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ů 4Fe²+ + O2(eq) + 4H+ → 4Fe³+ + 2H₂0 (1) eq (eq) eq 16.32 ), (H* (eq) = 0), (Fe³+ eq = −4.60), (H₂O(1) = −237.18) AG°f = (Fe2+ = -78.87), (02(eq) eq mg If there is 0.001 M of Fe²+, 0.002M of Fe³+ ,8. ofO₂. And 10-7M ofH*. Will the forward rxn occur? L Step 1: balance the reaction. Step 2 calculate AG° = -RT In k [C] [D]d [A]a[B]b = Step 3 calculate k = using AG° = -RT ln k. Alternative: step 3: Calculate AG using AG = AG° + RT In Q

ů 4Fe²+ + O2(eq) + 4H+ → 4Fe³+ + 2H₂0 (1) eq (eq) eq 16.32 ), (H* (eq) = 0), (Fe³+ eq = −4.60), (H₂O(1) = −237.18) AG°f = (Fe2+ = -78.87), (02(eq) eq mg If there is 0.001 M of Fe²+, 0.002M of Fe³+ ,8. ofO₂. And 10-7M ofH*. Will the forward rxn occur? L Step 1: balance the reaction. Step 2 calculate AG° = -RT In k [C] [D]d [A]a[B]b = Step 3 calculate k = using AG° = -RT ln k. Alternative: step 3: Calculate AG using AG = AG° + RT In Q

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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& 4Fe²+ + O2(eq) + 4H+ (eq) → 4Fe³+ + eq 3+ eq 120 (1) AG°f = (Fe²+ = -78.87), (02(eq) = 16.32 ), (H* (eq) = 0), (Fe³+ = -4.60), (H₂O(1) = -237.18) eq eq mg If there is 0.001 M of Fe²+, 0.002M of Fe³+, 8- ofO₂. And 10-7M ofH+. Will the forward rxn occur? L Step 1: balance the reaction. Step 2 calculate AG° = -RT In k [C]c[D]d [A]a[B]b Step 3 calculate k = Step 4 calculate Q Step. 5 compare K and Q. [C]c[D]d [A]a[B]b 0₂ mg = 8 X L 1mol 1g 1000mg 32.00g Step 2:AG° = [(4 x -4.60) + (2x −237.18)] - [(4 x −78.87) + (16.32)] = -193.60 kJ/mol X using AG -RT In k. Alternative: step 3: Calculate AG using AG = AG° + RT ln Q = 0.00025M
Transcribed Image Text:& 4Fe²+ + O2(eq) + 4H+ (eq) → 4Fe³+ + eq 3+ eq 120 (1) AG°f = (Fe²+ = -78.87), (02(eq) = 16.32 ), (H* (eq) = 0), (Fe³+ = -4.60), (H₂O(1) = -237.18) eq eq mg If there is 0.001 M of Fe²+, 0.002M of Fe³+, 8- ofO₂. And 10-7M ofH+. Will the forward rxn occur? L Step 1: balance the reaction. Step 2 calculate AG° = -RT In k [C]c[D]d [A]a[B]b Step 3 calculate k = Step 4 calculate Q Step. 5 compare K and Q. [C]c[D]d [A]a[B]b 0₂ mg = 8 X L 1mol 1g 1000mg 32.00g Step 2:AG° = [(4 x -4.60) + (2x −237.18)] - [(4 x −78.87) + (16.32)] = -193.60 kJ/mol X using AG -RT In k. Alternative: step 3: Calculate AG using AG = AG° + RT ln Q = 0.00025M
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