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Lem 7.90 - Enhanced WIlPart B 2C4Hu(e) + 13O2(g)-,8COz(g) + IOH,O(g) + 5315 kJ Gar' Ih" AH valle for the combuslion of bulane as shown In Ihe [eacllon pos...

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Lem 7.90 - Enhanced WIlPart B 2C4Hu(e) + 13O2(g)-,8COz(g) + IOH,O(g) + 5315 kJ Gar' Ih" AH valle for the combuslion of bulane as shown In Ihe [eacllon positive, do not include the sign in your answer: expres$ your answer using four significant figures. If the value isAEdSubmitBequest AnswctPar €Give the 4 H value for the formation 0f binary compounds aS shown the reaction Hz(s) Brz (g) 2HBr(s) + 36.3 kJ Express your Jnswer using three significant figures If the value is positive, do

lem 7.90 - Enhanced WIl Part B 2C4Hu(e) + 13O2(g)-,8COz(g) + IOH,O(g) + 5315 kJ Gar' Ih" AH valle for the combuslion of bulane as shown In Ihe [eacllon positive, do not include the sign in your answer: expres$ your answer using four significant figures. If the value is AEd Submit Bequest Answct Par € Give the 4 H value for the formation 0f binary compounds aS shown the reaction Hz(s) Brz (g) 2HBr(s) + 36.3 kJ Express your Jnswer using three significant figures If the value is positive, do not include the sign your answer: AEd TALACIA #Bc [Eudendc LcnLcr _ Eadcrg cn P Type here to search



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From given following equations and $\Delta H^{\circ}$ values, determine the enthalpy of reaction at $298 \mathrm{~K}$ for the reaction : $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{~F}_{2}(g) & \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) \\ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) & \longrightarrow 2 \mathrm{HF}(g) ; \quad \Delta H_{1}^{\circ}=-537 \mathrm{~kJ} \end{aligned} $$ $\mathrm{C}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g) ; \quad \Delta H_{2}^{\circ}=-680 \mathrm{~kJ}$ $\mathrm{ZC}(s)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) ; \Delta H_{3}^{\circ}=52 \mathrm{~kJ}$ (a) $-1165$ (b) $-2486$ (c) $+1165$ (d) $+2486$

So he was just considering a standard entropy off formation. We've been provided with some data for Sulphur Floren saw for Tatreau, fluoride and sulphur hexafluoride. They are as follows and so here we're just looking to determine the energy and the chemical bond of SF. So firstly, we write our equation which is us at four Afghans has s F four. When we have four s F bonds, we don't generate a figure off Negative 1369.8 which we divide by four because we have four sf bond and gives us a number off 342.45 So now we can compare this with the numbers that have been given. So SF four is 342.45 As we calculated on the value found from SF six, there's 3 to 7. Therefore the volley on the table was based on SF six and no in fact SF for So lastly we can look at the change of entropy, are formation values and discuss why they are not equal to zero. So it is the formation entropy of a given element in its most stable state. When we are considering standard conditions when we determine it to be zero so far is instead stable R S eight on flooring a stable at F two hands where they don't have figures at zero.

Problem. 81 says that given the following and a piece of formation, um, to predict wth ee bond energy for the bond between sulfur and foreign to compare a calculated bond energy a table 8.5 in your textbook and then to address why Sulphur and Florian have info piece of formation at all if they are elements. So the first part of this is to estimate the bond energy between sulfur and Florida. To do this, um, we have to use the, uh, the information that we're given a sort of creative way because it's not obvious how, um to form sulfur or florian gas from, um simpler reactant. So we can do instead is we can take the reverse reactions of either us four to sulfur and for Florian's or s f six to you sulfur and six florins. What we know from news is that the entropy of formation for S f force negative 775 collaterals Permal so we can reverse that, um, sign to positive 7 75 Since this is a reactive instead of a product in this particular, um, in this particular equation and then from this we're making sulfur gas, which the formation has a value of 278.8 kilograms. Killer jewels for more. Okay. And then we're making for florins, which has hit a formation of 79 collaterals promote. So Seattle of that together, Um, that gives us a total of 1369.8 killer Jules Total, which, um, killing joules per multiple, which, if you divide by four since s F four has four sulfur florid bonds, gets you a bond energy of 342.5. Kalid rules for more. Oops. She can do the same thing for S F six plugging in the appropriate, um, values for each of these either reacted so products. And this gives this slightly different answer. Um, 1961.8 Killer Jules per mole. Which, if we divide that by six, gives us 327 Killa Jules Permal Purcell for Florian Bond says you can see these two values are slightly different. So it's interesting about table 8.5. Is that the value listed there for a sulfur florian bond is 327 collaterals promote. So what this tells you is that the bond energy calculated for sulfur Florian bonds is most likely based off of, um, this reaction here and not SF for becoming sulfur and flooring. And then the reason, um, this question and parts. He also asks why sulfur gas and florian gas have heat of formation values all if their elemental. It's one thing to keep in mind about elements like Florian or Corinne, and oxygen or sulfur is that these particular gases exist is di atomic molecules. So chlorine gas a seal to florian gas is deaf, too. Oxygen gases, Oh, thio sulfur gases s too and sulfur gas isn't really existed room temperature or are very stable at all because sulfur is typically a solid at room temperature but sulfur gas that can be produced, um, from heating up solid sulfur takes on this die atomic form. So the reason that, um, sulfur and flooring at single atoms actually have good information is because, um, this is in fact, not their most seat

