Question

Given: N2 + 3H2 <-> 2NH3. At equilibrium [N2] = 0.55M , [H2] = 0.69M, and [NH3] = 0.52M
at a certain temperature in a 1.0Lvessel. Some N2 was removed and when equilibrium was established, [H2] = 0.72M. How much N2 was removed

Answer 1

so first I found the Ksp , which is 1.4966
and for the values after N2 was removed
is it N2 : 0.55-x
3H2 : 0.69 + 0.03 = 0.72
2NH3: 0.52 - 0.02 = 0.50

(3 mol of H2 = 1 mol of N2)
but doesnt need to add the 0.01 to N2? cuz when equilibrium is established, N2 will rises too right
so shouldnt it be like 0.55 -x + 0.01

Answer 2

"At equilibrium... [H2] = 0.69M"

"...when equilibrium was established, [H2] = 0.72M"

What?

Answer 3

that means the change is 0.03M
I 0.69
C 0.03
E 0.72

Answer 4

the reaction is moving to the reactant side, so the change is +ve