TopScore 1 - GC #56 Help

[Member 8586]

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When 14.250 mol of PCL5 gas is placed in a 3.00 liter container and comes to equilibrium at a constant temperature, 40.0% of the PCL5 decomposes according to the equation:

PCl5(g) <--> PCl3(g) + Cl2(g)

What is the Kc value of this reaction?



Can someone help me out with this please...I was trying to figure it out with Topscore's explanation but no luck really...Also, I've never seen a problem like this with the "40% decomposes"...is something like this even likely to land?
​

 

[SN1]

When 14.250 mol of PCL5 gas is placed in a 3.00 liter container and comes to equilibrium at a constant temperature, 40.0% of the PCL5 decomposes according to the equation:

PCl5(g) <--> PCl3(g) + Cl2(g)

What is the Kc value of this reaction?



Can someone help me out with this please...I was trying to figure it out with Topscore's explanation but no luck really...Also, I've never seen a problem like this with the "40% decomposes"...is something like this even likely to land?
​

14.25 mol / 3 L = 4.75 M

but only 40% of this actually decomposed so
40% * 4.75 = 1.9M

the mole ratio is 1:1:1 PCl5😛Cl3:Cl2 respectively so 1.9 M reacted means you produce 1.9M of Cl2 and 1.9M of PCl3

at equilibrium you have 4.75-1.9 = [PCl5]

then just plug in everything in the Kc expression(product/reactant thing).

 

[Member 8586]

14.25 mol / 3 L = 4.75 M

but only 40% of this actually decomposed so
40% * 4.75 = 1.9M

the mole ratio is 1:1:1 PCl5😛Cl3:Cl2 respectively so 1.9 M reacted means you produce 1.9M of Cl2 and 1.9M of PCl3

at equilibrium you have 4.75-1.9 = [PCl5]

then just plug in everything in the Kc expression(product/reactant thing).

ahh i took it as 40% decomposed so 60% reacted lol..fail ...

at equilibrium you have 4.75-1.9 = [PCl5]
Can you elaborate on that part? how do you know to do this?

and so the data table i get:
PCl5 <----> PCl3 + Cl2
1.9----------- 1.9--- 1.9
+x------------ (-x) --(-x)
2.85------- (1.9-x) (1.9-x)

ahh this seems so wrong..sorry im still lost on this...different from any of the data tables i did in destroyer :/

 

[SN1]

ahh i took it as 40% decomposed so 60% reacted lol..fail ... and so now that i have these numbers...just plug em into the little chart that you set up and form the Kc expression?

Yes sir. so

kc= ([Cl2][PCl3])/[PCl5]

and 60% of the initial value is actually what you will have LEFT in equilibrium


in all honesty, I don't think this kind of crap would show up on dat.

 

[Member 8586]

Yes sir. so

kc= ([Cl2][PCl3])/[PCl5]

and 60% of the initial value is actually what you will have LEFT in equilibrium


in all honesty, I don't think this kind of crap would show up on dat.

lol i really hope not ...i edited that post...how wrong is my chart? lol

 

[premac]

ahh i took it as 40% decomposed so 60% reacted lol..fail ...

at equilibrium you have 4.75-1.9 = [PCl5]
Can you elaborate on that part? how do you know to do this?

and so the data table i get:
PCl5 <----> PCl3 + Cl2
1.9----------- 1.9--- 1.9
+x------------ (-x) --(-x)
2.85------- (1.9-x) (1.9-x)

ahh this seems so wrong..sorry im still lost on this...different from any of the data tables i did in destroyer :/

PCl5 <------> PCl3 + Cl2
4.75 ------------ 0 -------- 0
-x ---------------+x-------+x
--------------------------------------
4.75-x ----------+x------- +x

you know x = 1.9

 

[SN1]

ahh i took it as 40% decomposed so 60% reacted lol..fail ...

at equilibrium you have 4.75-1.9 = [PCl5]
Can you elaborate on that part? how do you know to do this?

and so the data table i get:
PCl5 <----> PCl3 + Cl2
1.9----------- 1.9--- 1.9
+x------------ (-x) --(-x)
2.85------- (1.9-x) (1.9-x)

ahh this seems so wrong..sorry im still lost on this...different from any of the data tables i did in destroyer :/

Yes I did all of the destroyer gen chem and saw nothing like this either. but I think it should be :

PCl5 <----> PCl3 + Cl2
initial 4.75---------- 0 + 0
change -40%* 4.75 +40%*4.75 + 40% * 4.75


final [PCl5] = 4.75- 40%*4.75
final [PCl3] and [Cl2] = 40%*4.75 because 1:1 mole ratio

since they TOLD YOU 40% of the initial value have decomposed you don't need to put x on the change table. You only do that with weak acid and base when you say an "x" amount have dissolved and that is why they need to give you ka and kb.