chapter6_part1

# chapter6_part1 - Chapter 6 Energy and Chemical Reactions...

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Chapter 6: Energy and Chemical Reactions 235 End-of-Chapter Solutions for Chapter 6 Summary Problem Answer: (a) 2.08 × 10 13 J (b) 41.5% (c) 8.00 × 10 7 g MgO, 2.39 × 10 8 g MgSO 4 (d) Adds 386.4 kJ to the heat value of coal (e) first reaction: Δ H° = –296.830 kJ, second reaction: Δ H° = –98.89 kJ Strategy and Explanation: (a) Assuming the fuel value of coal is approximately the same as for graphite, determine the quantity of energy transferred to the surroundings by the combustion of a given mass of coal. Fuel value is identified by the enthalpy of combustion: C(graphite) + O 2 (g) CO 2 (g) Δ H° = –393.5 kJ Convert the mass to grams using the given conversion factor, use the molar mass of C to determine moles of C, then use the fuel value to determine the energy. 7000. tons C " 9.08 " 10 5 g 1 ton " 1 mol C 12.0107 g C " 393.5 kJ 1 mol C " 1000 J 1 kJ = 2.08 " 10 14 kJ (b) Calculate the percent efficiency, given the actual energy evolved and the theoretical energy output calculated in (a). 8.64 " 10 13 kJ 2.08 " 10 14 kJ " 100% = 41.5% (c) Given a fixed mass of a reactant, calculate the mass of another reactant required and the mass of a product produced. Convert tons to grams, then use the molar mass of the reactant to find the moles of reactant. 140. tons SO 2 " 9.08 " 10 5 g 1 ton " 1 mol SO 2 64.064 g SO 2 = 1.98 " 10 6 mol SO 2 Use the stoichiometry of the equation to determine the moles of the moles of the other reactant required and the moles of the reactant produced. Then use the molar masses to find grams. The balanced equation says: 1 mol SO 2 reacts with 1 mol MgO and produces 1 mol MgSO 4 . 1.98 " 10 6 mol SO 2 " 1 mol MgO 1 mol SO 2 " 40.3044 g MgO 1 mol MgO = 8.00 " 10 7 g MgO 1.98 " 10 6 mol SO 2 " 1 mol MgSO 4 1 mol SO 2 " 120.368 g MgSO 4 1 mol MgSO 4 = 2.39 " 10 8 g MgSO 4 (d) Given a balanced equation, calculate the heat energy transfer (i.e., the enthalpy of combustion) and indicate whether it adds or takes away from the heat energy transfer of coal combustion. Using molar enthalpies of formation is a common application of Hess’s Law. We will form all the products (using their molar enthalpies as written) and we will destroy – the opposite of formation – the reactants (using their molar enthalpies with opposite sign). That process is the basis of Equation 6.11. Δ H° = (moles of product) " # H f o (product) \$ % & ( ) * + (moles of reactant) " # H f o (reactant) \$ % & ( ) * Use the stoichiometric coefficient of the balanced equation to describe the moles of each of the reactants and products. Set up a specific version of Equation 6.11. Look up the Δ H f o for each; remember to check the physical phase and remember that Δ H f o for standard state elements is exactly zero. Plug them into the equation and solve for Δ H°. The product is MgSO 4 . The reactants are MgO, SO 2 and O 2 . O 2 is the elemental form for oxygen.

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Chapter 6: Energy and Chemical Reactions 236 Δ H° = (1 mol) × Δ H f o (MgSO 4 ) – [(1 mol) × Δ H f o (MgO) + (1 mol) × Δ H f o (SO 2 ) + ( 1 2 mol) × Δ H f o (O 2 )] Look up the Δ H f o values in Table 6.2.
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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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chapter6_part1 - Chapter 6 Energy and Chemical Reactions...

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