Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ToT)NaM( on January 27, 2009, 08:35:08 PM

Title: Equilibrium problems
Post by: ToT)NaM( on January 27, 2009, 08:35:08 PM
So the equation is CO + 3 H2 <--> CH4 + H2O
or in words: Carbon monoxide and hydrogen gas react to form methane and water.

The information given is that the equilibrium constant, K, is 4.0
The intial concentration of Hydrogen gas is 0.30M
and the concentration at equilibrium of hydrogen gas is 0.12M

What are the initial and final concentration of CO, or carbon monoxide?

I believe you have to ICE it but I cant quite make it work


and one more: PCl5 <--> PCl3 + Cl2

what is the equilibrium concentration of all 3 componets if the initial concentration of PCl5 is 1.00M and the K = 0.02??

I'm pretty damn sure that you have to ICE each problem. However, I've never had to do a problem where you dont start with all of the initial concentration.

For the second problem I think I have the right answer because it is much easier (you start with the initial concentration of all of the reactants. So i really would just like to check my answer with what you guys get.
Title: Re: Equilibrium problems
Post by: ToT)NaM( on January 27, 2009, 09:16:53 PM
Sorry for double post but I think I can only edit once. This is actually a question from my AP Chemistry high school class so should i actually put this in the undergraduate section or?
Title: Re: Equilibrium problems
Post by: vhpk on January 27, 2009, 11:13:31 PM
Some hints for u:
1/ Basing on the intial and equilibrium concentration of hydrogen, you can easily infer the difference of concentration of hydrogen and the one of carbon monoxide. Also, you can calculate the final concentration of methane and water. Using the Kc expression to calculate the concentration at equilibrium of CO. So you can calculate the intial concentration of CO
2/PCl5 <--> PCl3 + Cl2
Let the amount of PCl5 that has reacted be x
So you can calculate the concentration of reactant and products w.r.t x.
Using the Kc expression to find x
Title: Re: Equilibrium problems
Post by: Donaldson Tan on January 28, 2009, 06:27:51 AM
This is actually a question from my AP Chemistry high school class so should i actually put this in the undergraduate section or?

Don't worry about the categorisation as long as you don't cross-post. The moderators will move your post to the appropriate category if there is a need to.

Let's look at the 1st problem:
So the equation is CO + 3 H2 <--> CH4 + H2O
or in words: Carbon monoxide and hydrogen gas react to form methane and water.

The information given is that the equilibrium constant, K, is 4.0
The intial concentration of Hydrogen gas is 0.30M
and the concentration at equilibrium of hydrogen gas is 0.12M

What are the initial and final concentration of CO, or carbon monoxide?

CO + 3H2 <-> CH4 + H2O

Let make some assumptions to simplify the problem:
1. Let initial concentration of CH4 be 0.0M
2. Let initial concentration of H2O be 0.0M
3. Let this be a constant volume system whereby V is the volume of the system

Let x be the change in number of moles of CO.
From the chemical equation, if the amount of CO decreases by x, then the amount of H2 decreases by 3x.
=> 3x = [H2]initial.V - [H2]eqbm.V = 0.18V
=>   x = 0.06V
=> [H2O]eqbm = [CH4]eqbm = x/V = 0.06M

K = { [H2O]eqbm * [CH4]eqbm } / { [CO]eqbm * [H2]eqbm }
[CO]eqbm = { [H2O]eqbm * [CH4]eqbm } / { K * [H2]eqbm } = 0.062/(4.0*0.12) = 0.0075M
[CO]initial = (0.0075V + x)/V = 0.0675M
Title: Re: Equilibrium problems
Post by: ToT)NaM( on January 28, 2009, 02:45:41 PM
So for number 2 i got:

0.02 = x2/1
0.02 =x2

x= 0.141

so then the concentrations of both products is 0.141 and the concetration of the reactant is 1 - 0.141 = 0.859 is that right?
Title: Re: Equilibrium problems
Post by: ToT)NaM( on January 28, 2009, 02:55:17 PM
This is actually a question from my AP Chemistry high school class so should i actually put this in the undergraduate section or?

Don't worry about the categorisation as long as you don't cross-post. The moderators will move your post to the appropriate category if there is a need to.

Let's look at the 1st problem:
So the equation is CO + 3 H2 <--> CH4 + H2O
or in words: Carbon monoxide and hydrogen gas react to form methane and water.

