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Carbonyl fluonde COFz imporant intermediale used the production' Iuorine-containing compounds For instance #i Used make refrigerant carbon tetralluorde, CF vi...

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Carbonyl fluonde COFz imporant intermediale used the production' Iuorine-containing compounds For instance #i Used make refrigerant carbon tetralluorde, CF via the rezction2COF2(9) CO(g) CF (9), 7,.80 nt 850 K If only COF, present initally Plesbuie 200 kPn Khat is the partial pressure COF , ut equilibrium? You can convert the units 0f pressure (rom kPa to Iu , using the relation Ka 0,0 twr , Expres? your Answet with the appropruate units.

Carbonyl fluonde COFz imporant intermediale used the production' Iuorine-containing compounds For instance #i Used make refrigerant carbon tetralluorde, CF via the rezction 2COF2(9) CO(g) CF (9), 7,.80 nt 850 K If only COF, present initally Plesbuie 200 kPn Khat is the partial pressure COF , ut equilibrium? You can convert the units 0f pressure (rom kPa to Iu , using the relation Ka 0,0 twr , Expres? your Answet with the appropruate units.



Answers

Carbonyl fluoride, $\mathrm{COF}_{2}$, is an important intermediate for organic fluorine compounds. It can be prepared by the following reaction: $$ \mathrm{CO}_{2}(g)+\mathrm{CF}_{4}(g) \rightleftharpoons 2 \mathrm{COF}_{2}(g) $$ At $1000^{\circ} \mathrm{C}, K$ for this reaction is $0.50 .$ What are the partial pressures of all the gases at equilibrium when the initial partial pressures of $\mathrm{CO}_{2}$ and $\mathrm{CF}_{4}$ are 0.713 atm?

Question 80 is mostly a graphical question where you have to craft the data in order to determine the order of the reaction with respect to the reactant. To see four h six, then using the data that is provided in the graph, namely the slope of the Y intercept or the slope, in this case, determine the rate constant. However, most data is provided with respect to the reactant, and you can simply plot the pressure or the concentration of the reactant is a function of time. However, the data provided in this question is total pressure. So in order to prepare the graphs of the pressure of the reactant as a function of time, the natural log of the pressure of the reactant as a function of time and one over the pressure of the reactant as a function of time, we first need to determine what the pressure of just the reactant is from the total pressure. So to do this will define the total pressure as the pressure of the reactant initially at time zero minus the decrease in pressure as the reaction proceeds plus one half the pressure of the product. That's because the chemical reaction has historic geometry of one mole of reactant for every half full of product. So when we decrease the reactant by X will increase the, um product by one half X. So if we can figure out what X is that X will be the change or the decrease in pressure of the reactant. So the initial pressure of the reactant minus X is going to be equal to the actual pressure of the reactant at any particular time. The pressure of the reactant at time zero, which is what this value is, is given to us as 4. 36 millimeters of mercury. So if we solve for X, then we can solve for the pressure of the reactant at Time T simply as the 4. 36 this value here minus X So what I did is I put in this equation here, rearranged it to solve for X, knowing that this is 4. 36 this is whatever the value is that was given to us for the total pressure. And the only unknown is X and I solve for X. Once I get X, then I can solve for the pressure at time T and that's what these values are right here. Then I'll plot the pressure as a function of time. I'm sorry. Natural log of pressure is a function of time. One over the pressure is a function of time, and the pressure is a function of time in which everyone gives me the best straight line that will correspond to the order, namely first order. If natural log of pressure is a function of time, gives me a straight line Second order. If one over the pressure of the reactant is a function of time, gives me a straight line or zero order. If it's the pressure of the reactant plotted as a function of time, that gives me a straight line to determine which one gives me the best straight line. I look at the R squared value and the one that is closest toe one is going to be the one that is the best straight line, and that is one over the concentration, which means that this is second order because it is second order. Then I can determine the rate constant by simply looking at the slope. It's okay, then is going to be equal to two times 10 to the negative five. Because it's second, order the units on KR one over polarity time and the time units here are minutes.

