The reaction chamber in a modified Haber process for making ammonia by the direc

Question

The reaction chamber in a modified Haber process for making ammonia by the direct combination of its elements is operated at 575 degree C and 248 atm. The reaction mechanism, involving a heterogeneous catalyst, is believed to involve the following steps: The overall reaction is: N_2 (g) + 3H_2 (g) rightarrow 2NH_3 (g) How many liters of nitrogen, measured at these conditions, will react to produce 7.75 x 10^6 grams of ammonia? (Show all calculation steps.) The typical, actual yield for this reaction process is 15%. However, the gases can be recycled back into the catalytic bed to react some more. After four cycles, the yield is approximately 97%. If 7.75 x 10^6 grams of ammonia is needed, how many liters of nitrogen gas should be used if the actual yield is 97%?

Explanation / Answer

Q1.

liters of N2 at such conditions for:

7.75*10^6 g of NH3

MW NH3 = 17 g/mol

mol of NH3 = mass/MW = (7.75*10^6)/17 = 455882.35 mol of NH3

note that ratio is 1 mol of N2 per 2 mol of NH3

so

455882.35 mol of NH3 = 1/2*455882.35 = 227941.17 mol of N2 required

so...

PV = nRT

V = nRT/P

T = 575+273 = 848K

V = (227941.17)(0.082)(848)/(248)

V = 63911.762 Liters of N2 required

Q2.

yield is 15%

yield = 97% recycle

we just figured out that for 100% reaction -->  63911.762 Liters of N2 required

but for 97%...

we must relate

100 -->  63911.762 Liters of N2 required

97 = x

x =  63911.762/0.97 = 65888.414 Liter of N2 required for 97%

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