# Equilibrium Concentration CalculationsWatch

#1
Hi guys, just doing some summer work but I'm a bit stuck on method.

I've got a whole sheet of questions like this:

Work out the actual number of moles of each species present at equilibrium

N2 + 3H2 -----> 2NH3
Initial Moles 1 1 0 - respectively
Equ. Moles 0.8 (of the N2)

In terms of approaching this question using the method found in my textbook/notes, I would say the eq. moles of the H2 is 0.4 (1-3(0.2)) and the Eq moles of NH3 is 0.4 (2 x 0.2)

But then the moles don't add up...?

Do I use this method, or will the Eq. moles of NH3 be 0.8 as 0.8 moles are subtracted from the L.H.S?

Thanks!!
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9 years ago
#2
Yeah, the method described in the text book is correct - the total number of moles of molecules can change, but the overall mass can't change

For example you can split Cl2 with light to form 2Cl. so one mole of chlorine molecules gives two moles of chlorine atoms.
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#3
(Original post by EierVonSatan)
Yeah, the method described in the text book is correct - the total number of moles of molecules can change, but the overall mass can't change

For example you can split Cl2 with light to form 2Cl. so one mole of chlorine molecules gives two moles of chlorine atoms.
Ah of course!

Yeah, just thought I'd check that out!

Thanks!
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