A 14.58 g quantity of N2O4 was placed in a 1.000-L reaction vesse...

Question

A 14.58 g quantity of N2O4 was placed in a 1.000-L reaction vessel at 400 K. The N2O4 decomposed to an equilibrium mix- ture of N2O4 and NO2 that had a total pressure of 9.15 atm.
(b) How much heat (in kilojoules) was absorbed when the N2O4 decomposed to give the equilibrium mixture? (Stan- dard heats of formation may be found in Appendix B.)

Accepted Answer

Hello. In this problem we are told a one leader reaction vessel containing 13.9 g of tetra phosphorous was heated to 578 Kelvin, the tetra phosphorous associated into an equilibrium mixture of tetra phosphorous and die phosphorus with the total pressure of 6.71 atmospheres. We are asked how many kilograms of heat were released during the dissociation of tetra phosphorous. To produce the equilibrium mixture. Let's begin by determining the initial concentration of tetra phosphorous so that we can plug it into our ice table. Initial consultation of tetra phosphorous then is given by We have 13.9 g in one liter And make use of the more mass one more Of Tetra phosphorous has a mass of 123. g. We ensure that our units of grams cancels And we're left then with moles per liter. So our concentration is 0.1122 Polarity or Mueller P four. And now we'll make our ice table. So we put our reaction and then we have initial change and equilibrium. Initially we have 0.1122 Mueller tetra phosphorous and we have no di phosphorus are change then is minus X. We have one mole of tetra phosphorous and plus two X. We have two moles of phosphorus. The equilibrium. Then we are going to combine the initial and the change given that we have one liter of our solution. That means we have 0.1122 minus X moles of p four And we have two x smalls of p two adding these together. We get the total number of moles in our next step, then that's solved for X. Using the ideal gas equation to describe our total pressure. So our pressure total then is equal to our most total is the gas constant times temperature divided by the volume. Are moles total is given in the table above. We're told that the temperature is 578 Kelvin, Our volume is one l And we'll make use of the r value of 0.08206 leaders atmospheres per Kelvin mole. We want to ensure that our units cancel. So most cancels Kelvin cancels and leaders cancels. So we're left with units of pressure. So our total pressure we were told was 6. atmospheres and simplifying this equation, combining those things without variables. We get 47. Time 0.1122 Plus X. We are then going to solve for X. We will begin by Moving our 47.431 to the other side. And then we will move our 0.11-2 to the other side. And so our x. works out 0.0-9-7. If we refer back to our table Then we find that the change in tetra phosphorous is two X. So this tells us then that 0. moles of tetra phosphorous react. The next step then, since we're trying to determine the amount of heat that is released, we will calculate the entropy standard entropy or heat of reaction, recall then that our standard entropy or heat of reaction then is equal to the entropy or heat of formation for products minus the entropy of formation. For our reactant We have two moles of die phosphorus which has a entropy of formation of .9 Gold Jewels Per Mole. Recall that we can find these standard entropy ease of formation in the appendix of our text and then we have one mole of tetra phosphorous which has the entropy Information of 144.0 kill jules per mole. And so our MPP reaction works out to then negative 26.2 filled jewels for more. And our last step then we will calculate the heat we're going to make use of our moles of tetra phosphorous that react. And then the entropy or heat of reaction step four. Then they are calculating the heat. We have 0.0-9- nose of tetra phosphorous. Making use of the Empathy of reaction. We have -26.2 killed joules per mole. We set it up so that our units of moles cancel. And this works out to then negative 0.767 killed Jules. And so 0.767 killed jules of heat was released. So this is the amount of energy that was released when tetra phosphorous associated into an equilibrium mixture of tetra phosphorous and die phosphorous. Thanks for watching Hope. This helps.

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