4. Chemical Reaction Mathematicsg: Please provide complete correct detailed solu

Question

4. Chemical Reaction Mathematicsg: Please provide complete correct detailed solution. Preferentially done by hand or on paper with graphs and calculation.

Problem 4: Ammonia synthesis The gas phase synthesis of ammonia occurs via the following reaction: 3 N2 + H2 NH3 It is carried out isothermallyin a flow reactor with an equimolar feed of H2 and N2 a.) Write the stoichiometric table (for each species give an expression for In. Reacted, Out, and Outlet Concentration). Concentrations should only be in terms of conversion of the limiting reactant and the total initial concentration b.) If the feed is at 16.4 atm and 227 C, what are the concentrations of all species (as well as total concentration) when the conversion of H2 is 60% ? c.) The reaction rate has been fit to experimental data to yield the complex rate law 3/2 k3 + CH dm3 - s With rate constants that have been measured: 3 1/2 k1= 0.01 mol mol k2-0.5 mol 0.001 3. The inlet volumetric flow ratev-0.1 dm, s. Based on the 1nformation you have obtained in part (b), determine a reactor design scheme to minimize the total reactor volume needed to obtain a conversion of x = 0.95. (HINT: for this part think about whether you should use a PFR, CSTR, or some combination (specify the order) to achieve this desired conversion. Concepts from graphical reactor design will definitely help)

Explanation / Answer

a) The balanced reaction is :-

(1/2)N2 + (3/2)H2 --------> NH3

Let 1 mole each of H2 & N2 be present initially

Now, as per the balanced reaction N2 & H2 reacts in the molar ratio of 1:3

Thus, H2 will be the limiting reagent.

Now, moles of H2 reacted = 1

moles of N2 reacted = (1/3)

Moles of NH3 formed = 2/3

b) when 60% of H2 is converted , only 0.6 moles of it will react

Thus, moles of N2 reacted = moles of H2 reacted/3 = 0.6/3 = 0.2

Moles of NH3 formed = 2*moles of N2 reacted = 0.4

Now, total moles of all the species present after the completion of reaction = 0.8 + 0.4 + 0.4 = 1.6

Now, applying Ideal Gas Equation, i.e P*V = n*R*T ; we get

16.4*V = 1.6*0.0821*500

or, V = 4.05 litres

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