write-the-equilibrium-expression-for-each-of-the-following-reactions-a-2-mathrmo_3g-rightleftharpoon

Question

Write the equilibrium expression for each of the following reactions.
a. 2 $\mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)$
b. $\mathrm{CH}_{4}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
c. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g)$

Anatole Borisov · Accepted Answer

So here we have various reactions when they ask, Find master couple of reminders felt that if we have a product or like some species and there&#x27;s a tow fish in front of it in the mass action expression is going to be the concentration of that product raised to the power of its coefficient. Another things. Another thing we do consider gas and a qui ists species, but not liquid in solid species in our mass action expression. And finally, member, it&#x27;s going to the concentration of the products. I&#x27;m sorry. That&#x27;s a ugly line. It&#x27;s going to the concentration of the products over the concentration of the reactions. And so I&#x27;m gonna go work through these problems and explain what I&#x27;m doing. So this 1st 1 we have two molecules of ozone 03 in equilibrium with three molecules of oxygen part of me. And so our K here is going to be first, the concentration of the products 02 But we have this coefficient three and so&#x27;s race. The third power over the concentration of the reactions. We have 03 squared race, the power of to and so that&#x27;s going to be our equation for the 1st 1 Next we have Ah, right here we have They served Bazaar one? Um, yeah. This is written a little funny here, but we have ch four as a gas reacting with three oxygen&#x27;s to become CEO too gas and to water molecules also as a gas. Um, yeah, that&#x27;s correct. And so r k expression is going to be the product of the product of the concentration of the products. So, co two, there&#x27;s no coefficient, so there&#x27;s no power in the concentration of water squared. It is a bad over concentration of ch four and then times a concentration of 02 to the third. Um, actually, in the answers on numerator here it should. It says to the second here, but that&#x27;s long. It should be to the third because of this coefficient. All such colors to this last month, sir. Actually, friend shape. So you have C two h for Athen. Athene burned me. So you two each for gas and seal to gas becoming see to each for C l two gas. And so r k here is gonna be Did you mean no coefficients on no powers ones here And so you see, too a horse yell, too, over the concentration of C two age four over the concentration of CL, too. So that&#x27;s our answer.

Ronald Prasad · Answer

That&#x27;s right. Equilibrium Expressions for the following For a Larry of each 20 times Similarity of CEO divided by clarity of each to on C 02 for be expression is and oh to the fourth and 20 squared. Oh too. And for C C H. 30 h over CEO and H two squared.

Ronald Prasad · Answer

Let&#x27;s read equilibrium expressions for the following For a you got C H 30 h in the numerator over similarity of C O inch two squared for b expression is similarity of both to over the validity of coach or sorry, um, and oh, squared times 02 over the molar ity of n 02 squared. And for C, we&#x27;ve got PBR three p b or three to the fourth p four b r. Two to the sixth.

Anatole Borisov · Answer

So in this question, we are asked to write various equilibrium expressions, mass access expressions. And so I&#x27;m going to remind how that&#x27;s going to work in the No, just go through the chapel. So for generic reaction, a plus B becoming a B the mass action expression K is gonna be equal to the concentration of all the products. So a B over here, over the concentration of the reactors A and be now couple of things to know if there&#x27;s a coefficient in front of a reactant. So, for instance, let&#x27;s say we have three right here that&#x27;s put as a power like that. Um, obviously, it&#x27;s not like you magically right, But ignore that. And so also, the other thing is we only at gases or a quist solutions here. If it&#x27;s a liquid or solid, we ignore it in the mass axe expression. And so that&#x27;s the information you need. And let&#x27;s get to it. Our first reaction is age, too. Gas and groaning gas. Rumi is actually commonly a liquid at room temperature, becoming to HBR Vaz, and so archaic expression wouldn&#x27;t have the concentration of the products HBR right here squared because the coefficient over the concentration of age to right here times the concentration b r. Two. And that&#x27;s our answer. Our next one is going to be a minute change colors. Our reaction is to age two. The gas reacting with as to gas to become to H two s also has a gas. And so r k put e que Because I thought that that somehow people often do it Trappers at equilibrium in the case always a kilogram. So we have our concentration of our products. H two s squared because that coefficient and then we have aged two. But note it should be squared because the coefficient to in front of it times a concentration of s two and that there&#x27;s your answer for that one. And finally we have colors again. Here we have the reaction of age to as a gas reacting with C two and two as a gas become each CN gas. I think there should be two of those. Yes, that&#x27;s correct. It&#x27;s okay. There&#x27;s going to be concentration of h c n squared because that coefficient times a constellation of age to times a constellation of C two and two because of that. And, um, that&#x27;s our answer

