Prof Gilbert LECTURE 14 CHEM 1211 10 13 10 Announcements Reminders Mid Term Exam Average was 74 75 If you scored less than a 50 you should plan to see me in Chem Central today at noon or next Tuesday during my office hours Last time Chapter 5 Thermochemistry Units of energy Forms of internal energy in molecules Calorimeters as a class of thermodynamic systems First Law of Thermodynamics energy is not created or destroyed so Esys Esurr q w Change in Enthalpy Hsys heat flow into or out of the system at constant pressure qp Hreact 0 the reaction is exothermic Hreact 0 the reaction is endothermic Heating Curve Temp change vs heat flow during and between phase changes depends on values of Hfus Hvap and the molar heat capacity cp of substance in the solid liquid and vapor phases The heat capacity Cp of an object J C depends on how much there is of it and what it is made of The heat required to warm an object is q Cp T The heat needed to warm a mass m of a substance with a known specific heat cs is q m cs T The heat needed to warm a of moles n of a substance with a known cp is q n cp T Calorimetry measurement of heat flow generally used to determine the heat flow associated with a physical change or chemical reaction This time H0 calculate H0 of CH4 Calculating the standard heat enthalpy of a reaction H0 H0 f of the products and reactants rxn from standard enthalpies of formation f is the energy released or absorbed when a substance is formed from the free elements that make it up To f values of the products the reactants Consider the combustion rxn take the difference in the H0 CH4 g 2O2 g CO2 g 2 H2O g Substance CH4 g 2O2 g CO2 g 2 H2O g Hf 0 kJ mol 74 81 0 00 393 51 241 82 mol 1 2 1 2 74 81 kJ 0 00 kJ 393 51 kJ 483 64 kJ KEY Equation H0 rxn S H0 f products S H0 reactants f Applying it to this problem H0 rxn 393 51 483 64 kJ 74 81 0 00 kJ 802 34 kJ Inquiry is the value of H0 also 802 34 kJ rxn of the following reaction CH4 g 2O2 g CO2 g 2 H2O l NOTE Calculations involving the above equation are in effect tracking the energy changes that would have occurred if the reactants had been broken up into their individual elements and then those elements recombined into products Consider the enthalpy change associated with the combustion of propane C3H8 C3H8 g 5O2 g 3CO2 g 4H2O g Fuel Food Values heats of combustion expressed in kJ g instead of kJ mol H2 has the highest fuel value of all common fuels but alas a gram of it takes up a lot of volume so its fuel density kJ mL is much less than its fuel value even when compressed to 350 atm Transport and storage is a challenge Chapter 6 Gases Barometric Pressure is the force F exerted by the weight of the atmosphere over the surface area A of the Earth That is P Pa F N A m2 Where Pa is pascals the SI unit of pressure and N is newtons the SI unit of force It s more common to use kPa kilopascals where an atmosphere of pressure atm 1 atm 101 325 kPa 1013 6 mbar 760 mmHg That 760 mm value is the height of the column of Hg d 13 595 g cm3 at 0 C in this Hg pool barometer For any other liquid the height of the column above the pool is inversely proportional to the density of the liquid Why Working barometers don t rely on liquid mercury A popular design for making a recording barometer or barograph is shown on the left Variations in barometric pressure 1 meteorological Here for example is a satellite view of Hurricane Katrina as it intensified to a Category 5 hurricane with maximum sustained winds of 175 mph 280 km h and a minimum central pressure of 902 mbar Generally the lower the central pressure of a hurricane the higher the wind speeds Why 2 Pressure decreases with increasing altitude Why The Gas Laws Based on personal experience you know that The pressure of a volume of gas is proportional to the amount of gas in the volume as in inflating a bicycle tire P n The pressure of a volume of gas increases with increasing temperature P T The volume of a quantity of gas is inversely proportional to the applied pressure V 1 P 1 2 3 Combining these observations we have or PV nT Calculations Turning this expression into an equation by installing a constant of proportionality R we have the ideal gas law PV nRT Keep in mind that this equation works only if temperature values are on an absolute temperature e g Kelvin scale Also the value of R called the gas constant depends on the units Most of the time we use 0 0821 L atm mol K 1 2 Use PV nRT when 3 of the 4 variables are known and you seek the 4th Often the key variable is the number of moles n of a gas produced or consumed in a chemical reaction You may first need to convert a mass of gas into an equivalent number of moles by dividing by its molar mass Inquiry 1 The volume of helium in this balloon which was flown around the world by the late Steve Fossett in 2002 was about 1 56 107 liters at 20 C and 1 00 atm of pressure What mass of helium was needed to fill the balloon 3 When n is a constant then PV T is a constant and the following general gas equation is often used Determining molar mass M The number of moles n of a substance is equal to its mass m divided by its molar mass M Substituting this into fraction into the ideal gas law we have or M Inquiry 2 A balloon is inflated with 4 62 g of a gas to a volume of 2 46 L at 27 C and 1 05 atm What is the molar mass of the gas Also how would you determine the mass of gas in a balloon Gas density and molar mass A mole of any ideal gas occupies the same volume at a given temperature and pressure But the mass m of that volume and therefore the density of the gas is proportional to the molar mass M M or d M Inquiry 3 What is the density of propane C3H8 at STP Would propane leaking from a tank stored in the basement of a house tend to collect in the basement or rise into the floors above Inquiry 4 The tanks used in barbecue grills such as the one shown here contain about 20 pounds of propane If this much propane leaked out what volume would it occupy …