5

Question 1273Using the thermodynamic data below; calculate K for the conversion of oxygen tO ozone3 0,6g) $ 203at ZOOO.0K [Calculate AH",xn and_ 4S" frst_...

Question

Question 1273Using the thermodynamic data below; calculate K for the conversion of oxygen tO ozone3 0,6g) $ 203at ZOOO.0K [Calculate AH",xn and_ 4S" frst__]Substance AH (kJlmol) Ozlg) O3(3) 14235? (Jmol. K) 205.0 237.6291 *10-9267 - 10-8544 10-14 184 '10-152.04 < 107

Question 12 73 Using the thermodynamic data below; calculate K for the conversion of oxygen tO ozone 3 0,6g) $ 203 at ZOOO.0K [Calculate AH",xn and_ 4S" frst__] Substance AH (kJlmol) Ozlg) O3(3) 1423 5? (Jmol. K) 205.0 237.6 291 *10-9 267 - 10-8 544 10-14 184 '10-15 2.04 < 107



Answers

$$ \begin{aligned} &\text { Calculate free energy change for the conversion of oxygen to ozone }\\ &\text { at } 298 \mathrm{~K} \text { , if } K_{\mathrm{p}} \text { for the change } \frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g) \text { at } 298 \mathrm{~K} \text { is }\\ &2.47 \times 10^{-29}(\mathrm{~atm})^{-1 / 2} \end{aligned} $$

Looking to calculate the value of delta G. Of a reaction. The gibbs free energy. When we have delta gino is equal to delta H minus T. Delta Esnal. Where we can break this equation down so we have delta H not is equal to the sun. Delta H. Not of the products. Subtract the some delta H not of the reactant. We can also use the same form out of equation for delta S. Not. So the disorder where we have the some of the change in therapy on entropy of the products, subtract the sun change and peace of the reactant and so we can determine our values. So we have delta H. Is negative 75 kg joules per mole. So remember that units are very important here we have delta S. That is negative 83 joules per mole per kelvin. So we plug our values in to solve the delta G, which is negative 50 to 66 jewels per mole. While the temperature is 298 kelvin. Remember to use the unit of kelvin, not degrees Celsius. So because DELTA G is negative hit, this means that the reaction is in fact spontaneous and so we can determine the lowest temperature at which the reaction is spontaneous. So what we need to do is make one assumption. So we can assume that at equilibrium delta G not is equal to zero. So that allows us to rearrange our equation to solve the temperature, which is equal to delta H, not over delta S. Not. We plug in those values that we identified earlier, how we saw for temperature, we're left with units of Calvin and that temperature is zero Calvin and that is the lowest temperature that the reaction will be spontaneous.

