Q.
Nitrogen gas and hydrogen gas are added to a 5L vessel where they react to form ammonia. This is an equilibrium reaction, and there is initially 3 times the amount of hydrogen gas than there is nitrogen. At 100 degrees, the K value of this reaction is 0.13M^-2. Given that the amount of ammonia at equilibrium is 1.5mol, find the pressure of the vessel at equilibrium.
N2 + 3H2 <----> 2NH3
So, first up, K = [NH3]^2 / [N2][H2]^3 = 0.13 M^-2.
Since there is initially 3 times the amount of hydrogen gas than nitrogen, and there is 1.5mol of Ammonia at equilibrium (notice there was zero at the start), 0.75 mols of the N2 was lost to produce this, and 2.25 mols of H2 was used up to produce this.
Let x = number of moles of N2 intially.
Therefore
n(N2) final = x - 0.75
n(H2) final = 3x - 2.25
The only way I could think of doing this was by now putting this into the K equation.
[H2] = (3x-0.75)/5
[N2] = (x-2.25)/5
So 0.13 = [NH3]^2 / [N2][H2]^3 = (1.5/5)^2 / ((x-0.75)/5)((3x-2.25)/5)^3)
Solving this x = 2.75 mol
So the amount of moles of N2 final is x - 0.75 = 2 mol
Amount of mole of H2 final is 3x - 2.25 = 6 mol
Now, using ideal gas law, pV = nRT and knowing that it is at 373 K (100 degrees Celsius), and that the sum of the gas moles is 2 + 6 + 1.5 = 9.5 mol
p = nRT / V = (9.5)(8.31)(373)/(5) = 5.895*10^6 Pa
Honestly, I have no idea if that's right. Just wanted to have a crack at it. Could someone verify please?
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