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Published byKory Walters Modified over 8 years ago
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Chapter 6 Energy Law of Conservation of Energy Potential vs Kinetic Heat – transfer of Energy because of temp. difference
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State function System vs Surroundings Exothermic Endothermic
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Thermodynamics 1 st law – Energy of Universe is constant Internal energy E = q + w w = -p V
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Enthalpy H = q p H = H prod – H react Calorimetry Specific heat q= sm T Specific heat of water is 4.18 J/g 0 C
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What is the specific heat of 15 g of an unkown at 45 0 C if when mixed with 55 g of water at 25 0 C the final temperature is 32 0 C? - q hot = q cold - sm T unk. = sm T H 2 O - s(15g)(32-45 0 C) = (4.18J/g 0 C)(55g)(32- 25 0 C) s = - (4.18J/g 0 C)(55g)(7 0 C)/((15g)(-13 0 C)) s = 8 J/g 0 C
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Constant volume E = q v Bomb Calorimeter –q = C T C = heat capacity of the calorimeter
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Hess’s Law H total = H 1 + H 2 + …. If reaction is reversed change the sign If you multiply the reaction’s coefficients you also multiply the H N 2 + O 2 2NO H 1 = 180 kJ 2NO + O 2 2NO 2 H 2 = -112 kJ ----------------------------------------------- N 2 + 2O 2 2NO 2 H = H 1 + H 2 = 68 kJ
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C graph. + O 2 CO 2 H = -394 kJ C diam. + O 2 CO 2 H = -396 kJ C graph. + O 2 CO 2 H = -394 kJ CO 2 O 2 + C diam. H = +396 kJ --------------------------------------------- C graph. C diam. H = 2 kJ
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Standard Enthalpies of Formation – H f o –Values for formation of 1 mole at standard conditions –Reactants are all elements –Elements have a value of zero –From Appendix 4 on pg. A21 H rxn = n prod H f o prod - n react H f o react
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CH 4(g) + 2O 2(g) CO 2(g) + 2H 2 O (l) H = ? H = 1n(-394kJ/n)+2n(-286kJ/n) – 1n(-75kJ/n) – 0 H = -891 kJ
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