Equilibrium question Kaplan

[basophilic]

See attached image. Shouldn't the equilibrium NOT be affected? I mean even if more linear glucose forms for Fehling's rxn, it gets used up in that reaction; ultimately I wouldn't expect the linear to ring form ratio to change (since equilibrium is sensitive primarily to temperature).

---

Members don't see this ad.

 

[OrdinaryDO]

This is a very simple concept, but you have managed to make it more difficult than it should be. Since only the linear glucose can be oxidized, the Le Chatelier's principle would come into effect. If at first you have an equilibrium of linear and cyclic glucose and only linear can be oxidized (which is what is happening), then you would expect the cyclic glucose to shift in favor of linear glucose since you are losing the linear glucose to a redox reaction. I will demonstrate a better example...

C-Glucose ⇌ L-Glucose
L-Glucose becomes oxidized means--- C-Glucose → L-Glucose (in order to re-establish equilibrium)

make sense?

 

[basophilic]

Wait, really? The equilibrium constant shouldn't change since more L-glucose is formed because it's getting used up by the Fehling's reaction. Once the whole system equilibrates, shouldn't the new equilibrium constant (i.e. the ratio L-glucose/C-glucose) stay the same. Sure there might be lesser amounts of both C and L glucose, but their relative ratio should remain the same, no?

 

[OrdinaryDO]

The equilibrium constant does not change. The concentrations change due to losing L-Glucose to an oxidation reaction which would favor forming more L-Glucose to reach equilibrium once again. The K never changes.