All right, So the question is gonna ask us to calculate activation, energy and also frequency factor. So they're going to be two different steps here in doing this problem. We'll start with the activation energy. So to start out with, you're gonna need this equation on DSO. The information that we're gonna fill in in this equation is all coming from either Constance or from the table that they provided in the question. So, looking at the information to start out with, we're going to need to find out what k one and k two r the rate constants for one into. So you're going to choose whichever ones you want. It really doesn't make much of a difference as long as you know which one you've chosen. So in this case, we're going to use the values from temperatures to 75 250 degrees. So to 50 will be t 1 to 75 t two and their corresponding K values will be plugged. It's OK to will be five point negative and say 5.58 times 10 to the negative four and K one will be 2.64 times 10 to the negative for next. We have activation energy. That's what we're looking for. So we won't have anything to fill in. Are you? Should be familiar with is a constant. Um OK, so that's the gas constant. It's gonna be 8.31 for five. I'm not including units here. It's just run out of space. If I do so on your paper units would be appropriate thing to keep in mind. Um, moving on, we're gonna have a temperature. So again, the temperature to has toe match up with K two. So we got this. Que value from the temperature that was 275 degrees Calvin and then t one is 250 Calvin again. It matches up with the Katie one value. So next we're gonna pretty much do math. We're gonna go through into a little bit of algebra. So if you do the left hand part, the natural log, you should come up with that. It's equal to 7484 Still have that activation energy over 8.31 for five. That hasn't changed. That last part that we can calculate make our life a little bit easier. It's 0.36 So our next step is we're gonna divide. We're gonna get rid of that. 0.36 Divide that right hand side by 0.362 by the left hand side. Don't forget, that is a negative number. When you do that, you'll find that the left hand side is equal to negative 2058 Change. Now we've got E A along with 8.3 145 So the last step is we've got a multiply. We gotta get rid of that. Eat. Do not forget when you're doing this this negative sign right here. So give a couple of options. You can wait until the end and divide by negative one. Or you can, when you multiply, just multiply by. Negative eight. That's fine too. Okay, Well, when you do that once you've gotten rid of the negative sign. Arabia 17,113 0.7. It's going to be in what? Your units should end up this Jules Permal. Okay, so that's the first part. That's the part. We're calculating activation energy. So wrong near the problem is going to ask us to go ahead and calculate another fail. So from there, we have to go ahead and come up with the frequency factor. Now that's going to involve using what we just discovered. But it's also going to involve a bit of a different equation. So much of the same information that we've already used we're gonna use again, again becoming familiar with these equations. Really, really important. Our equation, in this case, great constant frequency factor E to the negative activation. Over. Yes, constant times temperature. Our Okay, so we've got to get rid of all of that. But once we do that, first we have to plug in some numbers. Okay? So again, you can choose whichever one you want to work with. That's totally fine. Come on. In our case, we're gonna I'm gonna use for me 325 degree Kelvin. I just chose that one. There's not any real reason that I chose that one. Someone start by plugging in the information on go step by step instead of doing it the shortcut way. So Che for that value is going to be 1.77 times tended negative. Three. You need to figure out what a is you calculated activation energy. Last problem. So that's the good news. You have that number one less getting ahead of myself. Looking Yoon are again. It becomes a problem space wise for me. But for you, you have your entire paper a little bit easier. All right, so that's the values all plugged in. Now we gotta figure out How the heck are we gonna get a by itself? So a couple of ways to do it, you can do out the math. Initially, you can just plug most of it into a calculator. It's really up to you. So the opposite of E to get rid of e would be to do a natural law. We have to do that for both sides. Okay, So what we will end up with? Well, im leaving behind parts. Well, we'll end up with is the natural log of 1.77 times 10 to the negative three. Careful when you during this, that's going to be equal to so again, we just made the numbers a little easier for ourselves to work with. When you plug that information in you should get 2702 point shoot. So now we're going to do to a few things to get a by itself so you can go out and do the math out. What you're gonna do is this national log you're going to multiply by 2702.2. That will bring that over. And we're also going to divide. Well, I 1000 or 17,000 113.7 teach. Got to do everything to both sides. Now you're gonna calculate that. So you're gonna put on a calculator when you do that, the value should come out to negative 1.1 Again. This last number might change just a little bit, depending on how you round in your units. And that's gonna be it. That's your number

So continuing on with the topic off thermodynamics. What we're looking at here is the standard and will be changed. So given our reaction, we're calculating Dr H No, on we calculate that as thesis. Um, the moles, Delta H no formation of the products. Subtract thes some the moles Delta h not formation after reactant. So our delta h not Here is 286 kg jewels. Then we can calculate our delta. Gino, Dr Gino is some of the moles Delta G of formation of our products. Subtract thes some the moles multiplied by the delta G formation of our reactant. So our Delta Gino in this instance, there's 3 to 6 kiddo jewels. So because we know that Delta G is equal to Dr G not add r t l n Q at equilibrium data G and is equal to zero. And cue is equal to K so we can rewrite this as L N. K is equal to negative. Delta Jeannot divided by r t. So we're calculating k here. So what we saw for is lnk. That is negative. 57.1 self. Okay, that is 7.22 times 10 to the minus 58 7.22 times 10 to the minus 58. So because the partial pressure of auction is one times 10 to minus three de partial pressure, P 03 that is equal to 3.3 times 10 to the minus 41 represents one molecule of ozone per 9.5 times 10 to the 17 liters of air. So therefore, the equilibrium is most likely not maintained under the conditions because the concentration of ozone is not large enough to maintain equilibrium.

