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Write the mathematical expression for the reaction quotient, $Q_{c}$ for each of the following reactions:(a) $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$(b) $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$(c) $\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$(d) $\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$(e) $\mathrm{NH}_{4} \mathrm{Cl}(s)=\mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$(f) $2 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)=2 \mathrm{PbO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$(g) $2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)$(h) $\mathrm{S}_{8}(g) \rightleftharpoons 8 \mathrm{S}(g)$

a) $Q_{c}=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{H}_{2}\right]^{3}\left[\mathrm{N}_{2}\right]}$b) $Q_{c}=\frac{[\mathrm{NO}]^{4}\left[\mathrm{H}_{2}\right]^{6}}{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{5}}$c) $Q_{c}=\frac{\left[\mathrm{NO}_{2}\right]^{2}}{\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}$d) $Q_{c}=\frac{[\mathrm{CO}]}{\left[\mathrm{H}_{2}\right]\left[\mathrm{CO}_{2}\right]}$e) $Q_{c}=\frac{[\mathrm{CO}]}{\left[\mathrm{H}_{2}\right]\left[\mathrm{CO}_{2}\right]}$f) $Q_{c}=\frac{[\mathrm{CO}]}{\left[\mathrm{H}_{2}\right]\left[\mathrm{CO}_{2}\right]}$g) $Q_{c}=\frac{1}{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}$h) $Q_{c}=\frac{[\mathrm{S}]^{\mathrm{s}}}{\left[\mathrm{S}_{\mathrm{s}}\right]}$

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

Shaun R.

December 28, 2020

E is wrong listed in the answer section

F is wrong also

B is listed in the answer as [H2]^3 but the explanation says [H20]^3

Rice University

University of Maryland - University College

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Brown University

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Well, everyone, this is Ricky. And today we're working on Problems 16 and will be writing a bunch of reaction potions. And so I'm not gonna be writing the reactions, but I will just be writing down. So starting with a we have Q is equal to of products over reacting. So N. H three swear because there's two of, um, each too cube because it's three h two times then, too. And, B um, we have what we have is h 2026 times, you know, forth over and H three to the floor. Times go to fifth. Next up we have a que is equal to and oh to square over and 24 he is D is Q is equal to carbon monoxide over each too. Co two. And remember, we don't include the concentration of pure solid or liquid. Where will we include them? But they're just the concept of their concentration is just considered one so we can skip out. Um, the next we have you. And in e we get Q is equal to and H three times hcl. Remember, we didn't include the solid because its concentration is just considered one next myth. Q is equal to and to to the fourth Arms lead, oxide squared times, oxygen. Um, next we had Oh, my bad action. Um P vo is actually a solid, so we will go thio and that it's just oh, too. Next we have G um, and Q is equal to Q is equal to one over h two square, multiplied by two. And last but not least, h we have to do is equal to us to the eighth over s eight. So I hope this video is helpful.

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