Chapter%2021%20Answers

Chapter 21 Answe - T = 25 Celsius(definition of using G values 4 What is the K of this reaction at 298 Kelvin ΔG = RT ln K 95920 = 8.314 298.15 ln

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CHEM 1212 Chapter 21 Practice 1. Balance the following redox reaction: Fe 3+ ( aq ) + I ( aq ) I 2 ( s ) + FeO ( s ) [acidic] e - + H 2 O + Fe 3+ FeO + 2 H + 2 I - I 2 + 2 e - 2 H 2 O ( l ) + 2 Fe 3+ ( aq ) + 2 I - ( aq ) 2 FeO ( s ) + 4 H + ( aq ) + I 2 ( s ) 2. Given the following values in kJ/mol, solve for ΔG 0 for the balanced redox reaction: G 0 (Fe 3+ ) = -10.5; G 0 (FeO) = -251.4; G 0 (I ) = -51.67; G 0 (I 2 ) = 0; G 0 (H + ) = 0; G 0 (H 2 O) = -237.19 ΔG 0 = [2 * -251.4 + 4 * 0 + 0] – [2 * -237.19 + 2 * -10.5 + 2 * -51.67] ΔG 0 = -502.8 + 598.72 ΔG 0 = 95.92 kJ/mol 3. What is the temperature for the ΔG 0 you solved in question 2?
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Unformatted text preview: T = 25 Celsius (definition of using G values) 4. What is the K of this reaction at 298 Kelvin? ΔG = - RT ln K 95920 = - 8.314 * 298.15 ln K K = 1.565 * 10-17 5. Solve for ΔG if there is initially 0.25 M of Fe 3+ , 0.15 M of I ‒ and 1.50 M of H + . Q = [H + ] 4 / [Fe 3+ ] 2 [I-] 2 = 1.50 4 / (0.25 2 * 0.15 2 ) = 3600 ΔG = ΔG + RT ln Q ΔG = 95920 + 8.314 * 298.15 * ln (3600) = 116218 J/mol = 116 kJ/mol...
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This note was uploaded on 11/29/2011 for the course CHEM 1212 taught by Professor Dockery during the Fall '08 term at Kennesaw.

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