Reaction that does not reach equilibrium

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SUMMARY

The reaction C(s) + CO2(g) = 2CO(g) does not reach equilibrium at 298K due to the positive Gibbs free energy associated with the reaction at this temperature. While all reactions theoretically reach equilibrium given sufficient time, the stability of CO at 298K is not favored, leading to minimal production of CO. The equilibrium condition is defined by the temperature, pressure, and composition where the change in Gibbs free energy is zero, which is not the case at room temperature for this reaction.

PREREQUISITES
  • Understanding of Gibbs free energy and its implications on chemical reactions
  • Knowledge of equilibrium concepts in thermodynamics
  • Familiarity with reaction kinetics and system types (closed vs. open systems)
  • Basic principles of chemical thermodynamics
NEXT STEPS
  • Study the derivation of the Gibbs equation and its application to chemical reactions
  • Learn about the conditions under which Gibbs free energy equals zero
  • Explore the differences between closed and open systems in chemical reactions
  • Investigate the stability of various chemical species at different temperatures
USEFUL FOR

Chemistry students, chemical engineers, and professionals involved in thermodynamics and reaction kinetics will benefit from this discussion.

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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