Given moles of H2 = 3 mol, moles of N2 = 9 mol for equation, 3H2 + N2 --> 2NH3, and the container containing both of them has 3L of volume
after Equilibrium NH3 gives 50% mole fraction(Means that after reaction is complete, the number of moles of NH3 formed from reaction occupies 50% of the total number of moles of all gases in the container) , the total volume of H2 and N2 in container before reaction is 3L.
1. Find mole of H2, N2 and NH3 after equilibrium
2. Calculate the heat. Given that delta H = -91 kJ
3. Calculate K
my solution :
3H2 + N2 ----> 2NH3
I 3 9
C -3x -x +2x
E 3-3x 9-x 2x
then 2x/(2x+12−3x−x) = 1/2 then x = 2
but it's impossible because moles of H2 after equilibrium will be -3
so what's my mistake?