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Question 42Provided the following information determine 4 H formation of CzHs(g):C(graphite) + Ozl(g) __ COz(g) 4H -393.5 kJ Hzlg) + {ozlg) HzO() 4 H -285.8 kJ CzHs...

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Question 42Provided the following information determine 4 H formation of CzHs(g):C(graphite) + Ozl(g) __ COz(g) 4H -393.5 kJ Hzlg) + {ozlg) HzO() 4 H -285.8 kJ CzHs(g) Zozlg) - 2COzlg) + 3H2O() 4H -1560.7 kJEdit View Insert Format Tools Table12ptParagraphB I Q 4y E~ T? v |

Question 42 Provided the following information determine 4 H formation of CzHs(g): C(graphite) + Ozl(g) __ COz(g) 4H -393.5 kJ Hzlg) + {ozlg) HzO() 4 H -285.8 kJ CzHs(g) Zozlg) - 2COzlg) + 3H2O() 4H -1560.7 kJ Edit View Insert Format Tools Table 12pt Paragraph B I Q 4y E~ T? v |



Answers

Given that $$ \begin{aligned} 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) & \Delta H=-571.6 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{~g})+4 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \\ \Delta H=-1937 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\ell) \\ \Delta H=-2220 . \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ determine the heat of the hydrogenation reaction $$ \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g}) $$

So there is a mistake in this question for that third reaction that say that the Delta h of the reaction is negative. Um, 100 90.4 killer jewels per mole. That would be if the chemical reaction were flipped and methane was reacting with oxygen forming carbon dioxide and water. So the Delta H for the chemical reaction as written is actually positive. 890 killed Jules for mold. So the first thing to do then is write the chemical reaction that describes what we desire the Delta H of formation of methane. So one mole of carbon reacts with two moles of hydrogen get gas to produce one mole of methane. Then the chemical reactions that are given to us are shown here below, and I'll just rewrite them so that we have them before way begin manipulating and using has a lot to get belt h of the reaction about. So you'll notice then that I have read written this here as positive. This is the only way you're able to get the answer in the back of the book. So the first thing that we need to do is multiply the second reaction by two. In doing so, we'll get to hydrogen is on the left hand side, which is what we need. And that's all we need to dio. You will notice that, um, the to oxygen's two moles of Oxygen's Here Castle, the two moles of oxygen's here. The two moles of liquid water canceled two moles of liquid water here and carbon dioxide dear will cancel the carbon dioxide here we can, then some up to chemical reaction in order to get the chemical reaction that we desire and then some about all the modified Delta H values. And we get negative 74.9 killer jewels.

Okay, So this question is asking us to compare our understandings of entropy of formation and entropy of Adam ization. And so to do that, it uses the example of phosphorus four. And here we're looking for the entropy of formation for the gaseous form of phosphorus port. So we're looking for the Delta H. Okay, so, SS to reference the entropy of Adam ization standard entropy of analyzation for gashes. Phosphorus four from Question seven. Question four, Dash seven if you haven't seen that video that is also available on new grade. Um, but the important thing here for this question is that we have the entropy of Adam ization for gaseous phosphorus poor as 1200 killer Gilles Permal and the other point that asked us to, uh, reference is from table 46 We need the standard entropy of Adam ization for, uh, phosphorus from the standard from for phosphorus from the standard a state of phosphorus as it is in, as you would find it in nature. So hear that, listed as also, phosphorus four white. Okay, so that might be a little bit confusing, but if you think about it, it says white because that's the color of it in its standard form. You could have that in a liquid, a gas or a solid. Any of those could be white. But if it's a liquid or a gas, you wouldn't really need to specify whether it's white or, um, black or red or whatever, because the organization of the molecules isn't going to affect it. It will always be just a random organization of gaseous and liquid molecules, so you would just say gas or liquid because it would always be the same color. But in solids they could be organized differently in different crystalline or a more fous structures. So we know that this is referencing that it's a solid. You can also check that in appendix 10. It will give you the same information, says Phosphorus four. White is a solid Okay, so those both point to the same thing. So we know we're talking here about ah, solid standard state. So we have the entropy of Adam ization for each of these. Oh, the table 46 It lists this as 315 kg Permal, but a very important point is that these were not talking about the same thing these air talking about. So we have our phosphorus molecule. Here, let me do p four real quick. Okay? Okay. That's what it looks like phosphorus for. And we're taking it into four phosphorus Adams with no bonds. So this mole 1200 kg Permal of gashes, Phosphorus four is talking about one mole of phosphorus four, whereas from table 46 this is talking about Adam ization on bits, giving it in terms of products. So it is one mole of phosphorus atoms, so we can see that if we want to make these equal, we need to multiply this by four. And I'll write that right under the other one. So it's clear, um, that that is 1260 kill jewels. Permal. Okay, Now, for both of these cases we're talking about, um, the the molecule per moles of phosphorus molecules. We're not talking about moles of atoms, so we can actually compare these numbers now. So if we think about how do we get the entropy of formation of gashes fosters for from these two numbers, it should be pretty. It should be pretty clear by by using Hess's law we know that every, um, entropy of formation of any, uh, molecular structure for any molecule is a linear combination of thean full piece of formation for any of the other linear structures. As long as you can go from the atomic a scale to any you can go however you want. As long as you end up at your final at the right molecule. Thean entropy formation will be correct. So we have our atoms and they are going to go to the solid, and then we're going to go to the gas, right? So that makes sense. So to go to the solid is actually going to be 1200 kill eagles, uh, perform als of atoms. That's four atoms toe, one solid molecule toe, one gashes, molecule or moles, I guess, Uh and then Sorry, that's 12 hundreds. 1260. Just solid. Excuse that. And then from solid to gas is going to be 1200 so we can actually just subtract thes two. And we can see that our final answer is just going to be 60 killer jewels, Permal. And that is again referencing moles of phosphorus four that are formed. Okay, so that's it. Thank you.

