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Topic: Major Problems With I-C-E charts  (Read 8310 times)

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Offline student8607

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Major Problems With I-C-E charts
« 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?

Offline Astrokel

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Re: Major Problems With I-C-E charts
« Reply #1 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?
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Offline student8607

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Re: Major Problems With I-C-E charts
« Reply #2 on: February 07, 2009, 01:45:24 PM »
1000 is the Kc?
Or do I not include that?


I did: products = Kc x reactants

Offline Astrokel

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Re: Major Problems With I-C-E charts
« Reply #3 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.

No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline student8607

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Re: Major Problems With I-C-E charts
« Reply #4 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?
« Last Edit: February 07, 2009, 07:08:25 PM by student8607 »

Offline Astrokel

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Re: Major Problems With I-C-E charts
« Reply #5 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)
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline student8607

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Re: Major Problems With I-C-E charts
« Reply #6 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.

Offline student8607

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Re: Major Problems With I-C-E charts
« Reply #7 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

Offline Astrokel

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Re: Major Problems With I-C-E charts
« Reply #8 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?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline student8607

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Re: Major Problems With I-C-E charts
« Reply #9 on: February 08, 2009, 09:45:12 AM »
Yeah I double checked?

Offline Astrokel

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Re: Major Problems With I-C-E charts
« Reply #10 on: February 08, 2009, 11:36:46 AM »
No idea, what you did are correct to me.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline enahs

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Re: Major Problems With I-C-E charts
« Reply #11 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?

Offline Astrokel

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Re: Major Problems With I-C-E charts
« Reply #12 on: February 08, 2009, 02:10:00 PM »
Ah yes enahs, i missed that.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline student8607

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Re: Major Problems With I-C-E charts
« Reply #13 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?

Offline Astrokel

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Re: Major Problems With I-C-E charts
« Reply #14 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!
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

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