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What are AHSrxn: 4S"rxn and 4G240.0 Kfor the following reactions at 240.0 K?Icompound HzS(g) Oz(g) HzOlg) SOzlg) SO3lg) HzO() HzSO4() S(g)AH?f (kJlmol) 20.17As...

Question

What are AHSrxn: 4S"rxn and 4G240.0 Kfor the following reactions at 240.0 K?Icompound HzS(g) Oz(g) HzOlg) SOzlg) SO3lg) HzO() HzSO4() S(g)AH?f (kJlmol) 20.17As? (Jlmol.K)205.6205.0241.8188.7296.8248.2395.7256.8 70285.8814.0156.9277.2167.8

What are AHSrxn: 4S"rxn and 4G240.0 Kfor the following reactions at 240.0 K? Icompound HzS(g) Oz(g) HzOlg) SOzlg) SO3lg) HzO() HzSO4() S(g) AH?f (kJlmol) 20.17 As? (Jlmol.K) 205.6 205.0 241.8 188.7 296.8 248.2 395.7 256.8 70 285.8 814.0 156.9 277.2 167.8



Answers

Classify each pericyclic reaction as an electrocyclic reaction, cycloaddition, or sigmatropic rearrangement. Indicate whether the stereochemistry is conrotatory, disrotatory, suprafacial, or antarafacial.

In this question we've been given a reaction in which two malls off and or two are creating with um all of F. Two to produce two malls off and all to act. So this is the reaction that we're looking at and the rate law for this reaction is going to be in the form of great is the quantum Okay The concentration of one or 2 raised to the power of em multiplied by the concentration of F. Troop raised to the power of end. So the main goal here is to determine the order and the right expression. But since we want to determine our um our end and finally our K. For us to have a full rate expression of this reaction. So for us to determine the quarter of reaction with respect to an autumn, We can substitute the two initial concentrations of n. 0. 2 and the corresponding right that is let's say we have afraid being a courtroom. Okay prime multiplied by the concentration of an auto is to the pope. And so here our K. Prime is representing K multiplied by the concentration of F. Troop. So we're going to leave that for a moment. So we want to determine the rate the order of the reaction with respect to an auto. In other words, we want to determine our camp. So let's take for example a certain condition where we have our right one Corresponding to a concentration number one and we have our late true corresponding to the concentration of an autumn. It points to raised to the power of em. We have been given all this information. For example When the rate is equal 0.0-6, remember our key. Okay prime this is the same. So when the rate is equal 0.0-6 The concentration is going to be equal to a 7.1. And we raise this to the power of um And again when the rate is 0.051 the concentration is going to be equal to self control again or raise this to the power of him. So what we here at the end of the day is an expression of 7.5 B equal to 7.5. Place to the power of em. And if we introduce logarithms we're going to have our law 0.5 Being equal to M Milk. So we can conclude that our aim is equal to one achieves this reaction is first order first order with respect to with respect to and or two then walking on we now need to determine the reaction order with respect to F two. And for us to do that we're going to apply the similar approach to say our right Let's look at .2 being equal to K. Prime. The concentration of F two raised to the power And at point #2 and our rate say it point blank it's going to be equal to K. Multiplied by the concentration of afternoon. 8.3 Ways to the power of end Here our K. Is going to be equal to care the concentration of an auto raised to the power of em. Now that we all have this information, remember this is the same reaction. So these kinds of So we know that when the rate is 0.051 concentration is 0.1 and we raised this to the power of and And when the rate is 0.103 the concentration of F two is equal to 0.2. And we also raise this to the power of mm. What we now have is we also have a situation where we have 7.5 billion equal to 7.5 raised to the power And And again if we introduce logarithms, the log of 7.5 is equal to end log officer 15. Again our L is equal to one. That is it is first order with respect to F two. So now that we have this information we can then look at our overall expression and if we look at the overall right expression we have the rate being equal to this is the overwrite expression. Now that we have our and our end this rate is going to be returned in the form of great is equal to k concentration of an autumn, Multiplied by concentration of F two. And the overall right over order of this reaction is M plus M which is one plus one and this is true. So the overall order of this reaction is while trump. Now we have to move on to determine our cake K. Part of this expression. So to determine the value of K. Let's look for example, we know that our K. Is going to be equal to right Divided by the concentration of N. or two, multiplied by the concentration of F Troop. So at any point let's just take any point The same .1 When the rate is zero, our concentration of one or two is going to be served with one And that of F two is also going to the airport one. So our K. here is there is 2.6 and let's take another point. The rate is 0.051. The concentrations are going to be equal to 0.2 by 7.1 and we have K two of 2.5. And let's take another point. If our rate is 0.103 the concentrations are going to be 0.2 by 0.2 and this is Equal Truth 2.5. Okay. And another point they say careful, The rate is equal to 7.411 Concentrations are going to be a 0.4 x 0.4 and this is equal to two seven and if we look at these these units are pay more could be lead to pick Tessa meter or later the second Or 2.7. I am yes, they can. Now, if we look at this they are relatively the same. So what we can do is to just take the average of these K values to get an average off K being equal to if we make the average here, They say 2.6 plus 2.55 Plus right up to 2.7 and we divide this by four. Our key Is going to be 2.61 am this again. So this is our value. Okay, now that we have this, we can then write the full rate expression to say the rate of this reaction. The rate is equal to 2.7, multiplied by the concentration of N or two. And the concentration of actually, so this is the final solution And over order of this reaction is 2nd order. That is the order is true.

