mr.pyro - 25-9-2006 at 13:41
Ive been stuck on this and cant find it anywhere in the textbook?
Can anyone offer some help? thanks in advance
You place 0.7366 mol of nitrogen, N_2, and 2.2098 mol of hydrogen, H_2, into a reaction vessel at a particular temperature and pressure.
N_2(g) + 3H_2(g) --> 2NH_3(g)
What is the composition of the equilibrium mixture if you obtain 0.04904 mol of ammonia, NH_3, from it?
chemoleo - 25-9-2006 at 15:29
Why do you need a text book for this?
If there is only 0.048 moles of ammonia, the rest will be N2 and H2. NH3 contains the 1.5 molar equivalent of H2, and 0.5 molar equivalent of N2.
Thus, to make 0.048 moles of ammonia, you require 0.048/2 moles of N2 and 0.048*1.5 moles of H2. Subtract this from the original quantities of N2/H2
and you know the composition of the equilibrium. You could even calculate Keq from it.