Question

In: Chemistry

(a) Write a stoichiometric table for the reaction N2 + 3 H2  2 NH3 for...

(a) Write a stoichiometric table for the reaction N2 + 3 H2  2 NH3 for an isothermal, isobaric flow system with equimolar (or equal molar) feeds of N2 & H2.

(b) If the entering total pressure is 16.4 atm and the entering temperature is 1727 oC, calculate the concentration of hydrogen and nitrogen entering the reactor.

(c) Plot the gas composition (molar fractions) as a function of the conversion. Is there anything worth noticing? Can you explain it?

What are the molar fractions of ammonia, nitrogen, and hydrogen (gas composition) when the conversion is 60%.

Expert Solution

(a): It is given that equimolar amount of H2 and N2 are taken.

Let the equimolar amount be 'N'

The stoichiometric table is shown in the figure below:

(b): Given the total entering pressure of the mixture of N2 and H2, Pi = 16.4 atm

Entering temperature of the mixture, Ti = 1727 ^oC = 1727 + 273 = 2000 K

Applying ideal gas equation

PiVi = nRTi

=> (n/Vi) = Pi / RTi

=> (n/Vi) = 16.4 atm / (0.0821 L.atm.mol-1.K-1 x 2000K)

=> (n/Vi) = 0.1 mol/L

Since we have taken equimolar amount of N2 and H2,

Concentration of H2 entering the reactor = 0.1 mol/L / 2 = 0.05 mol/L (answer)

Concentration of N2 entering the reactor = 0.1 mol/L / 2 = 0.05 mol/L (answer)

(c): 3 mol of H2 reacts when 1 mol of N2 reacts.

Since we have taken equimolar amount of H2 and N2, H2 is the limiting reactant.

Hence when the conversion is 60% there would be no more H2 remaining.

moles of N2 left = N - 0.6N = 0.4 N

Moles of NH3 = 2*0.6N = 1.2 N

Hence mole fraction of NH3 = 1.2 N / (1.2N+10.4N) = 0.75

mole fraction of N2 = 0.4 N / (1.2N+10.4N) = 0.25

queen_honey_blossom answered 2 years ago

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