October 17 2016 Chapter 6 HW due Sunday 10 23 SLW meets this week Lab meets this week Enthalpy H is thermodynamic quantity equivalent to the total heat content of a system State function H E P V Pressure change in volume Change in Internal energy 6 4 Stoichiometry of Thermochemical Equations A thermochemical equation is a balanced equation that includes the enthalpy change of the reaction or heat of reaction Hrxn 1 Sign The sign of Hrxn depends on whether the reaction is exothermic or endothermic The forward reaction has the opposite sign of the reverse reaction Example Decomposition of 2 mol H2O to its elements 2H2O l 2H2 g O2 g Hrxn 572 kJ Formation of 2 mol H2O from its elements 2H2 g O2 g 2H2O l Hrxn 572 kJ 2 Magnitude The magnitude of Hrxn is proportional to the amount of substance reacting Example Formation of 1 mol H2O from its elements H2 g O2 g H2O l Hrxn 572 kJ 286 kJ Sample Problem 6 7 The major source of aluminum in the world is bauxite mostly aluminum oxide Its thermal decomposition can be represented by Al2O3 s 2Al s 3 O2 g 2 Hrxn 1676 kJ If aluminum is produced this way how many grams of aluminum can form when 1 000 x 103 kJ of heat is transferred 1 From the balanced equation and H we see that 2 mol of Al is formed when 1676 kJ of heat is absorbed 2 mol Al 1676 kJ or 1676 kJ 2 mol Al 2 Convert heat transferred 1 000 x 103 kJ into moles aluminum produced and then to mass aluminum Mass of Al 1 000x103 kJ x 2 mol Al 1676 kJ x 26 98 g Al 1 mol Al 32 20 g Al Liquid hydrogen peroxide releases oxygen gas upon decomposition 2H2O2 l 2H2O l O2 g Hrxn 196 1 kJ How much heat is transferred when 21 519 moles of H2O2 decomposes A 2110 kJ is released by the reaction B 4220 kJ is released by the reaction C 2110 kJ is absorbed by the reaction D 4220 kJ is absorbed by the reaction E 8440 kJ is released by the reaction 21 519 mol H2O2 x 196 1 kJ 2109 9 kJ 2 mol H2O2 6 5 Hess s Law of Heat Summation One of the most powerful applications of the state function property of enthalpy H allows us to find H of any reaction for which we can write the equation This application is based on Hess s Law of Heat Summation the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps Hoverall H1 H2 Hn To apply Hess s Law 1 Imagine the overall reaction as the sum of individual reactions which have known H 2 Add together all the known H values for the steps to get the unknown H for the overall reaction May need to manipulate reverse equation change sign of H multiply equation multiply H equations to generate target equation Sample Problem 6 8 Two gaseous pollutants that form in auto exhaust are CO and NO An environmental chemist is studying ways to convert them to less harmful gases through the following equation CO g NO g CO2 g N2 g Hrxn Target equation Given the following information calculate the unknown H Equation A CO g O2 g CO2 g Hrxn 283 0 kJ Equation B N2 g O2 g 2NO g CO g O2 g CO2 g 2NO g N2 g O2 g Hrxn 180 6 kJ Hrxn 283 0 kJ Hrxn 180 6 kJ 90 30 kJ CO g O2 g NO g CO2 g N2 g O2 g Reversed and halved CO g NO g CO2 g N2 g Hrxn 283 0 kJ 90 30 kJ 373 3 kJ Use the following standard heats of formation N2 g O2 g NO g NO g Cl2 g NOCl g Hf 90 3 kJ Hf 38 6 kJ to calculate Hrxn for the following reaction 2NOCl g N2 g O2 g Cl2 g Hrxn A B C D E 257 8 kJ 103 4 kJ 257 8 kJ 103 4 kJ Insufficient information is given to calculate the H rxn 2x NOCl g Cl2 g NO g H f 2x 38 6 kJ 77 2 kJ 2x NO g N2 g O2 g H f 2x 90 3 kJ 180 6 kJ 2NOCl g 2NO g Cl2 g 2NO g N2 g O2 g 2NOCl g Cl2 g N2 g O2 g Hrxn 77 2 kJ 180 6 kJ 103 4 kJ