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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: student8607 on February 07, 2009, 12:03:01 PM

Title: Major Problems With I-C-E charts
Post by: student8607 on February 07, 2009, 12:03:01 PM
Working with equilibrium and ICE charts
Initial:
Change:
Equilibrium:


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

I: 0.11    0.02       0.2
C:   -x       -x        +x
E: 0.11-x  0.02-x   0.2+x

now we have to adjust for the 5.0L to get molarity

PCl3 = (0.110-x)/5 
Cl2 = (0.02-x)/5
PCl5 = (0.2+x)/5

I need to get a Kc of 998 & I know the correct value for x is 0.0095

But I can't figure out how to get there

.

(0.2+x)/5 = 1000[(0.110-x)(0.02-x)]
                                      5
0.2+x = 5000[(0.110-x)(0.02-x)]

0.2+x = 5000(0.0022-0.13x+x2)

0.2+x = 11-650x+5000x2

0 = 5000x2 - 651x + 10.8

x = 0.111 OR 0.0195


I must be making a silly mistake somewhere. We were given about 10 problems like this for homework and my answers are off on all of them?
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 07, 2009, 01:33:40 PM
Quote
(0.2+x)/5 = 1000[(0.110-x)(0.02-x)]
                                      5
What is this set up of equation? What's 1000?
Title: Re: Major Problems With I-C-E charts
Post by: student8607 on February 07, 2009, 01:45:24 PM
1000 is the Kc?
Or do I not include that?


I did: products = Kc x reactants
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 07, 2009, 04:05:27 PM
Is the Kc given and isn't it 998?

Anyway looking at your equation, there is something wrong with your math.

Quote
(0.2+x)/5 = 1000[(0.110-x)(0.02-x)]
                                      5
You right hand math is wrong when you combine both the concentration reactants.

Quote
0.2+x = 5000[(0.110-x)(0.02-x)]
This is wrong too.

Title: Re: Major Problems With I-C-E charts
Post by: student8607 on February 07, 2009, 06:38:30 PM
Yeah it's given as 1000, but I also calculated it to be 998 as well.

In any case, where did I go wrong with my math?
I just took EVERYTHING and multiplied it by 5.0 to get rid of the volume
Then I foiled the on the right side and then distributed
Combined - reduced to get quadratic?
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 07, 2009, 07:37:14 PM
[PCl5]eqm = Kc x [Cl2]eqm[PCl3]eqm
(0.2+x)/5 = 1000 [(0.02-x)/5] [(0.11-x)/5]
(0.2+x)/5 = 1000 [((0.02-x)(0.11-x))/25]
(0.2+x) = 200(0.02-x)(0.11-x)
Title: Re: Major Problems With I-C-E charts
Post by: student8607 on February 07, 2009, 07:48:23 PM
Ahhh I see now.

So then I get 0.125 and 0.0095
I'll use 0.0095

and then I get my Kc at 998.

Thanks.


I might have a few more questions with these ICE charts tomorrow as well. Your help is always appreciated.
Title: Re: Major Problems With I-C-E charts
Post by: student8607 on February 07, 2009, 08:10:28 PM
Did 3 problems correctly, but then ran into this one:

the Kc=5.0x10-3
in a 10.0L vessel
initially we have 1.0moles of N2, 1.5 of H2 and 0 of N2H4

N2(g) + 2H2(g)  ::equil::N2H4(g)

I solved for x and got 0.000748.
But that was not one of the choices?

.

x/10 = 0.005[(1.00-x)(1.50-2x)/100]
x = 0.0005(1.5-3.5x+2x2)
0.001x2-1.00175x+0.00075
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 07, 2009, 08:47:32 PM
Quote
So then I get 0.125 and 0.0095
I'll use 0.0095
0.125 is rejected because it is greater than 0.11 which is impossible.

Quote
x/10 = 0.005[(1.00-x)(1.50-2x)/100]

Quote
x/10 = 0.005[(1.00-x)(1.50-2x)/100]
x = 0.0005(1.5-3.5x+2x2)
0.001x2-1.00175x+0.00075
The setup of equation and math so far is correct. Did you solve the quadratic correctly?
Title: Re: Major Problems With I-C-E charts
Post by: student8607 on February 08, 2009, 09:45:12 AM
Yeah I double checked?
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 08, 2009, 11:36:46 AM
No idea, what you did are correct to me.
Title: Re: Major Problems With I-C-E charts
Post by: enahs on February 08, 2009, 01:24:29 PM
In an equilibrium expression, if you have a stochimetric coefficient in front of a chemical species, in the equilibrium expression you raise it to that power. I do not see that in your math?
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 08, 2009, 02:10:00 PM
Ah yes enahs, i missed that.
Title: Re: Major Problems With I-C-E charts
Post by: student8607 on February 08, 2009, 02:38:49 PM
Ahh i missed that also.

How would I do the math for the problem then?

If I square H2 I get (4x2-6x+2.25)/100

Then I still need to multiple that by (1-x)/10?
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 09, 2009, 07:12:43 AM
Then you will get a cubic equation which it's pretty tedious. There are another way to solve is through approximation. I'm not sure if your equilibrium constant is small enough for this approximation, as if Kc is very small, then the dissociation x will be very small too. Well wait for someone better knowledge in this than i do to reply, i'm sorry!
Title: Re: Major Problems With I-C-E charts
Post by: Borek on February 09, 2009, 04:49:55 PM
So, the equation is

Kc = 100 x/((1-x)(1.5-x)2) = 0.005

or

x/((1-x)(1.5-x)2) = 0.00005

This can be true only if x is much lower than 1 (check what happens if x gets close to 1). That means in turn that you can simplify substituting 1-x=1 and 1.5-x=1.5.

To be sure - substitute, solve for x, then check if calculated x is really much smaller than 1. If it is - your result should be OK. If it is not - you may have to solve cubic.

For pH calculation (which is nothing else but equilibrium calculation) there exist a so called 5% rule of thumb that says you can approximate x-y as x if y is smaller than 5% of x. Note, that after using this rule you should always check if the assumption (5%) is satisfied.

Quote from: Astrokel on February 09, 2009, 07:12:43 AM
I'm not sure if your equilibrium constant is small enough for this approximation, as if Kc is very small, then the dissociation x will be very small too.

It doesn't matter whether K value is small or large, what matters is whether summed values are comparable or not.
Title: Re: Major Problems With I-C-E charts
Post by: Astrokel on February 10, 2009, 03:35:56 AM
Thanks Borek for the reply! :)