To calculate the Delta G. Standard values for these reactions. We simply need to look at the Delta giza formation for the reactant and the products. And then we subtract, we sum up the delta jeez of formation for the products and subtract off the Delta jesus formation for the reactant. So we'll look at the Delta geo formation of H. Two, S. 04 And then subtract off that for s. 0. 3 And water. And we get negative 82 3 killer jewels for this reaction will look up the Delta G of formation of calcium chloride, 70 -7 50.2 and then add to that that of water and two times that of ammonia. And then subtract off the Delta geo formation of calcium oxide and two times that of ammonium chloride. And we get negative 8.8 killer jewels. And then last of all for this last reaction. Look up the Delta geo formation of calcium chloride and then that of sulfuric acid liquid. And then subtract off that of calcium sulfate And two times that of hydrochloric acid. And we get positive 70.74 killer jewels.

In this question are going to be looking into entropy and by definition entropy just gives a measure of the disorder nous, This wilderness offices and in terms of thermodynamics entropy gives the measure of the amount of thermal energy that is available to do actual work. So in this case or in this assignment we've been given Gas four systems or Garcia's combo. So what we're going to do is we're going to look at the amount of classes that are available in the initial state. That is the reaction. And we look at the amount of classes that are available in the final state that is the product. So to put this into useless for example, look at the first one. In the first case we've got C3 H eight reacting with five moles of oxygen to produce three moles of ceo. To end formals of It's true. So what we are seeing here, remember all of these are in gaseous form. What we're seeing is six moles of gas are reacting to produce seven miles off. Okay, so what is happening here is the entropy of this system is going to increase because we are increasing the dis ordinance of the system by increasing the amount of gaseous molecules available in the final stage. So as a result, Delta S. Of this system is going to be greater than zero. And if we are to look at the delta is of the surrounding this by definition is negative DELTA H over tea. And since our daughter H Over T is greater than zero, this whole thing is going to be or delta T. All daughter age is less than zero. So if this is less than zero it means this is negative, multiplied by this negative. This whole thing is going to become a positive. So the surrounding the changing and the entropy of the surrounding is going to be greater than zero. And if we are to have these two, if these two are positive, we can conclude that the change in entropy of the universe is also greater than zero and what this tells us is the reaction is spontaneous at all champions. And if we are to look at the second system, the second system, We have two moles of gas, two moles of gas that are reacting to produce two moles of the same gas. So we just have one species of gas in the final state. So we can tell that the entropy change entropy change. This system is going to decrease. That is the standard entropy change is going to be less than zero. And if we are to use this same equation to look at the entropy change of surroundings, we can tell that delta the standard and change is greater is greater than zero. And if it's greater than zero, it therefore means that the standard entropy change of the surrounding he's going to be less than zero and we have these two Being less than zero. Can conclude to say the standard entropy change of the universe is going to be less than zero. And what this again tells us is the reaction is non spontaneous old timbers. Now moving on to the next system, since we've got three moles of gas that are being converted into two moles of gas. So the entropy change is decreasing, the entropy is in entropy decreasing. So is the end the standard entropy change for this reaction is going to be less than zero. And since our end of change, our entire approach in the standard entropy change of this reaction is greater than zero. You can conclude that the entropy change of the surrounding Is less than zero. And since we put have these being less than zero, we can conclude that the entropy change of the universe is also going to be less than zero. And this tells us that the reaction is non spontaneous ad or temperature. And if we are to look at the last system We have nine moles of gas, nine moles of gas being completed into 10 malls of course. So the entropy of this system is increasing. Therefore the standard entropy change of this system is going to be return zero. And since we have our delta age Standard being less than zero. Standard entropy change of surrounding is also going to be Better than zero. And we can therefore conclude that the standard entropy change of the universe Is going to be greater than zero. That is the reaction is sponsoring us at all temperatures


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