Let's look at the reaction for the production of ozone. Just 302 equilibrium with 203 Yes. Using the data from Appendix four will calculate first of all, Delta each Not. So. You go to the an therapy of formation of two times infamy. Information of all three minus three times the entropy of formation of oh to gas to be equal to two times 143 Killah jewels. Permal minus three times zero. This work so too. 286. Kill a Jules Permal for Delta H not and Delta Jeannot with two Delta G not for hold three minus three. Felt that Jeannot for two. Again. Look at the data from Appendix four. This is 163 Killah jewels per mole minus 30 This work so too. 326 kill jewels per mole for Delta. She not now. Let's solve for the value of the equilibrium constant. So well, you see expression Lawn K is equal to negative Delta G, not over our tea. And this is equal to minus 300 26 killer jewels, sometimes 10 to the three jewels Permal over our just 8.314 Jules Permal Calvin temperature is to 98 Calvin and Lawn K is equal to negative 1. 31 point 573 And so K is equal to E to the minus 1 31 573 which works out to an equilibrium constant of 7 to 2 times 10 to the negative 58. Now let's estimate the partial pressure of ozone in equilibrium with oxygen at 30 kilometers above the earth's surface. So we need the value for the equilibrium constant at 2 30. So what, you can use our formula here lawn of K one over K two. Mm, it's actually switches up. K two over K one is equal to Delta each Not over our one over. Yeah, T one minus one over t to. So we need to solve for the equilibrium. Constant. It's 230. Um, Calvin. So let, um, Let's go lawn Now. K one is 7.22 times 10 to the negative. 58. We'll call this the equilibrium constant. At 230 Calvin Delta each not, is 286 que a Jules. So times 10 to the three jewels over 8.314 Jules Permal Calvin T. One is 230 Calvin minus t two, which is 2 98 Calvin and solving this we find that, um, mhm. We work at the right hand side. 7.22 times 10 to the negative 58 over the equilibrium constant to 30 is equal to e to the minus 34 13 There are positive 34.3. This is equal to 6.6 pence. 10 to the 14th and solving for the equilibrium constant to 30 Calvin It's equal the 1.1 times 10 to the minus 72. So the equilibrium constant at 230 Calvin is equal to 1.1 times 10 to the minus 72. And that's equal to the partial pressure of ozone squared over the partial pressure of 02 squared. This is equal tomb. The partial pressure of what's we write it down here. 1.1 times 10 to the minus 72 is equal to the partial pressure of 03 squared over 10 times 10 to the minus three atmospheres. We put the units here, cubed partial pressure of 03 works out to 3.3 times 10 to the negative 41 atmospheres. So this would be the partial pressure of ozone in equilibrium with oxygen at 2 30 Calvin, 30 kilometers above the earth's surface. In the last part, is it reasonable to assume the equilibrium between oxygen and ozone is maintained Under these conditions? Let's now calculate the volume occupied by one molecule of ozone. The volume is equal. The n r T overpay and so one molecule of ozone would be 1/6 0.22 times 10 to the 23 moles. That would be one molecule, 8.31 or just 0.8 to 06 leaders atmospheres per mole. Kelvin temperature is 230 Calvin, and the pressure here is 33 I'm 10 to the minus 41 atmospheres and selling for the volume. Here we find that this is equal to 95 times 10 to the minus seven are positive 17 leaders. So can equilibrium be maintained? Equilibrium is probably not maintained under these conditions when Onley to ozone molecules are in a volume of 9.5 times 10 to the 17 leaders. The reaction is not at equilibrium. Under these conditions, Q is greater than K, and the reaction shifts left. But with only two ozone molecules in this huge volume, it is extremely unlikely Uh huh, that they will collide with each other. All right. At these conditions, the concentration of ozone is not large enough to maintain equilibrium.


