The first step in the synthetic manufacture of nitric acid is the oxidation of a

Question

The first step in the synthetic manufacture of nitric acid is the oxidation of ammonia to nitric oxide. (NH_3 reacts with O_2 to form NO and H_2O). The commercial procedure is to feed the ammonia vapor at 25 degree C into a stream of air at 700 degree C before it passes over the catalyst. Sufficient air is used to provide 10% excess over that theoretically needed to oxidize all the ammonia. Approximately 80% of the NH_3 is oxidized to NO, and the gases leave at the reaction temperature of 920 degree C. Using a basis of 4 gmol of NH_3: Write the balanced chemical reaction equation for the oxidation of ammonia with stoichiometric oxygen. Write the balanced chemical reaction equation for the oxidation of ammonia using with 10% excess air and complete 100% oxidation of the NH_3. Write the balanced chemical reaction equation for the oxidation of ammonia using with 10% excess air and partial 80% oxidation of the NH_3. Assume that 4 gmol/min of NH_3 are fed to the reactor, calculate the gmols of all compounds entering and leaving the reactor. Using the basis of the elements at 25 degree C and 1 atm. calculate the enthalpy (kJ/gmol) of all compounds entering and leaving the reactor and the overall heat flow to or from the reactor. For simplicity, assume and use the following average (and constant) specific heats: NH_3 49.8 J/gmol-K; NO 32.3 J/gmol-K; N_2 31.2 J/gmol-K; O_2 33.0 J/gmol-K; H_2O 38.2 J/gmol-K To check your work complete the following Table.

Explanation / Answer

(a)

NH3(g)+1/0.21air (0.79 N2+0.21O2)(g)--------->NO(g)+H2O(g)+(0.79/0.21) N2(g)

NH3(g)+3.761 N2+O2(g)--------->NO(g)+H2O(g)+3.761 N2(g)

(b)

10% excess air

means (1.1/0.21)=5.238 moles of air

So

NH3(g)+4.138 N2+1.099 O2(g)--------->NO(g)+H2O(g)+4.138 N2(g)+0.099 O2(g)

(c)

10% excess air

80% oxidation

NH3(g)+4.138 N2+1.099 O2(g)--------->0.8NO(g)+1.2H2O(g)+0.2 NH3(g)+4.138 N2(g)+0.099 O2(g)

(d)

NH3=4 gmol/min

According to reaction

NH3(g)+3.761 N2+O2(g)--------->NO(g)+H2O(g)+3.761 N2(g)

N2 =3.761*4=15.044 gmol/min

O2=4 gmol/min

leaving

NO=4 gmol/min

H2O=4 gmol/min

N2=15.044 gmol/min

(d)

H=CpT

Cp=product-reactant=32.3+38.2+3.761*31.2-(3.761*31.2+33+49.8)=-12.3 J/gmol K

H=CpT=-12.3*(298.15)=-3667.245 J/gmol=-3.667 kJ/gmol

(e)

Reactant Product gmol/min gm/min kJ/gmol gmol/min gm/min kJ/gmol NH3 4 68 14.84787 N2 15.044 421.232 9.30228 15.044 421.232 9.30228 O2 4 128 9.83895 NO 4 120 9.630245 H2O 4 72 11.38933 617.232 33.9891 613.232 30.32186

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