Chapter20ISM

Chapter20ISM - CHAPTER 20 METALLURGY AND THE CHEMISTRY OF...

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CHAPTER 20 METALLURGY AND THE CHEMISTRY OF METALS Problem Categories Biological : 20.68. Conceptual : 20.36, 20.45, 20.50, 20.61, 20.65, 20.66, 20.73, 20.75. Descriptiv e: 20.13, 20.14, 20.15, 20.17, 20.18, 20.27, 20.28, 20.29, 20.34, 20.35, 20.37, 20.38, 20.39, 20.43, 20.47, 20.48, 20.49, 20.51, 20.52, 20.55, 20.56, 20.59, 20.62, 20.63, 20.64, 20.67, 20.69, 20.70, 20.71, 20.72, 20.74. Industrial : 20.11, 20.16, 20.53, 20.60. Difficulty Level Easy : 20.11, 20.33, 20.36, 20.38, 20.40, 20.45, 20.48, 20.49, 20.55, 20.58, 20.59, 20.67, 20.72. Medium : 20.12, 20.13, 20.15, 20.17, 20.18, 20.27, 20.28, 20.29, 20.30, 20.34, 20.35, 20.39, 20.43, 20.44, 20.50, 20.52, 20.53, 20.56, 20.60, 20.61, 20.62, 20.63, 20.64, 20.68, 20.70, 20.71, 20.74, 20.75. Difficult : 20.14, 20.16, 20.37, 20.46, 20.47, 20.51, 20.54, 20.57, 20.65, 20.66, 20.69, 20.73, 20.76, 20.77. 20.11 For the given reaction we can calculate the standard free energy change from the standard free energies of formation. Then, we can calculate the equilibrium constant, K p , from the standard free energy change. f4f f [Ni(CO) ] [4 (CO) (Ni)] Δ°=Δ − Δ GG G G DD D Δ G ° = (1)( 587.4 kJ/mol) [(4)( 137.3 kJ/mol) + (1)(0)] = 38.2 kJ/mol = 3.82 × 10 4 J/mol Substitute Δ G ° , R , and T (in K) into the following equation to solve for K p . Δ G ° = RT ln K p 4 p (3 . 8 2 1 0J /m o l ) ln (8.314 J/K mol)(353 K) −Δ ° − − × == G K RT K p = 4.5 × 10 5 20.12 The cathode reaction is: Cu 2 + ( aq ) + 2e ⎯⎯→ Cu( s ) First, let’s calculate the number of moles of electrons needed to reduce 5.0 kg of Cu. 2 1000 g 1 mol Cu 2 mol e 5.00 kg Cu 1.57 10 mol e 1 kg 63.55 g Cu 1 mol Cu ×× × = × Next, let’s determine how long it will take for 1.57 × 10 2 moles of electrons to flow through the cell when the current is 37.8 C/s. 2 96,500 C 1 s 1 h (1.57 10 mol e ) 37.8 C 3600 s 1mole × × = 111 h
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CHAPTER 20: METALLURGY AND THE CHEMISTRY OF METALS 611 20.13 Table 19.1 of the text shows that Pb, Fe, Co, Zn are more easily oxidized (stronger reducing agents) than copper. The Ag, Au, and Pt are harder to oxidize and will not dissolve. Would you throw away the sludge if you were in charge of the copper refining plant? Why is it still profitable to manufacture copper even though the market price is very low? 20.14 The sulfide ore is first roasted in air: 2ZnS( s ) + 3O 2 ( g ) ⎯⎯→ 2ZnO( s ) + 2SO 2 ( g ) The zinc oxide is then mixed with coke and limestone in a blast furnace where the following reductions occur: ZnO( s ) + C( s ) Zn( g ) + CO( g ) ZnO( s ) + CO( g ) Zn( g ) + CO 2 ( g ) The zinc vapor formed distills from the furnace into an appropriate receiver. 20.15 The trick in this process centers on the fact that TiCl 4 is a liquid with a boiling point (136.4 ° C), a little higher than that of water. The tetrachloride can be formed by treating the oxide (rutile) with chlorine gas at high temperature. The balanced equation is: TiO 2 ( s ) + 2Cl 2 ( g ) TiCl 4 ( l ) + O 2 ( g ) The liquid tetrachloride can be isolated and purified by simple distillation. Purified TiCl 4 is then reduced with magnesium (a stronger reducing agent that Ti) at high temperature. TiCl 4 ( g ) + 2Mg( l ) Ti( s ) + 2MgCl 2 ( l ) The other product, MgCl 2 , can be separated easily from titanium metal by dissolving in water. 20.16 (a) We first find the mass of ore containing 2.0 × 10 8 kg of copper. 81 0 100% ore (2.0 10 kg Cu) 2.5 10 kg ore 0.80% Cu ××= × We can then compute the volume from the density of the ore.
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Chapter20ISM - CHAPTER 20 METALLURGY AND THE CHEMISTRY OF...

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