Hello students in this question for the given values of delta H. Not we have to calculate the unhealthy of the reaction. That is DELTA H. Not for the reaction at temperature T. Equals 2 to 98 Calvin. Okay so DELTA H. Not for the reaction can be written for the particular. So tumor player by delta H. Not H. One not plus two delta H. Do not minus delta H. Three. Not. Okay so we have the given values. So tumor player by delta H. One not which is minus or 5 37 kg jal plus two. More players delta H. Two which is minus 6 80. Kill a jewel minus of delta X. Three which is 52. Okay so we can solve them. So this value will be minus of 1074 and this value is equal to minus of 1360 and -52. So after solving this value so we will obtain that this is completely after addition. This is minus of 2486 kg. Okay, so this is the and help for the reaction. Okay? Delta H. Are not. Okay. So from the given options, option B. Here is the correct answer for this problem. Okay. Thank you.

So we're given three thermo chemical equations. We're gonna try to use them to figure out the delta H. Associated with 1/4 equation. So the first thing I'm going to do is go ahead and write the equation that we're looking for. Okay, so we're gonna decompose B two H. Six into its elements. So we're gonna go ahead and write B. Two, age six. Okay? And that's gonna form two morons And three hydrogen because that's diatonic. And then we're gonna try to use the equations were given to add up to what we need. Okay, so the first one has some boron in it, but it's on the wrong side. So we're going to reverse that equation and I have twice as many as I need. So I'm going to reverse the equation and divided by two. Okay? So I'm gonna start by writing B-. 203 Gives us two morons And. 3/2.02. Right, so delta H. We're going to reverse the sign because we reverse the reaction. So we're going to make it positive 25 43.8. But I also cut the reaction in half. So I'm going to divide delta age by two. Mhm. The second equation actually going to jump to the third one. Okay, The third one has B two H. Six where we want it. So, I'm just gonna go ahead and write that just as it's given. Okay, Maybe two or 3 Than three H 20. And since I didn't do anything to the equation, I'm just gonna write delta age just as it was given a negative we go, Mhm. So, so far I've got Some stuff that I want, right? I've got the B to H6. I've got the boron. Okay, But I don't have any hydrogen. So, I'm gonna use that middle reaction to get hydrogen on the right side and to get rid of some of these waters. Okay, So what I'm gonna need to do is take that reaction and reverse it. Okay, The one that was given because I want the water on the left side and I'm going to Multiply it by three. So three waters Give me three hydrogen and three halves. Oh two. All right. And so I reversed that and I multiply it by three. So I'm gonna take three times positive to 41 8. Okay? So if we look now the waters are canceled out, Okay, I've got the three hydrogen is on the right as I want to And 1.5 and 1.5 is 302. So these guys will cross out with that and I've got what I'm looking for. So, since those reactions add up, I can just go ahead and add up the delta H. Is now. So when I combine all of those, I'm gonna get negative 35 0.6. Kill the jewels


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