[Harlz420]

This is a independent study course, the materials I have barely help me at all, I've gotten this far however, Id love for some input, advice, suggestions! Thanks a lot guys! I can't get help anywhere else



Question:

After 4.00 Mol of C2H4(g) and 2.50 mol Br2(g) are placed in a sealed 1.0 L container, the reaction reaches equillibriu, and is written following.

C2H4(g) + Br2(g) --> C2H4Br2(g)

calculate the equillibrium concentrations for all three substances


Answer:
       C2H4       Br2       C2H4Br2
I       4.0 M     2.5 M      0.00 M
C        -x          -x            +2x   ( why + 2 x? explain?)
E       2.5 M       1.0 M       3.0 M

4.0-x= 2.50
x=1.50 mol/L
 ^ dont understand why I set 4.0-x=2.50 ( conc of br2; explain?)



closed system so input=output  I assume, and the only way to satisfy that is to use 2x under C2H4Br2

but I can get my in to = out



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Question: In a gaseous 2L reaction system, 2.00 mol of methane, CH4(g) is added to 10.00 mol of chlorine, Cl2(g).  At equilibrium the system contains 1.40 mol of chloromethane, CH2Cl(g), and some hydrogen chloride, HCl(g)

a) write out the balanced chemical reaction
b) calculate the amount of each substance at equilibrium


Answer: 
2 mol CH4 in 2 L = 1 M CH4
10 mol CL2 = 5 M CL2
Equillibrium achieved at 1.40 mol or 0.7 M


       CH4       Cl2       CH3Cl       HCL

I      1.0 M    5.0 M     0.00 M     0.00M
 
C      -x           -x           +x         +x

E       0.3 M     4.3 M     0.7 M     0.7 M


inital conc of CH3CL = 0.00
Conc at equillibrium = 0.7 M

therefore change  (x) is equal to +0.7
x=0.7 M

use x to fill out table

then 0.3 M CH4 x 2L = 0.6 mol CH4

0.6 mol CH4 x  16.042 g       =     9.625 g
                    1 mol CH4



Basically does my work, resulting concentrations, answers look correct?

Please *delete me*

[DevaDevil]

Answer:
       C2H4       Br2       C2H4Br2
I       4.0 M     2.5 M      0.00 M
C        -x          -x            +2x   ( why + 2 x? explain?)
E       2.5 M       1.0 M       3.0 M

it is not 2x; you make 1 mole of C₂H₄Br₂ for each mole of C₂H₄ that reacts, so it should be +x (with x in moles)

4.0-x= 2.50
x=1.50 mol/L
 ^ dont understand why I set 4.0-x=2.50 ( conc of br2; explain?)

is the end point of the reaction given? The math basically says: Start - change = End, that End is equal to the concentration of Br₂ is probably coincidence

[Harlz420]

but If  I dont use 2x Input =/= output and its  a closed system so it should I think



As for the other question, everything that is given to me is given to you, so what are you implying? Im a little confused sorry...

[DevaDevil]

my question to you is:

is the problem is given as:

After 4.00 Mol of C2H4(g) and 2.50 mol Br2(g) are placed in a sealed 1.0 L container, the reaction reaches equillibriu, and is written following.

C2H4(g) + Br2(g) --> C2H4Br2(g)

calculate the equillibrium concentrations for all three substances

?

[Harlz420]

the problem is given exactly as I wrote it apart from the one spelling error on equilibrium but really that is just semantics..

[DevaDevil]

this means that the end concentrations (aka E       2.5 M       1.0 M       3.0 M
) were not given.

How did you come to them?

I ask this because the equation does not have the equilibrium sign, which would imply the reaction is continuing until one of the reactants runs out. (this is then the end "equilibrium" state)


as far as moles go, it is basic stoechiometry. For each mole of bromine and ethylene that reacts, you get one mole of dibromoethane.

[Harlz420]

its in a closed container, it can't go until one runs out, it has to reach equillibrium....


I showed my work, to get 2.5 1.0 and 3.0 I used

4.0-x (ethane) = 2.50 mol (br2 initial)

-x=-1.50
x= 1.50


sub x into equation, I know how to do simple stoic.. I dont know if that will help me here.. Im not really sure what to do.. would br2 be the limiting reagent at 2.5 moles and just calculate from there?

[Harlz420]

yes it has the double arrows im sorry I forgot that, seems obvious since its in a closed container... now can you help me please?

What about # 2?

Also sorry for double posting, Im new 

[DevaDevil]

4.0-x (ethane) = 2.50 mol (br2 initial)

this does not make sense.
The only way this would make sense is if bromine would run out, then you have 1.5 moles of ethene left over.


Basically what I am trying to say is that the end point values for the concentrations (assuming the volume is kept at 1L) do not make sense in any way. Also, it says you have to CALCULATE your end point concentrations. The only thing you have shown us is that you ARE GIVEN the end point conditions and calculate the change. this is very different from calculating the final concentrations.


this is what would be my answer:

Ethene: Start = 4.0 moles Change = -x, End = 4.0-x moles
Bromine: Start = 2.5 moles Change = -x, End = 2.5-x moles
2-bromoethane: Start = 0 moles Change = x, End = x moles

To determine x you have to be given an equilibrium constant or an outright number.

Remember, the question was to calculate end point concentrations.

[DevaDevil]

For Q2:

Question: In a gaseous 2L reaction system, 2.00 mol of methane, CH4(g) is added to 10.00 mol of chlorine, Cl2(g).  At equilibrium the system contains 1.40 mol of chloromethane, CH2Cl(g), and some hydrogen chloride, HCl(g)

a) write out the balanced chemical reaction
b) calculate the amount of each substance at equilibrium

First of all, chloromethane is CH₃Cl


a) I do not see your answer or attempt, but I can see from the rest of your answer that you must have reasoned this out correctly.

b) your table is best filled out in moles; this is more practical, just in case the volume would change during reaction. The answer is correct.

[Harlz420]

4.0-x (ethane) = 2.50 mol (br2 initial)

this does not make sense.
The only way this would make sense is if bromine would run out, then you have 1.5 moles of ethene left over.


Basically what I am trying to say is that the end point values for the concentrations (assuming the volume is kept at 1L) do not make sense in any way. Also, it says you have to CALCULATE your end point concentrations. The only thing you have shown us is that you ARE GIVEN the end point conditions and calculate the change. this is very different from calculating the final concentrations.


this is what would be my answer:

Ethene: Start = 4.0 moles Change = -x, End = 4.0-x moles
Bromine: Start = 2.5 moles Change = -x, End = 2.5-x moles
2-bromoethane: Start = 0 moles Change = x, End = x moles

To determine x you have to be given an equilibrium constant or an outright number.

Remember, the question was to calculate end point concentrations.

Yea I wasn't given any end point number, I figured its a 1:1 reaction 2.5 mols br2 reacts with 2.5 mol ethene

if I dont have an end point number I can't solve?

[DevaDevil]

Yea I wasn't given any end point number, I figured its a 1:1 reaction 2.5 mols br2 reacts with 2.5 mol ethene

if I dont have an end point number I can't solve?

okay, so now I know your table was wrong.

if you have a full reaction, indeed 2.5 moles of Br reacts with 2.5 moles of ethene. This means that the change aka, x is 2.5 moles. Not the end concentration of ethene.