write-the-equilibrium-expression-for-each-of-the-following-heterogeneous-equilibria-a-4-mathrmals3-m

Question

Write the equilibrium expression for each of the following heterogeneous equilibria.
a. $4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) ightleftharpoons 2 \mathrm{Al}_{2} \mathrm{O}_{3}(	ext { ) }$
b. $\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) ightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)$
c. $2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) ightleftharpoons 2 \mathrm{MgO}(s)$

Shazia Naz · Accepted Answer

Yeah, We have to write a equilibrium expression for heterogeneous equilibrium. So here, heterogeneous actually bringing in World Special State are not of the same face with each other. So here, for about a mm. For a general reaction here, a plus b see bliss The equilibrium, constant expression will be here. Yeah, is equal to concentration of a product of the concentration of re obtained here. Yeah, right. Oh. Mm. Uh huh, Yeah, yeah. However, we should not include the solids in pure liquids in the equilibrium Constant expression because their concentrations are assumed to be constant. So therefore, the equilibrium constant expression here is equilibrium. Constant expression When divided by a concentration of or to power tree notice. Their soldiers and liquids are not included here. Okay, for part B here, equilibrium constant expression is equal to went out of concentration of mhm NH four into concentration of it&#x27;s here. Notice the soldiers and the liquids are not included here. So this is heterogeneous equilibrium involved Species state are not of the same face. Yeah, this is the equilibrium. Constant expression for the given reaction. So next we have four part c. Yeah. For part C, we have Yeah. Yeah. It&#x27;s the Librium. Yeah. Constant expression is equal to one of the concentration of turmoil. Hugh. Notice their solids and liquids and not included here. Yeah. Wow.

Anatole Borisov · Answer

So in this question here were asked to write the people really expression or mass ach expressions, um, for these different rations. So let&#x27;s work through this. And as we go, we&#x27;ll note the important thing. So the K is always K values, always product of the concentrations of products off the reactor. So but I&#x27;m gonna need it&#x27;s gonna hit us right away. Whenever we have solids or liquids, we ignore them in the mass act expression sofas first, to react reverse direction. We have to, um, species that we need to ignore, and so that&#x27;s gonna cancel those out. So the first thing we need to dio is we need to have the concentration of water in the numerator times a concentration of CO two. And since there&#x27;s nothing in the denominator, it&#x27;s really just over one, and that&#x27;s our mass tax expression. Next one again, we ignore solids so these two species get ignored, and stark a value is really simple. You just have the concentration of CO two, and for this last one again, we&#x27;re going to solids living him on aluminum oxide, and so our mass action expression is going to be. There&#x27;s nothing on the product is just gonna be one over concentration of 02 Now, remember, whenever there is a tow efficient 100 you need to Ah, take the concentration to the power of us. We have three right here. Memory goes three right there. So that&#x27;s our answer. Um, yeah, that&#x27;s it.

Ifeoluwa Bolujo · Answer

I want to start with a year to write the equilibrium expressions for these problems. So if you remember, solid and liquids are not used in equilibrium expressions because their concentration does not affect the expression, totally you&#x27;re not going to use it or this, and the products will be the new Marie. Tears and the denominator would be the reactors. So lighting the expression here have each team here Yes, each meal, Nuria and that moving on to be Let&#x27;s go through and take out the solids and liquids. So here&#x27;s a solar, and this is actually also the small would. So take that as well. So again, if the there&#x27;s no products will use you run, that&#x27;s the new normal and then oxygen and then three moles of oxygen&#x27;s Who will use that? So that is the equilibrium expression for the 2nd 1 For the last one, we&#x27;ll go through first, find all the liquids and solids. Good. Then start with the products. We&#x27;ll use HCL formals of it here, including onto the products. Well, has C H one no, only one more, and then seal to. And then we have formals of that here that rules that you were with my expression for certain one

Ifeoluwa Bolujo · Answer

Oh, her The equilibrium expression for thes 50 problems started after the 1st 1 First thing is to go to really put the solids and liquids because they don&#x27;t contribute to the expression, they have a constant concentration. So going through, you can knock off this one right here. That&#x27;s the only solid or a liquid. Shouldn&#x27;t they expression? We start off with the products, the products will be the numerator. So go s too. It&#x27;s a product and then going over to the reactant because Iraq into to have three moles of CEO put three. Everyone to the 2nd 1 will do the same thing. Start off looking for solids or liquids. We have a solid for a product so will cost off. So next when we write the actual equation, we&#x27;ll start off with the critics. You don&#x27;t have a product because this was a solid. So we&#x27;re going to use that any expression. So we&#x27;re just gonna put one as the numerator and they you there. We have three moors have to so three there. Last but not least, I won&#x27;t do the same thing here. Let&#x27;s look for the solids or the liquids. So product in this reactive. So starting off, we don&#x27;t have any products. Will just put one there and then going to the reactions. We just have this one Three moles of oxygen We&#x27;ll put to I&#x27;m entering here. Okay, that&#x27;s it.

