Question

One way to remove harmful “NO_x” (NO and NO₂) atmospheric pollutants from flue gas is the “thermal deNOx” process, in which NH₃ is used to reduce the NO to N₂:

__4__ NO (g) + __1__ O₂ (g) + ____ NH₃ (g) → ____ N₂ (g) + ____ H₂O (g)

The research lab has several gas cylinders available for studying this reaction, including one cylinder containing a mixture of 3.0 wt% NO and 15 wt% O₂ in an N₂ diluent, and a second cylinder containing a mixture of 4.5 wt% anhydrous ammonia in N₂ diluent. Assume that all the gases behave ideally.

Balance the thermal deNOx reaction.

You introduce 8.0 g from the NO cylinder and 3.5 g from the NH₃ cylinder into an isothermal, constant volume (20 L) batch reactor. What are the initial molar concentrations of NO, NH₃, and O₂?

What are the initial NO, NH₃ and O₂ mole fractions?

What is the total initial reactor pressure (in bar) at 250°C? What are the partial pressures (in bar) of NO, NH₃, and O₂?

What is the limiting reagent and the final partial pressures of NO, NH₃, and O₂, assuming the reaction goes to completion? Note: the limiting reagent is the reactant that, due to stoichiometry, limits the extent to which the reaction can proceed.

What molar concentrations of NO and NH₃ remain when 75% of the limiting reagent is consumed (i.e., 75% conversion)?

Answer 1

NO + 0.5O₂ + (4/3)NH₃ ------> (7/6)N₂ + 2H₂O

Cylinder 1: 3 wt% NO, 15 wt% O₂, 82 wt% N₂ --- Taken 8 g

Cylinder 2: 4.5 wt% NH₃, 95.5% N₂ --- Taken 3.5 g

Thus, quantities in reactor:

NO = 0.24 g = 0.008 moles

O₂ = 1.2 g = 0.0375 moles

NH₃ = 0.1575 g = 0.00926 moles

N₂ = 9.9025 g = 0.3537 moles

Total moles = 0.40842 moles

For mole fractions, we just need to divide individual no. of moles by the total moles.

Mole fractions:

NO = 0.01959, O₂ = 0.09182, NH₃ = 0.3856, N₂ = 0.503

For molar concentration, just divide individual number of moles by total volume

Molar concentration:

NO = 0.0004 mol/L, O₂ = 0.001875 mol/L, NH₃ = 0.000463 mol/L, N₂ = 0.01769 mol/L

For total reactor pressure, we use ideal gas law using total moles as n=0.40842 moles and T= 523 K

P = nRT/V = 0.40842 mol x (8.314 J/mol-K) x 523K / (0.002 m³ ) = 887950 Pa = 8.88 bar - Reactor pressure

Partial pressure : Multiply total pressure with mole fraction

NO = 0.174 bar, O₂ = 0.8153 bar, NH₃ = 3.424 bar, N₂ = 4.467 bar

NH₃ is the limiting reagent

Overall conversion = 75%

Thus reactant concentrations of LR become 25% of initial

Final concentrations:

NH₃ = 0.0001158 mol/L

From which

NO = 0.0001395 mol/L

O₂ = 0.00174 mol/L

N₂ = 0.000304 + 0.01769 = 0.01799 mol/L