Similar Solved Questions

5 answers
Create Swnthusis hox_tu_fllaWing Lorupaund in one veadtion Step
create Swnthusis hox_tu_fllaWing Lorupaund in one veadtion Step...
5 answers
Ihetme (insecends) fera free -folling objeet +o #ll d 52.15 quutnoh cpnstruction Wcrier dm s anailand ebxrtts 9trin4? @ Puadle atier seconds. EefimaltIh neqhiof Ihabulling
Ihetme (insecends) fera free -folling objeet +o #ll d 52.15 quutnoh cpnstruction Wcrier dm s anailand ebxrtts 9trin4? @ Puadle atier seconds. EefimaltIh neqhiof Ihabulling...
5 answers
Describe how a change in the amino acid sequence of a protein affects the mutated protein in cells of patients with CF,and that protein'$ function
Describe how a change in the amino acid sequence of a protein affects the mutated protein in cells of patients with CF,and that protein'$ function...
5 answers
Q 5. Findl (hc fuuctions (a) fog, (b) yof (c) fof and (d) gog andl their <lomain (a) f(r) = Vi
Q 5. Findl (hc fuuctions (a) fog, (b) yof (c) fof and (d) gog andl their <lomain (a) f(r) = Vi...
5 answers
13) The number of cows that can graze 0n mnch approximated by C(xy) 9x + 5y - 2, where x is the number of acres of grass and the number acrc? of alfalfa. If the ranch has 40 acres of alfalfa and 90 acres of grass, how many cows may graze? A) 1010 cows B} 810 cows 0)808 cows D) 1OORcows
13) The number of cows that can graze 0n mnch approximated by C(xy) 9x + 5y - 2, where x is the number of acres of grass and the number acrc? of alfalfa. If the ranch has 40 acres of alfalfa and 90 acres of grass, how many cows may graze? A) 1010 cows B} 810 cows 0)808 cows D) 1OORcows...
3 answers
II:Prove or disprove: Let T be a biagraph Kn,n mt n then L(T) is regular:
II: Prove or disprove: Let T be a biagraph Kn,n mt n then L(T) is regular:...
4 answers
The number of scratch in metal sheet during its manufacturing is assumed to follow a Poisson distribution with a mean of 0.1 scratch per square meter_ Use manual calculation to solve this problem.(a) What is the probability that there are at most 2 scratches in 1 square meter of the metal sheet?(b) What is the probability that there is one scratch in 10 square meters of the metal sheet?
The number of scratch in metal sheet during its manufacturing is assumed to follow a Poisson distribution with a mean of 0.1 scratch per square meter_ Use manual calculation to solve this problem. (a) What is the probability that there are at most 2 scratches in 1 square meter of the metal sheet? (b...
1 answers
Find the least squares line. Graph the data and the least squares line. $$\begin{array}{ll}x & y \\1 & 1 \\2 & 3 \\3 & 4 \\4 & 3\end{array}$$
Find the least squares line. Graph the data and the least squares line. $$\begin{array}{ll}x & y \\1 & 1 \\2 & 3 \\3 & 4 \\4 & 3\end{array}$$...
1 answers
Find each product. Recall that $a^{2}=a \cdot a$ and $a^{3}=a^{2} \cdot a$. $$ 3 p^{3}\left(2 p^{2}+5 p\right)\left(p^{3}+2 p+1\right) $$
Find each product. Recall that $a^{2}=a \cdot a$ and $a^{3}=a^{2} \cdot a$. $$ 3 p^{3}\left(2 p^{2}+5 p\right)\left(p^{3}+2 p+1\right) $$...
5 answers
The dlagram below shows right angle bend, current of 2.98 flows through the wlre Determlne the magnetlc fleld (In unit vector notatlon) at the polnt whlch from the corner the bend; (Assume that the +X aris directed the right; the +% axls directed up and the aris toward You,) 32edistance2.26 cm
The dlagram below shows right angle bend, current of 2.98 flows through the wlre Determlne the magnetlc fleld (In unit vector notatlon) at the polnt whlch from the corner the bend; (Assume that the +X aris directed the right; the +% axls directed up and the aris toward You,) 32e distance 2.26 cm...
5 answers
What types of intermolecular forces exist in a crystal of ice? How do these forces differ from the types of intermolecular forces that exist in a crystal of solid oxygen?
What types of intermolecular forces exist in a crystal of ice? How do these forces differ from the types of intermolecular forces that exist in a crystal of solid oxygen?...
5 answers
Define reflexive closure and symmetric closure by imitating the de finition of transitive closure_Use yur definitions to compute the reflexive and symmetric closures of examples in the text What are the transitive reflexive closures of these examples? Convince yourself that the reflexive closure of the relation On the set of positive integers P is
Define reflexive closure and symmetric closure by imitating the de finition of transitive closure_ Use yur definitions to compute the reflexive and symmetric closures of examples in the text What are the transitive reflexive closures of these examples? Convince yourself that the reflexive closure o...
1 answers
Solve each system of inequalities $$\left\{\begin{array}{l} y-x^{2} \leq-16 \\ y^{2}+x^{2}<9 \end{array}\right.$$
Solve each system of inequalities $$\left\{\begin{array}{l} y-x^{2} \leq-16 \\ y^{2}+x^{2}<9 \end{array}\right.$$...
5 answers
Question ZZ of 32At 4 certain temperanire, the Kp for the decomnposition 0f HS Is 0.701HSte) = Hs(g) + Stg)Initially. only H,S is present at prccsune of 0.188 atm in a closcd container Whut is the tOlal pressure in the containet equilibrium?Firu
Question ZZ of 32 At 4 certain temperanire, the Kp for the decomnposition 0f HS Is 0.701 HSte) = Hs(g) + Stg) Initially. only H,S is present at prccsune of 0.188 atm in a closcd container Whut is the tOlal pressure in the containet equilibrium? Firu...
5 answers
Let A = { 2 +( 1/ n) : n ∈ N }. Show that inf A = 2.
Let A = { 2 +( 1/ n) : n ∈ N }. Show that inf A = 2....
5 answers
Aluminum' sphere 8.05 cI in diameter1 Part A L 5 change 1 is V 3eviouSnmlan 1
aluminum' sphere 8.05 cI in diameter 1 Part A L 5 change 1 is V 3 eviouSnmlan 1...
5 answers
14) Show that the civatne ol' Ghe involute ( = 0 + (c _ s)ttol' a is given by Sm& k? + T2 (c T 8)2k215 Show that the unit binormal of the involute 0 =0 + (c _ s)tt of c = a(8) is given bykb Ttt 6 = K(c F s)k k"
14) Show that the civatne ol' Ghe involute ( = 0 + (c _ s)ttol' a is given by Sm& k? + T2 (c T 8)2k2 15 Show that the unit binormal of the involute 0 =0 + (c _ s)tt of c = a(8) is given by kb Ttt 6 = K(c F s)k k"...

-- 0.020363--