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10) Reaction of 0.1050 mol PCIs w/ 0.0450 mol Clz & 0.0450 mol PCIz in 0.5000 flask- 4.2* 10 2

Question:

10) Reaction of 0.1050 mol PCIs w/ 0.0450 mol Clz & 0.0450 mol PCIz in 0.5000 flask- 4.2* 10 2 at 250"C PcIs(g) PCis(e) Clz(x) a) Which direction will rxn proceed? b) At equil, if [PCIs] 0.2065 M, what are [PCI] & [Ch]?


Answers

13.20 What products would you expect from the reaction of 1 mol of 1.3-butadiene and each of the following reagents? (If no reaction would occur, you should indicate that as well.)
(a) $1 \mathrm{~mol}$ of $\mathrm{Cl}_{2}$ (b) $2 \mathrm{~mol}$ of $\mathrm{Cl}_{2}$ (c) $2 \mathrm{~mol}$ of $\mathrm{Br}_{2}$ (d) $2 \mathrm{~mol}$ of $\mathrm{H}_{2}, \mathrm{Ni}$ (e) $1 \mathrm{~mol}$ of $\mathrm{Cl}_{2}$ in $\mathrm{H}_{2} \mathrm{O}$ (f) Hot KMnO $_{4}$ (excess) (g) $\mathrm{H}_{2} \mathrm{O},$ cat. $\mathrm{H}_{2} \mathrm{SO}_{4}$

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Beautiful and serene is treated with the variety of religions. Project supporters. A Pulls one More a shell in water each year we'll put away the oxygen. It will open up introduce car back in time which will react with water and debris to make. And as a result, oh two hydroxy groups will be introduced.

Our reaction be start similarly bob Carlberg italian we get attacked by ruth um all and after the detonation it will produce either at this position. Mhm. Mhm. Mhm. Yeah.

Yeah. Yeah. See it like this block mod, we'll open up a park side and it will attack more strictly uh available. Primary carbon at the second recover and Hancock settled the form and after press automation it will give an alcohol. Uh that's uh group introduced from the group really urgent and that the final patinated alcohol.

Mhm. .1 More shooting hydroxide. It will open up the back said produce a cock side at the secondary carbon. And after detonation it will be the same die all. Mt you first said will open up.

Uh huh. Okay. Uh huh. Primary problem And the secondly run will have a cock side and after contamination it will produce an alcohol. Now we have been trackside here.

Um alcohol at the secondary carbon of course. Yeah. Mhm.

So in this problem we're going to compare Q and Kr equilibrium constant. Okay, when we put our pressures or uh concentrations into our equilibrium constant, they have to be at equilibrium. Okay. But when we put them into Q, they can be at any conditions, Right? So Q. Which is called the quotient is the same exact expression as K products.

Overreactions raised to their coefficients. And if Q equals K, we know our systems at equilibrium. So let's go ahead and write our expression for our equation. Okay, so that's gonna be the partial pressure of a mhm products Overreacted. It's raised to their coefficients partial pressure of cube squared.

And then we'll just plug in the numbers that were given for those. Okay, 0.5. Um And we'll divide that. Bye. 0.3 Well, square that.

So we only have A and Q. Because our is a solid and be as a liquid. So we don't include those. So this gives us a value for Q. Of 56.

Well, we're told that K is 9.4, so we know that we're not at equilibrium because they're not equal. So now let's predict which way it will shift since Q is greater than K. That means we have too much product. Okay, so these are products overreacted. It's so if it's too big, we know that we have too much product.

So if we have too much product, that means our reaction is going to need to shift to the left to return to equilibrium. Okay, So the reverse reaction will occur until we reach equilibrium again..

Okay to figure out which direction the reaction proceeds, we need to calculate Q. The reaction quotient, which will have the same form as the equilibrium expression. So, um it's going to be the let's see concentration of the gas A. B. As a liquid.

So we admit it and then we have the concentration of Q. And that's going to be squared. So if we have 0.3 moles and 10 leaders for Q and 100.5 moles and 10 leaders for A, the B and the are are irrelevant. So we have .5 Divided by 10, divided by .3 divided by 10. And that's gonna be squared.

It's important thing with the volume, not just the mold because it doesn't automatically cancel out. Um All right, So .5 divided by 10 is .05 and the other one's point of 3.05 divided by .03 Squared come out to 55.6. So Q is obviously greater than K, which is 9.4, which means that you have too many products relative to react. And so it's going to shift to the left.

Okay, so our reaction here is ceo gas reacts with two H two gas to give us methanol And we're told that we take .1 more and we're gonna divide that by two leaders. Here, we take .2 moles and that's going to be in two leaders And at equally room or we have .5 malls here. We don't know for an equilibrium yet. The two leaders. Okay, so this is going to be .050 moller 0.1 more 1.25 Mauler.

So our expression for Q is the concentration of our methanol divided by the concentration of our ceo And our H two squared. So if we plug that in, we're going to get .25, Try to buy .050 And .1 Squared. So our cue comes out to be 500. So Q is definitely bigger than K. So this is going to shift to the left or to the reverse.

The reverse reaction will occur..

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