Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitrogen pentoxide and has a half-life of 2.81 h at 25 °C. If a 1.5-L reaction vessel initially contains 745 torr of N₂O₅ at 25 °C, what partial pressure of O₂ is present in the vessel after 215 minutes?

Video transcript

All right. Hi, everyone. So this question says that di nitrogen pent ide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in Dion nitrogen pent oxide and has a half life of 2.81 hours at 25 °C. If a 1.5 L reaction vessel initially contains 745 tor of N 205 at 25 °C. What partial pressure of 02 is present in the vessel after 215 minutes? And here we have four different answer choices labeled a through D proposing different partial pressures in units of torque. So let's begin by drawing out or writing out the reaction in question. Here, we have di nitrogen pent oxide or N 205 as a gas decomposing to form nitrogen dioxide. That's no two and 02 or oxygen gas. Now, because there is a correlation between partial pressure and concentration. What we're first going to do is use the initial pressure of and 205 to find the initial concentration of di nitrogen PTX side. So to do this, recall, the ideal gas law where PV equals N RT. Now using the partial pressure or the pressure of the nitrogen pent oxide. Initially, we're going to find the molar concentration, meaning that we're going to solve for N divided by V or moles divided by the volume and this is going to be equal to P divided by RT. Now recall that R is the gas constant 0.08206 leader atmospheres per mole. Kelvin. This means that for the purposes of this question, we're going to have to use a conversion factor to convert tor into atmospheres. So in my equation, the initial pressure of die nitrogen pide is 745 tour. No recall that one atmosphere is equivalent to 760 tour, which is the conversion factor that I'm going to use. And so I divide this quantity by the gas concert divided by the temperature. But because our temperature is in units of degrees Celsius, we're going to have to convert this into Kelvin to keep our units consistent. So to do that, we're going to add 273 to the value of our temperature in degrees Celsius. So after I solve for this expression, the initial concentration of di nitrogen pent ide is equal to 0.400862 molar. Slight correction because it is actually 0.0400862 molar. So at this point, we have calculated the initial concentration of di nitrogen pide. So we're going to use this and the integrated rate law or this process to find the concentration of D nitrogen PTX side after 215 minutes. And using that information, we can then find the concentration of oxygen gas which we can then use to find its partial pressure. Now, recall it before we can go ahead and apply the first order rate law, we have to find the rate constant or K. But we are given the half life, recall that the half life or the T one half is equal to 0.693 divided by the rate constant K. This means that K is equal to 0.693 divided by the half life. Let me just go ahead and label the initial concentration of di nitrogen pent oxide there. All right. So here to solve for the rate constant, that's 0.693 divided by the half life which was 2.81 hours. And so 0.693 divided by 2.81 equals 0.2466 inverse hours. So now we can apply this towards the first order rate law or die nitrogen pent oxide. In the case of D nitrogen and oxide, the natural logarithm of the concentration of die nitrogen ptx side after a time T is equal to negative KT add it to the natural logarithm. Of the initial concentration of nitrogen pide. So we can use the data determined previously to solve or the concentration of nitrogen pide after 215 minutes. Now, one quick note here because we're given the time in minutes, but our half life was expressed in hours, right? So T is equal to 215 minutes divided by 60 minutes per one hour. And so this equals approximately 3.58 hours. This is to make sure that units remain consistent. So here, the natural logarithm of the concentration or final concentration of nitrogen pent oxide is equal to negative K. That's negative 0.2466 inverse hours multiplied by 3.58 hours. Add it to the natural logarithm of the initial concentration which was 0.0400862 M. So here, the natural logarithm of the final temperature of dentox side is equal to negative 4.10044. And so the final concentration of N 205 is equal to E to the power of negative 4.10044 which equals 0.0165 654 muller. All right. So now our next step is to calculate the concentration of oxygen gas after those 215 minutes. So that is equal to first the difference between the initial temperature or excuse me, initial concentration of di nitrogen pent oxide. And the natural logarithm of the final concentration of no nitrogen pide. And then I'm going to multiply this by one half moles of oxygen gas, divided by one mole of N 205. And the reason for this, if I scroll up here to the very beginning is because earlier, I did not balance the equation that I wrote though, in actuality to balance out the number of nitrogens and oxygens. On either side, I would have two moles of nitrogen dioxide and one half moles of oxygen gas. This ensures that I have five oxygens on both sides of the equation or five oxygen atoms as well as two nitrogen atoms on both sides. So here we're using the stoichiometry ratios to relate di nitrogen pent oxide to oxygen gas. So here, but actually, instead of the natural logarithms, you're simply taking the difference between the initial and the final concentration of Dion nitrogen pent oxide. So no natural logarithm, right. So the initial concentration was 0.0400862 molar subtracted by the final concentration of nitrogen p oxide, which was 0.016 5654 moller multiplied by one half moles of 02, divided by one mole of N 205. So this gives us a final concentration of oxygen gas equal to 0.0117604 molar. So now we can use this information to yield or to find the partial pressure of 02 after 215 minutes. So recall once again, that PV equals NRT, which means that the pressure is equal to N divided by VRT. So N divided by V is the final concentration. So that's 0.0117604 molar or moles per liter multiplied by R that's 0.08206 leader atmosphere. Her mole Kelvin multiplied by the temperature in units of Kelvin. Now recall earlier that it was 25 °C which after adding 273 yields 298 Kelvin. So after evaluating this expression, the pressure is equal to 0.2 8757 atmospheres which I would then convert into Tor by multiplying by 760 because 760 tour equals one atmosphere. And so my final answer in units of Tor after rounding 23 significant figures is 219 Tor and there you have it. So our answer is 219 tour which matches option A in the multiple choice and there you have it. So if you watch this video until the end, thank you so very much for watching. I appreciate it. And I hope you found this helpful.