5

Ldo?locator-assignment-take&takeAssignmentSessionLocatornassign ` Use the References to access Important values K needed for this question:The equilibrium cons...

Question

Ldo?locator-assignment-take&takeAssignmentSessionLocatornassign ` Use the References to access Important values K needed for this question:The equilibrium constant, Kc for the following reaction is 7.00x10-5 at 673 KNHAI(s) --NHz(g) + HI(g)Ifan equilibrium mixture of the three compounds in a 7.78 L container at 673 K contains 3.32 mol of NHAI(s) and 0.383 mol of NHy; the number of moles of HI present moles.Submit AntwerRetry enune Groupl 3 more group attempts remaining

Ldo?locator-assignment-take&takeAssignmentSessionLocatornassign ` Use the References to access Important values K needed for this question: The equilibrium constant, Kc for the following reaction is 7.00x10-5 at 673 K NHAI(s) --NHz(g) + HI(g) Ifan equilibrium mixture of the three compounds in a 7.78 L container at 673 K contains 3.32 mol of NHAI(s) and 0.383 mol of NHy; the number of moles of HI present moles. Submit Antwer Retry enune Groupl 3 more group attempts remaining



Answers

Consider the reaction:
$$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$
A reaction mixture in a 5.19 $\mathrm{L}$ flask at a certain temperature contains
26.9 $\mathrm{g} \mathrm{CO}$ and 2.34 $\mathrm{g} \mathrm{H}_{2} .$ At equilibrium, the flask contains 8.65 $\mathrm{g}$
$\mathrm{CH}_{3} \mathrm{OH}$ . Calculate the equilibrium constant $\left(K_{\mathrm{c}}\right)$ for the reaction at this
temperature.

Problem, 97. More percent it calls to place represent Since that total pressure is 180. Um Therefore More percent by % 100 equals two mole fraction. That is equals two partial pressure. Sure, keep equals two partial pressure of carbon monoxide to the power to divided by partial pressure. Up She or two guys. So mm we can calculate K. P. In the form of table. Okay, So we live off PP. We can simply put this value in this formula and get the answer. So for this Cap equals 2 14 0.1 73 point it 22 point seven into 10. to the power two and two and 2 approximate to into 10 to the power three. This is our answer. Apart 1st value of keeping and because K grower larger with increasing the temperature, we clearly see here when temperature is increases than the value of K P. Is also increases. So the reaction must be in the thermic. This is our final answer, last point.

Our expression for K. P. Will be the partial pressure of SEO which is our product squared. This is a coefficient of two over the partial pressure of co two There were 850°. The partial pressure of co two is our total pressure of one atmosphere times the percentage it's more percentage. So that gives us When 0623 ATM and the partial pressure of the ceo. Similarly you can see that our pressures are just going to end up being our percentages. So 937,780 M. So k. p at 850° Is going to be 93 77 squared Divided by .0623. So that will be 15.1. Okay so rather than do this math out every single time I think we can see the pattern that whatever our percentages is going to be our pressure. So at 950° R K. P is 0.98 68 because it was 98.68% squared Divided by .0132. So that will be 73.8. That 10 50 degrees 1,050°C K. P. Is going to be 9963. It's 99.63% of one. So Over .0037. So that's going to come out to be 268 or to to sig figs to seven Times 10 to the second and then finally At 1200° go ahead and put in our percentages which are now pressures Mhm. So now we have our case and we can see that as the temperature increases, the pressure increases. Sorry, the KPM increases. So this tells us the reaction must be end a thermic. Okay, So if I write it that way, that means the heat would be over here on the left. If we increase the temperature, it's like adding heat and it's going to shift the reaction to the right, forming more product, right? And increasing the value of K. P. So this reaction must be and a thermic based on our experimental evidence.

