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Equilibrium moles question

N2 + 3H2 --> 2NH3


For nitrogen and hydrogen the initial moles is two, and none for ammonia.

At equilibrium there is 2X amount of moles of ammonia, how many moles of hydrogen and nitrogen are there at equilibrium?


I worked it out as there being 0 moles of hydrogen and one mole of nitrogen, although hydrogen would lose three moles you can't get negative quantities of moles.
(edited 10 years ago)
Original post by JackTheNerd
N2 + 3H2 --> 2NH3


For nitrogen and hydrogen the inital moles is two, and none for ammonia.

At equilibrium there is 2X amount of moles of ammonia, how many moles of hydrogen and nitrogen are there at equilibrium?


For 2X moles of ammonia to be produced they must come from reaction between X moles of nitrogen and 3X moles of hydrogen ....
Reply 2
Original post by charco
For 2X moles of ammonia to be produced they must come from reaction between X moles of nitrogen and 3X moles of hydrogen ....


I'm quoting this directly from chemsheets so I don't think it's written incorrectly. But there is three moles of hydrogen and one mole of nitrogen, hence the N2 and 3H2?

The confusing thing is that only the equilibrium moles of ammonia is given in terms of X, not any other value...
Original post by JackTheNerd
I'm quoting this directly from chemsheets so I don't think it's written incorrectly. But there is three moles of hydrogen and one mole of nitrogen, hence the N2 and 3H2?

The confusing thing is that only the equilibrium moles of ammonia is given in terms of X, not any other value...


If X moles have reacted and you started with 2 moles you must have 2-X moles left at equilibrium ...
Reply 4
Original post by charco
If X moles have reacted and you started with 2 moles you must have 2-X moles left at equilibrium ...


I asked at school, I realised that the answer is just 2 - 3x moles, since there's three times as many moles lost in hydrogen compared to ammonia. It's a little bit tedious since the answer is in terms of X, it's a completely different answer compared to all of the other questions.

Thank you for actually replying thought

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