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Consider Ihe following rexction where Ke 0.159 # 72} K:Nz(g) JHz(g) = ==ZNIy(g)FAcon mLXIUTC Liter contuinCT-tound t0 contain 41J*1O 2 moles of N1(g). 4J3*IO moles ...

Question

Consider Ihe following rexction where Ke 0.159 # 72} K:Nz(g) JHz(g) = ==ZNIy(g)FAcon mLXIUTC Liter contuinCT-tound t0 contain 41J*1O 2 moles of N1(g). 4J3*IO moles of Hz(g) and 6.2J*1O ' moles of NHygh. in & [0QIndicli Iug([LElse (ELL schollebllowing:In oruer to reach equilibrium NA;(E) MUST be COnsuned; In order Trach cquilibnum Ke must Incre Int Orcer to reuch equilibrium must be pFoduced Qc is less than Kr" The feaction @` equilibrium. No {urthet = nucuon will occurJubmti cnzwor

Consider Ihe following rexction where Ke 0.159 # 72} K: Nz(g) JHz(g) = ==ZNIy(g) FAcon mLXIUTC Liter contuinCT- tound t0 contain 41J*1O 2 moles of N1(g). 4J3*IO moles of Hz(g) and 6.2J*1O ' moles of NHygh. in & [0Q Indicli Iug([LElse (ELL schollebllowing: In oruer to reach equilibrium NA;(E) MUST be COnsuned; In order Trach cquilibnum Ke must Incre Int Orcer to reuch equilibrium must be pFoduced Qc is less than Kr" The feaction @` equilibrium. No {urthet = nucuon will occur Jubmti cnzwor Rouy Enun Group more group Rutempt remiining



Answers

Consider the reaction $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$ for which $K_{\mathrm{c}}=278 \mathrm{M}^{-1} .0 .001 \mathrm{~mole}$ of each of the reagents $\mathrm{SO}_{2}(g), \mathrm{O}_{2}(g)$ and $\mathrm{SO}_{3}(g)$ are mixed in a $1.0 \mathrm{~L}$ flask. Determine the reaction quotient of the system and the spontaneous direction of the system : (a) $Q_{c}=1000$, the equilibrium shifts to the right (b) $Q_{c}=1000 ;$ the equilibrium shifts to the left (c) $Q_{c}=0.001$; the equilibrium shifts to the left (d) $Q_{c}=0.001$; the equilibrium shifts to the right

23 The reaction given is B r he is We'll call br Those powerful feel good went to help with the standard free energy change that ideology Nord which is no which is n for the liturgy, not off all the problems you don't have off all the products minus and their stuff, do you not my older The evidence It is half Delta Gino D f Well, we are too plus half their touch. Do not have feel too minus day rejecting that ideology. Note of off the RC two They are They're little spheres in your home. So after putting all the values using Appendix Day, we're going to that 0.54 killer jewels five months. So with a relation between equilibrium, that is K and standard free energy, we're going to calculate K. Uh, okay, then care is started. Stuff she not by rt when she is 2.54 Kill a jewel Farmers. My art is a 0.314 going to tend to the power minus three. Just remove the Calvin. His killer Joel farm. All Calvin on there is to 90 days. He killed them. So it was a 1.2 So, Allen case Okay, will be Ellen inwards. Ellen, Inverse off. One point. There are two. Which will we eat? The bar minus 1.2 or chill turns out the base 0.36 This is a value off carriages. The equilibrium constant. The initial concentration on we are sale is obtained by taking ratio of the moles and volume. The little would be RC else. One point it is little moles. Well, they are CIA. My 10.0 liters, which will be 0.100. Come. They live in constant expression for the reaction. Can Britain s care we are to, huh? Sale due? Uh huh. By older? Yes, sir. That is their serial. They are still sealed, so Yeah, so we need to find a change in the concentration eggs. So when it is of strategic kilogram concentration in this reaction okay, will be point receive. Have my to that's where to? To the bar house. My 0.1. Double zero minus six. Therefore, from here we can find as physical stool 0.41 name to convert most from the majority will take lie. We need to multiply the morality with this volume in letters. So 0.419 moles into 10 point zero letters will be 0.4 to 0. Well, the initial amount off reaction the initials amount. Well, for the yes Turned is point for one little little M in tow. 10 liters, which is equals two one. Well, the equilibrium, all the equilibrium, most off B R C l is one point. There is a minus 1.0.42 more. They are cycles 2.58 more off B. R. C s. So another collision concentration off the are too. His 0.42 moles. No, they are my two, which is 20.21 Mole? No, the arto. Now they live them constant off the alto, which is 0.42 moles off sale toe by two. Which is you don't point to one. No field. Therefore the amount off their cell tow is Montel be are safe. How's this point? Five Admin. The amount off they are to and point to one mall and the amount of sales. So this point to one more