Our expression for K. P. Will be the partial pressure of SEO which is our product squared. This is a coefficient of two over the partial pressure of co two There were 850°. The partial pressure of co two is our total pressure of one atmosphere times the percentage it's more percentage. So that gives us When 0623 ATM and the partial pressure of the ceo. Similarly you can see that our pressures are just going to end up being our percentages. So 937,780 M. So k. p at 850° Is going to be 93 77 squared Divided by .0623. So that will be 15.1. Okay so rather than do this math out every single time I think we can see the pattern that whatever our percentages is going to be our pressure. So at 950° R K. P is 0.98 68 because it was 98.68% squared Divided by .0132. So that will be 73.8. That 10 50 degrees 1,050°C K. P. Is going to be 9963. It's 99.63% of one. So Over .0037. So that's going to come out to be 268 or to to sig figs to seven Times 10 to the second and then finally At 1200° go ahead and put in our percentages which are now pressures Mhm. So now we have our case and we can see that as the temperature increases, the pressure increases. Sorry, the KPM increases. So this tells us the reaction must be end a thermic. Okay, So if I write it that way, that means the heat would be over here on the left. If we increase the temperature, it's like adding heat and it's going to shift the reaction to the right, forming more product, right? And increasing the value of K. P. So this reaction must be and a thermic based on our experimental evidence.

So let's figure out what the equilibrium concentrations are for each of these substances by setting up an ice table. Now if we know what the initial concentrations are of the S. 03 and um the N. O. We can plug that into the initial line. But we plug in polarities. So we're told that we have 0.240 moles of each of these in a two liter container. So when we take molds and divide by leaders or 20.240 divided by two. Each of the polarities are .1-0. So make sure you put polarities into your ice table and we don't have any of our right side. So we're gonna have to make some right side. Which means that the right side is going to go up, the left side is gonna go down all of our coefficients in our chemical reaction or one. So let's look at the N. 02 1st I have to make some I don't know how much X. But I know for every N. 02 I make the same amount of S. 02 So that's gonna be X. I also know that for each N. 02 I make I have to use one N. O. Use the same amount and for the S. 03 of the same thing I use the same amount. So in equilibrium I started with .120. And it went down by a certain amount. Started with .120 went down by a certain amount started with none. Wind up by a certain amount started with none when up by a certain amount. Now I can write the law of mass action expression. So I can take my products and divide by my reactant since yeah I can now put those in X times X is X squared and 0.12 oh minus X times 0.12 oh minus X. 6.12 oh minus X Squared. And we know the value of the equilibrium constant is .5. All right. So why don't I put this to look like squares? Because now it's easy to just square root both sides. To make the math easy. So if I square root both sides, I get X Over .120 -1 Equals the Square Root of .5, which is .77. Now we'll go ahead and I'll cross multiply and then I'll collect life terms and solve for X. so .120 times .707 is point oh eight for eight And then .707 -1 is negative .707 x. And then when I cross multiply x times one his ex. So I'll collect like terms bring the X to the other side, Divide by 177. And I get X. Is equal to 004 nine seven. Okay so if I want to know what the concentrations are an equilibrium I have to go back to my ice table, N. 02 and S. 02 are represented by X. So in my polarity I have my concentrations of those two substances. So I know The concentration of n. 0. 2 And the concentration of S. 02. If I want the other ones I have to take .1-0 -1. Yeah. Yeah. Yeah. Mhm. Six. And that's point 0703 moles per liter. And that's the concentration of both My reactant. So is the concentration of S. 03. Mhm. And the concentration Havana. Yes.