Anatole Borisov · Answer

this question here asked us to go through various reactions and write the mass action expressions. So our 1st 1 is before in the solid state reaction with five oxygen Gash estate, which is in equilibrium with P for 0 10 in the solid state. And so remember, in mass acting expressions Hey, we&#x27;d ignore solids and liquids so we don&#x27;t have those. And so it&#x27;s actually a really simple massive compression. Nothing in the products and on Lee Oxygen raise the fifth power because of this coefficient. And so that&#x27;s our mass acts expression for this first question. Next react. We have co two gas reacting with, uh, sodium hydroxide as a solid to be in equilibrium with, ah, sodium carbonate, also as a solid. And so remember we ignore solids in our mass action specials. So nothing here, nothing there. And so we have one over there in the denominator. We have the concentration of CO two. Nothing here. And so you got. It&#x27;s a simpler one than it looks. Don&#x27;t be caught by always just every single thing in there. Um, actually, part of me miss something, this question who is? There&#x27;s also another product, which was water. Just seen practice. Where is that hard you&#x27;re going? We actually have. Ah, now we have aged 20 here. And so in this in a gaseous form in. So we do actually have something in the room. There we have the concentration of aged. Well, parliament. Um, Now, finally, we have our last reaction. We have manage for and of three solid, which is gonna be any equilibrium with, I know, to gas and to water molecules in a gaseous form so lasting north solids. But then the rest is just gonna be simple. We have concentration of n 02 from here concentration H 20 but raised the second power and nothing that number is so it&#x27;s just this

Ifeoluwa Bolujo · Answer

so it will be calculating the equilibrium expression for these equations here, starting off with a well first go through and we&#x27;ll knock off the solids and liquids because again they don&#x27;t contribute to the reaction they their their concentration just remains the same. So it&#x27;s no need this well, so the rest of gases, which is good so writing equilibrium cost the products will be the numerous. So we&#x27;ll go h two and C o And then the reactant will be the denominator. So that would be the equilibrium constant for the 1st 1 from separate one. Let&#x27;s go through and do it again. Yes, I actually Miss wrote it. This is actually gas. So going through it again, we again miracle the solids in the luckless because the concentration stays constant. So we&#x27;ll just have eight to a gas so age too low because it&#x27;s a product was in the new Maria. And then there&#x27;s nothing is the denominator because we don&#x27;t have any rapid socialist with, So they&#x27;re actually just h 20 is equilibrium question. So going up the last one, what&#x27;s a gun? Go through work or solids and liquids. Okay, so we don&#x27;t have any products, so we&#x27;ll put one as our numerator and then for the nominator we have go to. But there&#x27;s three moles of it. It&#x27;s all right. Go to then in parentheses. Three right there and then that&#x27;s it.

Problem

What is the effect on the position of a reaction …

Problem 100 Medium Difficulty

Answer

a. $K_{e q}=\frac{1}{\left[O_{2}\right]^{3}}$
b. $K_{e q}=\frac{1}{\left[N H_{4}\right][H C l]}$
c. $K_{e q}=\frac{1}{\left[O_{2}\right]}$

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a. $K_{e q}=\frac{1}{\left[O_{2}\right]^{3}}$b. $K_{e q}=\frac{1}{\left[N H_{4}\right][H C l]}$c. $K_{e q}=\frac{1}{\left[O_{2}\right]}$

Introductory Chemistry

Stephanie C.

Nadia L.

Allea C.

Morgan S.

Liquids - Intro

Comparison Of Gases Liquids And Solids

Steven S. Zumdahl, Donald J. DeCosteIntroductory Chemistry

Stephanie C.

Nadia L.

Allea C.

Morgan S.

a. $K_{e q}=\frac{1}{\left[O_{2}\right]^{3}}$
b. $K_{e q}=\frac{1}{\left[N H_{4}\right][H C l]}$
c. $K_{e q}=\frac{1}{\left[O_{2}\right]}$

Steven S. Zumdahl, Donald J. DeCoste

Introductory Chemistry