The rate constant for the first-order decomposition of gaseous
N2...

Question

The rate constant for the first-order decomposition of gaseous
N2O5 to NO2 and O2 is 1.7 * 10-3 s-1 at 55 °C.
(a) If 2.70 g of gaseous N2O5 is introduced into an evacuated
2.00 L container maintained at a constant temperature
of 55 °C, what is the total pressure in the
container after a reaction time of 13.0 minutes?

Accepted Answer

Well, everyone's in this video want to calculate for the total pressure and the reaction vessel after 10 minutes being told that this is a first order reaction decomposition. So for our first order integrated rate law, that's the natural log of the concentration of R. A. And that's just the final concentration here equal into negative K. Times T. Plus the natural log of the initial concentration of A. Alright, so first we can go ahead and calculate for the initial concentration of c. four H. 8. So I just put this here. So that's c. H8 initial. So I just put that as not. And of course we want the units for this to be equaling to mole per leaders because that's what the polarity is and that's what concentration is. So one thing to notice is that the molar mass O. c. four H is equal to 56.104 g per mols. Alright, now we're gonna go ahead and start with the calculations of C four H eight. Were given that we have an eight g sample of this. We want to go ahead and convert this into moles by using the molar mass. So for every one mole of this we have 56.104g. And that's just the moles since we know that the g will cancel. And since it's per leaders will just multiply one per the leader of a solution that we're given is 5.00 L. So then we'll get moles over liters which is just the concentration. So putting this into my calculator. I'll get the value of 0.02852 most per leader. And that's just capital M. Here. Now calculate for the final concentration of c. four H. 8. So brackets here c. four H. 8. Is that if not or the zero will have f. Or in this here we'll just have it as tea. So you know that the time here we said is 10 minutes we want to go ahead and cover this in two seconds. So for every one minute we have 60 seconds We see here that the units of minutes will cancel giving us a value of 600 units being seconds. Now we can go ahead and plug this in to calculate for our final concentration. Alright so using the first order integrated rate law that we just wrote here we can go ahead and solve for the final concentration of our c. four H. Shaped. So the natural log of the concentration of C4 H is equal to negative cape. So negative and the K. Is 3.12 times 10 to negative three. And then our T. We just calculate for that and that's 600 seconds. And then adding the natural log of the concentration of our initial C. Four H. Eight. When we saw that to be 0. to moller Again we're here we're solving for the natural log of c. four H. 8. So then here we get negative 1. plus negative 3.5571. Now doing more simplification here. Once we add these two negative values we get the sum to be negative 5.4291. I was gonna go ahead and do this here. So taking the E. Of the natural log of the concentration or the final concentration of C. Four H. Eight equaling to the E. To the negative 5.4291. So here this E. In this natural log will cancel leaving us with just this here and isolate that. So then continuing our simplification. We get that the concentration or the final concentration of C4 H8 will be equal to 4.387 times 10 to the - Moller. Alright so now we'll have to go ahead and find the moles of the c. H8. Using what we just calculated for the polarity. So the moles of C4 H8 is equal to all. We said we have five liters of the solution Using this concentration that we just solved for. So 4.387 times 10 to the negative three moles her leader. We see here that the Leaders will cancel leaving us with just the moles of c. H8. So once we put this in the calculator I get the value of 0.02194 moles. Now cackling for the initial moles of C four H 8. We have the starting eight g. And then using the molar mass. So that's for everyone more. We have 56.104g. You can see here that the grounds will cancel leaving us with the units of the moles and a numerical value to be 0.14 to six moles. And we need this because we are going to do the I. C. F. Table. So just scrolling down to give a space for the I. C. F. Table. So I stands for initial sti stands for concentration and F stands for final. Our chemical equation is going to be that C H8 Will yield and give us two moles of C two H 4. Now the initial values are just stuff that we have all we know. So for the initial value of C four H eight, that 0.14 to six moles. And that's what we saw for in green. And then for our product here we have zero of this for starting then for concentration here, we don't know this. And then for our products we'll add two X. Because this is also unknown. The reason why we have a coefficient of two here is because we have two moles of our product. So then our final here Is going to be 0.14 26 moles minus X. And their product to be just two X. All right now, for the moles of C four, H eight is equal to 0.14 to six moles times X. Very calculated for the moles of C four H eight, that's 0.2194 moles. That again, which is equaling to 0.14 to six moles minus X. We see here that we can easily isolate this X. That's our unknown. So if we just add both sides and then divide by negative one, we get that X is equal to 0.1 to one moles. So what we can see here is that after Smooth this little lower. So after 10 minutes the total moles is equal to the moles of a C four H eight plus the moles Of our C two H 4. So basically our exes here. So for the most of C four H eight, that 0.1 to one moles and then the most of C two H four is two X. So two times 0.1 to one moles. So you put that into a calculator, You get that, a total moles is equal to 0. moles. Now, finally using the prevnar to creation here. So that equation is PV equals N. R. T. And this equation resolving for api so we can isolate RP by dividing each by both sides by V. Because it's the new equation to be an R T over V. Alright, so for a temperature were given that to be 480 degrees Celsius, we're gonna convert this into kelvin. So we're gonna go ahead and add to 73 point 15th. That gives us 753.15 kelvin's. So now we have all the values to plug into this ideal gas law equation, I apologize. It's this one right here, this new one here. Alright, so p is equal to n is equal to moles And we just said that to be 0.26394 moles. And then for our that's our constant and that's 0.8206 units to being laters times a T. M per mole times kelvin. And then our tr temperature, we just calculate that in red right above and that's 753.15 kelvin's. This all over V and V is just our five liters. Just make this line a little bit straighter. Alrighty. And we can see here that for the units cancelation moles will cancel, kelvin's will cancel and leaders will cancel leaving us with the final unit to be a T. M. And that's what we want for pressure. So putting all of these numerical values into the calculator, I get the final value of 3.26 and units of course being just a T. M. And this right here is going to be my final answer for this problem

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