Similar Solved Questions

4 answers
3jor_! Jiw SE cundl $ E" #e(xy+2t+(4v-xf+yZE~uIVGn
3jor_! Jiw SE cundl $ E" #e(xy+2t+(4v-xf+yZE ~uIVG n...
5 answers
04.(15 pts) Let the initial value problem {vc)-[;] be given where Y(t) = L8l andA = [2 %1 (e) Show the coefficient matrix A = [2 is nilpotcnt: (b) Solve the IVP by using matrix exponential eAt
04.(15 pts) Let the initial value problem {vc)-[;] be given where Y(t) = L8l andA = [2 %1 (e) Show the coefficient matrix A = [2 is nilpotcnt: (b) Solve the IVP by using matrix exponential eAt...
5 answers
[-/2 Points]DETAILSEvaluate each expression based on the following table:~3-2f(x)-1.50.25(a) f(o)(b) f(2)Need Help?Read ItSubmit Answer
[-/2 Points] DETAILS Evaluate each expression based on the following table: ~3 -2 f(x) -1.5 0.25 (a) f(o) (b) f(2) Need Help? Read It Submit Answer...
5 answers
The following forms of packed DNA are in the correct order from highest packing compacted) to lowest: (most a) scaffold loops; nucleosomes, beads on a string, 30 nm fibre b) chromatin, 30 nm fiber, nucleosomes, B-DNA c) B-DNA, nucleosomes, 30 nm fibre, higher order chromatin d) higher order chromatin, beads on a string, nucleosomes, 30 nm fibre e) scaffold loops, chromatids, nucleosomes, B-DNA
The following forms of packed DNA are in the correct order from highest packing compacted) to lowest: (most a) scaffold loops; nucleosomes, beads on a string, 30 nm fibre b) chromatin, 30 nm fiber, nucleosomes, B-DNA c) B-DNA, nucleosomes, 30 nm fibre, higher order chromatin d) higher order chromat...
5 answers
A7.50 kg bowling ball moving 6.42 mls strikes a 1.60 kg bowling pin at rest. After; the ball moves 5.43 mls at a 12.08 angle. What is the X-component of the pin's final velocity?X-component (m/s)Enter
A7.50 kg bowling ball moving 6.42 mls strikes a 1.60 kg bowling pin at rest. After; the ball moves 5.43 mls at a 12.08 angle. What is the X-component of the pin's final velocity? X-component (m/s) Enter...
5 answers
Need Find Submit Answer an 2 Help? 5 points RI I Practice Another Version the point Answers and SCalcET8 12.5.024 with normal vector 8itek
Need Find Submit Answer an 2 Help? 5 points RI I Practice Another Version the point Answers and SCalcET8 12.5.024 with normal vector 8itek...
5 answers
IncorrectKcut GeaaIFCansGraph the solution to the followlng etem inequalities~3r+4vz-I6 X<4 "z-8
Incorrect Kcut Geaa IFCans Graph the solution to the followlng etem inequalities ~3r+4vz-I6 X<4 "z-8...
5 answers
R-5, fr x <-3 Jsiven that f(x)={ 2x+3, for -3<x<0 finda) f(-1), b) f(-3) ad c) f(6) . 2* jor r > 0
r-5, fr x <-3 Jsiven that f(x)={ 2x+3, for -3<x<0 finda) f(-1), b) f(-3) ad c) f(6) . 2* jor r > 0...
5 answers
Let8 0+3d det] Sd 6+3 3 7Suppose det _Find the fcllowing detenninants.
Let 8 0+3d det] Sd 6+3 3 7 Suppose det _ Find the fcllowing detenninants....
5 answers
Duh Houu 7_Lm eat[7aru 072 54o7 ebuulur { 0440) Cslcuske E78 IObtortanerye Wt aTLEETTNend Helpr
Duh Houu 7_Lm eat [7aru 072 54o7 ebuulur { 044 0) Cslcuske E78 IObtort anerye Wt aTLEETT Nend Helpr...
3 answers
For Ax = b where A = ~2and ? =Construct the normal equations for the least squares solution:Find the least squares solution %
