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Q Jndtics Axd EduLibrim Calculating cquilbrium composition from an cquilibrium constantSuppose 500. mL flask Is filled with 1,6 mol of NO andmol of - NOz: The folla...

Question

Q Jndtics Axd EduLibrim Calculating cquilbrium composition from an cquilibrium constantSuppose 500. mL flask Is filled with 1,6 mol of NO andmol of - NOz: The follawilng rcaction bccomes possible:NOj(g) + Nokg) ~ ZNO,(g) The equillbrium constant K for this reaction 8.30 at the temperature of the flask-Cslculete the cqullibrum molarty OfNO Rcund vour answer to two decImal places.

Q Jndtics Axd EduLibrim Calculating cquilbrium composition from an cquilibrium constant Suppose 500. mL flask Is filled with 1,6 mol of NO and mol of - NOz: The follawilng rcaction bccomes possible: NOj(g) + Nokg) ~ ZNO,(g) The equillbrium constant K for this reaction 8.30 at the temperature of the flask- Cslculete the cqullibrum molarty OfNO Rcund vour answer to two decImal places.



Answers

A flask with a volume of 1.50 $\mathrm{L}$ , provided with a stopock, contains ethane gas $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ at 300 $\mathrm{K}$ and atmospheric pressure $\left(1.013 \times 10^{5} \mathrm{Pa}\right) .$ The molar mass of ethane is 30.1 $\mathrm{g} / \mathrm{mol}$ . The system is warmed to a temperature of $380 \mathrm{K},$ with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

So is asking us for the final pressure of the ethane and the the final pressure of the ethane and the total and the total mass that is left of the ethane. So we should write our givens one point five leaders and then we're going to automatically convert to point zero zero one five meters cube, given that one thousand leaders are in one cubic meter and then we have the Mueller Mass of ethane. Ethane, by the way, is six C two eight six and this is going to be equal to point zero three zero one kilograms. Parimal, we have, ah temperature the initial temperature of five hundred fifty Calvin and then the final temperature of three hundred, three hundred Calvin And again this is Ah if the system is being warmed to five hundred fifty Calvin. But Thebes, the question is asking what happens when it's cooled. So we're going to define the initial temperature as five fifty Kelvin and then the final temperature as three hundred Calvin and then the pressure of this Ah, the pressure of this is going to be one hundred and one three to five Pascal's and that will be our initial pressure. And that's asking us, of course. What is our final pressure? Well, are we can write the ideal gas constant P vehicles and Artie and let's see what this constant. So the actual thing that air constant is going to be p over t equals and are over V. So V is gonna stay constant and the number of moles within the container stays constant. And, of course, the idol gas Constant is a cop constant. So of course, that's constant. And we can then say we can create a relationship and say p over one p one better by T suborned equals piece of two. There are my teeth, too. And then we're trying to find peace of tomb. So the piece of two equals a piece of one t's up to the head by a piece of one. This is going to be equal to, um, the pressure of the NFL pressure one o one, three to five, and then the final temperature of three hundred divided by the initial temperature five fifty. This is going to give us fifty five thousand two hundred and sixty eight point one eight to Pascal's, so that would be our final pressure. This would be our answer for apartment and there's asking us. Okay, what's the temperature? What's the Ah, Mass of this entire? What's the mass of this entire ethane gas? So again, let's Peavy Conner's TV get her ideal gas law. PV equals and Artie. The number of malls equals the mass of the FAA and divided by the Mueller mass of the ethane. So we can read, write this and say that PV equals M Archie over capital. And then let's ah, isolate them. So am is going to be equal to P V over Artie. And then at this point, you can choose final initial Doesn't matter, because, of course, they're constant. Pressure over the temperatures will be staying constant throughout the entire process. And so we can say m equals piece of we'LL Just use two V m on divided by t's up to our. This is simply going to be equal to again fifty five thousand whether they want us to find the final pressure, the final final mass. However, the final mass and the initial mass are the same, but we'LL just use the stuff script to anyway. So fifty five thousand two hundred sixty eight point one eight two and then times point zero three zero one kilograms per mall, point zero zero one five meters cubed all divided by the temperature final temperature of three hundred Calvin and then ideal gas, constant of a point three one four. And our mass is going to be equal to point zero zero one zero zero or one gram. Exactly. See, too and six What? So that would be our final mass of ethane And that would be your answer for Part B. And that's the end of the solution. Thank you for watching.

