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Topic: Kc and heterogeneous equilibria  (Read 942 times)

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Offline Aldebaran

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Kc and heterogeneous equilibria
« on: January 31, 2022, 07:31:12 AM »
An exam question states: peroxyethanoic acid can be prepared by reacting hydrogen peroxide with ethanoic acid. It also states this is a heterogeneous equilibrium giving this equation:
  H2O2(aq) + CH3COOH(aq)   ::equil:: CH3COOOH(aq)+ H2O(l)

It also gives Kc as 0.37 dm3+mol-1 and having also given concentrations for hydrogen peroxide and ethanoic acid it asks the student to calculate the amount in mol of peroxyethanoic acid.

The units given for Kc are consistent with the statement that it is a heterogeneous equilibrium and thus if you leave the H2O out of the calculation you get the answer in the mark scheme. However my question is in what sense is this a heterogeneous equilibrium since all components are present in an aqueous solution. Where is the phase boundary?

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Re: Kc and heterogeneous equilibria
« Reply #1 on: January 31, 2022, 09:39:00 AM »
For me it looks like a homogeneous.

Concentration of water is often considered to be constant so water is not included in the equilibrium constant, that will explain the units.
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Offline Corribus

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Re: Kc and heterogeneous equilibria
« Reply #2 on: January 31, 2022, 09:56:55 AM »
I think they're calling it a heterogeneous reaction because H2O(l) is formally considered a different phase than the other three aqueous species. Subtle distinction that I wouldn't worry too much about it. You just have to remember that pure liquids and pure solids (at standard state) are assigned an activity of 1 and therefore do not contribute to the equilibrium expression.

« Last Edit: January 31, 2022, 11:37:03 AM by Corribus »
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Offline Aldebaran

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Re: Kc and heterogeneous equilibria
« Reply #3 on: January 31, 2022, 10:12:45 AM »
Thank you both for your replies. I agree with both comments but for my own  interest I calculated the molar amounts present from the other info they gave in the question and then worked out what the Kc would be if all were present as pure liquids i.e. no added water, only that generated from the actual reaction. In this case you get a different Kc (because it seems right to include the water as a product if its not added as a solvent medium). In this case whilst it is essentially the same equilibrium you get a different Kc. I do take the point about the activity of pure solids but does that actually apply to pure reacting liquids? I am confusing myself now so any further comment would be most welcome.

Offline Corribus

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Re: Kc and heterogeneous equilibria
« Reply #4 on: January 31, 2022, 11:28:40 AM »
Activities simply don't work like that. A signature failure of most general chemistry courses is that they don't properly explain why concentrations are used for determining equilibrium constants in solution-phase reactions. The fact is that concentrations are often used as surrogates for the chemical activity values because the activity of species in solution is usually proportional to the concentration. But there are lots of occasions where that substitution simply does not work because of how chemical activity is defined. "Pure" substances (liquids and solids) are some of those occasions.

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« Last Edit: January 31, 2022, 11:39:23 AM by Corribus »
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