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Quest: (15 pts). The value ofK for the reaction below is 0.342.2 HS (g)2 H2(g) S2 (g)The concentration of the gases are a5 follows: H.S (g). 0.050 molL; H: (g) 0.10...

Question

Quest: (15 pts). The value ofK for the reaction below is 0.342.2 HS (g)2 H2(g) S2 (g)The concentration of the gases are a5 follows: H.S (g). 0.050 molL; H: (g) 0.100 molL; Sz (g) 0.100 molL; respectively Write the Equilibrium expression and then do calculation to determine the direction in which the equilibrium will move?Complete the concentration summary tableInsert the values from your table into the equilibrium expression but do not solve the equation (do nor even simplify the equation)

Quest: (15 pts). The value ofK for the reaction below is 0.342. 2 HS (g) 2 H2(g) S2 (g) The concentration of the gases are a5 follows: H.S (g). 0.050 molL; H: (g) 0.100 molL; Sz (g) 0.100 molL; respectively Write the Equilibrium expression and then do calculation to determine the direction in which the equilibrium will move? Complete the concentration summary table Insert the values from your table into the equilibrium expression but do not solve the equation (do nor even simplify the equation)



Answers

The equilibrium constant for the reaction
$$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$
has the value $9.2 \times 10^{-6}$ at a particular temperature. The system is analyzed at equilibrium, and it is found that the concentrations of $\mathrm{NOCl}(g)$ and $\mathrm{NO}(g)$ are
0.44 $\mathrm{M}$ and $1.5 \times 10^{-3} \mathrm{M},$ respectively. What is the concentration of $\mathrm{Cl}_{2}(g)$ in the equilibrium system under these conditions?

In this given chemical reaction, three molecules of a reacts with one molecule, er B to produce two molecule of C and one molecule of D. One thing we cannot hear that is in gaseous state. B is in pure solid estate, while C and D are inadequate state. Now, after the, after this reaction has achieved equilibrium, their concentrations in terms of mood or more clarity is given in the problem. So is 2.48 mol B is 2.41 mole C has the concentration 1.13 Mueller, and D has the concentration 2.27 Golden. So these are the datas are the equilibrium values Now in the problem, it is also mentioned that volume of the flask in which the a gas has filled. The volume is 1.87 liter. And we have to find equilibrium constant for this lee Jackson. So first of all, we will we will convert all the concentrations in a single unit, since C. And they are already in Mueller units. Therefore we will convert the concentration of a also in Mueller unit. So how Mueller unity is achieved for that. We divide the moral values with the uh most by volume. So here for a the mole is 2.48 moles divided by volume is 1.87 later. So this will be the modularity of I guess now the question is how we will write expression for Casey or equilibrium constants. We know that while we write the expression for equilibrium constants, we neglect the pure solids and pure liquids here. Pure solid is present and that is at um be therefore we will neglect its concentration since its concentration will not change with time, it will be always unity. Now, first we enumerated, we write the product concentration 1st. So what is the water? The products that are C&D. Now the power of these concentrations will be their corresponding coefficient. Now, coefficient of CS two, Therefore, concentration of C will be squired and coefficient of D is one. Therefore there will be one in the power of fish, in the power of the concentration of D. So this will be divided by the concentration of reactant and there is only one reactant since we are neglecting the concentration of be it, since it's a pure solid. Therefore, concentration of A. And its coefficient is three. Therefore it will go into the power of A. So this is equal to now we can put the data's of their concentration values at the equilibrium. So C. Is one point 13 Mueller police square and two D. Has the concentration 2.27 Mueller Divided by concentration of a. That is 2.48 divided by 1.87 whole cube. So from here we get the value of Casey that is equals two, one point 242 So the correct option is option E. That is one point 24 so this will be the answer for this problem.

Okay, so the reaction here is three a. Which is a gas because B. Which is a solid gives us to see which is a chris and D. Which is Aquarius, I'm not gonna worry about are solid. And let's let's work with everything in polarity here at 1.1, 3 more And 2- seven more. And for a, I'm told that I have 2.48 moles And that's going to be in our volume of 1.87 L. That's gonna be 1.33 moller. And that's true at equilibrium. So we'll go ahead and write RK for this. It's going to be the concentration of C squared time's D. All over the concentration of a cube. Okay, We're gonna ignore because it's a solid. So we'll plug in our concentrations at equilibrium. We need them all to be in polarity. So we're going to get a K. c. here equal to 1.23. So that's going to be answered. A.

For this question, we can first start down the equilibrium constant expression for this reaction. K equals concentration of Eno to power of two times concentration of BR to over concentration of N o B r to the power of to And we can then rearrange this equation so n o B r Concentration square equals and no concentration square br to concentration over the liberal constant. So the nlb our concentration equals their roots square of right inside. So then we plug in numbers given by a question. We know that at a cooler beom, the concentration of Eno and they are two are both point a move so we can plug in the numbers. So it's point a well for both and no concentration and b r two concentration over the equipment Constant, which is given asked 25. And that gives us 0.20 for eight, which equals 0.14 wolf. Therefore, at a gloom, the concentration off a no PR is 0.14 months

Okay, so we're giving the equation. We have two moles of die nitrogen monoxide reacting with oxygen gas that's in equilibrium with the creation of four moles of night nitrogen monoxide. Now, we're also told that the equilibrium constant for this equation is 1.7 times 10 to the minus 13th and were given the concentration of die nitrogen monoxide and oxygen at equilibrium. So we need to figure out what would be the concentration of nitrogen monoxide at equilibrium. So the first thing we're going to do is write out How would we express the equilibrium? Constant K. So remember, we're doing concentration of products divided by concentration or reactions, and those are raised to the power of the coefficient. So, for example, there are four moles of nitrogen monoxide. So this is the concentration of nitrogen monoxide is going to be raised to the fourth power. Um, so we have night die nitrogen monoxide, and that's gonna be raised to the second power because of this coefficient of two. And then that's going to be multiplied by the concentration of oxygen, which would just be raised to the first power because it has a coefficient of one So let's plug in what we are get. We are told that K is equal to 1.7 times 10 to the negative 13. We don't know the concentration of nitrogen monoxide, so that's gonna be our X. And again it's raised to the fourth power. The concentration of into Oh, we are told in the problem is 0.35 and again, we'll have to square this. And then the concentration of oxygen zero point 00 Well, 27 And we need to solve for X in order to figure out the concentration of nitrogen monoxide. So, in order to get exploit itself to start, I'm gonna multiply both sides by this term here, so we're gonna multiply both sides by 0.35 squared times 0.27 on both sides. So these air going to cancel out of the denominator on the right hand side. And so what we're going to plug into our calculator is 0.35 squared times 0.27 times 1.7 times 10 to the negative. 30. When we do that, we end up with 5.62275 and that is times 10 to the negative 21st power. Now we're still left with X to the fourth power. So in order to get acts by itself, we have to take the fourth root of both sides. So when I take the fourth root of 5.6 2275 times tense and negative 21st we get X is equal to 8.7 times. Turn to the negative six. Mohler And remember, we're on Lee reporting this to two significant figures. So we round to 8.7 because our original K value is given to two sick things as well as the concentrations. So the concentration of nitrogen monoxide equilibrium is 8.7 times 10 to the negative six Mohler.


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