for-the-reaction-mathrmco_2gmathrmh_2g-rightleftharpoons-mathrmcogmathrmh_2-mathrmog-the-equilibrium

Question

For the reaction
$$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) ightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$
the equilibrium constant, $K$ , has the value $5.21 	imes 10^{-3}$ at a particular temperature. If the system is analyzed at equilibrium at this temperature, it is found that $[\mathrm{CO}(g)]=4.73 	imes 10^{-3} M,\left[\mathrm{H}_{2} \mathrm{O}(g)ight]=5.21 	imes 10^{-3} \mathrm{M}$ , and $\left[\mathrm{CO}_{2}(g)ight]=3.99 	imes 10^{-2} \mathrm{M}$ . What is the equilibrium
concentration of $\mathrm{H}_{2}(g)$ in the system?

Ifeoluwa Bolujo · Accepted Answer

So looking at this problem, we&#x27;re gonna be looking at how to find equilibrium, concentration of hydrogen. And I&#x27;ve written out the problem here. An ever not the respective concentrations for each is was the casar you for the reaction. So really solving for this concentration starting off were to write out a K Q equation and we look for solids and liquids. Let me see that there are none. So moving on will have the products in the numerator. So it&#x27;s rate, See oh, and H 20 and the reactions of the denominator and booth water. So putting you are the values we have My point, too, to negative three is equal to 4.73 times 10 to predictive three times 5.21 times 10 to the negative three. And in the denominator, we have great way 99 times 10 to the negative, too times. So, sultan crags, we&#x27;re gonna multiply these two now I used together and doing that should get. And we&#x27;re also gonna multiply these $2 together just a simplify. So doing nothing should get 2.8 times 10 to the negative four x this equal to 2.46 times 10 to the negative. Five. Got that? Because we did 3.99 times 10 to the negative two times X. And then we&#x27;re gonna cross multiply, so kind of act like there&#x27;s one over here. But I&#x27;m also quite times this and multiply and multiply these two together. And those two together you will come up with this. So 2.8 times 10 the negative for X her to 2.46 times sent a five those sulfur x we&#x27;re gonna defied inside times to quit. We&#x27;ll eat incentive to get it for and doing that X should be one 0.119 And that&#x27;s how you saw for

Ifeoluwa Bolujo · Answer

mystically Hey, for this reaction, written out the reaction here at the top as well as the corresponding concentration. So first we looked through the reaction to look for any solids and liquids because those concentrations aren&#x27;t used to find the value of K because their concentration remains constant. So now that we see that there&#x27;s no solids and liquids been going to the next step, which is writing out r k e que equation. So we have our products as the numerator. Shall we have co two? And then because there&#x27;s two moles of O Reilly&#x27;s she was exported and then we&#x27;re gonna use Go to are reacting as our denominator. And then we have CEO to molars of that. So the twos is her and exponents and then multiply that times two, she lets r. K. So when we saw for K, we&#x27;re gonna basically put thes concentrations in this equation right here. So rewriting it, we have 1.1 times. Let me actually read this 1.1 times 10 to the negative one. And that is to second power over 2.7 times 10 to the negative four times. Oh, going back. That&#x27;s actually to the second power because there&#x27;s two moles of that. No, that&#x27;s multiplied times Your monthly, my time, stent. Negative three. So to simplify that we get it. I want you one over 1.39 tons to to the negative 10. So when we did or was me multiply this times itself and then multiplied it by this to get this value right here and then this is simplified to this value. So get him a final answer. We basically divide the two, and you should get 8.74 times 10 to the seventh as the value of K for this reaction.

