Back Exam IV - Answers

Back Exam IV - Answers - BACK EXAM IV ANSWERS 1 Given the...

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BACK EXAM IV – ANSWERS 1. Given the table of thermodynamic data for the reaction: B 2 O 3 (s) + 6 HF(g) Æ 2 BF 3 (g) + 3 H 2 O(l), answer the questions that follow. B 2 O 3 (s) HF(g) BF 3 (g) H 2 O(l) Δ H o f (kJ/mol) –1273 –271 –1137 –286 S o (J/mol.K) +54 +174 +254 +70. (a) Is this an exothermic or an endothermic reaction? Δ H o rxn = [(2 x –1137) + (3 x –286)] – [(1 x –1273) + (6 x –271)] = –233 kJ Exothermic (b) Is the entropy change for this reaction positive or negative? Δ S o rxn = [(2 x +254) + (3 x +70)] – [(1 x +54) + (6 x +174)] = –380 J/K Negative (c) Calculate Δ S o surr at a temperature of 25 o C. Δ S o surr = – T H 0 rxn Δ = – K 298 J ) 10 x 233 ( 3 = +782 J/K (d) Based upon your answers to questions (a) , (b) , and (c) , what can you say about the spontaneity of the reaction at 25 o C? Δ S o universe = Δ S o rxn + Δ S o surr = –380 + 782 = +402 J/K Positive, so reaction is spontaneous at 25 o C. or: Δ G o rxn = Δ H o rxn – T Δ S o rxn = –233 – (298)(–380 x 10 –3 ) = –120 kJ Negative, so spontaneous. (e) Show that the reaction is non-spontaneous at a temperature of 350 o C. Δ S o surr = – T H 0 rxn Δ = – K 623 J ) 10 x 233 ( 3 = +374 J/K; Δ S o univ = –380 + 374 = –6 J/K Negative, so non-spontaneous at 350 o C or: Δ G o rxn = Δ H o rxn – T Δ S o rxn = –233 – (623)(–380 x 10 –3 ) = +4 kJ Positive, so non-spontaneous. (f) Will increasing the temperature above 350 o C make this reaction spontaneous? Explain. No. Increasing T will make the –T Δ S o rxn term increase (more positive) so that Δ G o rxn will always be positive, and therefore non-spontaneous at temperatures above 350 o C. (g) What is the range of temperatures for which this reaction is spontaneous? Δ G o rxn = 0 = Δ H o rxn – T Δ S o rxn ; T = 0 rxn 0 rxn S H Δ Δ = J/K 380 J 10 x 233 3 = 613 K = 340 o C Spontaneous between 0 K and 612.99 K, or between –273.15 o C and 339.99 o C
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2. There are two forms of tin: White powdery Sn, Δ G o f = 0 kJ/mol; Grey metal Sn, Δ G o f = +0.13 kJ/mol (a) Which form is the most thermodynamically stable form of tin? Why? White powdery tin is the most thermodynamically stable form.
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This note was uploaded on 03/26/2011 for the course CHEM 1032 taught by Professor Guenard during the Spring '06 term at Temple.

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Back Exam IV - Answers - BACK EXAM IV ANSWERS 1 Given the...

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