5

(10%) 5. From the following mechanism, find the rate of the reaction: NO(g) +0x(g) ki NOs(g) fast k NQ (g) NQ(g)kz 2NQ_(g) slow 2NO(g) +0,(g) 2NO-(g) overall reacti...

Question

(10%) 5. From the following mechanism, find the rate of the reaction: NO(g) +0x(g) ki NOs(g) fast k NQ (g) NQ(g)kz 2NQ_(g) slow 2NO(g) +0,(g) 2NO-(g) overall reaction(10%) 6. For the reaction H,S(g) Hx(g) + S(g) Kp = ? (find the value for Kp) The equilibrium concentrations are Puzs 0.020 atm; Puz 0.045 atm; Ps 0.030 atm

(10%) 5. From the following mechanism, find the rate of the reaction: NO(g) +0x(g) ki NOs(g) fast k NQ (g) NQ(g)kz 2NQ_(g) slow 2NO(g) +0,(g) 2NO-(g) overall reaction (10%) 6. For the reaction H,S(g) Hx(g) + S(g) Kp = ? (find the value for Kp) The equilibrium concentrations are Puzs 0.020 atm; Puz 0.045 atm; Ps 0.030 atm



Answers

If the value of the equilibrium constant $K_{c}$ for the following reaction is $5 \times 10^{5}$ at $298 \mathrm{K},$ what is the value of $K_{\mathrm{p}}$ at $298 \mathrm{K} ?$ $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$

So we're looking for the equilibrium constant for the reaction of H C n gas in equilibrium with H plus and C N minus. So we're giving a couple equations, the first one we're going to leave just the same so H C N was O H minus will give us CN minus Was H 20. And the second one we're changing our H 20 into H plus and L H minus. So we haven't changed anything about those who are casar the same. Okay, here is 49. I'm sending the 4th from here, it is one times 10 to minus 14 cause that's kW So if we can show that these reactions add up to what we're looking for or H 20 cancels and R O H minus cancels and we're left with the reaction we're looking for. So if we can add up equations, we multiply the case. So okay is simply going to be K1 times K two. So RK is going to be 4.9 Time to 10 to the 4th Times 1.0. I'm sent to the -14. So our K for the reaction we're looking for is 49. I'm standing the -10

And this problem, we are given an overall reaction and for each part we need to sell for the change in Gibbs Free Energy at various conditions in part a. We need to sell for Delta G at standard conditions, we are given a value for the equilibrium, constant K And so this top equation is the one that we need to use to solve for that Beltagy value at standard conditions. Given that value of K so we can use that equation to directly software Delta G eight Standard conditions apart A so that is going to equal to negative our which is eight 0.314 joules per mole Kelvin and we typically saw for Delta G and units of Kill Egil so he can divide that constant by 1000 to get it in units of killer jewels and we multiply that by the temperature. And we noticed that we were given that value of K at a temperature 25 degrees Celsius which corresponds to 298 Kelvin and then we want to play that by the Ln the natural log of that value of K that we are given, which is 2.26 times 10 to the power of four. And now, when we want to play those three values together, you should find that Delta G at standard conditions for this reaction comes out to negative 24 0.8 killer jewels. And in Part B, we want to again sell for the Change and Gibbs Free Energy Delta G at equilibrium when we are at equilibrium, remember that that means that the value of the equilibrium constant K is one. And based on this equation, we know from mathematics that any time we take the natural log of one that comes out to be zero in any time we multiply any other quantity By zero, it comes out to be zero. So that means that the changing Gibbs free energy at equilibrium for this reaction is zero killer. JAL's finally in part C. We are solving for a non standard Delta G value, and so use the bottom equation in order to help us sell. For that, we need the Delta G at standard conditions, which is what we saw. Foreign Port A, and we also need the value of the reaction quotient. Q. Based on the reaction that were given. We conform the expression for the reaction quotient. Q. In terms of the partial pressures that were given, that would be the partial pressure of methanol divided by the partial pressure of carbon monoxide times, the partial pressure of hydrogen gas squared. We pull again, those personal pressures that were given into that expression. And when we do that, we have 1.0 atmosphere for methanol. We have zero 0.10 atmosphere for CEO, and we also have zero 0.10 atmosphere for H two. But we square that one, so the reaction quotient comes out to a million. Now. When we put that all into this bottom equation, we can sell for that Delta G value non standard conditions. So Delta G is equal to Delta G at standard conditions, which from part A, we found was negative 24.8 killer jewels. And this time we add are Tee Ellen of Q Are is 8.314 times 10 to the negative. Third killer jewels. For Kelvin Times, a temperature is still 298 Kelvin times the Ellen of Q. We saw for Q to be one million. You plug that all in. And for part C, we should find that changing Gibbs Free Energy comes out to nine 0.4, killing JAL's.

