Question about using the ICE table

  • #1
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Homework Statement:

The following equilibrium reaction is known as the 'water gas shift' reaction:
$$CO(g) + H_2O(g) \Leftrightarrow H_2(g) + CO_2(g)$$
When initially 0.40mol of CO(g) and 1.00 mol of $$H_2O(g)$$ are present in a vessel at $$1200°C$$, the pressure is 2 atm. The amount of $$CO_2(g)$$ is 0.225 mol at equilibrium.
a) Calculate the total equilibrium amount, n, of the gas mixture. Hint: Use the ICE table
b) Calculate the extent of dissociation, α.
c) Calculate the number of moles of each gas, except the carbon dioxide gas in the mixture at equilibrium.

Relevant Equations:

pV=nRT
This is how I started my ICE table:
$$CO(g)$$$$H_2O(g)$$$$H_2(g)$$$$CO_2(g)$$
Initial (mol)0.401.0000
Change (mol)-0.40α-1.00α1.4α1.4α
Equilibrium (mol)0.40-0.40α1.00-1.00α1.4α1.4α (or should I use 0.225)
I am not sure if my table is correct. When I work out the total equilibrium amount, n, using the 1.4α I get: $$n = 0.40 - 0.40α +1.00 - 1.00α +1.4α +0.225$$
and that gives $$n = 1.625 mol$$
Rounding to the correct significant figures: n = 1.6 mol
I then used the 1.6 mol to get the extent of dissociation, α, but my answer does not make any sense when I substitute it back into the 'n' equation.
Can anyone give me hint to help me to see where I am making a mistake?
Thank you very much.
 

Answers and Replies

  • #2
BvU
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Hi,
I don't understand the second line of your table: if 0.4##\alpha## of CO reacts, then: how much H2O reacts ?
 
  • #3
Borek
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In other words: your changes don't follow stoichiometry.

While technically it is possible to express changes in terms of α, it will be much better if you use x (preferably defined as amount of H2 produced) as the unknown. Once you find the equilibrium α will be trivial to calculate.
 
  • #4
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In other words: your changes don't follow stoichiometry.

While technically it is possible to express changes in terms of α, it will be much better if you use x (preferably defined as amount of H2 produced) as the unknown. Once you find the equilibrium α will be trivial to calculate.
So for the Change Part:
If 0.40x of CO reacts, then 0.40x of $$H_2O$$ also reacts?
Then, for the hydrogen gas and carbon dioxide, would it also be 0.40x (under the Change row)?
 
  • #5
Borek
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No idea why you write 0.40x instead of just x, perhaps you are somehow mistaken about what the change really is?

Change line should look like


$$CO(g)$$$$H_2O(g)$$$$H_2(g)$$$$CO_2(g)$$
Change (mol)-x-xxx

What it means is that hen x moles of CO are consumed, there are also x moles of water consumed, and x moles of hydrogen and carbon dioxide produced. This is just the stoichiometry of the reaction.
 
  • #6
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No idea why you write 0.40x instead of just x, perhaps you are somehow mistaken about what the change really is?

Change line should look like


$$CO(g)$$$$H_2O(g)$$$$H_2(g)$$$$CO_2(g)$$
Change (mol)-x-xxx

What it means is that hen x moles of CO are consumed, there are also x moles of water consumed, and x moles of hydrogen and carbon dioxide produced. This is just the stoichiometry of the reaction.
I was following the example in my book.
So then:
$$CO(g)$$$$H_2O(g)$$$$H_2(g)$$$$CO_2(g)$$
Initial (mol)0.401.0000
Change (mol)-x-xxx
Equilibrium (mol)0.40 - x1.00 - xxx

a) The total equilibrium amount would then be n=(0.40 - x + 1.00 - x +x +x) mol = 1.40 mol
b) Would the extent of dissociation then be α = 0.225?
 
  • #7
Borek
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ICE table looks OK.

a) The total equilibrium amount would then be n=(0.40 - x + 1.00 - x +x +x) mol = 1.40 mol
Yes. Hardly surprising - 2 moles of products are made for 2 moles of substances reacting, 1:1, no change in the total number of moles.

b) Would the extent of dissociation then be α = 0.225?
How would you define extent of dissociation for this reaction?

This is a bit tricky IMHO, as in a metathesis reaction technically nothing dissociates.
 
  • #8
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ICE table looks OK.



Yes. Hardly surprising - 2 moles of products are made for 2 moles of substances reacting, 1:1, no change in the total number of moles.



How would you define extent of dissociation for this reaction?

This is a bit tricky IMHO, as in a metathesis reaction technically nothing dissociates.
I'm struggling to grasp the whole part about extent of dissociation. I know it should be the molecules that have dissociated, but this question is way different from what my book is saying. For the Change part in the ICE Table, my book states that the reactants should be -nα where n is the number of moles. I'm really lost about defining the extent of dissociation for this one.
 
  • #9
Borek
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It is not your fault that you are confused, question is confusing and IMHO tries to apply wrong concept to the problem at hand.

Normally, assuming we have an acid HA of initial concentration C0 and it dissociates into H+ and A-:

HA ↔ H+ + A-

extent of dissociation would be

[tex]\alpha = \frac {[A^-]}{C_0}[/tex]

then what we called x in the ICE table (amount of products that appeared) would be just αC0. Technically using αC0 in such calculations (or αn, which - with some trivial assumptions - is equivalent) is correct, and can be applied to simple cases like an acid dissociation. But I don't see any value in this approach. First, using just x is much easier and much more universal (works perfectly for any equilibrium problem), second, once we know the equilibrium, calculating α is trivial. And as you have just found this approach can be quite confusing, especially when we have a situation where it is not entirely clear what α is intended to mean.

My bet is that the idea was that you can assume CO2 "dissociates" into CO and an oxygen atom that is consumed by hydrogen, then "extent of dissociation" would be "number of CO2 molecules that disappeared" over "initial number of CO2 molecules". But I don't like it, it is just overcomplicating things.
 
  • #10
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It is not your fault that you are confused, question is confusing and IMHO tries to apply wrong concept to the problem at hand.

Normally, assuming we have an acid HA of initial concentration C0 and it dissociates into H+ and A-:

HA ↔ H+ + A-

extent of dissociation would be

[tex]\alpha = \frac {[A^-]}{C_0}[/tex]

then what we called x in the ICE table (amount of products that appeared) would be just αC0. Technically using αC0 in such calculations (or αn, which - with some trivial assumptions - is equivalent) is correct, and can be applied to simple cases like an acid dissociation. But I don't see any value in this approach. First, using just x is much easier and much more universal (works perfectly for any equilibrium problem), second, once we know the equilibrium, calculating α is trivial. And as you have just found this approach can be quite confusing, especially when we have a situation where it is not entirely clear what α is intended to mean.
Thank you very much. It makes more sense now.
 

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