For the thermal decomposition of nitrous oxide, 2 N2O1g2S 2 N21g2 + O21g2, values of the parameters in the Arrhenius equation are A = 4.2 * 109 s-1 and Ea = 222 kJ>mol. If a stream of N2O is passed through a tube 25 mm in diameter and 20 cm long at a flow rate of 0.75 L/min at what temperature should the tube be maintained to have a partial pressure of 1.0 mm of O2 in the exit gas? Assume that the total pressure of the gas in the tube is 1.50 atm.

Video transcript

Hello. Everyone is on This. Problem being asked at one temperature is you need to be kept in order for 02 in the exit gas to have a partial pressure of 20.95, we're here assuming that the gas and the to the total pressure of 1 M. So this problem we are given to us is first the balanced chemical reaction. So that's two moles of R N 205 gas. That goes ahead to yield four moles of our end to gas and one mole of 02 gas. So here that the pressure of the 02, so the exit gas is you go to 0. mila meters of mercury. So then the total pressure Is equal to 80 M. It was just equal to 760 mm of Mercury. And that equals to the pressure of the end 205 at its initial state. Alright, so we're gonna go ahead and use some stark eom a tree. So the pressure of the n. 0 2, the exit gas is equal to four Times the pressure of our 02 and that's equal to 3.8 mm of mercury. And now for the pressure of the end 205, which is just the exit gas, he goes to one more or one times the p totals or total pressure minus the pressure of the N. 02 minus the pressure of the 02, which gives us 7 60 millimeters of mercury minus all the pressure of an 02 is 0.95 millimeters of mercury. And then lastly, the pressure of the 02 is equal to 3. millimeters of mercury. So if we do that, we get that. The pressure of N is equal to 755.25mm of Mercury. Now we're assuming the constant pressure in the tube now. Alright, it's just scrolling down for more space here. Now we can go ahead and calculate the volume of the tube. So the equation for the volume of the tube is that volume is equal to pi R squared times height. So our here is just the diameter divided by two or or D divided by half. Unless it's equal to roll, the diameter is 30. If we're multiplying it by one half or dividing it by two, we get that still our units in millimeters, but we want this in meters And then two cm. So he was doing some dimensional analysis. So for every one millim we have 10 to the negative three m. And now we can go ahead and see that the millimeters will cancel. And now we go ahead and convert meters into centimeters. So for every one centimeter we have 10 to the negative two m. And now we can see that the meters will cancel. All right, So putting all this into my calculator, I get that my radius. So my r is 1.5 centimeters. Using again. This volume equation for our tube because the volume is pi and are we can plug that in 1. centimeters squared and our height is 25 centimeters. So then we get the final value for volume to be 0.17671 for six units being Leaders. And what we did here is of course we see that the units is going to be centimeter cubed once we multiply this and this and the conversion into leaders is well we know that one centimeters cubed is equal to one mL. And then we just want to convert milliliters into leaders and that's how we got this final value. So now calculate for all the time the gasses are in the tube. So just T. T. is equal to all the volume is 0.1767146 L. Gonna go ahead and convert this into minutes first. So for every one minutes we have the rate to be one minute for every 0.80 L. And now converting simply the minutes in two seconds is that for every one minute we have 60 seconds. So you see here that leaders will council and municipal council giving us that the T. Value is going to be 13.2536 units being seconds. Using the Iranian equation, that equation is K. Is equal to a E. To the negative E. E. Sub a divided by R. T. So the units of K is going to be for our frequency factor. And that goes by this capital A. Here. And that's equal to us to a negative one. Which gives us that this reaction is a first order reaction. Alright? So then calculating for when the time is at 13.2536 seconds. Again, that's what we saw for in this blue color here. So using the first order integrated rate law says we are determined that we have a first order reaction. That rate law is the natural log of the final concentration of the molecule that were interested in equaling two negative K. Times T. Plus the natural log of the final or the initial concentration of the molecule that we're interested in. So you go ahead and get some values here. All right, so again the natural log of the concentration of a. Alright, so what we're doing here first is isolating this value here. This negative K. Of T. So what we're gonna do is subtract the natural log of the initial concentration of the molecule that are interested on both sides. And that gives us two equal to negative K. F. T. So then plugging in the actual molecule instead of just having our ace here, We just get the concentration of n. Again, that's the Final concentration over the concentration of our initial end 205 molecule and hear how we're able to do this as well. First this negative means division in our long dramatic roles. And here I'm just plugging in the actual molecule. So that's how we got this here again. That equals to negative K. T. And that's just equals 755.25 millimeters of mercury Over 760 of Murky. And again I'm just plugging values that we got for here and here. So once we do so we get that this equals to 0.99375, squirrel down. Alright, so we're gonna go ahead and calculate for T. R. K. Now the constant. Alright so K. Is equal to the negative natural log of the final concentration of the molecule that are interested over the initial concentration of the market that were interested in this all over T. Alright. And basically we're just rewriting sort of this right here. Alright so plugging in some values now We get that. This here is 0. from this calculation here And then T. is 13 points 2536. That gives us a value as we put everything into a calculator. That gives us 4.73 times 10 to negative four S. Two negative one. So for the linear form of our Rania's equation, that's just Ln of K. You're going to the negative E. Of A over R. Times one over T plus Ln of eight. So here we already said that the E. Of A. As you go to one oh 3.26. Kill jules per month. We'll go ahead and convert this into jewels per mole. So we'll multiply this by 10 to the three jewels for everyone. Kill a jewel that gives us 10 30 to 60 jewels per remote. Now we'll go ahead and plug this into this equation here. So the Ln of K. And K is 4.73 times 10 to the negative four as two negative one. They're going to well E. Of A Is 103, jules Permal. You forgot this negative here. Alright. And it's divided by the R. That's 8. joules per mole times calvins Multiplied by one over T. And the virgin of hook. Go ahead and add this will Ln of eighth. So natural log of 4.27 times 10 to the 13th as two negative one. So once we go ahead and simplify this, we'll get that again. Of course we see that everything has a numerical value. So for one over T. So we're solving for one over T. Once we do so we get that one over T. Is equal to 3.143 times 10 to the negative three K two negative one. Again we're finally solving for T now and we get that T physical to 318 Calvin's. And this right here is going to be my final answer for this problem