# In Finkelstein reaction, how does precipitate formation favor forward reaction? [duplicate]

In my textbook it's stated that:

$\ce{NaCl}$ and $\ce{NaBr}$ thus formed is precipitated in dry acetone. It facilitates the forward reaction according to Le Chatelier's principle.

Is this related to common ion effect in any way? I'm not familiar with Le Chatlier's principle. Please explain this in simple words as I'm in high school.

How does formation of $\ce{NaCl}$ and $\ce{NaBr}$ (both insoluble) prevent backward reaction from taking place? Does it mean $\ce{NaI}$ will be produced again?

• If NaCl and NaBr are insoluble in the reaction medium then they are not available for the backward reaction because they are not in solution which is a prerequisite for reaction – Waylander Feb 21 '18 at 14:24
• @Waylander Explained it without Le Chatelier. Nice! – Gaurang Tandon Feb 21 '18 at 14:25
• chemistry.stackexchange.com/questions/6947/… – Mithoron Feb 21 '18 at 21:10

$\ce{NaCl}$ and $\ce{NaBr }$formed as products are polar compounds which are vey less soluble in non- polar solvents like dry acetone, but $\ce{NaI }$due to high polarizability of $\ce {I^-}$ is soluble in acetone. So the products precipitate.
If a reaction is in equilibrium in constant temperature and it is distorted somehow, it goes in the direction to nullify the effect of distortion. This is the Le-Chattelier's principle. If some of the products are removed to keep the $\ce{K_c}$ constant , some more reactants will react and convert into products. Thus constant removal of products by precipitation helps to dissociate reactants into products and thus favour the forward reaction.