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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: kirushanth on August 01, 2012, 07:48:19 AM

Title: Reversible or not
Post by: kirushanth on August 01, 2012, 07:48:19 AM

Why is nitration of benzene not reversible?

Generally, when we study about organic mechanisms, what type of mechanisms determine whether the reaction is reversible or not?

I guess any sort of rearrangement is not reversible.

Title: Re: Reversible or not
Post by: Babcock_Hall on August 01, 2012, 01:05:58 PM

I usually use the words reversible or irreversible to describe the thermodynamics of a reaction, not the kinetics.  Reactions that proceed with a large value of the equilibrium constant are irreversible.

Title: Re: Reversible or not
Post by: kirushanth on August 01, 2012, 04:17:23 PM

Which sort of mechanisms have large equilibrium constant?

Title: Re: Reversible or not
Post by: Babcock_Hall on August 02, 2012, 08:55:40 AM

It might be better to substitute the words "class of reactions" for mechanisms.  Oxidations often proceed with large equilibrium constants, but isomerizations (rearrangements) can have either large or small equilibrium constants, to take two examples.  I am not especially good at predicting equilibrium constants, speaking only for myself.  However, many (perhaps most) of the reactions one encounters in synthetic organic chemistry must be somewhat favorable thermodynamically, or they would not be useful transformations.

Title: Re: Reversible or not
Post by: kirushanth on August 02, 2012, 12:13:40 PM

Thanks, now what reason can one give for the fact that nitration of benzene to be non reversible?

Title: Re: Reversible or not
Post by: fledarmus on August 02, 2012, 02:18:27 PM

One good way to estimate whether a reaction will be reversible is to see how much energy is required or released by the reaction. If the free energy released by the reaction is high, then the reaction is probably not reversible.

Can you calculate the thermodynamic properties for this reaction?

Title: Re: Reversible or not
Post by: camptzak on August 03, 2012, 02:13:37 AM

you would use Gibbs equation right?
ΔG°=ΔH°-T·ΔS°

you would subract the reactant enthalpy values from the products enthalpy values. If there is a decrease in enthalpy then there is a release of that energy that was trapped in the system right?

The Entropy value is a factor of the temperature in which the reaction is taking place correct? If the overall ΔG° is negative then the reaction will be spontaneous and your K value will depend on the amount of free energy released correct?

I dont understand the Entropy value. How do we assign these values?