole ratio. Alright, so we know how many moles of our gaseous halogen we have because we just calculated that. So now that we've calculated the moles of gas. Now we're going to calculate the moles of metal. All right. So we'll start off with our moles of our gas. And we said that was 0.03184 moles of our gashes halogen. So now we're going to use the multiple ratio, which we need to refer to our equation that we wrote up at the top. So looking at our multiple ratio, we see that for every one mole of metal there is yes, for everyone a mole of metal, there is half a mole of gas. Okay, so we'll use that to calculate our molds of metal. Alright, So for every one mole of metal, there is half a more of the gas. So our units for moles of the gas canceled. And now we're left with molds of the metal. So when we do this calculation, what we're going to get is 0. six, 0. 368 moles of metal. So now we can calculate the molar mass of the metal. So we're going to calculate the molar mass of the metal by in the problem we're told what the grams of the alkali metal are. And remember molar mass is grams per mole. Alright, so in the problem we're told the grams and we just calculated the moles. Okay, so our molar mass is equal to grams over mold. And then the problem, we're told what the grams is, we're told it is 8.4634g. And the moles we just calculated is zero 063 moles. So this is grams per mole. And so once we do this calculation, we see that the value we're going to get is 100 and 32. g per mole. If we look at our periodic table, this molar mass is near the molar mass of cesium. Okay, so now we have everything we need to determine the identity of our halogen Using the given unit cell edge. Alright, so the ionic radius for cesium which is a cat eye on the ionic radius of cesium is equal to 181 And this is a value that was not given in the problem. We would have to look this up. It's 181 PICO meters and we know that the edge length of the unit cell As a whole is 696 people meters. This value was given in the problem. So what we can do here is calculate the unit cell edge using that information. And the way that we're going to write this out is the unit cell edge is equal to two times the radius of cesium. Okay plus two times the radius of x minus. That's our an ion. And this is going to be equal to the total which is PICO m. Alright so what we want to isolate here is this radius times x minus. Alright. So if we isolate that, what we're going to get is the radius times X minus Is going to equal 696 km minus two times the radius of cesium Cat eye on divided by two. Okay, so what we're going to get is if we simplify this 696 PICO meters minus two times the ionic radius of cesium which we said is 181 PICO meters and this is all over two. Alright so the value we're going to get is 100 and PICO meters is the unit cell edge. So looking at this value, the ionic radius of 167 PICO meters is closest to chlorine. Alright, so right here closest to chlorine. Alright, since the ionic radius of the cesium, cat ion and chlorine and ion are relatively comparable. So we have 181 PICO meters and 167 PICO meters. Since they're relatively comparable, there's no contact between each an ion. So our binary ionic compound is going to be cesium chloride and it will have a simple cubic unit cell. Okay, so let's go ahead and draw this out. So we'll have cesium be represented by blue and we'll have our chlorine represented by green. Alright, so now we can draw the detailed illustration and what it's going to look like is this? And then the problem we're told Yes. So it's gonna be Alright, so this is going to be the unit cell of the binary ionic compound and the and ions do not have contact with each other because the cat ion and an ion have comparable ionic radius is Okay, so that is it for this problem? I hope this was helpful

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