The information given is that the equilibrium constant, K, is 4.0
The intial concentration of Hydrogen gas is 0.30M
and the concentration at equilibrium of hydrogen gas is 0.12M

What are the initial and final concentration of CO, or carbon monoxide?

CO + 3H2 <-> CH4 + H2O

Let make some assumptions to simplify the problem:
1. Let initial concentration of CH4 be 0.0M
2. Let initial concentration of H2O be 0.0M
3. Let this be a constant volume system whereby V is the volume of the system

Let x be the change in number of moles of CO.
From the chemical equation, if the amount of CO decreases by x, then the amount of H2 decreases by 3x.
=> 3x = [H2]initial.V - [H2]eqbm.V = 0.18V
=>   x = 0.06V
=> [H2O]eqbm = [CH4]eqbm = x/V = 0.06M

K = { [H2O]eqbm * [CH4]eqbm } / { [CO]eqbm * [H2]eqbm }
[CO]eqbm = { [H2O]eqbm * [CH4]eqbm } / { K * [H2]eqbm } = 0.062/(4.0*0.12) = 0.0075M
[CO]initial = (0.0075V + x)/V = 0.0675M



I think it might be because im not used to having to read these problems on a computer but i didnt manage to follow that >.<

I understand that because of the chemical reaction that for every amount that the CO went down the H2 went down three times that. So that means since the hydrogen concentation went down 0.18 the Hydrogen went down a third of that? But wouldnt that make the equilibrium concentration of CO a number minus 0.06 and not just 0.06? So in this case it would be the equilibrium concentration 0.075 plus x, or how much it lost, which is 0.06 so the intial would be 0.135M?
Title: Re: Equilibrium problems
Post by: Astrokel on January 28, 2009, 03:45:09 PM
Quote
But wouldnt that make the equilibrium concentration of CO a number minus 0.06 and not just 0.06?
Yes you are right that [CO]eqm is a number minus 0.06. As you are using [H2]eqm, [CH4]eqm, [H2O]eqm to calculate, your result should give an equilibrium concentration of CO as well and not its initial.

Since you mentioned, [CO]eqm = [CO]initial (number you're talking about) - 0.06, and rearrange it gives the same as geodome's equation.
Title: Re: Equilibrium problems
Post by: ToT)NaM( on January 28, 2009, 04:25:20 PM
Quote
But wouldnt that make the equilibrium concentration of CO a number minus 0.06 and not just 0.06?
Yes you are right that [CO]eqm is a number minus 0.06. As you are using [H2]eqm, [CH4]eqm, [H2O]eqm to calculate, your result should give an equilibrium concentration of CO as well and not its initial.

Since you mentioned, [CO]eqm = [CO]initial (number you're talking about) - 0.06, and rearrange it gives the same as geodome's equation.

Oh my decimal play was off by one on the 0.0075. Ok so can anyone confirm my answer to #2 is correct?
Title: Re: Equilibrium problems
Post by: Astrokel on January 28, 2009, 04:51:17 PM
Quote
So for number 2 i got:

0.02 = x2/1
0.02 =x2

x= 0.141

so then the concentrations of both products is 0.141 and the concetration of the reactant is 1 - 0.141 = 0.859 is that right?

Method is correct, but you should not approximate 1-x to x as the equilibrium constant is not small.
Title: Re: Equilibrium problems
Post by: ToT)NaM( on January 28, 2009, 05:05:52 PM
What do you mean aproximate 1-x?? what would you get if you "didn't approximate"?

I guess I dont understand what about what i did involved "approximating"
Title: Re: Equilibrium problems
Post by: Donaldson Tan on January 28, 2009, 07:16:18 PM
I think it might be because im not used to having to read these problems on a computer but i didnt manage to follow that >.<

It is alright. Just build an ICE table using the following assumptions:
Let make some assumptions to simplify the problem:
1. Let initial concentration of CH4 be 0.0M
2. Let initial concentration of H2O be 0.0M
3. Let this be a constant volume system whereby V is the volume of the system

Everything will fall nicely in place.
Title: Re: Equilibrium problems
Post by: Astrokel on January 29, 2009, 02:29:35 AM
       PCl5 ::equil:: PCl3 + Cl2
I         1        0         0

C        -x      +x        +x

E       1-x       x          x