December 08, 2022, 05:30:04 PM
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Topic: Gaseous equilibrium problem that seems to elude my ICE tables!  (Read 988 times)

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Offline slyasafax

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I've been stuck with this problem:

The following reaction has Kc = 2.8 · 102 at T=1000 K:

C2H4 + H2 ::equil:: C2H6

A reactor with volume V = 10 L initially contains 0.32 mol C2H4, 0.16 mol H2 and 0.68 mol C2H6.

Calculate the number of C2H6 moles after:
a) the addition of 1.00 mol of C2H6 in the reactor.
b) the volume of the reactor is increased by 20 L.

I know the answer to a) must be 1.515 moles, while b) must be 0.645 moles.

I can see these numbers make sense by LeChatelier principle. By adding 1 mole to 0.68 mol you get 1.68 mol of C2H6, but a part of it gets converted back to its reactants; by adding 20 L to the volume of the recipient, you shift the equilibrium position towards the reactants (because they are gaseous, so they are entropically favoured)

I tried to solve this exercise with ICE tables, resulting in
Numerator: 1.68 + x

Denominator: (0.32 - x) × (0.16 -x)

But when I solve for Numerator/Denominator = Kc, I do not get the answers above. I also do not know how to include volume in my calculation. I've tried using the ideal gas law and converting into partial pressures, but I still do not get the correct answer.

Offline slyasafax

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Re: Gaseous equilibrium problem that seems to elude my ICE tables!
« Reply #1 on: April 09, 2022, 07:24:21 AM »
Might the first result (point a) contain an error?

The way I've gone about this problem is this: define Kp in terms of Kc. Use the fact that, for a reaction with two reagents and one product, all of them with stoichiometric coefficient = 1:

Kp = Kc/RT

So you get

Kc/RT = [nCRT/V] / [nA·nB·(RT/V)2]

by eliminating all of the RT's, this massively simplifies to

Kc/V = nC / nA·nB

at this point you can calculate what happens to the ratio if you increase V by 20 L. Simply substitute nC for nC-x,  nA·nB for  (nA+x)·(nB+x), and 10 for 30 in the denominator of Kc. By doing this you get  nC-x ≈ 0.645.

In point a), it is de facto impossible for the moles of H2 to decrease by 0.165 (1.68 - 1.515 of the alleged result) because they are 0.16 to begin with.

Offline mjc123

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Re: Gaseous equilibrium problem that seems to elude my ICE tables!
« Reply #2 on: April 09, 2022, 04:07:52 PM »
Quote
In point a), it is de facto impossible for the moles of H2 to decrease by 0.165 (1.68 - 1.515 of the alleged result) because they are 0.16 to begin with.
Moles of H2 don't decrease, they increase. As you said, Le Chatelier.

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