For Ax = b where A = ~2 and ? = Construct the normal equations for the least squares solution: Find the least squares solution %...
5 answers
Mtdr *1'16J PakL RrerJ] (l98 ds+ib-ton Ur! pfe^ V= 9 ire^ 2}JnDL F L`164FL?UG!'er6 7V =u& Vc ) Cs"46.7' 50l+2 ke +-6o+ ;Ilowue-er ~k^I =^ < =Jvz % 6+ + 6Fj-%k @ 2 <9 &704 4777LchE9-Jhr
Mt dr *1'1 6J PakL RrerJ] (l98 ds+ib-ton Ur! pfe^ V= 9 ire^ 2}Jn DL F L` 164 FL? UG!'er6 7 V = u& Vc ) Cs "46.7' 50l+2 ke +-6o+ ; Ilowue-er ~k^ I =^ < = Jvz % 6+ + 6Fj-% k @ 2 < 9 & 704 4777 Lch E9-Jhr...
5 answers
Iaa Oran-Oranti ve Problemler X:y:z=3:4:5 2x - 3y+2 = - 2 olduguna gore, x in degeri kactir? A) 6 B) 3 C) 1 D) - 3E)-6
Iaa Oran-Oranti ve Problemler X:y:z=3:4:5 2x - 3y+2 = - 2 olduguna gore, x in degeri kactir? A) 6 B) 3 C) 1 D) - 3 E)-6...
5 answers
6Q.53.a) f dt (4t2+1)2 ett b) f _ dt e2t+3et+52c) Find the value of sin(2sec-13) + sin(cos-13
6 Q.53.a) f dt (4t2+1)2 ett b) f _ dt e2t+3et+5 2 c) Find the value of sin(2sec-13) + sin(cos-13...
5 answers
Point) Consider the ordered bases B = d2 2] [8 3] [a of upper triangular 2 x 2 matrices Find the transition matrix from C to B.andc= ([0 '] [s 2] [6 for the vector space VT8Find the coordinates of M in the ordered basis B if the coordinate vector of M in C is [MJc =[MB =E Find M.M =
point) Consider the ordered bases B = d2 2] [8 3] [a of upper triangular 2 x 2 matrices Find the transition matrix from C to B. and c= ([0 '] [s 2] [6 for the vector space V T8 Find the coordinates of M in the ordered basis B if the coordinate vector of M in C is [MJc = [MB = E Find M. M =...
4 answers
Prove the following statement: If fn Z 0, then J Efdu = X J sdg
Prove the following statement: If fn Z 0, then J Efdu = X J sdg...
5 answers
ILS(u,v)=(3t,0) S 92at ~u2+v2 . Then Y-3v du ZeXy-e*siny, X=uv, Let
ILS (u,v)=(3t,0) S 92at ~u2+v2 . Then Y-3v du ZeXy-e*siny, X=uv, Let...
5 answers
21 - 5 Find the inverse of f (2) for r > -2/3. Further, determine its domain and range inverse. 31 + 2 Please note that this is simple algebra problem and no calculus is involved)_Prove that(sin ~ 1 1) =Find the equation of the tangent line to graph of inverse of the function f(c) = 2 +1+ at € = 1.
21 - 5 Find the inverse of f (2) for r > -2/3. Further, determine its domain and range inverse. 31 + 2 Please note that this is simple algebra problem and no calculus is involved)_ Prove that (sin ~ 1 1) = Find the equation of the tangent line to graph of inverse of the function f(c) = 2 +1+ at ...
5 answers
In the game of Bingo number is selected at random Ifit is number between and 15,then the letter B" has been chosen. Ifit is number between [6 and 30_ then the letter "T" has been chosen. Similarly, ifit is a number between 3[ and 45.46 and 60.or 6/ and 75, then the letters "N" G" Or 0" have been chosen; respectively. Find the probability that either a "G" an even number, or multiple of 3 is selected as the first Imber in Bingo_
In the game of Bingo number is selected at random Ifit is number between and 15,then the letter B" has been chosen. Ifit is number between [6 and 30_ then the letter "T" has been chosen. Similarly, ifit is a number between 3[ and 45.46 and 60.or 6/ and 75, then the letters "N&quo...

-- 0.070278--