Reaction that does not reach equilibrium

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Discussion Overview

The discussion centers on whether the reaction C(s) + CO2(g) = 2CO(g) will reach equilibrium at 298K. Participants explore concepts related to chemical equilibrium, Gibbs free energy, and the conditions under which reactions stabilize.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • Some participants assert that all reactions reach equilibrium given enough time, questioning the validity of the book's assertion that the reaction will not reach equilibrium at 298K.
  • Others suggest that the stability of CO at 298K may influence the reaction's ability to reach equilibrium, referencing Gibbs free energy and its positive value for the reaction in the forward direction.
  • A participant emphasizes that equilibrium is defined by the condition where the change in Gibbs free energy is zero, and that this condition may not be met at 298K for this reaction.
  • There is mention of the distinction between reaching a state of stabilization and achieving true equilibrium, particularly in relation to the Gibbs free energy and the nature of the system (closed vs. open).

Areas of Agreement / Disagreement

Participants express differing views on whether the reaction will reach equilibrium at 298K, with some arguing that it will eventually stabilize while others contend that the conditions do not favor equilibrium. No consensus is reached regarding the interpretation of the book's answer.

Contextual Notes

Participants note that the equilibrium state is dependent on temperature, pressure, and system composition, and that the reaction may behave differently in open versus closed systems. The discussion highlights the complexity of defining equilibrium in the context of Gibbs free energy.

kasse
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Homework Statement



Will you expect the reaction C(s) + CO2(g) = 2CO(g) to reach equilibrium at 298K?

The Attempt at a Solution



Yes, because all reactions reach equilibrium given enough time.

Am I wrong? My book says so. The answer is simply "no". But why?
 
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All reactions reach 'an' equlibrium at a certain temperature.

Imagine mixing hydrogen and oxygen, at room temperature the equilibrium is for it to form water - on the surface of a star it won't.
 
So the answer is no because CO is not stable at 298K? And that can be seen because the gibbs free energy for the reaction is positive from left to right?
 
kasse said:
Yes, because all reactions reach equilibrium given enough time.

I'm with you on this one. It's too bad the answer couldn't have been more specific (e.g., a reasonable amount of time). It makes it hard to understand what's expected of you.

Even if the production of CO isn't favored, the reaction should still reach equilibrium--just with very little CO in the system.
 
kasse said:
Am I wrong? My book says so. The answer is simply "no". But why?

Change the book.

It may take eons, but finally reaction will reach the equilibrium.

Could be they meant something else than they asked - but if so, change the book.



 
This question is concerns the equilibrium as pertaining to free energy and at room temperature the free energy for the reaction may not be zero meaning that either the forward or reverse mechanism is favored ; this does not mean that things are not going to stabilize at some point however this latter point is different from the actual concept of equilibrium.


Remember that the equilibrium is at the temperature where the free energy is zero. Use the a derivation of the Gibbs equation along with the standard Gibbs potential value to solve for the temperature.
 
GCT said:
Remember that the equilibrium is at the temperature where the free energy is zero.

Equilibrium is at the temperature, pressure, and composition where the change in Gibbs free energy is zero.
 
I was referring to this question in particular which is soley with respect to the temperature.

Also remember that we're not necessarily referring to a closed system here - in such a system the rate of the reverse increase with successively to eventually equal the forward assuming that the rate constants are comparable. Also we're dealing with gases - in an open system things may simply go to completion unless we're at the exact temperature where the Gibbs is zero where things are at an equilibrium state.
 

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