Charotte Mcadam · Answer

so the equilibrium constant of a chemical reaction is the value of the reaction quotient at equilibrium. So estate approached by a dynamic chemical system after enough time has passed at which its composition has no measurable tendency towards father change. So here we will just be writing out our equilibrium expressions for the following reactions. So we have to and 02 generates to end 204 so that reaction is up on the screen. So we generate RK equilibrium equation from the formula just over on the right of the screen on then that generates end 204 to the power one over N 02 to the power to that is because we multiply are concentrations by austro que metric coefficients. So then I&#x27;m doing the exact same for the following three reactions in order to generate okay, equilibrium values, which is our equilibrium constant value again, just using the same equation where we have our products on the top are starting materials on the bottom in terms of their concentration, where they are all multiplied by their strike you metric coefficients

Ronald Prasad · Answer

sorry. Equilibrium expressions Reach of the following reactions For a We will not include a solid HCL over Learned you c two, h six and seal too for be We&#x27;ve got clarity N o to the Fort H 2026 in each three to the fourth go to to the fifth. And for C equilibrium constant is P c 03 c 02 over p c 05

Problem

At high temperatures, elemental bromine, $\mathrm…

Problem 97 Hard Difficulty

Answer

a. $K_{e q}=\frac{\left[O_{2}\right]^{3}}{\left[O_{3}\right]^{2}}$
b. $K_{e q}=\frac{\left[C O_{2}\right]\left[H_{2} O\right]^{2}}{\left[C H_{4}\right]\left[O_{2}\right]^{2}}$
c. $K_{e q}=\frac{\left[C_{2} H_{4} C l_{2}\right]}{\left[C_{2} H_{4}\right]\left[C l_{2}\right]}$

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Steven S. Zumdahl, Donald J. DeCoste

Introductory Chemistry

Lizabeth T.

Karli S.

Jacquelin H.

Jake R.

Liquids - Intro

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a. $K_{e q}=\frac{\left[O_{2}\right]^{3}}{\left[O_{3}\right]^{2}}$b. $K_{e q}=\frac{\left[C O_{2}\right]\left[H_{2} O\right]^{2}}{\left[C H_{4}\right]\left[O_{2}\right]^{2}}$c. $K_{e q}=\frac{\left[C_{2} H_{4} C l_{2}\right]}{\left[C_{2} H_{4}\right]\left[C l_{2}\right]}$

Introductory Chemistry

Lizabeth T.

Karli S.

Jacquelin H.

Jake R.

Liquids - Intro

Comparison Of Gases Liquids And Solids

Steven S. Zumdahl, Donald J. DeCosteIntroductory Chemistry

Lizabeth T.

Karli S.

Jacquelin H.

Jake R.

a. $K_{e q}=\frac{\left[O_{2}\right]^{3}}{\left[O_{3}\right]^{2}}$
b. $K_{e q}=\frac{\left[C O_{2}\right]\left[H_{2} O\right]^{2}}{\left[C H_{4}\right]\left[O_{2}\right]^{2}}$
c. $K_{e q}=\frac{\left[C_{2} H_{4} C l_{2}\right]}{\left[C_{2} H_{4}\right]\left[C l_{2}\right]}$

Steven S. Zumdahl, Donald J. DeCoste

Introductory Chemistry