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Write the mathematical expression for the reaction quotient, $Q_{\mathrm{c}}$ for each of the following reactions:(a) $\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g)$(b) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$(c) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$(d) $\mathrm{BaSO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{SO}_{2}(g)$(e) $\mathrm{P}_{4}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10}(s)$(f) $\mathrm{Br}_{2}(g)=2 \mathrm{Br}(g)$(g) $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g)=\mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$(h) $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(g)$

a) $Q_{c}=\frac{\left[\mathrm{CH}_{3} \mathrm{Cl}\right][\mathrm{HCl}]}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{Cl}_{2}\right]}$b) $Q_{c}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{N}_{2}\right]\left[\mathrm{O}_{2}\right]}$c) $Q_{c}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}$d) $Q_{*}=\left[\mathrm{so}_{2}\right]$e) $Q_{c}=\frac{1}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}$f) $Q_{c}=\frac{\left[\mathrm{CO}_{2}\right]}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{O}_{2}\right]^{2}}$h) $Q_{c}=\left[\mathrm{H}_{2} \mathrm{O}\right]^{5}$

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

Rice University

Drexel University

University of Maryland - University College

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So for this problem, we're looking at a set of eight equilibrium reactions, and we just want to write out the reaction quotient for these. So just very briefly. Keep in mind that the reaction quotient is equal to a relationship between the products and reactions. So you either have the concentration or partial pressures of the products of the reactions reactions. In this case, we're talking about QC, so we're going to be looking totally concentrations. And it's also important to remember for these, the the only species that will be included in the reaction Kwo Shen are going to be gassed, your species and a curious species, mainly because solids and liquids in terms of changes in concentration and partial pressures, are going to be negligible because their volumes are not going to change substantially within the context of the equilibrium mixture, let's just jump right into it. The first equation is fairly straightforward, so to write out the Q C. For this, we're just going to have the concentrations of all of the products multiplied by each other, divided by the concentrations of the reactant CE. Fairly straightforward. This is because all of the terms and our acquaintance or gases. So they all get included with respect to each other in terms of their concentrations. And again, we have products of the reactions. So we look at number two. Number two's a little bit more complicated because we start to involve coefficients in front of our terms. So the important thing to remember is any time you have a coefficient that is going to get included in your equilibrium expression or, in this case, your reaction quotient expression as an explanation. So for this particular equation, all of our terms are still going to be included because we're dealing with gases. But we're going to produce two mattress oxide, uh, for every, um, nitrogen gas and oxygen gas molecule that react together. For that reason, we end up with a squared in the top of our numerator. So moving forward, we keep on going forward with this, um, question. Our equation, see, is very much going to operate under the same principle. The only difference being in this equation. We actually have two terms that have a cohen on one coefficient, so our product is going to be square as well as one of our reactions. So that our reaction question will look like that. D is where things start to get interesting even further. So D actually only has one term in the entire equation. That's a non solid or liquid. So so for dioxide is the only gas or equals species in this equation, because it's a product that's going to go in the numerator. So essentially, what we end up with is that the reaction quotient is just going to be the concentration of sulfur dioxide we keep going forward. Toa letter e letter E is an interesting case where we only another interesting case where we only have one side of the equation that actually has usable species with respect to our equation. So in this case, just like before, we're only going to have one side of the equation contributing to our reaction question. The only difference being this time. Those terms are going to be on the bottom. So the concentration of phosphorus relative to the concentration of oxygen. The only other thing with this one is because the coefficient in front of oxygen is five. It's going to be going to correspond to oxygen reused to the fifth power within the reaction in question. Letter F is relatively straightforward. We're just going to take the reaction. The two terms in the equation. We're going to raise the broom Ian Adams, to the power of two. Because of their two coefficient within the equation, we will include the gas. Yes, roaming on the bottom. Finally, if we take a look at letters G and H, we look quickly. Letter G. Ah, water is included. This equation is a liquid. So we're just going to cut that out of our reaction quotient such that our reaction quotient is going to be equal to carbon dioxide on the top, divided by I think the concentration of methane gas times the concentration of oxygen square in the denominator. That's going to be that reaction question. And for letter H, we're taking a look at, um, a salt that coordinates toe water, copper, Sophie, and actually what's going to be happening? As we look at this correctly, there's an equilibrium between copper. So if they coordinated toe water and gaseous water molecules, and in this case, because the the water molecules are the only gas yesterday qui of species, we have another case where the Q C is actually only going to involve one of the species in our equation, and in this case we raise it to the fifth because of the coefficient in front of the water.

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