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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 159b

Consider the sublimation of mothballs at 27 °C in a room having dimensions 8.0 ft ⨉ 10.0 ft ⨉ 8.0 ft. Assume that the mothballs are pure solid naphthalene (density 1.16 g/cm3) and that they are spheres with a diameter of 12.0 mm. The equilibrium constant Kc for the sublimation of naphthalene is 5.40⨉10-6 at 27 °C. C10H8(s) ⇌ C10H8(g) (b) How many mothballs are required to saturate the room with gaseous naphthalene?

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hi everyone for this problem it reads a dry ice container with a capacity of 20 U. S. Liquid quarts was used to hold cylindrical dry ice pellets, assuming that each pure cylindrical dry ice pellet has a density of 1.6 g per cubic centimeter a diameter of 1.6 centimeters and a length of six centimeters. Find the number of pellets required to saturate the container with gaseous carbon dioxide. The equilibrium constant for the sublimation of carbon dioxide is 1189.2. So for this problem, the question that we want to answer is the number of pellets required to saturate the container. Alright, so that's our goal. And let's take a look at what we're given. We know that we have 20 U. S. Liquid quarts. Okay. We know that the density of a dry ice pellet is 1.6 g per cubic centimeter with a meter diameter of 1.6 centimeters and a length of six centimeters. And were also given the equilibrium constant is the following. So let's take a look at our reaction. We have solid carbon dioxide at equilibrium to produce gasses. Carbon dioxide. We are given the equilibrium constant K. C. And what this equates to is the concentration of product over the concentration of reactant and only gashes and acquis substances are considered in the equilibrium constant expression. And so for this we see that we have a solid here. So this solid is not going to be included in our equilibrium expression. So our gaseous carbon dioxide is the product. Okay so we know that the concentration of product based off of the K. C. Or the equilibrium constant that was given. So our gaseous carbon dioxide, That concentration is equal to 1,889.2 moles per liter. Because concentration is moles per liter. So we know what our concentration of gasses. Carbon dioxide is based off of the equilibrium constant given. So now that we know that we need to calculate our moles of carbon dioxide. Alright, so our moles of carbon dioxide, we're going to calculate that based off of what is given. And we're going to start off with the 20 U. S. Liquid quarts. Okay, so there's 20 quarts and we're first going to go because we want to go from quartz to moles. Okay, so let's write down that pathway here. Our goal is to go from quartz to moles. All right, so the first thing that we're going to do is go from quartz to Leaders. Alright. And one leader, There is 1.057 quarts. Okay, so our units for quartz cancel And we're in Leaders and now we want to go from Leaders to moles and we can do that because we know what the MMA larry T is because we just wrote that out here. Okay, So we're going to use this. So in one leader there is 1889.2 moles. So our units for leaders cancel and we're left with moles. Now we can calculate our moles of carbon dioxide gas and that is going to be 35,000 746 .45222 malls. Okay, so that's our moles of carbon dioxide. So now let's go ahead and calculate our mass of carbon dioxide. Our mass. We're gonna take this moles per leader. Yeah, we're going to take the molds and convert it to grams. Okay, so we have 35,000 746.45222 moles. And we want to go from moles to grams. So we're going to need the molar mass of carbon dioxide And one mole of carbon dioxide. The mass is 44 009g. So our units for moles cancel and we're left with grams. And now we can do this calculation and what we're going to get is a large number. So let's just write this out. 1573165.616 g. Okay, so that's our mask. So now we have what we need to calculate the what we're looking for, which is the number of pellets required. The way that we're going to do this is by solving first for the volume of a cylindrical dry ice pellet. So the volume of cylindrical dry ice pellet. This is equal to pi R squared times H. Okay, So let's go ahead and plug in our values volume is equal to pi the radius is going to be let's take a look at what we're given. The radius is 1.6 cm divided by two. Okay. Because we're told what the diameter is, we're told the diameter is 1.6. Okay, so we're going to take 1.6 and divide that by two. And that's squared times height. Okay. And the problem, we're told A length of six cm. So that length is the height. Okay, so actually let's write the units here. So we know what we're writing. So 1.6 cm divided by two. And that squared times our height, which is 6cm. All right, so that means our volume is going to equal 12.0637 15 79 cubic centimeters. So that's the volume of a cylindrical dry ice pellet. Now we're going to calculate the mass of cylindrical dry ice pellet. So this is going to be our density. So we're gonna use our density here. So our mass is going to equal So let's write down our volume as our starting point. The volume of the cylindrical dry spell. It is 12.6371579 cubic centimeters. And the density that was given in the problem is 1.6 g per cubic centimeter. Alright, so our cubic cm cancel. And we're left with g. So this answer is what we're going to get for the mass of our cylindrical dry ice pellets. So when we do this calculation, the mass that we're going to get is 19.301 94 5-6 g. Okay, so now the last step is our number of dry ice pellets. What we're going to do here to calculate this number is we're going to take the mass of our carbon dioxide gas. Okay, So we're going to take the mass of carbon dioxide that we solve for and divided by the mass of the cylindrical dry ice palette. Okay, so let's go ahead and do that. So, our mass for the carbon dioxide gas that we saw for was Following 157316 5.616g. And the mass of the driest pellet that we just solved for is 19.301 94 5-6 g per palette. Okay, so once we do this calculation, what we're going to get Is 8.2 times 10 to the four. Okay, so our final answer Is the number of dry ice pellets is equal to 8.2 times 10 to the 4th. Dry ice pallets. And that is it for this problem. I hope this was helpful
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Textbook Question

Consider the sublimation of mothballs at 27 °C in a room having dimensions 8.0 ft ⨉ 10.0 ft ⨉ 8.0 ft. Assume that the mothballs are pure solid naphthalene (density 1.16 g/cm3) and that they are spheres with a diameter of 12.0 mm. The equilibrium constant Kc for the sublimation of naphthalene is 5.40⨉10-6 at 27 °C. C10H8(s) ⇌ C10H8(g) (a) When excess mothballs are present, how many gaseous naphthalene molecules are in the room at equilibrium?

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