GC question

[UnimaasMED]

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When 14.250 moles 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 value of Kc for this reaction?


My set up:
PCl5(g) <-> PCl3 (g) + Cl2 (g)
I 14.250/3; 0 ;0
C -x ;+x; +x
E 4.75-x; x; x

Since 40.0 % decomposes, reactant= (0.40) (4.75)=1.9
since reactant= 2x= 1.9, x=0.8
Kc= 0.8^2/ 3.95

What am I doing wrong?

 

[mashinka]

Ok, so Kc= 0.8^2/ 3.95...I am not quite sure where you are getting 3.95.
For the Equil.[PCl5] I got 2.85 because I did 4.75 - 1.9, since that is what you will have left. The rest of it, I have the same.

 

[RCT PC CRN]

When 14.250 moles 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 value of Kc for this reaction?


My set up:
PCl5(g) <-> PCl3 (g) + Cl2 (g)
I 14.250/3; 0 ;0
C -x ;+x; +x
E 4.75-x; x; x

Since 40.0 % decomposes, reactant= (0.40) (4.75)=1.9
since reactant= 2x= 1.9, x=0.8
Kc= 0.8^2/ 3.95

What am I doing wrong?

I'm not 100% sure but this is what I would do....

Products are x and x which is 40% each - 1.9
Reactant - 60% which is 2.85

Kc= x^2/reactant = 1.9^2/2.85 = 1.266.....is this the answer?

 

[gigawatt]

When 14.250 moles 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 value of Kc for this reaction?


My set up:
PCl5(g) <-> PCl3 (g) + Cl2 (g)
I 14.250/3; 0 ;0
C -x ;+x; +x
E 4.75-x; x; x

Since 40.0 % decomposes, reactant= (0.40) (4.75)=1.9
since reactant= 2x= 1.9, x=0.8 ...THIS IS THE MISTAKE: x = 1.9
Kc= 0.8^2/ 3.95

What am I doing wrong?

Initial [PCl5] = 14.25 mol / 3 L = 4.75 M ...correct

40% of this decomposes, and for every molecule of PCl5 that decomposes you get one molecule of PCl3 and one molecule of Cl2.

4.75 M x 40% = 1.9 M reacted

This gives 1.9 M PCl3 and 1.9 M Cl2 and leaves 2.85 M PCl5 (= 4.75 M - 1.9 M).

Kc = [1.9][1.9]/[2.85]

 

[UnimaasMED]

Initial [PCl5] = 14.25 mol / 3 L = 4.75 M ...correct

40% of this decomposes, and for every molecule of PCl5 that decomposes you get one molecule of PCl3 and one molecule of Cl2.

4.75 M x 40% = 1.9 M reacted

This gives 1.9 M PCl3 and 1.9 M Cl2 and leaves 2.85 M PCl5 (= 4.75 M - 1.9 M).

Kc = [1.9][1.9]/[2.85]

Thanks 👍

 

[mashinka]

shouldn't [PCl3] and [Cl2] be 0.8 each, because the PCl5 decomposes 1.9 all together.

PCl5(g) <-> PCl3 (g) + Cl2 (g)
I 4.75 0 0
C -1.9 +x +x
E 2.85 x x

So, 2x = 1.9 and x= 0.8

 

[RCT PC CRN]

shouldn't [PCl3] and [Cl2] be 0.8 each, because the PCl5 decomposes 1.9 all together.

Reactant [ ] goes down by 40% that means both the products [ ] goes up by 40%

If you want to do pure 40% decompose then convert 14.25 mols in to grams and then take 40% of that and divide those grams between PCl3 and Cl2 accordingly..now turn those back in to mols of PCl3 and Cl2 ...divide by 3 to get [ ] ...you will have same [ ] i.e. 1.9 each. This is of course the longer way but it will clear your doubt.

make sense?

 

[mashinka]

word. thanks!