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'0.971 mol N2O4 0.0580 mol NOz 0.750 L 25 P2.25 L 25 %C...

Question

'0.971 mol N2O4 0.0580 mol NOz 0.750 L 25 P2.25 L 25 %C

'0.971 mol N2O4 0.0580 mol NOz 0.750 L 25 P 2.25 L 25 %C



Answers

Calculate $K_{P}$ for the following reaction at $25^{\circ} \mathrm{C}$ : $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=2.60 \mathrm{kJ} / \mathrm{mol} $$

For this question we have a reaction of two moles nitrogen dioxide decomposing. We're actually combining to form one mole die nitrogen tetroxide K. P. Is going to be equal to the pressure of die nitrogen tetroxide divided by the pressure of nitrogen dioxide squared. And we were given a value of 67. So to determine if it's at equilibrium, we need to solve for Q. Q. Is going to be the concentration of diet nitrogen tetroxide which can be calculated using the ideal gas law. The pressure will be equal to mhm. The moles that are provided multiplied by are multiplied by T divided by the this is the pressure of N 204. Well then divide that by the pressure squared of two which will be its moles multiplied by are multiplied by T divided by the volume and will square it And we get a value of 249. So no, it's not an equilibrium. In order for you to achieve equilibrium we need to increase the K value. The way to increase it is to have it shift to the right

Continuing on with thermodynamics. We've got a repeat of the previous video but just with different values. So what we're looking to do is calculate the absolute entropy pommel of ozone this time. And so what we start with is the delta H. F. No. To the entropy of formation. Okay so we have delta H not is equal to the sum of the malls. Entropy of formation of the product. Subtract the some of the most of the entropy of formation of the reactant. And so the value we should get is negative 613.4 kg jewels. So its negatives. That tells us that the reaction is exhaust ceramic and so a temperature and units of Calvin is 298.15 kelvin. So we can rearrange the gibbs ham halt equation from delta G. Note is equal to delta H. Nought minus T. Delta S. Not to find delta S. Not is equal to delta H. Not minus delta G. Not divided by T. So for Delta S. Not, which is 1 8.1 job Calvin answer. Then what we can do is plug that into the following equation while we have delta S. Not is equal to the sum of the malls. The disorder of the product subtracted some of the malls, multiplied by the disorder of the reactant. And what we're looking to do is solve for X. So all of our valley's plugged in as two on 3.7 at 188.7 take away 219.0 had to over three X. Where X. Is the absolute entropy of ozone self racks And we should get 248.0 jewels hummel per kelvin.

Yeah, for each of the following. That's calculate how many moles of N O to form with each quantity completely reacts. He started with 2.5 moles and 25 from the start geometry. Moles bend to five on four moles of 02 and this would yield an answer of five points. Your moles n 02 for be starting with 6.8 moos of into a five times two moles. Eventual five 24 moles of any two and to the correct number six pigs to Sig figs will get an answer. 14 moles of 02 for C, starting with 15.2 grams and 25 Let's convert this two moles. Molar Mass is 108.0 grams into a five. One more been to a five two moles of into five for more zaventem to and 23 Sig figs. This would work out to 0.281 moles of 02 and lastly for deep. We're starting with 2.7 kilograms and 25 pictograms. 1st 1 kilogram of 1 to 1000 grams molar mass 108.7 grams of into five in one model into a five in our strike geometry two moles of end to five to formals of N 02 and we'll find that this will be equal to three sig figs 53.1 moles of n 02

Okay, So to start this question, we're gonna use our equation here. Felt g equals rt Ln KP So we will be solving for KP. So let's rearrange this equation. So we have itself for KP. So as you can see, I divided by a negative rt and then we're going to get rid of this natural log by using E. And we come out with this final equation. KP equals e raised the Delta G over rt. All right, so moving on, we're gonna look at our equation and I put our delta G values here, so we know rt, you know, are which is the constant It's eight 0.314 in temperature were given as to 98. So if we confined, Delta G will be able to find KP by plugging in values here in solving for KP. So, in order to solve four Delta G, we're going to use our products minus reactant since again you can find the Delta G values online if you just Google Delta G A formation. So here's our equation written out so we know exactly what we're looking for. And again always remember that if there is a coefficient. This also needs to go into the equation. So right here, we're going to have to times the delta g of I C L minus the Delta G of I two plus the Delta G of CO. Two. So from here, it's just a plug and drug. We're gonna plug the use numbers, and here Seal to is going to go over there, and this one will plug in right here. And don't forget to multiply this value by two. Since we have a cowfish and right here. All right, so if you want a positive view and do that calculation, it should come out to be negative 31 0.52 killer jewels. But we're gonna want to put this in jewels, since the units of our our value are in jewels. So we're gonna multiply that by 1000 which will give us negative. Do you 15 to 0. And that will be in jewels. Not right writing the units just because it takes me a while. I'm not very good of the pen. All right. So moving on since we have our delta G value right here, come off the negative sign. We have our delta G value right here. Let me make that a little more clear. We can plug that into our first equation. So I've written out the skeleton right here. So we're gonna plug or adult Aegean right here. 315 to 0. And then it's just plug and chug again. So we are going to come out with 3.35 times 10 to G fit.


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