Ammonia is an important commodity chemical used for the manufacture of the fertilizer urea and many — Analytical Chemistry Chemistry Question
Production of ammonia
Ammonia is an important commodity chemical used for the manufacture of the fertilizer urea and many other products. The production of ammonia takes place according to the equilibrium reaction:
N2 + 3 H2 2 NH3
The hydrogen in the ammonia plant is obtained from methane and water by the reaction:
CH4 + H2O CO + 3 H2
Nitrogen is taken from the air, whereby oxygen is removed by the reaction with CO as follows:
O2 + 2 CO 2 CO2
In the air the nitrogen content is 80%. The reactions are performed in a catalytic reactor, the diagram of which is shown below.
The respective flows are numbered in the arrows. Assume that the reactants are converted completely. Take as flow for ammonia at position ; = 1000 mol s-1.
Calculate the following flows in the plant in mol s-1
n[H2 ], for hydrogen at position
n[N2 ], for nitrogen at position
n[CH4 ], for methane at position
n[H2O ], for water at position
n[CO ], for CO at position
n[O2 ], for oxygen at position
n[CO ], for CO at position
Model Answer
n(H2, ) = 3 × ½ × 1000 = 1500 mol s–1
n(N2, ) = ½ × 1000 = 500 mol s–1
n(CH4, ) = ½ × 1000 = 500 mol s–1
n(H2O, ) = ½ × 1000 = 500 mol s–1
n(CO, ) = ½ × 1000 = 500 mol s–1
n(O2, ) = ¼ × ½ × 1000 = 125 mol s–1
n(CO, ) = n(CO, ) – 2 n(O2, ) = 250 mol s–1
In real practice the ammonia formation is an equilibrium reaction, converting only a part of the reactants. The ammonia unit thus must be equipped with a separator and a recycle unit, as shown below.
Suppose the recycle of N2 + H2 that leaves the separator is two times the NH3 flow.
Calculate the flow of N2 at position and the flow of H2 at positions .
Model Answer
n(N2, ) + n(H2, ) = 2 n(NH3, ) = 2 n(NH3, )
n(N2, ) = 500 mol s-1
n(H2, ) = 1500 mol s–1
At a temperature T = 800 K, the Gibbs energies of the three gases are:
G(N2) = – 8.3×10^3 J mol-1
G(H2) = – 8.3×10^3 J mol-1
G(NH3) = 24.4×10^3 J mol-1
Calculate the change in the Gibbs energy (ΔGf) for the conversion of one mole of N2.
Model Answer
ΔGr = 2 G(NH3) – G(N2) – 3 G(H2)
ΔGr = (2 × 24.4 + 8.3 + 3×8.3) ×10^3 = 82×10^3 J mol-1
Calculate the equilibrium constant Kr, for the NH3 formation, using ΔGrf (see 1.3).
Model Answer
ΔGr = – RT ln Kr
Kr = 4.4×10–6
The gas constant equals to: R = 8.314 J mol-1 K-1. Equilibrium constants can also be expressed in partial pressures of the reactants, thus:
Kr = p(NH3)^2 / ( p(N2) * p(H2)^3 )
The partial pressure of ammonia at position is a fraction x of the total pressure:
p(NH3) = x * ptot where x is also expressed by the flow ratio n(NH3) / ntot.
Derive the equations for the partial pressures p(N2) and p(H2) at position .
Model Answer
p(N2) = 1/4 (1 – x) ptot
p(H2) = 3/4 (1 – x) ptot
Insert the partial pressures in Kr and simplify the formula thus obtained as much as possible.
Model Answer
Kr = (x ptot)^2 / [ (1/4 (1 - x) ptot) * (3/4 (1 - x) ptot)^3 ]
Kr = (4^4 * x^2) / ( 27 (1 - x)^4 * ptot^2 )
Calculate x when p0 = 0.1 Mpa and ptot = 30 MPa. (Hint: Kr has been calculated in 1.4).
Model Answer
x^2 / (1–x)^4 = 0.0418 x / (1–x)^2 = 0.204
– 0.204 x^2 + 1.408 x – 0.204 = 0 x = 0.148