[student8607]

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 K_c 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+5000^x2

0 = 5000^x2 - 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?

[Astrokel]

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

What is this set up of equation? What's 1000?

[student8607]

1000 is the Kc?
Or do I not include that?


I did: products = Kc x reactants

[Astrokel]

Is the K_c given and isn't it 998?

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

(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.

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

This is wrong too.

[student8607]

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?

[Astrokel]

[PCl₅]_eqm = K_c 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)

[student8607]

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.

[student8607]

Did 3 problems correctly, but then ran into this one:

the K_c=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

[Astrokel]

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.

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

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?

[student8607]

Yeah I double checked?

[Astrokel]

No idea, what you did are correct to me.

[enahs]

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?

[Astrokel]

Ah yes enahs, i missed that.

[student8607]

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?

[Astrokel]

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 K_c 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!