Discussion in 'Pre-Medical - MD' started by NubianPrincess, Apr 14, 2002.

1. ### NubianPrincess Perpetually Bored 7+ Year Member

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In the reaction CO(g) + CL2(g) &lt;--&gt; CoCL2(g),
Kc = 1.2*10^3 at 395 degrees Celcius. What is the equilibrium value of [COCL2] if at equilibrium [CO]=2[CL2] = 1/2[COCL2] ?
Any help would be greatly appreciated.

2. ### imtiaz i cant translate stupid Moderator Emeritus 10+ Year Member

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Easy.

Kc = products/reactants

Kc = [COCl2]/[CO][Cl2]

[CO]= 1/2[COCl2]

2[Cl2] = 1/2[COCl2] so
[Cl2] = 1/4[COCl2]

[COCl2] equals [COCl2] (duh)

1.2e3 = [COCl2]/[CO]*[Cl2]
1.2e3 = [COCl2]/(1/2[COCl2])*(1/4[COCl2])
1.2e3 = [COCl2]/1/8[COCl2]^2
1.2e3 = 1/[COCl2]*1/8
(1/8)*1.2e3 = 1/[COCl2]
1/((1/8)*1.2e3) = [COCl2]

There you go.

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3. ### Mr.D insipidus maximus 7+ Year Member

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You're going to have do some substitution to get the anser since they give you a relationship between the reactants and products and the K. With this in mind try to substitue a value for [COCl2], such as x, and then use the same variable to get the other 2 equilibrium concentrations (ie. [Cl2]=4x, [CO]=2x). Now set your equilibrium expression K= [products]/[reactants] using the substituted values from above, and simply solve for x (recall [COCl2] = x). The answer I got was 1.04E-4. One more thing, PURE solids and liquids have activities equal to 1, therefore you don't have to worry about incorporating their values into an equilibrium expression. Hope this helps.

4. ### imtiaz i cant translate stupid Moderator Emeritus 10+ Year Member

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Everything is in the gas phase, so it will be partial pressures instead of concentrations, but it's the same deal. If you did have liquids and solids, it wouldn't be a Kc, but a Keq.

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5. OP

### NubianPrincess Perpetually Bored 7+ Year Member

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Thank you. I tried the substitution method but got really large values, so I'll try it again.
On another note, I HATE the last two weeks of class.
Thanks again
Nubian

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