A 0.500 L reaction vessel equipped with a movable piston
is fill...

Question

A 0.500 L reaction vessel equipped with a movable piston 
is filled completely with a 3.00% aqueous solution of 
hydrogen peroxide. The H2O2 decomposes to water and O2 
gas in a first-order reaction that has a half-life of 10.7 h. 
As the reaction proceeds, the gas formed pushes the piston 
against a constant external atmospheric pressure of 
738 mm Hg. Calculate the PV work done (in joules) after a 
reaction time of 4.02 h. (You may assume that the density 
of the solution is 1.00 g/mL and that the temperature of the 
system is maintained at 20 °C.)

Accepted Answer

Hello. Everyone in this video, we're dealing with a whole lot of things. We're dealing with rate, constant integrated rate laws. ICF tables and hurt. So first let's go ahead and start with our given 2.5% solution. So from this information, first of all, we need to know the molar mass of our ammonium nitrate. So the molar mass of NH four N is 64.52 g per milliliters. Now, we need to first offer the mass of our ammonium nitrate. So how we do this is while we have one liter of solution. So the mass of the NH four N 02. Alright, again we have one liter of solution. We can convert this into milliliters. So we do a direct conversion. So for every one liter here we have 1000 mL. And then from there we can go ahead and take the density of the solution which is one g for every one millim. And then here we can continue with the 2.50 over 100. And that's just the percentage of the agree solution. Alright, after we do that, we can see here for our units, the leaders will cancel the male leaders council. Then our final units will be in grams for the numbers I'll put this into my calculator. I get the value of 25.0 g of ammonium nitrate. So NH four N 02. Alright. Now calculating for the moles of NH four N 02. So now we know that we have 25 g of NH four N. 02. We go ahead and use the molar mass that we wrote in the first line here to calculate and go from grounds into malls. So for every one mole of this We have 64.052 g. And we can see here that the grams will cancel. So putting this into my calculator, I get the value of 0.3903 Units being moles of NH four n. 0 two. So then the polarity When we know that clarity is moles over L. So we're taking the moles calculated which is 0. moles. Leaders is just one leader. That gives us the malaria T to be 0.3903 units being moles over leaders. Alright let's just go ahead and scroll down. So next we can do is calculate the rates constant. So the half life here is equal to the L. one, so natural log of two over cake. And we every rearrange this we get that K. is equal to the natural log of two Over t. half which is just the half life. So then our constant then is the Ln of two divided by while we're given that the time here is 2.54 hours. And that just gives us a value of 0.2729. Alright now we go ahead and calculates using the first order integrated rate law. So for that equation that's natural log of the concentration of NH four n. O. T. And that's already something that we calculated and that equals to negative K. Times T. Plus the natural log Of the concentration of NH four n. 0 two. And that's for the initial. All right, let's go ahead and sell for this. So, again, the natural log of the concentration of NH four N. 02. That gives us just plugging in the values here. K. Is 0.2729. T. is going to be 1.50. And we're adding the natural log of the concentration that we calculate for to be 0.3903. Alright, so I'm just putting everything into my calculator. The natural log of the concentration of NH four, N. 02 gives us a value of -1.3502. Now finally isolated the concentration of NH four and 2. We're going to take the E. Of the power of both sides. So we do this, we just basically cancel out the natural law here and we take this here. So basically what we're doing here, I'll just write in a different color is E. To the power of the natural log of the concentration of N H four, N. 02. Equalling to the e. to the negative 1.3502. And we know that these two will cancel each other out. Alright, so again putting this into the calculator. I get the value of 0.2592, you need to be moles per liter. And then we're gonna multiply this by The leader of the solution which is 1.00 leaders. And this leaves us with just 0. moles. Alright now again, just scrolling down, we can use our I. C. F. Table now. So I stands for initial C stands for concentration and F stands for final. So we have I. C. And F. Of course we have the chemical equation to be N. H. Four And 02. This is Chris and this yields and two a gas and two moles of H 20. And its liquid state. Alright, so for initial We have 0.3903 moles. And we know this because we calculated for this and two is just zero, same thing for H20. Now for our concentration we don't know this but call this as X. Same thing for N two, We do not know this and that's just plus X. And then for H +20, that's plus two X. And the reason why we have this two coefficient is because we have two moles of H +20. Alright. And lastly for our final we don't know our X. So that just gives us 0.3903 moles minus X. For N two. That's just X. And then our final for H 20. Is just two X. Alright. Again, just scrolling down to focus on our I. C. F. Table. So then our concentration of the N H four And 02 is equal to 0.3903 moles minus X. And this just this right here. So then we get 0. moles equaling to the 0. moles minus X. And this again is just something that we already saw for where somebody's plugging that in. So now if we just isolate our X. Here, we get X to be equal to 0. units being most. So next hour moles of our end to then, since we already calculated for X is equal to 0.1311 Most. All right. So now we're gonna go ahead and use the pivot equation. All right. So of course we want to do some unit conversions. So, for pressure here we have 746 millimeters of mercury cover this into a. T. M. So for every one A. T. M. We have 760 millimeters of mercury. We see here that the units here will cancel leaving us with just a T. M. And I guess it's a value of 0.9816 A T. M. R. Temperature were given this to be 25 degrees Celsius. We want the temperature to be in kelvin's. We can just convert this to kelvin's by adding 273.15 Which gives us the temperature to be 298.15 Kelvin's. All right now we have all the factors to go ahead and use our pivot equation. So PV equals N R. T one himself for volume here, which is just V. So we get V equaling two N. R. T over P. Now plugging everything in here for our end. That's our moles. And we set that to be our X. so 0.133 or malls are are just are constant. So that's 0.8 to 06. He was being leaders times A T. M. Over moles, times kelvin. And then lastly for our temperature, that's 298 0. kelvin's. All right. And we'll have that all over our pressure and we just calculate that to be 0.9816 ATM. Let's go ahead and move this over a little. Alright, so we see that for our units here can cancel out moles Kelvin's and ATM leaving us with just the units of leaders right over here from our constant and exactly what we want for our volume. So once you put all the numerical values into a calculator, we get that value to be 3.27. And again units being leaders. Just one last step here that will go ahead and squeeze and bread on top of our work. So we need to solve for W. So W. Equals to negative p delta v. So W. It goes to negative P. Which is all four to be 0.98168 AM in the volume. Again, we just calculate this that is 3. liters. Once you get that you can put this into a calculator again, let's just put this a little here. Alright, so that's negative 3.21 Leaders times A. T. M. We're gonna convert this into jewels. So multiply this With 10 1.325 jewels per every one leader times A. T. M. So you see here that those units will cancel leaving us with just jewels and this gives us a final value that W is equal to negative 25 jewels. And this right here is going to be my final answer for this problem.

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