Prof Gilbert LECTURE 13 CHEM 1211 Fall 10 Announcements Reminders Mid Term Thursday bring a 2 pencil and your calculator Note on your answer sheet the letter A B C or D preceding the page numbers on your exam Last time Chapter 4 Chemical Reactions in Solution Titrations Ion Exchange CHAPTER 5 Thermochemistry definitions o Kinetic Energy o Thermodynamic systems This time o Potential Energy chemical energy electrostatic energy are forms of potential energy Units of energy one calorie cal is the quantity of heat required to raise the temperature of 1 g of water from 14 5 to 15 5 C The SI unit of energy is the joule J 1 cal 4 184 J One dietary Calorie Cal with a capital C kilocalorie kcal 1000 cal Internal Energy E of a System the sum of kinetic and potential energies of the substances in the system Hard to quantify Easier to measure the change in internal energy Esys There must be a balancing change in the energy of the system s surroundings so that This is a way of expressing the First Law of Thermodynamics energy is not created or destroyed only changes from one form to another Calculating Esys two ways to increase the internal energy of a system Esurr Esys 1 2 Add heat q to it Do work w on it Doing work on a gas could mean compressing it into a smaller volume Esys q w Esys q P V So why is there a sign on the P V term Inquiry This balloon the first to fly around the world nonstop contained 550 000 cubic feet of He How much work was done to inflate it at 1 00 atm of pressure Given 1 L atm 101 34 J Change in Enthalpy Hsys heat flow into or out of the system at constant pressure qp Most reactions including the biochemical reactions in the body occur at constant P For these reactions Hsys Hreact Hreact 0 the reaction is exothermic Hreact 0 the reaction is endothermic H of phase changes such as Hfus and Hvap have values The reverse processes liquid freezing or vapor condensing have H values 0 Heating Curve of water Heat flow during phase changes The slope of line CD depends on the specific heat cs of water 4 185 J g C which is related to the molar heat capacity cp of water 75 5 J mol C The length of line BC is proportional to the value of the molar heat of fusion Hfus and the length of line DE is proportional to the value of the molar heat of vaporization Hvap These are intensive values characteristic of a particular substance The heat capacity Cp of an object J C depends on how much there is of it and what it is made of An extensive property The heat required to warm an object is The heat to warm a mass of a substance with a known specific heat is The heat to warm a quantity of a substance with a known molar heat capacity is Inquiry Consider the following table of molar heat capacities and other properties of several metals Metal Atomic Number Molar Mass g mol Density g mL Molar Heat Capacity J mol C q Cp T q m cs T q n cp T 0 534 2 70 8 94 7 87 5 73 19 2 19 3 24 8 24 4 24 5 25 1 24 6 24 8 25 4 Li Al Cu Fe As W Au 3 13 29 26 33 74 79 6 94 26 98 63 55 55 84 74 92 183 84 196 97 How do cp values of these metals relate to their molar masses densities and other properties Calorimetry measurement of heat flow generally used to determine the heat flow associated with a physical change or chemical reaction Calorimeter apparatus that measures heat flow This one is a bomb calorimeter used to determine heats of reaction such as combustion The heat from the reaction warms the water and insulated container producing a T If we also know CP of the calorimeter then If the reaction happens at nearly constant pressure then qP Hrxn 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 H0 calculate H0 of CH4 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