A mixture of 14.0 g of N2 and 3.024 g of H2 in a 5.00 L container...

Question

A mixture of 14.0 g of N2 and 3.024 g of H2 in a 5.00 L container is heated to 400 °C. Use the data in Appendix B to calculate the molar concentrations of N2, H2, and NH3 at equilibrium. Assume that ∆H° and ∆S° are independent of temperature, and remember that the standard state of a gas is defined in terms of pressure.

Accepted Answer

Hello. Everyone in this video, we want to go ahead and calculate for the equilibrium molar concentrations of N 204 were under the assumption that delta H. And delta S do not vary with temperature and we have to recall that delta G. Is related with the pressure equilibrium constant. So the reaction that goes on here is that we have N 204 A gas which is an equilibrium with two moles of N 02 gas. Alright so first step here is to calculate from my delta age of reaction. So this is just the entropy of my products minus the entropy of my reactant. So starting off we have two moles of the product. We're gonna go and multiply this by the given delta age of our product. So that's 33.2 killer jewels per mole subtract this with let's see here for my starting materials, we have one mole. I'm gonna multiply this by the given Delta H. So that is 11.1 killer jewels per mole. We can see here by multiplying with the mole amount, we can go ahead and cancel out the mole units. Alright, so my delta asia reaction. If I put all these numerical values into my calculator, I get the value to be 55.3 killer jewels. Alright, now calculated for my delta S of the reactions. The same exact concept started with my products. So we have two moles for the entropy value that's going to be 2 40.1 jules, per mole times kelvin training. This will again we have one mole of my starting materials but the entropy values different. This is 304.4 joules per mole times kelvin's. Alright so again we can see here that the moles unit will cancel. Alright so now after every I put all these values into my calculator, we get the value for my delta X. Of the reaction to be 1 75.8 jewels per kelvin. Let's go ahead and convert our jewels unit into our killer jewels. So we can multiply this. So for everyone kill a jewel that we have, we have 10 to the three jewels. So slashes 1000 jewels. You see here that the jewels unit will cancel leaving us with the killer joules per kelvin. So once I put everything into the calculator again we get the value of 0.1758 killed Jules per Kelvin's. Alright so for my delta g of reaction this is equal to the delta H. Of my reaction minus T. For temperature multiplied by or with delta S. Of direction. The delta G of reaction can also equal to negative R. Times T. Times natural log of R. K. Constant. So now kind of using this equation here, Starting at the age of reaction which we've already solved for now go ahead and do this in green. So we have 55.3 killed jules per mole subtracting from The T. For temperature. This is at 350 Kelvin's multiplying this by the Delta reaction. Which again, we are soft for this is 0.1758. Killer joules per kelvin. We can set this equal to the negative rt natural log of K. So negative. And our value this is our gas constant. This is 8.314 times 10 to the negative three killer jewels, per mole times kelvin. When I multiply this by T. Again that's 3 50 kelvin's. And then we have the natural log of K. So now we're isolating the natural log of Cape. So to do this We can get then negative 6.23 kg jewels. Her mole is all divided by -2.9099. Again, units being killed Jules Permal. You see then that the entire unit will go ahead and cancel. This gives us a numerical value Of 2.14097. I'm Gonna go ahead and scroll down for more space here. Now that we have the natural log of K. P. Giving us this value. We're gonna go ahead and finally isolate R. K. P. So we can do this is that we go ahead and raise E. To the power of the answer of 2.14. So 2.14097. And this value is then 8.50769. So we're taking the E. To the power of this. We're basically canceling out the natural log here and isolating our K. P. All right, so continuing on with our calculations, R K. P. Is equal to the K. C. Constant. Multiplied by R. T. To raise the power of delta N. We know that the change in moles here. It's going from 2 to 1 or one or two. But we're doing final minus initial. So let's 2 to 1 again. I'm gonna go ahead and scroll down for more space here. So not plugging on all the american values or the equation I wrote above. We're gonna go ahead and sell for of course our K. C. Actually this binary software K. P. So get that K. C. Is equal to K. P. Times are T raised the power of the natural for the negative delta again, we're gonna go ahead and fill this equation here. So then we have 8.507692 B. R. K. P. The R value here is going to be 0.8 to 06 A. T. M. Times. Leaders over mold times, kelvin's Alright. And then we're multiplied this by team which were established earlier to be 350 calvins. This raised to the power of negative delta N. Said delta N. Is already equal to one. So embrace the power of negative one. Once we put everything into the calculator received that. My K. C value is equal to the 0.29622. Continuing our calculations here, The concentration of N 204 initially is equal to 46 zero g of N 204 divided by the two leaders that we have. And I'm gonna go ahead and multiply this with the molar mass of N 204. So for everyone move N 204, we have 92.011 g of N 204. So once you put everything into the calculator we get that the initial concentration of N 204 to go to 0.24997 molars. Alright, so we can go ahead and use this in our ice table then. Alright, so again we have the Equation chemical equation to be N 204 in grams. Or in this case a state. Sorry. And the product then is two moles of N. 02 and its gaseous state. Of course we have our eye for initial C. For training equilibrium or change in a concentration and E. For equilibrium. So initially were calculated for the concentration of N 204. So that is 0.24997 moller. And then for my product here, that's just zero. So for a change we don't know this. So of course we're gonna do this as X. For most other materials this will always be a minus sign. And for my products will always have a plus, we also want to go ahead and include two X. The coefficient of our chemical reaction. We can see here for the starting materials that we have a coefficient of one. So we don't need to include this so we have minus X. And then for my products we have a coefficient too. So it's plus two X. And now for the equilibrium part we're just gonna go ahead combined with above. So starting off with my certain materials that is 0.24997 minus X. And then for my N. +02 here that's just simply two X. So my K. C. Value is equal to the concentration of N. 02 raised to a power of two. Since we have that other coefficient divided by the concentration of N 204. We go ahead and plug in some values here. So from A. K. C. We know this value that's 0.29622. And then we don't know the concentration of R. N. 02. But from the ice table that is just two X ray Starcraft two. Still since we have this at the formula then for my denominator here that is just 0.24997 minus x. And doing some certifications here we get four X squared plus 0.29622 X minus 0.7405. Equaling to zero. You can see that we can go ahead and solve this using the quadratic formula. Once we do so we get that X. Can either equal to 0.10398. They can also equal to negative 0.17803. We're gonna go ahead and take a positive value of X. Because the concentration can't be negative. So we have this here again. I'm gonna go ahead and scroll down for a little bit more space. Alright, so for my concentration of N 204 is equal to 0.24997 m minus X. Were established that X is equal to 0.10398 molars. And once you take that difference, we get that. The value is 0.146 molars. So that's from my concentration of N. 204. Now, solving for my concentration of N. 02, this is two times X. And X is again 0.10398 molars. So once I put that into my calculator, I get the value of 0.20796. Let's write the six a little bit better here. So we have 96 molars writing those 23 significant figures. We have 0.208 molars. So this is my last answer for this problem. Thank you all so much for watching

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