Kevin Zaborsky · Answer

We&#x27;re given the value of the equilibrium constant Casey for this reaction and were also given the values of the equilibrium Total number of moles for both h 20 and CH 30 h gas. We&#x27;re told what the volume is, so we can we can calculate the equilibrium concentrations for both of the species. We want to use all of this information to determine the equilibrium concentration of CEO gas. We can begin by reading out the expression for the equilibrium concentration constant K C, which is the concentration at equilibrium of ch three. Hu h divided by equilibrium concentration of H two squared times the equilibrium concentration of CEO which is what we&#x27;re solving for and so we can rearrange that expression to sulfur the equilibrium, concentration of CEO and that will come out to equilibrium. Concentration of methanol divided by the equilibrium constant K C times the equilibrium concentration of H two squared. And now we&#x27;re told that for H two, at equilibrium, we have 1.17 times 10 do the negative two moles and this is in a volume of five 0.60 leaders. And so that comes out to 2.9 times 10 to the negative third Moller for the equilibrium concentration of H two and that equilibrium for methanol. We are told that we have 3.46 times 10 to the negative third moles again in that volume of 5.60 leaders. And so that comes out to about 6.18 times 10 to the negative Fourth Moller for the equilibrium concentration of methanol and now having to sell for the equilibrium concentration of carbon monoxide. So that is equal to based on that expression that we wrote earlier. The equilibrium concentration of methanol, which we just saw four to be 6.18 times 10 to the negative fourth Moeller divided by the value of K C, which were given to be 1.6 times 10 to the negative, too times the equilibrium concentration of H two squared. We solved for that equilibrium concentration to be about 2.9 times 10 to the negative Third Moeller. We square that term. And now when we played that all in, we should find that the equilibrium concentration or CEO, she come out to about eight 0.84 times 10 to the third. Moeller

Charotte Mcadam · Answer

So as always, thought contextual information is at the top of the screen. We&#x27;ve run over it, perform previous podcasts. So we&#x27;re just gonna move straight onto the core material. So what we&#x27;re looking at is a reaction. Here we have CEO at each tour generates co two and H two. We&#x27;ve got a number off father used given to us that what we need to do is simply plug into the equation that I&#x27;ve listed over on the right hand side of the screen. So we have A, B, C and D. The lower case values are striking. Metric coefficients C and D are the concentrations off our products. A and B are the concentrations of our starting materials, so that is all on the left on site. So here is my line of equation. More if somebody plugged in the numbers that are all listed here. So what I&#x27;m doing is plugging numbers into equations. They&#x27;re already in the correct unit form, and we can generate a number off 17.9 for our cake. Librium. Constant value

Evey Z · Answer

For this question, we can first right out the equilibrium expression K equals concentration of age to square times concentration of 02 over concentration of age, slow square. And if we rearrange this equation, we will get the concentration of H 20 square equals concentration of age to square times concentration of 02 over the equilibrium constant and the concentration of H 20 would be the square root of the right hand side. And now we can plug in the numbers given by the questions. The H your concentration is 1.9 times 10 to the minus two mole square and 02 concentration is a times 10 to the minus two more and the equilibrium constant here is 8.7 times 10 to the third and after calculation are answers should be 5.8 homes, 10 to the minus five month. So this is the concentration for water at equilibrium

Samantha Grieco · Answer

we have this equilibrium expression where we have hydrogen gas reacting with carbon dioxide gas to produce water vapor and carbon monoxide gas and were asked to determine the equilibrium constant for this reaction. So remember the equilibrium constant que is going to be represented by the concentration of your products divided by the concentration of your reactant since. So in this case, our products are water vapor and carbon monoxide, and our reactant are hydrogen gas and carbon dioxide gas. So this is the equation that we&#x27;re going to be working with. And then we are given the concentrations of each of these four substances at equilibrium. And so we&#x27;re simply going to plug those values into this equation and come up with our K value. So for thus were given that the concentration of water at equilibrium was 0.11 Moeller the concentration of carbon dioxide a equilibrium waas 0.14 Moeller the concentration a hydrogen at equilibrium was 0.61 Moeller on the concentration of carbon dioxide was 0.16 Mohler equilibrium. So we&#x27;re going to plug this into our calculator. We&#x27;re going to multiply these two terms in the numerator thes two terms in the denominator and then divide numerator by denominator. When we implode this into our calculator, what we get is 1.5 seven 78 a bunch of other numbers. But in order with keeping with significant figures, we have do because Ra multiplying and dividing round our answer to the least number of sick bags. And all of these given concentrations were only given to Sig Figs. So our answer should also be reported to to sick banks. So that would be one point in this case. The five is going to round two a six, and that is our equilibrium constant.

Problem

The equilibrium constant for the reaction $$\mat…

Problem 48 Hard Difficulty

Answer

$\left[H_{2}\right]=0.119 M$

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$\left[H_{2}\right]=0.119 M$

Introductory Chemistry

Stephanie C.

Lizabeth T.

Karli S.

Jake R.

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Comparison Of Gases Liquids And Solids

Steven S. Zumdahl, Donald J. DeCosteIntroductory Chemistry

Stephanie C.

Lizabeth T.

Karli S.

Jake R.

Steven S. Zumdahl, Donald J. DeCoste

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