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Chapter 6 Thermochemistry

Chapter 6 Thermochemistry - Chapter 6 Thermochemistry...

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Chapter 6 Thermochemistry : Energy Flow and Chemical Change 6.1 E = q + w The sign of the energy transfer is defined from the perspective of the system. 6.2 No. An increase in temperature means that heat has been transferred to the surroundings, which makes q positive. 6.3 E = q + w = w , since q = 0. Thus, the change in work equals the change in internal energy. 6.4 Increase: eating food, lying in the sun, taking a hot bath. Decrease: exercising, taking a cold bath, going outside on a cold day. 6.5 a) electric heater b) sound amplifier c) light bulb d) wind turbine e) battery (voltaic) 6.6 E universe = E system + E surroundings = 0 E system = -∆ E surroundings The E for the heater and air conditioner are the same magnitude. either heater or room air conditioner 6.7 heat energy; sound energy (impact) kinetic energy (falling text) potential energy (raised text) mechanical energy (raising of text) chemical energy (biological process to move muscles) 6.8 E = q + w = 425 J + ( - 425 J) = 0 J 6.9 E = q + w = - 255 cal + ( - 428 cal) = - 683 cal 6.10 E = q + w = - 675 J + J 10 x 52 1. = cal 1 J 4.184 x cal 25 5 3 6.11 E = q + w = cal J 4.184 x kcal cal 10 x kcal 47 2 0. + kJ 1 J 10 x kJ 15 6 0. 3 3 = 615 J + 1.03 x 10 3 J = 1.65 x 10 3 J 112
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6.12 C( s ) + O 2 ( g ) CO 2 ( g ) + 3.3 x 10 10 J (1.0 ton) a) E (kJ) = 3.3 x 10 10 J x J 10 kJ 1 3 = 3.3 x 10 7 kJ b) E (kcal) = 3.3 x 10 10 J x cal 10 kcal 1 x J 4.184 cal 1 3 = 7.9 x 10 6 kcal c) E (Btu) = 3.3 x 10 10 J x J 1055 Btu 1 = 3.1 x 10 7 Btu 6.13 CaCO 3 ( s ) + 9.0 x 10 6 kJ CaO( s ) + CO 2 ( g ) (5.0 ton) a) E (J) = 9.0 x 10 6 kJ x kJ 1 J 10 3 = 9.0 x 10 9 J b) E (cal) = 9.0 x 10 6 kJ x J 4.184 cal 1 x kJ 1 J 10 3 = 2.2 x 10 9 cal c) E (Btu) = 9.0 x 10 6 kJ x J 1055 Btu 1 x kJ 1 J 10 3 = 8.5 x 10 6 Btu 6.14 E (J) = 4.1 x 10 3 Calorie x cal 1 J 4.184 x Calorie cal 10 3 = 1.7 x 10 7 J E (kJ) = 1.7 x 10 7 J x J 10 kJ 1 3 = 1.7 x 10 4 kJ 6.15 t (h) = 4.1 x 10 3 Cal x 4.184 J 1 hr x 9.3hr 1 cal 1850kJ = 6.16 The system does work and thus its internal energy is decreased. 6.17 E = q v Measuring the heat transfer at constant pressure is more convenient than measuring the heat transfer at constant volume. H = q p 6.18 E < 0; exothermic 6.19 a) exothermic b) endothermic c) exothermic d) exothermic e) endothermic f) endothermic g) exothermic 6.20 The internal energy of a substance is the sum of kinetic (E k ) and potential (E P ) terms. k(total) k(translational) k(rotational) k(vibrational) = + + E E E E 113
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P(total) P(atom) P(bond) = + E E E E P(atom) has nuclear, electronic, positional, magnetic, electrical, etc. components. 6.21 H = E + P V (const P) a) H < E , P V is negative b) H = E , a fixed volume means P V = 0. c) H > E , P V is positive for the transformation of solid to gas. 6.22 Reactants H H = ( - ) Products 6.23 Reactants H H = (+) Products 6.24 a) CH 4 ( g ) + 2O 2 ( g ) CO 2 ( g ) + 2H 2 O( g ) + heat CH 4 + 2O 2 H H = ( - ) CO 2 + 2H 2 O b) H 2 O( l ) H 2 O( s ) + heat H 2 O( l ) H H = ( - ) H 2 O( s ) 6.25 a) Na( s ) + ½ Cl 2 ( g ) NaCl( s ) + heat Na + ½ Cl 2 H H = ( - ) NaCl b) C 6 H 6 ( l ) + heat C 6 H 6 ( g ) C 6 H 6 ( g ) H H = (+) C 6 H 6 ( l ) 114
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6.26 a) C 2 H 6 O( l ) + 3O 2 ( g ) 2CO 2 ( g ) + 3H 2 O( g ) + heat C 2 H 6 O + 3O 2 H H = (-) 2CO 2 + 3H 2 O b) 1 / 2   N 2 ( g ) + O 2 ( g ) + heat NO 2 ( g ) NO 2 H H = (+) ½N 2 + O 2 6.27 a)
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