Driver formula. Casey is equal to K f Re concept for forward. We're re concept for reverse. Let's plug in what we're told in the question. Um, great concept for the Ford reaction is six times 10 to the negative six. And the equilibrium casa is one times 10 to the 16th. And we're going to solve the re concept for the reverse reaction, which is going to be equal to six times 10 to the negative 22.

Reaction of sulfur dioxide and oxygen gas reacting to forms sulfur try oxide. So in order to find the equilibrium constant, which is our KP, we need to use the equation that are standard free energy of reaction is equal to the negative of the universal gas. Constant times the temperature that this reaction has run out times the natural log of our equilibrium constant. So let's go ahead and do some algebra here because we're solving for art. Okay, so the first thing that we can do is we can divide both sides by the negative are times t and that will give us that Delta G standard negative doctor to standard over our times T is equal to the natural log of Okay, so we can raise both sides to the E to allow this natural log to cancel and we're left with okay is equal to e to the negative Delta G standard divided by our times t. So let's go ahead and plug in some numbers. We know that mm to the negative and then from the previous problem, we know that our delta G standard of reaction is negative. 141 0.8 killer jewels divided by are but his 0.0 8314 Jules Permal Calvin Times 298. Calvin, there we go. Okay, then we can go ahead and put that in our calculators, and we find that our K of this reaction is equal to 7.1 times 10 to the 24th power. All right, so okay does not have any units because it is a constant and a couple of tips and tricks for solving these types of problems. It's always good to make sure you keep track of your negative signs here. So you see that our delta G standard of reaction here is negative, but this entire term is also negative. So make sure not to lose your negative signs here. And then you could easily plug this into your calculator and get an answer for your KP


