# A relation to Le-chatelier's principle?

1. Apr 30, 2015

### mooncrater

1. The problem statement, all variables and given/known data
$Fe+2HCl(g) \longrightarrow FeCl_2+H_2$. In my textbook it's written that liberation of $H_2$ prevents formation of $FeCl_3$.
But why?And how?

2. Relevant equations

3. The attempt at a solution
I think it is related to the Le-Chatelier priciple . As $H_2$ is liberated more, the reaction equilibrium is shifted towards the products . So more $FeCl_2$ will be produced. But for $FeCl_3$, I think there should be an alternative path for the reactants to produce $FeCl_3$ along with the path producing $FeCl_2$ and $H_2$. I think this path is neglected when the reaction equilibrium is shifted towards the products of the other path.Am I correct , or is there any other reason for that?

2. May 4, 2015

### leafjerky

I think you're right. As you take away product, more product is formed. As far as an alternate route goes, the reaction could go Fe + 3HCL = FeCl3 + 3/2H2. If you are constantly taking away H2 then the reaction should constantly be shifting to the right and keeping the FeCl3 from forming.

3. May 4, 2015

### mooncrater

I think the reaction I am talking about forms $FeCl_2$ not$FeCl_3$.
Though $H_2$ is liberated in both the reactions why $FeCl_2$ formation is preferred? Is it due to the reason that $Fe^{+2}$ is the most stable o. S. Of $Fe$. I am leaving the Le-chatelier approach, as it is working for both