Ammonia is produced by reacting nitrogen and hydrogen according to the reaction

Question

Ammonia is produced by reacting nitrogen and hydrogen according to the reaction below. N, +3H 2NH, A feed stream consisting of 1.700% argon (by mole) and stoichiometric proportion of the reactants is fed into the system at the rate of 100 molmin. The components enter a reactor, and then all ammonia is separated from the other components and leaves the process. The other components are recycled back to the feed stream, with a portion being purged from the system. The mole percentage of argon in the purge stream is and the conversion per pass in the reactor is 12.50%. What is the production rate of ammonia? Number mol NH min 32 What is the overall process fractional conversions of nitrogen and hydrogen? Number 0.65 What is the flow rate of the recycle stream? Incorrect. Using the extent of reaction (E 15.96 mol min) calculate the flow rate of each species in the purge stream Number 380 mol/min

Explanation / Answer

Feed to the reactor contains 100 mole/min , Argon in the feed= 1.7 moles/min

Combined stream of N2 and O2= 100-1.7= 98.3 moles/min

The reaction is N2+3H2 --->2NH3

They are present in stoicometric ratio of 1 :3

So moles of N2= 98.3*1/4= 24.575 and H2= 98.3*3/4= 73.725 moles/min

The mole percentage of Argon is 4.7% in the purge. All the Argon in the feed will be removed through purge under steady state conditions

Let P= purge, P*4.7/100 = 1.7, P= 1.7*100/4.7 =36.2 moles/min

Let R= recycle stream which contains 100-4.7= 95.3% of N2 and H2.

Every 4 moles of N2 and H2 gives 2 moles of NH3 for 100% conversion

but the conversion per pass is only

hence total moles of N2 and H2 entering = 98.3+R*0.953

moles of N2 and H2 reacted is only 12.5%. Hence moles of NH3 formed = (98.3+R*0.953)*0.125/2

moles of N2 and H2 unreacted= (98.3+R*0.953)*0.875

this is divied into 36.2moles/min of purge containing 100-4.7= 95.3% N2 and H2 and Recycle (R) contaiing 95.3% of N2 and H2

hecne (98.3+R*0.953)*0.875= (36.2+R)*0.953

98.3+R*0.953= (36.2+R)*0.953/0.875 =1.089*(36.2+R)

98.3+R*0.953= 1.089*36+1.089R

59.096= R*(1.089-0.953)

R= 434.5 moles/min

Moles of N2 and H2 in purge= 36.2*0.953= 34.5 moles/min

Total moles of N2 and H2 reacted = 98.3-34.5= 63.8 moles/min

For every 4 moles of N2 and H2, 2 moles of NH3 is formed.

Moles of NH3 formed= 63.8/2= 31.4 moles/min

If all the N2 and H2 react, moles of NH3 formed =98.3/2= 49.15 moles/min

% overall conversion =100*31.4/49.15= 63.88%

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