Similar Solved Questions

5 answers
Section 3.5 Undetermined Coefficients: Problem 10PreviousProblem ListNextpoint) Findfunciion of I ifIly" 28y 36e'4 (0) 26,_v" (0)v(O) y(I)PreviewAnsweisSubrnil AnswersYou have allempted Inis problem tirnes. You have unlimited altempts remaining
Section 3.5 Undetermined Coefficients: Problem 10 Previous Problem List Next point) Find funciion of I if Ily" 28y 36e' 4 (0) 26,_v" (0) v(O) y(I) Preview Answeis Subrnil Answers You have allempted Inis problem tirnes. You have unlimited altempts remaining...
5 answers
ProveFor constants a.b,€ and d,and random variables X and Y , Cov(a + bX,c + dY) bdCov(X.Y) (b) For random variables -1. Y,and Z,Cov(X,Y+Z) = Cov(X,Y)+Cov(X,Z)
Prove For constants a.b,€ and d,and random variables X and Y , Cov(a + bX,c + dY) bdCov(X.Y) (b) For random variables -1. Y,and Z,Cov(X,Y+Z) = Cov(X,Y)+Cov(X,Z)...
5 answers
Consider Alamouti: transmit diversity scheme. Week 14 lecture notes showed how construCT the diversity combiner for the case of 2 x and 2 x 2 diversity: Extend the concept and construct the combiner for the case of 2 x L diversity:
Consider Alamouti: transmit diversity scheme. Week 14 lecture notes showed how construCT the diversity combiner for the case of 2 x and 2 x 2 diversity: Extend the concept and construct the combiner for the case of 2 x L diversity:...
5 answers
Find the moment of inertia about its axis of an ellipsoid of revolution.
Find the moment of inertia about its axis of an ellipsoid of revolution....
1 answers
Simplify each radical. See Example 2. $$ \sqrt{75} $$
Simplify each radical. See Example 2. $$ \sqrt{75} $$...
4 answers
Provide an IUPAC name for the structure shown_ball & sticklabels(Do not use stereochemical terms in YOu name_
Provide an IUPAC name for the structure shown_ ball & stick labels (Do not use stereochemical terms in YOu name_...
5 answers
LnalnPlinetxhas Ueeunes thc mass Limth thtauttoce Planet X is
Lnaln Plinetxhas Ueeunes thc mass Limth thtauttoce Planet X is...
5 answers
What is the rate at which the sun is converting hydrogen into helium?Please give your answer in kilograms per second
What is the rate at which the sun is converting hydrogen into helium?Please give your answer in kilograms per second...
5 answers
Make a combinatorial circuit for a hallway light fixture controlled by two switches $x$ and $y$ . Assume the light is off when both switches are.
Make a combinatorial circuit for a hallway light fixture controlled by two switches $x$ and $y$ . Assume the light is off when both switches are....
5 answers
Suppose a CI for a difference of two means (mean1-mean2) is(-0.1, 0.6).Which of the following is the correct conclusion? (Note, in allclaims below, "mean" refers to the respective population mean, notsample mean.)Group of answer choicesWe claim there is no difference between the means.We claim the second mean is larger.We are not sure which mean is larger. It is possible the twomeans are equal.We claim the first mean is larger.
Suppose a CI for a difference of two means (mean1-mean2) is (-0.1, 0.6). Which of the following is the correct conclusion? (Note, in all claims below, "mean" refers to the respective population mean, not sample mean.) Group of answer choices We claim there is no difference between the mean...
1 answers
8. The position of a particle moving in a circular path is given by r(t) = < ~4sin(3t),4cos(3t) > Find the speed v of the particle at any time t.
8. The position of a particle moving in a circular path is given by r(t) = < ~4sin(3t),4cos(3t) > Find the speed v of the particle at any time t....
5 answers
Question 1 (0.6 points) Find the area under the standard normal curve that lies outside the interval between 1 =-1.15 and 2 = 2.14.Write only number as vour answer: Round to decimal places (for cxample 0.6148). Do not write as a percentage:Your Answer;
Question 1 (0.6 points) Find the area under the standard normal curve that lies outside the interval between 1 =-1.15 and 2 = 2.14. Write only number as vour answer: Round to decimal places (for cxample 0.6148). Do not write as a percentage: Your Answer;...
5 answers
Find the Taylor polynomial Tz for the function f(c) In(x) at the number a = 1OA ( _ 1)+d6 _ 1)2+ {6 _ 1)8B. (1 _ 1)+3(_ 1)2+ 3(_ 1)3(x -1) - Z(2 _ 1)2 + 3( _ 1)3none of theseE. (x +1) - %(2+1)2+ 3(+1)3
Find the Taylor polynomial Tz for the function f(c) In(x) at the number a = 1 OA ( _ 1)+d6 _ 1)2+ {6 _ 1)8 B. (1 _ 1)+3(_ 1)2+ 3(_ 1)3 (x -1) - Z(2 _ 1)2 + 3( _ 1)3 none of these E. (x +1) - %(2+1)2+ 3(+1)3...
5 answers
The wall and 5 N floor square? hotfon the H wall carpenter (pocheck; "Cbexputatiroment } U from point A 2 Is marked wall and point B 'doofon thefooorFLOOACeroane Leamnrq 2113
the wall and 5 N floor square? hotfon the H wall carpenter (pocheck; "Cbexputatiroment } U from point A 2 Is marked wall and point B 'doofon thefooor FLOOA Ceroane Leamnrq 2113...

-- 0.018600--