Yes. For this question, you will be complete each of the following nuclear equations. These are bombardment reactions, bombardment nuclear reactions where we take one nucleus and we bombard it with something such as a neutron or a small nuclei. So if we take a neutron and we bombard bore on with it again for all nuclear equations to make sure their balance to some of the mass numbers and atomic numbers need to be the same on both sides. So if we have one and 12, that gives us 13 and zero and five, that gives us five an element five is boron. So we just get bore on 13 from boron 12. And for the next one flooring nineteen's bombarded with the neutron one plus 19 gives us 20. Perhaps we have this positron. So one plus 19 gives us 20 so 20 plus zero is 20 and then zero plus nine gives us nine and eight plus one gives us nine and eight is oxygen. A new trial bombard something producing these two products something back up. So 2 31 plus four gives us 2 35 minus one, gives us 2 34. And then two plus 89 gives us 91 and zero plus 91 gives us 91 91 is element palladium. And for the next one we've got a helium nucleus, an alpha particle bombarding potassium 40 producing a proton. So four plus 40 gives us 44 so 43 plus one, gives us 44 and then two plus 19 is 21 and 20 plus one gives us 21. An element 20 is calcium.

So in this quite a question, we're looking at these three different reactions and classifying them as the, you know, molecular by molecular or trying Molecular. So you know, molecular, Basically, this means one reactant molecule by molecular is to reactant molecule molecules and then try Molecular, as you probably guessed by now, is three reactant molecules so and a we can see that we have two molecules of Eno and one molecule of br to ah, colliding. So that means a is gonna be trying molecular b. We have just one molecule of ch three n c. And it's ah kind of being rearranged so that b is gonna be, you know, molecular and then see, we can see we have one molecule eso and one molecule of co two So that's to be by molecular

Other fallen reactions. Press it in the commentator or dIsa territory manner and under which conditions, thermal or for the chemical. Okay, this double bunk breaks and its form certainly industry and double bar We fell. Orbital's just sledded deviated from the plane, and this part of the molecule is spread. A symmetric are so this area off orbital's is formed by rotation off those friends in the opposite direction. It's, Ah, dis Rotator Way that six alert journals participated in the Iraqi injury two from double bunk from double bond and to fund its syndrome bomb. So that's six electrons. And for the district attorney mechanism A there no additional lords. So it's an automatic condition state. So the reaction goes by the thermal mechanism. Damn, they're six electrons. Those balls are participating and cottages and after being under opposite sides off the born, so the mechanism is, ah, coming up day. Today there are six intervals and the camera territory mechanism. It introduces additional note for the orbiter so six electrons will respond to the end of our magic transition stayed and in order for the nation to proceed, we need to change symmetry off homer and make it again and our magic transition state, and it's achieved by irradiation, which promotes one electoral to the next senator Orbital.


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