So here we're giving information about reaction or hydrogen gas reacts with iodine gas to form two molecules of H I. This reaction has an equilibrium constant with respect concentration of 50 and we have to remember that everything we're going to define has to be in terms of concentration even though we have gases Were given information that we have a volume of a container of 20 leaders. So we have to convert everything in moles to concentration units. So 0.75 moles of H.I Will be zero. So in terms of concentration will be .752 by by 20, which is 0.0375 moller In turn, iodide for iodine and hydrogen, we have 0.025. So we have to divide that by 20. So this becomes 0.00125 Mueller, which is the same for our ID. So for checking the concentration of current species, we always use reaction quotients in this case to be the concentration of H. I squared over the concentration of hydrogen times the concentration of ID. So we should substitute all our information into our equation here to see whether or not we're at equilibrium. And based on the information we have based on the information we have here and making sure we have used uh plugging in everything properly. We find that the reaction question here is equivalent to 900. So from this information, we can basically guess which direction the reaction will shift. In this case Q. Is greater than the equilibrium constant with respect the concentration. So that's important because it shows that with a greater Q. That means you have too much product in this case. So in order to eliminate that extra product, the reaction would have to shift to the left. Okay, so the concentrations the concentration of H. I. Would decrease while the concentration of our separate hydrogen gas and ideal gas would increase. So you would have zero 0.037, Molars. And we have 0.00 125 Molars, which is the same for id. Yeah, so the concentration would decrease by a factor of two X. Since two more two moles of H. I would have to be consumed to form one mole of hydrogen gas and one mole of iding gas. And the amount that would increase by would be X. Since there's one mole of ah The monitor, one mole of H two formed per two moles of H. I. So now we can write our equilibrium expression which is 50 is equivalent to 0.375 minus two X squared over 0.125 plus x square And now we can take the square root of both sides of the positive square root. We never use negative concentrations. So the square 50 Is equivalent to about 7.07 equals 0.0375 minus two X Over 0.00125 plus X. And now we should rearrange and essentially solve for equation here. So we can find that we have 9.07 x on one side. So then we have 0.00125 times 7.07 0.375 minus that, Which is equivalent to 0.0287. And we divide this by 9.07 and we can find that X in this case is equivalent to 0.00 32 molars. So now we have to find the concentration of all our species in this case. So the concentration of hydrogen is always going to be equal to the concentration hiding in this case, Which is equivalent to 0.0044 molars. And now we have to solve for other value here. So we remind us two times x. In this case X once again is 0.0032. And we can find that the concentration HI is equivalent now to 0.031 molders. And these are our final answers.

First we will write the equilibrium constant expression K p. It's going to be equal to the pressure of the product. N O C L squared because of the coefficient divided by the pressure of Eno, also squared And then times the pressure of Cl two. And then we can plug in the values to calculate the value of the constant. So um NOCL is 1.2 atmospheres, So it's 1.2 squared. We omit the units for the K expression you have, N O is Um 0.05. Okay. And we'll square that. Yeah. And then cl two is 0.3. So then we just have to calculate 1.2 Squared, Divided by .05 Squared And then divided by .3 Is 2, 2 significant figures, 1.9 Times 10 to the 3rd. Okay. And that is the value of K p.

So this question is asking us to write or solve K a Q for the chemical equation and 204 reacts into two and two. We've also been given the concentration of and 204 which is equal to 2040.185 miles per leader. And we've been given the concentration of Eno to which is 0.6 to 7 moles per leader. And we're trying to find cake you Now. First, we have to recognize the equation for K E Q. Which in this case is the concentrations of the products raised to their coefficients over the concentrations of the reactant SSRIs to their coefficients. So first, I'm just going to plug in our respective react in some products into this equation. So to do that we have Que que is equal to the concentration of the product, which is N 02 is our only product, and it's squared because as a coefficient of two over the concentration of N 204 and I don't right and exponents there because it's only raised to the first power. And so this is our equation that we're working with. But we also have numbers for these concentrations that we can plug into this equation, so I'm going to go ahead and do that. I'm going to say that K E Q is equal to the concentration of Eno to gas, which is 0.6 to 7 moles per leader and that is squared, divided by the concentration of into 04 which is 0.18 five moles per leader. And so I'm gonna go ahead, pause this video. Now, if you have a calculator and plug that in, you'll do 0.6 to 7 times itself, and then you'll divide that by 0.185 And if you do that, you're actually going to get zero point 213 for your answer. And it's not going to have any particular units because you've divided most per liter over most pre Leader, so those will cancel out and so will end up with a unit. Lists K e Q equals 0.213 for this expression,


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