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May be solved by the molar volume method (Section 14.7 ) or by the ideal gas equation method (Section 14.8 ). In the answer section, the setups are given first for the molar volume method, printed over a tan background. Then, over a green back- ground, the answers are given for the ideal gas equation method. Check your work according to the section you studied. The reaction chamber in a modified Haber process for making ammonia by the direct combination of its elements is operated at $575^{\circ} \mathrm{C}$ and 248 atm. How many liters of nitrogen, measured at these conditions, will react to produce $9.16 \times 10^{3}$ grams of ammonia?

For this next question, We have the reaction of into with 02 to produce nitrogen monoxide. The first task is to solve for the moles of nitrogen dioxide per volume. Ultimately, we want to get the number of molecules per centimeter cubed using the ideal gas law with pressure and temperature that are provided, we can at least solve for a number that gives us the moles of gas per leader of solution and this will be the total moles of gas, oxygen, nitrogen and whatever else is found in the atmosphere. Yeah, we can then convert this into number of molecules of gas per leader by multiplying by avocados number and then we can go from moles of gas, I'm sorry, number of molecules of gas per leader to a number of molecules of gas per centimeter cubed, Recognizing there's 1000 cm cubed in a leader, We get 2.46 times 10 to the 19 molecules per centimeter cubed. If 78% of them are nitrogen, then we have 1.92 times that of the 19 nitrogen molecules per centimeter cubed. And if 21% of them are oxygen, then we have point to one times are total molecules per centimeter cube, giving us 517 times 10 to 18 oxygen molecules per centimeter cubed. We are provided the K. P value which we will use as an estimate for the equilibrium constant for this reaction. We have gone to a concentration in molecules per centimeter cubed as opposed to a pressure and it's not polarity. So I'm really not certain if it's going to be required to convert K. P into Casey. So I'm just gonna leave it as K. P. For now. But it is possible that you would get a more accurate result if you converted it to Casey because we really have concentration values here now to convert it to K. C. We would recognize that we have two moles of gas product and two moles of gas reactant. So essentially K. P equals K. C. So then we'll plug in our concentrations for oxygen and nitrogen and all we have left is the concentration of nitrogen monoxide, which we can calculate using a little bit of algebra To be 6380 nitrogen monoxide molecules per centimeter cubed. The question then asks, how would you expect this concentration to change in an automobile engine in which combustion is occurring? Well, combustion is occurring. The equilibrium can be reached more quickly because there is a large amount of energy, so there's an equilibrium and then there's kinetics and although the equilibrium might suggest that we have more molecules of nitrogen monoxide um in pure air then within the combustion engine, because in the combustion engine oxygen is being consumed if oxygen is consumed, and according to the Shotty lawyers principle, the reaction shifts to the left and we should get fewer molecules of nitrogen monoxide per centimeter cubed. So, although this might be suggested by the equilibrium, because the kinetics are so much quicker, we may end up with a net increase in the nitrogen monoxide molecules because of the higher temperatures. So answering this, just based upon the equilibrium, the equilibrium inside of the combustion engine should result in fewer nitrogen monoxide molecules because oxygen is being consumed, shifting the equilibrium to the left.

The Great Wall Decriminalization isn't the equation that relates the rate of the reaction. So the product off the rate constant multiplies modified by one or more rating concentration each raised to the power appropriate power with which is called order of Direction. For the reaction we have energy three, a general stool, an itch for and all to on H two, image four and or school. Oh, yeah. E told you the storage toe liquid. But the reaction is first order with respect to NH three and seven. Order with respect to Asian does a little over. The reaction is neat. What? Because Okay, the next, please. A general too. From rate law. Weaken right. Expression for the constant and okay is close to Mm hmm. By N h t A. General tools for me. We know that the unit for the concentration is morality. And four times Second, now the unit off. Great constant case. It constant care This them part second him. I'm good. Um, okay, I am too So on and get it online, Istres. But because I'm on the stores now for the unit off. Okay? Are this right? The little expression and were given us, J. It was a case. Yeah, and is for the final two, not a span gold. Cool. This is this expression is equivalent to rate. This expression is equivalent to me is a ghost. Okay? And if please age and not told Yeah, the overall direction is given below, which is an 83 They've plus a turn. Old bull. That girl's returns and door. My story still or from the night reaction we have, there's a not our reaction is a little plus two in substitute all the values. Well, my next 46.1 killer jewels. Find my voice minus what did today? 0.1 your joy for most? Yeah, because minus 5 71.6. Electoral caramel minus nine point. Oh, the large Obama minutes for 82.4 Joe profile diagram can be given us is a nice cheese paragraph off the action in village duel image for and not do. And it's a blast. And our door. The energy profile is drawn by taking energy on the Y axis Andi program on stuff. The progress of the reaction on the excesses, the demonization shown on intermediate, that energy profiles that can contain to pick where it is. It's better than a me

For part A. Since the equilibrium constant for in a church three b. C. H 33 is the largest. This favours e products the greatest. Therefore, ye largest partial pressure uh, B C h 33 will be obtained from in each three C h 33 part beat. Let's, um, solve for the partial pressure of B N. H three b ch 33 Initial First, which will be for the end r t over V. We're told that we have 0.10 moles. Ideal gas constant is equal to 0.8 to 06 leaders atmosphere is more Calvin. 100 degrees Celsius is 373 Kelvin and 100 millilitres is 1000.100 liters. This works out to 3.6 atmospheres. We use this most separates table into initial change. Equilibrium 3.6 atmospheres 00 minus x plus X plus x 3.6 months X x and x Mr. Robert KP expression here would be the partial pressure of B. C H 33 partial pressure of and H three over the partial pressure of an H three. B c. H. 33 We're told that the Librium constant here is equal to you. 4.62 secrets You picks x three points your six minutes X going head and solving for X, and you can go ahead and verify that the two routes air to put with zero and minus 6.72 Gonna reject the negative root and therefore using the value of X equal to 2.10 The polarity of N. H three b c. H. 33 at equilibrium is 3.6 minus 2.10 which works out to 0.96 atmospheres. The polarity of the B. C H 33 at equilibrium is equal to similarity of N H three, an equilibrium which is on those are both equal to 2.10 atmospheres. The total pressure is equal to 0.96 atmospheres, 10 atmospheres close to put one's their atmospheres of equilibrium and this is equal to 5.16 atmospheres and lastly percent dissociation. For the in age three, bean see 33 is equal to 2.10 about it by three points, or six times 100% and we get a percent association equal to 68.6%

Live on this Is Ricky it today? Working on problem number 30 from chapter 12. So we're given a set of three reactions, of which A and C are fast and be a slow. So we really only care about the second equation when it comes to rates of the reaction. So Torno plus 02 is equal to two and no square right, too. 202. And if we're going to write down our rate long we get rate is equal to K, and we're second order for N O and first order for 02 And then what we also know is that the rate constant K is equal to 5.8 times tends to be negative. Six leaders swear to moles squared per second. All right. So using what we know, we have to calculate the rate of formation of a note too. Yeah, when we have a go to concentration equal 2.50 muller and a n o concentration equal 2.75 All right, so we're giving the concentrations man. Now you can just hold you so great. It will too. Okay, plug this in. Sometimes a concentration of Eno square times a concentration of 02 We get that our rate ISS 1.631 times 10 to the negative. Six Mueller per second. Now to get our rate of formation, we have to look at the fact that one point 63 we have to see that because we're making to know, too. We'll be forming twice a cZ, much so that the rate that it's produced is two times as fast. So our rate is going to be 3.262 times 10 to the negative sex. More percent soaps videos helpful. See you in the next.