{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter%2021%20Answers

Chapter%2021%20Answers - T = 25 Celsius(definition of using...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}