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Thread: Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber...?

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    Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber...?

    Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber process? If we increase the concentration of one reactant, then this would shift the equilibrium to the right which favours the production of ammonia (forward reaction)...

    The textbooks (i.e. Jacaranda and Conquering Chemistry) explicitly mention that the ratio must remain 1:3, but they don't really mention why.

    Can someone please explain this to me?
    Thank you
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    Re: Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber

    because if you write out the chemical equation:
    N2(g) + H2(g) → NH3(g)

    and then balance it, you get:
    N2(g) + 3H2(g) → 2NH3(g)


    i.e. the ratio 1 N2 : 3 H2 (1:3)
    Last edited by PaterzAttack; 30 Jun 2012 at 10:14 PM. Reason: edited to add clarification to my post
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    Re: Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber

    Sorry, I don't think my question was very clear ... In Jacaranda, for example, it says:

    "Although LCP predicts that the equilibrium can be shifted to the product side by increasing reactant concentration, it is important that the 1:3 mole stoichiometry is maintained". (Page 248)

    Can someone please explain to me why this mole stoichiometry must be maintained and why we cant just increase the concentration of one reactant and shift the equilibrium...
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    Re: Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber

    You don't have to be too specific with it, all it's saying to you is that it needs to be constantly monitored so that the ratio is always 1:3 to maintain the highest yeild possible. The key thing about the Haber bit is that you have to know that the process has to be kept efficent. Refer to Le Chatliers principle, if you still don't understand post back here and i'll give a detailed explanation
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    Re: Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber

    From what I've been taught, you maintain the stoichiometric ratio as a higher concentration of one other will act to clog the catalyst, reducing effective surface area.

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    Re: Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber

    1. The stoichiometric ratio is kept because if you have an excess of one reactant then you are effectively only creating the yield of the limiting reagent (the one that was not in excess).
    2. Industrially, they recycle the unreacted nitrogen and hydrogen gases multiple times while removing ammonia each time to effectively allow a yield of 98% but as said in (1), this can only be done until it hits a limiting reagent.
    3. Having excess nitrogen (example) will just be wasting gas space since all the hydrogen will be used up but there will be a whole lot of excess nitrogen left over that would have money to produce since it was obtained through cooling to about -190C.
    --> similarly for production of H2, it was required to be clean and not using it all would be a big waste of money.
    4. Also note that the equilibrium is significantly to the left, making an excess will mostly likely need to be very significant, let's say 1 of N2:1000 of H2
    -->you will get 2 molecules of ammonia but be left with 997 of H2 making it uneconomical and technically still poor yield.
    Last edited by someth1ng; 8 Jul 2012 at 11:50 PM.
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    Re: Why do we need to ensure that the mole ratio (N2:H2) remains 1:3 during the Haber

    Quote Originally Posted by someth1ng View Post
    1. The stoichiometric ratio is kept because if you have an excess of one reactant then you are effectively only creating the yield of the limiting reagent (the one that was not in excess).
    2. Industrially, they recycle the unreacted nitrogen and hydrogen gases multiple times while removing ammonia each time to effectively allow a yield of 98% but as said in (1), this can only be done until it hits a limiting reagent.
    3. Having excess nitrogen (example) will just be wasting gas space since all the hydrogen will be used up but there will be a whole lot of excess nitrogen left over that would have money to produce since it was obtained through cooling to about -190C.
    --> similarly for production of H2, it was required to be clean and not using it all would be a big waste of money.
    4. Also note that the equilibrium is significantly to the left, making an excess will mostly likely need to be very significant, let's say 1 of N2:1000 of H2
    -->you will get 2 molecules of ammonia but be left with 997 of H2 making it uneconomical and technically still poor yield.
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