One way to remove harmful “NO x ” (NO and NO 2 ) atmospheric pollutants from flu
ID: 700863 • Letter: O
Question
One way to remove harmful “NOx” (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)?
Explanation / Answer
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
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