chapter18_part1

# chapter18_part1 - 822 Chapter 18: Thermodynamics:...

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Chapter 18: Thermodynamics: Directionality of Chemical Reactions 822 End-of-Chapter Solutions for Chapter 18 Summary Problem Answer: (a) 583.3 kJ, 532.54 kJ, –321.050 kJ, –565.968 kJ; steps 3 and 4 are exothermic, steps 1 and 2 are endothermic (b) Steps 1, 2, and 3 increase. Step 4 decrease. (c) Steps 1 and 2 are reactant-favored at room temperature, Steps 3 and 4 are product-favored at room temperature; Steps 1,2, and 3 are product-favored at high temperatures; Step 4 is reactant-favored at high temperatures. (d) Entropy changes: 261.9 J/K, 144.9 J/K, 178.73 J/K, – 173.01 J/K; Gibbs Energy change: 1239.28 kJ, 490.24 kJ, – 274.336 kJ, – 514.382 kJ (e) – Δ S°, more negative Δ G°: Steps 1 and 2; more positive Δ G°: Steps 3 and 4; yes (f) Reactions 1 and 2; phase change seen in FeO at about –1370°C (g) 190. kJ/mol, 315 kJ/mol, – 489.1 kJ/mol, – 306.5 kJ/mol (h) Fe 2 O 3 , C(s), Steps 1 and 2 (i) yes (j) See reactions below, 467.8 kJ, 560.3 J/K, at 25°C: 1036.7 kJ, at 1500 K: – 3073.4 kJ (k) 3 × 10 4 kJ destroyed, 7 × 10 3 kJ stored, 80% wasted Strategy and Explanation: (a) To find Δ H°, use Equation 6.11: Δ H° = (moles of product) " # H f o (product) \$ % ( ) * + (moles of reactant) " # H f o (reactant) \$ % ( ) * Step 1: Δ H° = (4 mol) × Δ H f o (FeO) + (1 mol) × Δ H f o (O 2 ) – (2 mol) × Δ H f o (Fe 2 O 3 ) Δ H° = (4 mol) × (– 266.27 kJ/mol) + (1 mol) × (0 kJ/mol) – (2 mol) × (– 824.2 kJ/mol) = 583.3 kJ Step 2: Δ H° = (2 mol) × Δ H f o (Fe) + (1 mol) × Δ H f o (O 2 ) – (2 mol) × Δ H f o (FeO) Δ H° = (2 mol) × (0 kJ/mol) + (1 mol) × (0 kJ/mol) – (2 mol) × (– 266.27 kJ/mol) = 532.54 kJ Step 3: Δ H° = (2 mol) × Δ H f o (CO) – (2 mol) × Δ H f o (C) – (2 mol) × Δ H f o (O 2 ) Δ H° = (2 mol) × (–110.525 kJ/mol) – (2 mol) × (0 kJ/mol) – (1 mol) × (0 kJ/mol) = – 221.050 kJ Step 4: Δ H° = (2 mol) × Δ H f o (CO 2 ) – (2 mol) × Δ H f o (CO) – (1 mol) × Δ H f o (O 2 ) Δ H° = (2 mol) × (– 393.509 kJ/mol) – (2 mol) × (–110.525 kJ/mol) – (1 mol) × (0 kJ/mol) = – 565.968 kJ Steps 3 and 4 are exothermic. Steps 1 and 2 are endothermic. (b) As described in Section 18.3, page 877, predicting entropy changes involves examining the physical state (solid, liquid, gas) of the reactants and products. Solids have much lower entropy than gases. A reaction that has more gaseous products than gaseous reactants will be predicted to have an increase in entropy. A reaction that has fewer gaseous products than gaseous reactants will be predicted to have a decrease in entropy. Step 1: no gas reactants one gas product Entropy increases. Step 2: no gas reactants

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## This note was uploaded on 07/15/2009 for the course CHE 131 taught by Professor Kerber during the Fall '08 term at SUNY Stony Brook.

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chapter18_part1 - 822 Chapter 18: Thermodynamics:...

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