Dinitrogen pentoxide decomposes in the gas phase to form nitrogen...

Question

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?

Accepted Answer

Welcome back everyone. We have a two liter reaction vessel with 780 tour of hydrogen peroxide at 25°C. We're told that the peroxide decomposes into water and oxygen gas. And we need to determine the partial pressure of oxygen that is present in the vessel after 350 minutes. So we're going to recall that we need to find our initial concentration of hydrogen peroxide. And so we want to recall that we can use our ideal gas law pressure times volume equal to the most of our gas times the rate constant or sorry, times the gas constant. R times our temperature in Kelvin. And according to the prompt, we have a temperature of 25°C which we will add to 73.152. So that we get to 98.15 Kelvin as our temperature for the formula. And because we need to find the concentration of hydrogen peroxide initially, we should recall that this is expressed in terms of polarity and recall that polarity is equivalent to moles divided by leaders. And so we can go ahead and reorganize our ideal gas law so that we have moles of n divided by our volume and leaders equal to the pressure divided by r gas constant. R times temperature. And so plugging in what we know, we would say that our pressure according to the prompt is 780 tour which we want to convert to a T. M. So recall that to go from tour to a T. M. We would recall that we have 760 Tour for 1 80 M. That's a conversion factor there, which allows us to cancel out tours and we're left with a T. M. In the numerator and denominator. We have our gas constant R. Which we should recall is 0.8 to 05 Leaders times a. T. M's, divided by moles times kelvin. And then this is multiplied by our temperature which above we converted to to 98. kelvin. And so canceling out our units will get rid of kelvin's will get rid of A. T. M. And we're left with leaders per mole which the inverse will give us our concentration in polarity. And when we type this into our calculators, we're going to get a concentration of 0.419 molar. And so now we have our concentration of hydrogen peroxide initially And now we want to go ahead. And our next step is to recognize that the prompt says that we have a first order reaction with a half life of 50 minutes. So we should recall that for a first order reaction. We have the following integrated rate law where we have the natural log of our concentration of our reactant which in this case is hydrogen peroxide at a given time is equal to negative one times a rate constant K. Times time. Which is added to the natural log of our initial concentration of our reactant. In this case hydrogen peroxide. We also want to recall the half life for a first order reaction, which is that our half life T one half is equal to the constant value 0.693 divided by the rate constant K. And so solving for rate constant K. Because we are given our half life in the prompt, we can say that we can take 0.6 93 divided by our half life of 15 minutes. And so this gives us our value for our rate constant. For our first order reaction equal to 0.01386 in verse minutes. So we're going to use this rate constant and our integrated rate law to find our concentration of hydrogen peroxide after 350 minutes. And so what we would have and we'll just do that below here, is that our natural log of, sorry, our natural log Ln of hydrogen peroxide At a given time is equal to negative one times a rate constant. Which above we found to be a value of 0.01386 in verse minutes. Which is then multiplied by our time, which is given in the prompt above as 350 minutes. And then added to the natural log of our initial concentration of hydrogen peroxide, which above we stated is equal to 0. Molar. So simplifying this into our calculators, we would find that our concentration of hydrogen peroxide After the 350 minute time frame is equal to a value of 3.27 times 10 to the negative fourth power. And we have units of polarity here. So now that we have our initial and final concentrations of hydrogen peroxide, we want to calculate how much hydrogen peroxide is left over after the reaction occurs. And to write out our full reaction, we have two moles of hydrogen peroxide Which produced two moles of water And one mole of oxygen. So Taking the difference between our initial concentration of hydrogen peroxide, which above we found as 0.0419 Moller Subtracted from our final concentration of hydrogen peroxide being above as 3.27 times 10 to the negative 4th power polarity. We would get a difference equal to 0. Moller where this represents our hydrogen peroxide that is left over after the reaction. And we should make note of our ratio between hydrogen peroxide and our product oxygen 02. Where we see we have a coefficient of two in front of peroxide and a coefficient of one in front of oxygen. So we have a 2-1 ratio And we're going to use this as a conversion factor to go from our moles of peroxide. So 0.00415 moles of peroxide Which is for one liter of solution. And we're going to multiply it by our molar ratio to go from moles of peroxide, Two moles of oxygen. And so plugging in our molar ratio, we said that we have a 2-1 molar ratio here. And this will allow us to cancel out moles of peroxide leaving us with moles of oxygen per liter of solution, which gives us our concentration of oxygen equal to 0.0208 moller. So again, this is our concentration of oxygen that we have just found here. And so now we want to go ahead and to solve this answer or this pumped for our final answer, we have to calculate the pressure. So recall again our ideal gas law where we would solve for pressure by saying that that is equal to the molds of our gas, which is then divided by our volume and multiplied by r gas constant R times temperature. And so we can say that our pressure is going to equal our moles divided by volume, which we understand is another way to say molar itty. So we can just plug in our polarity of oxygen. And above we said that that is equal to 0.208 moller which we can also write as 0. moles per one liter of solution. Since we calculated this value here and so we would multiply now by r gas constant R. So recall that that is a constant value of 0.82065. And we have units of leaders, times A T. M's divided by moles, times kelvin. So this constant is multiplied by our temperature in Kelvin, which is going to be above. We converted to 298.15 Kelvin For 25°C. And so now that we have everything plugged into our formula, we can calculate our pressure and we would get a value of 0.508 and canceling out our units. We can get rid of moles, we can get rid of leaders as well as Kelvin and we're left with a T. M. So we have units of a tm for pressure. However, in the prompt were given units of tours. So we're going to go back to converting to tours. So we're going to multiply by the conversion factor to go from a. T. M. To tour and recall that for 1 80 M, we have 760 tour. And so multiplying by 7 60 we have a final pressure of 387 tour. And so for our final answer, we would say that our pressure of oxygen gas After 350 minutes Is equal to 387 tour. And so this would be our final answer here to complete this example. So I hope that everything that I reviewed was clear. If you have any questions, just leave them down below and I will see everyone in the next practice video.

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