Help me find the percentage of the volume of nitrogen gas N2 (g) and hydrogen gas H2(g)?
The reaction for the Haber process, the industrial production of ammonia, is:
N2(g) + 3H2(g) 🡪 2NH3 (g) When the equilibrium arrived, the NH3(g) has a 25% volume of the Haber. Find the percentage of the volume of nitrogen gas N2 (g) and hydrogen gas H2(g), respectively.
- Trevor HLv 76 months agoFavorite Answer
The balanced equation:
N2(g) + 3H2(g) → 2NH3(g)
1 mol N2 reacts with 3 mol H2 to produce 2 mol NH3
For gases , moles = volumes
1 volume N2 reacts with 3 volumes H2 to produce 2 volume NH3
Now you ask for the % volumes when equilibrium has been achieved:
Let us say that you have Haber reactor that has a volume of 100 L
You load the reactor with 25 L of N2 and 75 L of H2
In going to completion :
25 L N2 reacts with 75 L H2 to produce 50 L NH3
The reaction proceeds until an equilibrium is achieved , when the NH3 is 25% of the volume of the reactor.
This means that 25 L of NH3 has been produced.
We know that 25 L N2 produces 50 L NH3 at completion
Therefore 12.5 L N2 produced 25 L NH3 at equilibrium
and 37.5 L H2 produce the 25 L NH3
Therefore , remaining unreacted :
25L - 12.5 L N2 = 12.5 L N2
75 L - 37.5 L H2 = 37.5 L H2
% composition of the gases in the reactor at equilibrium:
NH3 = 25 L = 25%
N2 = 12.5 L = 12.5%
H2 = 37.5 L = 37.5%
- oeman50Lv 76 months ago
At a constant final pressure in the reaction vessel, the volume % is equal to the mole %. Since the NH3 is occupying 25%, the N2 and the H2 are occupying 75%, with 1 mole of N2 and 3 moles of H2 for a total of 4 moles.
For the N2:
1/4 X 75% = 18.75%
For the H2:
3/4 X 75% = 56.25%