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A relation to Le-chatelier's principle?

  1. Apr 30, 2015 #1
    1. The problem statement, all variables and given/known data
    We know about the equation:
    ##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. jcsd
  3. May 4, 2015 #2
    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.
     
  4. May 4, 2015 #3
    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
     
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