September 19, 2024, 02:18:35 PM
Forum Rules: Read This Before Posting

### Topic: chemical equilibrium  (Read 11062 times)

0 Members and 1 Guest are viewing this topic.

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### chemical equilibrium
« on: July 18, 2007, 07:08:02 AM »
Does anyone know how to set-up an equation if the concentrations are already in equilibrium, then an amount of the reactant was added. Kc is also given?

Thank you.

#### Borek

• Mr. pH
• Deity Member
• Posts: 27790
• Mole Snacks: +1806/-411
• Gender:
• I am known to be occasionally wrong.
##### Re: chemical equilibrium
« Reply #1 on: July 18, 2007, 07:25:50 AM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### Re: chemical equilibrium
« Reply #2 on: July 18, 2007, 08:02:56 AM »
thank you. I have read it.

Here's the problem:
For the process below, K = 1025 at some temperature. The system is initially at equilibrium with 0.0400 M CO, 0.100 M Cl2, and 4.10 M COCl2. If 0.060 mol CO is added, and the volume of the container is subsequently doubled, what is the molarity of CO after equilibrium is reestablished?

CO(g) + Cl2(g) <==> COCl2(g)

a.0.0052 M
b.0.050 M
c. -0.46 M
d. 0.045 M

#### firzzy87

• Regular Member
• Posts: 53
• Mole Snacks: +2/-13
• I'm a mole!
##### Re: chemical equilibrium
« Reply #3 on: July 18, 2007, 09:01:06 AM »
this should be belong to physical chemistry forum

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### Re: chemical equilibrium
« Reply #4 on: July 18, 2007, 09:46:41 AM »
Well, it can also be in inorganic chemistry. Some topics in inorganic chemistry are also in physical chemistry.

• Regular Member
• Posts: 16
• Mole Snacks: +5/-0
• Gender:
• it's up to you
##### Re: chemical equilibrium
« Reply #5 on: July 18, 2007, 11:40:47 AM »
I calculate the concentration of CO for V= 1litre initially. I find it 0.045M. Is this correct? Do you have the answer?

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### Re: chemical equilibrium
« Reply #6 on: July 18, 2007, 08:41:13 PM »
Yes, that's correct. How did you get it? Can you show your solution? thank you very much.

• Regular Member
• Posts: 16
• Mole Snacks: +5/-0
• Gender:
• it's up to you
##### Re: chemical equilibrium
« Reply #7 on: July 19, 2007, 03:12:56 AM »
Here is the solution:

Kc=[COCl2]eq/[CO]eq[Cl2]eq. If you have a initially volume of 1l, double it and put into the system  an additional amount of 0.060 mol CO then the equilibrium concentrations must be changed to balance the system.
Then Kc=(2,05-x)/(0,05-x)(0,05-x). The initial concentrations must be semidoubled because the volume doubled. If  you solve this equation you must find x=0.0052. [CO]eq=0.05-x=0.0448=0.045M

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7978
• Mole Snacks: +555/-93
• Gender:
##### Re: chemical equilibrium
« Reply #8 on: July 19, 2007, 03:55:20 AM »
Then Kc=(2,05-x)/(0,05-x)(0,05-x)
??
Kc=c(COCl2) / (c(CO) x c(Cl2) ) = V x n(COCl2) / (n(CO) x n(Cl2) )
c - molar concentration, n - moles, V - volume in liters

After addition of CO and volume change
Kc = 2V x (n(COCl2)+X) / ( (n(CO)+0.06-X) x (n(Cl2)-X) )
capital X - change of reagents (in moles) to reach an equilibrium
AWK

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### Re: chemical equilibrium
« Reply #9 on: July 19, 2007, 04:07:29 AM »
Thank you very much. But I still can't get the right answer. I used the quadratic equation. I'll just try again. Maybe something is wrong with my computation.

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### Re: chemical equilibrium
« Reply #10 on: July 19, 2007, 04:13:08 AM »
I got it. I just forgot to divide my answer to 2 L.

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7978
• Mole Snacks: +555/-93
• Gender:
##### Re: chemical equilibrium
« Reply #11 on: July 19, 2007, 04:23:30 AM »
I got 0.0495. If your result is the same, this should be a printing error
AWK

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### Re: chemical equilibrium
« Reply #12 on: July 19, 2007, 05:59:51 AM »
I got 0.045 M.  I also followed your equation.  Thank you very much sir.

• Regular Member
• Posts: 16
• Mole Snacks: +5/-0
• Gender:
• it's up to you
##### Re: chemical equilibrium
« Reply #13 on: July 19, 2007, 08:33:52 AM »
Sorry, the equation is wrong. It is 2,05+x. My calculations are with the right equation but i write the wrong because I am very busy today. I am sorry.

#### kimi85

• Full Member
• Posts: 133
• Mole Snacks: +0/-12
##### Re: chemical equilibrium
« Reply #14 on: July 19, 2007, 10:46:11 AM »
That's okay. Thank you very much.