Question

problems. Ask the Instructor to make sure you and your group all achieve the purposes. Ammonia, NH3, is one of the most important industrial substances in the world. Its synthesis is accomplished taking advantage of our understanding of the thermodynamic and kinetic factors that affect the chemical process. Synthesis and Thermodynamics: Most industrial Synthesis Process ΔΗ ammonia is produced by combining N2(g) and AS H2(g). Relevant data for this synthesis reaction is (ld mol") K'mol") listed in the table: 1/2 N,(g) + 3/2 H,(g) =NH, (g) -45.9 -99.1 Fun 1. Justify the signs of AH and AS based on the analysis of the nature of reactants and products. 2. Express the equilibrium constant for the process ke in terms of the pressures of reactants and products. Calculate its value under standard conditions. 3. Build a qualitative graph for AGºran (T) as a function of temperature T and discuss whether the reaction will be favored at low or high temperatures. Thermodynamics versus Kinetics: The activation energy for the synthesis of NH3(g) from H2(g) and N2(e) in the absence of catalyst is high, E, = 325 kJ/mol. The rate of the reaction may be increased by carrying out the synthesis at a high temperature (~500 °C). 4. Build a potential energy diagram for the synthesis of ammonia 5. Determine how many times faster the reaction is at 500 °C than at 25°C 6. Determine how many times smaller Kvis at 500 °C than at 25 "C.

Answer 1

Solution for Question 1:

The Reaction given here is:

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

This Reaction is taking Place at a very High Temperature of 450°C in the presence of Fe Catalyst otherwise it is impossible for Formation of NH3 Gas and therefore this Reaction releases a lot of Heat Energy and is Exothermic in Nature

If a Reaction is Exothermic,the Value of Enthalpy Change (∆H°rxn) is Negative and this Overall Decrease in Enthalpy is achieved by the Generation of Heat.Therefore;

The Value of ∆Hrxn is -ve = -45.9KJ/mol Itself.

In this Reaction, the Total Number of Moles of N2 and H2 in the Reaction as per Stoichiometry = (1÷2) + (3÷2) = 0.5 + 1.5 = 2moles.

The Number of Moles of NH3 = 1 Mole.

So there is a Decrease in the Number of Moles in the Reaction,and so there is a Decrease in Degrees of Freedom and Distortion.

Therefore the Entropy of the Reaction(∆S°rxn) Decreases and becomes Negative.

The Value of ∆S°rxn is -ve = -99.1 J/mol.K itself.

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