Web viewTHERMODYNAMICS FREE RESPONSE II. 2009 part B, question #5, form B. Answer the following...

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THERMODYNAMICS FREE RESPONSE II 2009 part B, question #5, form B Answer the following questions about nitrogen, hydrogen, and ammonia. (a)In the boxes below, draw the complete Lewis electron-dot diagrams for N 2 and NH 3 . (b)Calculate the standard free-energy change, ΔG°, that occurs when 12.0 g of H 2 (g) reacts with excess N 2 (g) at 298 K according to the reaction represented below. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 298 = -34 kJ mol -1 (c)Given that Δ298 for the reaction is −92.2 kJ mol -1 , which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain. (d)The value of the standard entropy change, Δ298 , for the reaction is −199 J mol -1 K -1 . Explain why the value of Δ298 is negative. (e)Assume that ΔH° and ΔS° for the reaction are independent of temperature. (i)Explain why there is a temperature above 298 K at which the algebraic sign of the value of ΔG° changes. (ii) Theoretically, the best yields of ammonia should be achieved at low temperatures and high pressures. Explain.

Transcript of Web viewTHERMODYNAMICS FREE RESPONSE II. 2009 part B, question #5, form B. Answer the following...

Page 1: Web viewTHERMODYNAMICS FREE RESPONSE II. 2009 part B, question #5, form B. Answer the following questions about nitrogen, hydrogen, and ammonia

THERMODYNAMICS FREE RESPONSE II

2009 part B, question #5, form BAnswer the following questions about nitrogen, hydrogen, and ammonia.(a) In the boxes below, draw the complete Lewis electron-dot diagrams for N2 and NH3 .

(b) Calculate the standard free-energy change, ΔG°, that occurs when 12.0 g of H2(g) reacts with excess N2(g) at 298 K according to the reaction represented below.

N2(g) + 3 H2(g) 2 NH3(g) ∆G˚298 = -34 kJ mol-1

(c) Given that ΔH˚298 for the reaction is −92.2 kJ mol-1, which is larger, the total bond dissociation energy of the reactants or the total bond dissociation energy of the products? Explain.

(d) The value of the standard entropy change, ΔS˚298 , for the reaction is −199 J mol-1K-1. Explain why the value of ΔS˚298 is negative.

(e) Assume that ΔH° and ΔS° for the reaction are independent of temperature.(i) Explain why there is a temperature above 298 K at which the algebraic sign of the

value of ΔG° changes.(ii) Theoretically, the best yields of ammonia should be achieved at low temperatures and

high pressures. Explain.

Page 2: Web viewTHERMODYNAMICS FREE RESPONSE II. 2009 part B, question #5, form B. Answer the following questions about nitrogen, hydrogen, and ammonia

2008 part B, question #6, form BUse principles of thermodynamics to answer the following questions.

(a) The gas N2O4 decomposes to form the gas NO2 according to the equation below.

(i) Predict the sign of ∆H˚ for the reaction. Justify your answer.(ii) Predict the sign of ∆S˚ for the reaction. Justify your answer. (b) One of the diagrams below best represents the relationship between ∆G˚ and temperature

for the reaction given in part (a). Assume that ∆H˚ and ∆S˚ are independent of temperature.

Draw a circle around the correct graph. Explain why you chose that graph in terms of the relationship:∆G˚ = ∆H˚ - T∆S˚.

(c) A reaction mixture of N2O4 and NO2 is at equilibrium. Heat is added to the mixture while the mixture is maintained at constant pressure.

(i) Explain why the concentration of N2O4 decreases. (ii) The value of Keq at 25°C is 5.0 10-3. Will the value of Keq at 100°C be greater than,

less than, or equal to this value? (d) Using the value of Keq at 25°C given in part (c)(ii), predict whether the value of ∆H˚ is

expected to be greater than, less than, or equal to the value of T∆S˚. Explain.