chapter22 - 958 Chapter 22: Chemistry of Selected...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
958 End-of-Chapter Solutions for Chapter 22 Summary Problem Part I Answer: (a) See reactions below (b) –780.6 kJ, –1703.9 kJ (c) – 599.5 J/K, – 383.1 J/K, the second reaction is more product-favored. (d) –257.2 kJ, –1369.5 kJ, the second reaction is more product-favored. Strategy and Explanation: (a) In Section 22.3, page 1073, roasting is described as the reaction of a metal sulfide with oxygen to form a metal oxide and SO 2 . Balance the equations for the reactions: 2 FeS(s) + 7 O 2 (g) 2 Fe 2 O 3 (s) + 4 SO 2 (g) 4 Fe, 4 S, 14 O 3 FeS(s) + 5 O 2 (g) 2 Fe 3 O 4 (s) + 3 SO 2 (g) 3 Fe, 3 S, 10 O (b) To find Δ H°, use Equation 6.11: Δ H° = (moles of product) " # H f o (product) $ % ( ) * + (moles of reactant) " # H f o (reactant) $ % ( ) * Reaction 1: Δ H° = (2 mol) × Δ H f o (Fe 2 O 3 ) + (4 mol) × Δ H f o (SO 2 ) – (2 mol) × Δ H f o (FeS) – (7 mol) × Δ H f o (O 2 ) Δ H° = (2 mol) × (– 824.2 kJ/mol) + (4 mol) × (– 296.830 kJ/mol) – (2 mol) × (– 101.67 kJ/mol) – (7 mol) × (0 kJ/mol) = – 780.6 kJ Reaction 2: Δ H° = (1 mol) × Δ H f o (Fe 3 O 4 ) + (3 mol) × Δ H f o (SO 2 ) – (3 mol) × Δ H f o (FeS) – (5 mol) × Δ H f o (O 2 ) Δ H° = (1 mol) × (– 1118.4 kJ/mol) + (3 mol) × (– 296.830 kJ/mol) – (3 mol) × (– 101.67 kJ/mol) – (5 mol) × (0 kJ/mol) = – 1703.9 kJ (c) To get Δ G°, Use Equation 18.3: Δ G° = Δ H° –T Δ S°. It is possible to assume that Δ H° and Δ S° have nearly constant values as long as their physical state doesn’t change. To find Δ S°, use the equations given in Section 18.4: Δ S° = (moles of product) " S o (product) # $ % ( ) * (moles of reactant) " S o (reactant) # $ % ( ) Reaction 1: Δ S° = (2 mol) × S°(Fe 2 O 3 ) + (4 mol) × S°(SO 2 ) – (2 mol) × S°(FeS) – (7 mol) × S°(O 2 ) Δ S° = (2 mol) × (87.4 J K –1 mol –1 ) + (4 mol) × (248.22 J K –1 mol –1 ) – (2 mol) × (82.81 J K –1 mol –1 ) – (7 mol) × (205.138 J K –1 mol –1 ) = – 599.5 J K –1 Reaction 2: Δ S° = (1 mol) × S°(Fe 3 O 4 ) + (3 mol) × S°(SO 2 ) – (3 mol) × S°(FeS) – (5 mol) × S°(O 2 ) Δ S° = (1 mol) × (146.4 J K –1 mol –1 l) + (3 mol) × (248.22 J K –1 mol –1 ) – (3 mol) × (82.81 K –1 mol –1 ) – (5 mol) × (205.138 J K –1 mol –1 ) = – 383.1 J K –1 Now use Δ H° and Δ S° to calculate Δ G° at 25°C. T = 25°C + 273.15 = 298 K Reaction 1: Δ G° = (–780.6 kJ) – (298 K)(–599.5 J K –1 ) " 1 kJ 1000 J = –601.9 kJ Reaction 2: Δ G° = (–1703.9 kJ) – (298 K)(–383.1 J K –1 ) " 1 kJ 1000 J = –1589.7 kJ The second reaction is more product-favored. (d) Use
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 31

chapter22 - 958 Chapter 22: Chemistry of Selected...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online