We give in the reaction between hydrogen and sulfur to form age to us, and we're given the starting, um there, among the starting materials and also at the ICO TBM condition. So we have to figure out the key peele toe we Asher. And at 1670 K Ah, we've the pressure for expressing and most fear. All right, so if I'm here ah, were given their mass and also the number more so Ah, we're going to first final. Casey. So that's why I wore Casey over here and then fell KP. All right, so you onto the vowel, Casey Red took over everything to concentration. So the concentration off hydrogen, we have one gram. So and, ah, volume of the flask. This Ah, yes. I actually skipping this cereal in five, see Rosile litter. So we're going to foul the concentration of hygiene first. So we have one gram divide by the molar mass and then the wired by the concentration, and we I'll find Yeah, we kind find as he was 21 border, and then we're going to do the same thing we have. Ah, hydrogen. Uh, the high that had you know, sell If I and we should be able to find as he goes a serial 0.6 to 2 smolder. All right, so and then we go to ur convert again. The lumber more sulfur Teoh, Um, to ah, the concentration. So we have eight point overtime tender power of s six, divided by soup on five. So we're 1.6 time tent in there, Five motor. All right, so yeah, we already used up those information. And also, the concentration is the most important. All right, so ah, from here, But we don't know the k. So we don't know where the reaction is going to proceed on the left or right, So it doesn't matter. So we just assume that we go in a positive, right? All right. So, Ray, we have ah, how did you So hydrogen is going to be con su aches. And then the same time force Ah, as to gas Because going to consume two X because his, um, went to two ways show and then the formation off. It's too, as we re closer, plus sex at the end off the reaction where one point at equilibrium, my steaks and then we have ah minus ah, to aches. And then we will have, um plus six at the final. Um, you couldn't concentration off h two s if you pay attention that we've assumed that the reaction go to the right. But we don't have any says to gas and we assume they were going constant to be left to eggs. So as a resell, we we have cereal minus name to accept the equilibrium and concentration cannot be let If so, that's fine. From here, we can find that yes, method medically is right. But chemical, it is not right. So that's why we ah wi way. We should be able to see that the reaction should be going to the left. That mathematically That's fine. It is. That's fine. We still can't solve for X. Okay, so from here, we can see that, um, we did You could even concentration off hours as to gases equals to 1.6 until they're five. So therefore, silver serum as to a C, goes through 1.6 time tension and then the five. And if we are going Teoh So for that we should be able to find that takes its secrets. Teoh Cedar 0.8 Time. 10 to 5 off course here will be minus. It will be minus. Okay, So you know the world We are going. Teoh, Um, we are going to plug you, Xing. Ah, um equal labor expression over here. Ah, but when we go back, we can extend aged O K. C. We will have X square divided by the concentration off Ah h two one point. Oh! Oh, my sakes. But it would be square time. Negative two x two a powerful one. All right, so from here, we already know that excess minus syrah pointing time Tell f five. So we're going to play aches over here, and then we will have Casey. So we have so point Any time T ah minus your point. Intend to five and then we have her. Ah, square. We have one check on, sir. Square times. We have minded too. That's quick. Here, Um and then we should be a pretty physical to 4.0, all the time. Chandon power. That is six to be the Casey. All right, so the next step is we're going to final KP KP goes Casey signed are tee debtor end and data and will be goes to that's foreign field. We have to market. Right. Look your left. Okay, so we just plug in all K c and ah, I guess Constance. And then the temperature is keeping this 1670 Calvin. And then we will have this to and that's one and we should be able to find that KP goes to 1.79 for the

This question asks us to calculate the pressures of Eno. Seal to an end of C. L in an equilibrium mixture produced by the reaction off a starting mixture with four atmospheres and no two atmospheres. Seal too. Here's that reaction to Eno with seal two forms two anos e l with the equilibrium constant with respect to pressure of 2.5 times 10 to the third and the initial partial pressure of N. O s four atmospheres and of Seal two is two atmospheres with no and no see out. Let's finish our ice table. We have minus two acts here minus X here, plus two x here and now expressions for the equilibrium Sze would be four minus two x to minus X and two ex And so when we plug these in, we would have to ex squared invited by for minus two X squared times two minus X And if you expand it and solve for X, you would find that X equals one point. Eat not and then we can plug it in to each of these expressions. When we would have draw little arrows here we would have, um, 0.2 two atmospheres of NL, we would have 0.11 atmospheres of C l, too, and we would have three point about three point eight atmospheres of N O. C L.

So in this problem, we are asked about this equilibrium system in a two liter vessel. What we're given Is this this reaction here were given that we have 0.177 atmospheres of carbon monoxide at 25°C and 0.391 g of water. So The vessel is then heated to 400°C. And we're asked to find the partial pressure Of this HDH. Now, we might first, of course, consider the partial pressures of each of these two gases are two reactant gases in this case presented as a product um At 400°C. So first we'll start with the carbon monoxide. So we'll just apply some ideal gas laws will say P one over T one equals P two Over T two. So we might say that 0.177 atmospheres over 298.15 Kelvin is equal to Ex atmospheres over 673.15 Kelvin, Which we get X is equal to 0.400 atmospheres. And then for of course water we might do the exact same thing but slightly different because we have a set amount of grams. First we're going to convert to moles which is going to be 0.39 1/18 0.2 Which is An amount of moles equal to zero 0- 1 7 moles of water. Now that we have a number of moles, we can apply the real ideal gas law PV equals NRT. So the pressure is what we're solving for. So P times V of two leaders equals number of moles. 0.217 moles times are zero point zero Yeah. Uh 8 to area three Times the temperature. 673.15 Kelvin. And if we solve this out we'll get p. is equal to 0.5599 atmospheres. Okay, now that we have these two, we can start solving our K. P expression so we'll write our K. P expressions. Okapi is equal to products. Overreacting. So the partial pressure of carbon oxide times the partial pressure of water over the partial pressure Of HDH. 02. And so we know some of these values. We know the value of K P. Which is here uh well, solve for, We solved for the partial pressure of carbon monoxide 0.4. We've solved for the partial pressure of water 0.599. And notice that I'm making the X's small assumption here because of course this will be over X. The reacted amount. And theoretically each of these would be minus X. But we're assuming that X is far far less than 0.4 because K p is so large. And we'll we'll we'll have a look at that assumption at the end. So if we solve this we can get X is equal to 0.4 times 0.599 over K P. Which is going to give us 1.498 times 10 to the minus seven. Which if we apply significant figures of which we have three, we can get a number of 1.50 times 10 to the minus seven atmospheres. And we we might then of course evaluate the excess small assumption. We might notice that any number of times 10 to the -7 is four orders of magnitude Smalling than 0.4. And so the excess small assumption holds true, and this will be our final answer.


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