The rate constant for the decomposition of gaseous NO2 to NO and O2 is 4.7>1M ~ s2 at 383 °C. Consider the decomposition of a sample of pure NO2 having an initial pressure of 746 mm Hg in a 5.00 L reaction vessel at 383 °C. (c) What is the mass of O2 in the vessel after a reaction time of 1.00 min?

Video transcript

Hello. Everyone in this video. I want to determine the mass of diatonic brewing produced after two minutes has passed. So for our balanced chemical reaction we have two moles of no Trachsel brain. Mind. So that's N. O. B. R. That's in this gaseous state that decomposes into two moles of our nitric oxide and And one mole of our di atomic grooming. Alright, so the rate constant which is given to us in the problem is said to be 1.064 per Polarity time seconds and that just equals to 1.064 M two negative one times seconds two negative one. And that shows that the reaction is second order. So first we can go ahead and calculate for the initial hilarity of NLB are by using our ideal gas law equation and that's just PV equals N. R. T. So permanent. So we get that volumes over our moles is equal to P over R. T. Here is doing some rearrangements. I'm gonna see here that this right here is just moles per leaders and that's just polarity and that's exactly what we want. So then m familiarity is equal to P over R. T. So first we do some unit conversions before plugging into this equation. We're at this tea a little bit better as well. Alright, so first information that we're given is the temperature. So T. Were given that it is 50 degrees Celsius. We want this in kelvin's so add 273.15. That gives us the numerical value to be 3 to 3.15 minutes kelvin's for pressure here. We're given that the pressure is 785 millimeters of mercury, convert this into A. T. M. We do a direct conversion. So for everyone a. Tm we have 760 millimeters of mercury. You see here that the unit of mm H. G. Will go ahead and cancel. And that gives us a final value of 1. A. T. M. Not going ahead to utilize this equation. Then he which is again when we solve for that is 1.03289 A. T. M. And that's over while ours are constant and that's 0.8206 A. T. M. Times leader Per mole times Cape. Go ahead extend this line here and then we'll go ahead and multiply this by a temperature which is 3-3.15 Kelvin. So for the unit cancelation we see the A. T. M. Will cancel and kelvin's will cancel leaving us with just mole and leader. So once I put everything into my calculator I get the Final value to be 0.03895. And units being more per leader which is just something as concentration. So capital M. We'll just go ahead and rewrite this 3895 M. Alright now calculate for the final concentration of the N. O. B. R. Go ahead and utilize the second order integrated rate law and that great law is just one over the concentration or the final concentration of the molecule that we're interested in. You're going to K. Times T. Plus one over the initial concentration of the molecule that we're interested in. So t here we know that T. Is two minutes. We want this in seconds. So we'll do a direct conversion. So for every one minute we have seconds and then we can see here that the minutes will cancel leaving us with the final value of 1 20. S. Alright now we can go ahead and plug in everything for the second order integrated rate law. So just growing down for more space here so the molecules are interested in of course is N. O. B. R. And R. K. Value the constant. That is 1.64 per mole time. Second Times R. T. Which is 1/22. And then plus one over this initial concentration to be 0.03895. And that's just what we saw for here earlier in the ideal gas law equation. Alright, so once we have that we just need to go ahead and simplify this. So again one over the final concentration of N. O. B. R. That's equal to 1 27.68. Forever more time. Second. I apologize. The unit calculation here is that the second will cancel Since this s here is at the bottom and this is on top. So these two will cancel. So let's go ahead and race this here and just have it per moller. Alright, and then we'll add this with 25.6739 per moller. We see here that these two units do match and we go ahead and just combine those and we can simplify this to equal to 1:53. 39 per lola. And just isolating this here, I'm taking the reciprocal basically we get that the concentration of N O B R. This final concentration is equal to 6. times 10 to the negative three moller. Alright, so finally we can go ahead and do the I C. F. Table. So again, the balanced equation we said to be is to N O B R. And that yields two moles of N And one mole of RBR. two. So we have I for initial C for concentration and F for final. So for initial We solve that to be 0. capital M for concentration, initial for products both is zero then for concentration that's unknown. So for our certain materials that's negative two X. Products, that is just plus two X and B R. Is plus X. The reason why we have X. Because our concentrations are unknown. And then the coefficients here is because we have two moles of each of those three agents. So then the final then for R. And R B R. So are only starting material. That is 0.3895 m minus two X. For our first product R N O. That's just two X. And our final last product is just X. Alright. The main goal of course, is just to solve for X. Being first solve for the Products are the starting material. So that's the concentration of RNO. is equal to 0.03895 - x. So we already know the concentration of the starting material and that is 6.52 times 10 to the -3. And that's just from this calculation here again, we'll just rewrite the right side of this equation. Well first go ahead and subtract both sides by this and multiply it by negative one which is about two X. Is equal to 0.3243. And I'll divide both sides by two. To isolate our X. We get that X. Is equal to 0. Moeller. And that is also the concentration of R. BR two. Alright, now the molar mass of br two Physical to 79.9 g Permal by this by two. So this is here, it's just the atomic mass of the element itself says be art and has two atoms here. So almost have this by two giving us the molar mass of BR two to be 1 59.8 g per moles. That calculating finally for the mass of our di atomic bombing, that's just taking our X value that we just offer from the I. C. F. Table. So 0.16215 and capital M. Is just polarity. And that's something as moles per liters. We'll go ahead and multiply this by one leader and then finally kept going from moles to grams. Will use the molar mass of B. R. Two and that's 1 59.8 g per one more. You can see here that the leaders were canceled and the most were canceled. Giving us a final value for the mass of the BR two to be 2.59 g. And this right here is going to be my final answer for this problem.