Chapter 6 Energy Law of Conservation of Energy Potential vs Kinetic Heat – transfer of Energy because of temp. difference

State function System vs Surroundings Exothermic Endothermic

Thermodynamics 1 st law – Energy of Universe is constant Internal energy  E = q + w w = -p  V

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

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)( C) = (4.18J/g 0 C)(55g)( C) s = - (4.18J/g 0 C)(55g)(7 0 C)/((15g)(-13 0 C)) s = 8 J/g 0 C

Constant volume  E = q v Bomb Calorimeter –q = C  T C = heat capacity of the calorimeter

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

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

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

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