Ch 101 1st Edition Exam 5 Study Guide Chapters 10 11 Lecture 1 Chapter 10 Standard Enthalpies of Formation Define standard conditions Understand standard enthalpies of formation Calculate H rxn from standard enthalpies Pgs 370 374 Reminder H is the change in enthalpy for a chemical reaction Standard Conditions The standard state is the state of a material at a defined set of conditions o For a gas Pure gas at exactly 1 atm pressure o For a liquid or solid Pure solid or liquid in its most stable form at exactly 1 atm pressure and termperature of interest Usually 25 C o For a solution Substance in a solution with concentration of 1 M Standard Enthalpy Changes o Change in enthalpy for a process when all reactants and products are in their standard states Standard enthalpy of formation o For a pure compound the change in enthalpy when 1 mole of the compound forms from its constituent elements in their standard states o For a pure element in its standard state H formation 0 Formation Reactions Reactions of elements in their standard state to form 1 mole of a pure compound Elements in standard state have H formation 0 H formation kJ mol Writing Formation Reactions ex C O CO g ex 2Al 3 8 S 8 6 O 2 Al 2 SO 4 3 s Calculating Standard Enthalpy Change for a Reaction Any reaction can be written as the sum of formation reactions or the reverse of formation reactions for the reactants and products Important Note reactants elements H f sum of reactants elements products H f sum of products reactants products H rxn H 1 H 2 Ex Ca CO3 s CaO s O 2 g Ca CO3 1207 6 kJ mol O2 0 CaO 534 9 kJ mol Soo 1207 6 634 9 572 7 kJ mol Special notes when coeffecients are involved must multiply H f by the coefficient H rxn np H f products np H f reactants Lecture 2 Chapter 10 Hess s Law Utilize Hess law to calculate the H of a reaction Relationships Involving H rxn When reaction is multiplied by a factor H rxn is multiplied by that factor If a reaction is reversed then the sign of H is changed o For equation H rxn 393 5 kJ o For adding coefficient of 2 to equation H rxn 787 0 kJ o For equation in other direction H rxn 393 5 kJ If a reaction can be expressed a series of steps then the H rxn for the overall reaction is the sum of the heats of reaction for each step Hess Law the change in enthalpy for a stepwise process is the sum of the enthalpy changes of the steps o This is like previous lecture Lecture 3 Chapter 11 Into To Gases The molecular description of the behavior of gases The laws of Boyle Charles and Avogadro The structure of a Gas Gases are composed of particles that are flying around very fast in their container s Gases are constantly in motion and their collisions with the surface creates pressure o The pressure exerted by a gas can cause some amazing and startling effects o Ex Difference in air pressure results in weather and wind patters o The higher in the atmosphere you climb the lower the atmospheric pressure is around you o Difference in pressure across the eardrum the membrane is pushed outward causing a popped eardrum o Units of pressure Boyle s Experiment Pressure and volume are inversely related Found this by using a j tube and adding more mercury to the tube decreasing the volume of the air in the tube increased the air pressure Applied can be used as o P 1 V 1 P2 V 2 Charles s Law Volume is directly proportional to temperature o Constant pressure and amount of gas o Graph of volume vs temperature is a straight line Temperature is in kelvin Kelvin Celsius 273 V 1 T 1 V 2 T 2 Avogadro s Law Volume is directly proportional to the number of gas molecules assumes constant pressure and temperature V 1 n1 V 2 n2 Lecture 4 Chapter 11 Ideal Gas Law The ideal gas law Apply the ideal gas law to calculate molar mass If gas is heated it will EXPAND If gas is put under higher pressure it will CONTRACT 2 moles of a gas is how many times bigger than 0 5 moles of gas 4 Standard Conditions Because the volume of a gas varies with temperature and pressure chemists have agreed on a set of conditions to report measures so comparison is easy this is called standard conditions o STP is Pressure 1 atm Temperature 273 K 0 C Applying all Boyle s Charles s and Avogadro s law together forms ideal gas law o PV nRT o This is the ideal gas law o R is a gas constant 0 08206 atm L mol K o P pressure in atm V volume in L n mols T temperature in kelvin Molar Volume Solving the ideal gas equation for the volume of 1 mol of any gas at STP gives 22 4 Liters Density is the ratio of mass to volume and at standard conditions you divide the molar mass of an element by 22 4 the volume of the gas to find density D helium 4 00g mol 22 4 L mol 0 179 g L Molar Mass of a Gas One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it is a gas Measure the temperature pressure and volume and use the ideal gas law o n PV RT Be able to manipulate the ideal gas law to find different measurements Lecture 5 Chapter 11 Partial Pressures Calculate partial pressure of gas mixtures mole fraction Pgs 407 411 1 mole of O 2 in a 22 4 L container at 273K has a pressure of 1 atm 2 moles of O 2 in a 22 4 L container at 273K has a pressure of 2 atm 1 mole of N 2 in a 22 4 L container at 273K has a pressure of 1 atm Mixture of Gases When gases are mixed together their molecules behave independently of each other o In a container with a mixture of gases each gas occupies the same volume o Each gas has the same temperature Therefore in certain applications the mixture can be thought of as one gas Partial Pressure The pressure of a single gas in a mixture of gases is called its partial pressure o Ptotal P1 P2 P3 P4 o Only when all volume and temperature are the same The partial pressure of each gas in a mixture can be calculated using the ideal gas law o P nRT V Lecture 6 Chapter 11 Effusion Calculate the molecular weight of unknown gas with Graham s Law of Effusion Pgs 420 421 Diffusion and Effusion The process of a collection of molecules spreading from high concentration to low concentration is called diffusion The process by which a collection …