Kp = 5.3E
5 for the reaction N
2 (g) + 3H
2 (g) 2NH
3 (g)when a certain partial pressure of ammonia gas is put into an otherwise empty rigid vessel at the same temperature, equilibrium is reached when 50 percent of the original ammonia has decomposed. what was the original partial pressure of ammonia before any decomposition occurred?
I wasn't sure how to approach the problem, so I set up an equilibrium expression and tried to express all partial pressures in terms of P
NH3.
I made the following assumptions:
-The partial pressure of nitrogen gas must be half of the equilibrium partial pressure of ammonia
-The partial pressure of hydrogen gas must be one and a half times the equilibrium partial pressure of ammonia
-let x equal to the equilibrium partial pressure of ammonia
5.3E
5 = x
2/[(x/2)(3x/2)
3]
the pre-equilibrium pressure of ammonia using this assumption is 2.24 E
-6, which is far from the answer given which is 2.1E
-3 ATM. what's the correct way to approach this type of problem?
Thanks in advance for any input.