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Equilbriumn 373 are favored in the following reactcnts , products Keq 1.8 x 10 s State whether CHjCOo "(aq) CH;coOHlaq) = H"(aq) - Keq 2.6 X10 "2 HzO...

Question

Equilbriumn 373 are favored in the following reactcnts , products Keq 1.8 x 10 s State whether CHjCOo "(aq) CH;coOHlaq) = H"(aq) - Keq 2.6 X10 "2 HzOz(aq) H(aq) + HOz(aq) Keq 1037 CuSO_(cq) - Zn(s) =Culs) ZnSO-(aq) Kq 1.0 10 * HzO() H(ag)+ OH" (aq) and find the value of the equilibrium constant for this reactior Write the equilibrium constant expression concentrations are: [Hz] = 12x 10 ' M: [Iz] = 3.4x Hz (g) + Iz (g) 2 HI (g) given that the equilibrium 10 * M: [HI] = 4

equilbriumn 373 are favored in the following reactcnts , products Keq 1.8 x 10 s State whether CHjCOo "(aq) CH;coOHlaq) = H"(aq) - Keq 2.6 X10 "2 HzOz(aq) H(aq) + HOz(aq) Keq 1037 CuSO_(cq) - Zn(s) =Culs) ZnSO-(aq) Kq 1.0 10 * HzO() H(ag)+ OH" (aq) and find the value of the equilibrium constant for this reactior Write the equilibrium constant expression concentrations are: [Hz] = 12x 10 ' M: [Iz] = 3.4x Hz (g) + Iz (g) 2 HI (g) given that the equilibrium 10 * M: [HI] = 4.0 x10 8 M (Answer: K 3.9 X 10-19)



Answers

Calculate the value of the equilibrium constant $K_{\mathrm{p}}$ at $298 \mathrm{K}$ for the reaction $$ \frac{1}{4} \mathrm{S}_{8}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ from the following $K_{\mathrm{p}}$ values at $298 \mathrm{K}$ $$\frac{1}{4} \mathrm{S}_{8}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ from the following $K_{\mathrm{p}}$ values at $298 \mathrm{K}$ $$\frac{1}{8} \mathrm{S}_{8}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g) \quad K_{\mathrm{p}}=4.0 \times 10^{52}$$ $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) \quad K_{\mathrm{p}}=7.8 \times 10^{24}$$

For this question, we have several different K values for the generic reaction A. Goes to be. And first well actually have one K value, we have different coefficient values. If both coefficients are one, then it's going to be the concentration of C. A. B divided by the concentration of a. Both raised to the one power. If we're starting out with one moller of a, then it's going to have to shift to the right will increase be by X and will decrease A by X. We solve for X&X 0.80 Mueller. So that means a is going to be 1 -18 or point to Mueller For the next one. They're both too. So we're gonna have to square both of them. We solve for X annex ends up being The concentration of be still in its .67 Mueller. And then a is going to be still one -X. So it'll be .33 Mueller For the last one. It's a little bit more challenging. We have two on B. I'm sorry, Yeah, two on B and a one on A. So as this shift to the right B is going to increase at twice the rate as a will decrease. So if we're defining the increase in B is X, which we're going to have to square because there will be a two here on little B, then the decrease in A is going to be one half X. So it will be one point oh minus one half X and the coefficient here is one. So we have a superscript one here, then we do a little bit of algebra and solve for X. And we get 1.24 Mueller for be for X. And then A is going to be one Mueller minus one half X or 10.38 Mueller.

To calculate the equilibrium constant for the new overall reaction from the two reactions that are given to us. We need to first figure out how to manipulate these chemical reactions in order to get the overall reaction and then determine what to do with the equilibrium constants for these two reactions to get the overall equilibrium constant. Here is the target chemical reaction. And to get this chemical reaction, we're going to have to multiply this second chemical reaction by two so that our co two that our two CEOs will cancel because Seo is not part of this overall reaction. If we multiply this chemical reaction by two, we need to square its equilibrium constant. Then we can sum these two reactions together, and when we some the reactions, we multiply the equilibrium constants in order to get the equilibrium constant. For the overall reaction 4.7 times 10 to the negative nine

So first equilibrium here we have end to plus O two gas equilibrium with to an O gas. And we're told that the room cousin here is equal to you 4.4 times 10 to the negative 30 one. And second equilibrium of God is to an old gas bless 02 gas. It will be equal to two and o to gas. And if what? The equilibrium cosy here is 2.4 times 10 to 12. Let's add these two equations together and we will get been to gas add to go to gas in equilibrium with two and two gas. Now to calculate the equilibrium constant here we're going to multiply the two equilibrium constants together to yield 1.1 times 10 to the negative 18. But I

For question 13.2 will use the equilibrium expressions that we wrote in 13.1 and substitute the values in the Miller d of S 03 We'll leave out units is 5.0 times 10 to the negative too Squared little come down here 4.0 times 10 to the negative three squared The polarity of S 02 is three 0.0 times 10 to the negative three, 3.0 times 10 to the negative three. And that squared 3.0 times 10 to the negative. Three squared and the polarity of 02 is 3.5 times 10 to the negative. Three. When we compute the answers here for a we'll get an equilibrium constant. That is 7.9 times 10 to the fourth. And for B, we're going to get in equilibrium constant.


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