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Topic: Equilibrium: Heterogeneous Reaction  (Read 3799 times)

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Offline Winga

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Equilibrium: Heterogeneous Reaction
« on: March 28, 2005, 07:29:56 AM »
When a solution mixture reacts to give a solid product, will the equilibrium shift completely to the product side?

If not, from which step that governs the reaction equilibrium?
« Last Edit: March 28, 2005, 07:32:17 AM by Winga »

Offline Organic_lover*

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Re: Equilibrium: Heterogeneous Reaction
« Reply #1 on: January 21, 2012, 07:44:58 PM »
To understand how a reaction of this type occur, you have to remember that:

-To occur the formation of the product, the system itself have to present a negative Gibbs free energy. The :delta: G it's a function that express the spontaneity of the reaction, in other words, if the reaction is energetically favorable or not.

So, assuming that we have the reagents (A and B )and mixing this two it's formed a product (C), we can determine if the reaction is occuring in differents times calculating* :delta: G/ :delta: t.

Considering the idea that the mechanism involved in a energetically favorable reaction may take differents steps (and some of this steps are yet studied by cientists), to tell which step governs the  equilibrium you have to talk about a specific reaction (and still would require a detailed study on the reaction in question).

In the equilibrium (you have to remember yourself) that the quantity of reagents and products are nearly constant. So, even in a heterogeneous reaction the equilibrium have the interconversion of C to A e B. The product will be obtained in the system if it's energy is lower than the energy of the reagents.

I hope this helps. If I told you something wrong someone will came here and answer your question better :)

Bye-bye o/
"Nothing in life is to be feared, only understood. Now is the time to understand more to fear less." {Marie Curie}

Offline orgopete

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Re: Equilibrium: Heterogeneous Reaction
« Reply #2 on: January 21, 2012, 10:49:58 PM »
When a solution mixture reacts to give a solid product, will the equilibrium shift completely to the product side?

If not, from which step that governs the reaction equilibrium?

That is the essence of the Finkelstein reaction (RCl + NaI  :rarrow: RI + NaCl(s)). Iodide is a good leaving group and the reverse reaction is possible. However, since sodium chloride is less soluble and precipitates, its low concentration in the solution drives the equilibrium toward the iodide.
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