This next problem is going to require some significant calculations because we need to convert the masses and volumes into concentrations. So our chemical reaction is going to be a church two plus I two goes to two H two. And we have Initial masses which we need to convert into concentrations will take the massive H2 divided by its smaller mass to get moles and then divide that whole thing by the volume Of the reaction. 367 leaders to get concentration initial concentration of H. two. We'll do the same thing with I. two. We have 96.9 g of it. We'll divide by the molar mass of I. Two in order to get the molds of I. Two and then divide that bye 367 leaders to get concentration. Then um we are told that there's nothing of H. I. To begin with. Then the reaction occurs and we end up producing 94 g of H. I. So again we need to calculate a concentration will take the 90.4 g of H. I, divide by the molar mass of H. I. In order to get moles of H. I. And then divide that by the same volume 3.67 So because the strict geometry is 1-1-2, then the change in concentration of H. I. Is going to be twice the change in concentration of H. Two and I. Two. The change in concentration of H. I. Is going from zero to this final concentration. It is this final concentration. Mhm. So Mhm. The decrease in concentration of H. Two and I. Two. Because they have half the coefficient of H. I will be one half the change. Or namely one half of the 90.4 g converted into molds. By dividing by the molar mass of H. I. Divided by the volume 3.67 So that will be the decrease one half of that will be the decrease in concentration of H. two and I too. So now what we need to do is calculate this concentration so we can use it to determine Casey This right here, calculates 2.1926. Then we need to calculate this this whole expression minus 1/2 this whole expression and when we take this and minus one half this, We get .00774 Mueller. Same thing over here, we'll take this minus one half that and we get .00683. So now Casey is going to be this concentration squared because of the coefficient to divided by each of these concentrations and we get a K c value Of 702.

So we have the ideal gas law T. V. Equalling Nrt however, we can say that the number of moles N. Would be equaling then the mass divided by the molar mass. Of course this would be the number of moles and so for part of a we have volume constant, the number of moles constant, it's a volume number of moles, it was constant. And therefore we can say that piece of one over piece of two would be equaling two t someone over T sub tube solving for piece of to this would simply be equaling then piece of one multiplied by T sub two Divided by Teeth of one. And so this would be 300 Calvin Well divided by rather 380 Calvin Multiplied by piece of one being 1.013 times 10 To the 5th Pascal's. Okay, Piece of two as an equaling Approximately 8.00 times 10 to the fourth pascal's. And so that would be the final pressure in the tank for part B. Now we can use that other formula PV equals N. But instead of em will say um R T mass over and molar mass. And so the mass of the gas lower case m would be equal and then pressure volume Mueller mass divided by ideal gas constant, multiplied by temperature And the mass is then equaling to 8.00 Times 10 to the 4th pascal's multiplied by the volume of 1.5 L. 1.5 Times 10 to the negative 3rd cubic meters, multiplied by The molar mass of 30 0.1 times 10 To the negative 3rd kg. For more. This would all be divided By the ideal gas constant 8.314 jules per more. Her Calvin multiplied by time. Uh multiplied by temperature rather of 300 Calvin. And we find that the masses equaling then 1.45 Times 10 to the negative 3rd kg. Or we can say 1.45 g. That is the end of the solution. Thank you for watching.

So here we're giving information about specific reaction where we have all gash of species and we have natural so chloride associating with nitric oxide and chlorine gas. And were given that we have initial concentration of the following of 2.00 molars. And were given that equilibrium there's 0.66 molars of nitric oxide. So this means that the amount of nitric oxide increased by 0.66 molars. The amount of chlorine increases by half as much due to destroy key metric coefficients. So we have our final concentration of chlorine And this means since Nitro, so chloride and nitric oxide or in the same style geometric ratio, the amount of nitrate chloride would have to decrease by the same amount, which is 1.34 molars. And now we can use this to determine the equilibrium constant with respect to concentration, which would be given by the following quantity. And essentially substituting for our certain quantities. We can see the equilibrium constant is essentially about zero 080. And this gives our final answer


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