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Unformatted text preview: Topics for the day Administrative stuff Constant pressure calorimetry Work Enthalpy Where were we? H is a state function, so it doesnt matter what path you take to get there, the value of H is the same. This is the basis of Hesss law . We used three different ways to calculate H rxn , the Frst of which is a general use of Hesss law, and the other two are speciFc applications of the law Addition of a set of reactions following a different pathway from reactants to products Difference between the sum of the enthalpies of fusion H f for the products and the sum of H f for the reactants Difference between the sum of bond energies of the reactants and the sum of bond energies of the products (approximation, for gas phase only) Calorimetry is an experimental way to measure H. A. Yes B. No C. Is this a trick question? iClicker Time The Frst law of thermodynamics says that the energy of the universe is a constant. So do we really have an world energy crisis? The crisis is not in the Quantity of energy available, its in the Quality . e.g., fuel is made up of chemicals with a lot of potential energy stored up in their bonds. Once its burned, that energy has been converted into thermal energy transferred to the surroundings and mechanical work done pushing pistons, turbines, etc. When we use energy, the amount of energy is unchanged, but its new form is less useful than its original form. 1 2 3 Given these reactions: C(graphite) + O 2 (g) CO 2 (g) H = 393.5 kJ mol 1 H 2 (g) + 1 / 2 O 2 (g) H 2 O(l) H = 285.8 kJ mol 1 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2 H 2 O(l) H = 2598.8 kJ mol 1 calculate the standard heat of formation for acetylene C 2 H 2 (g) Calculation time 1. Determine balanced target reaction. Acetylene comes from C and H. Standard states of these elements are C(graphite) and H 2 (g): 2C(graphite) + H 2 (g) C 2 H 2 (g) 2. Find each of the products and reactants from the target reaction in the supplied reactions and multiply each of them appropriately. First reactant is C(graphite). Two of them on the reactant side of the target reaction. One of them on the reactant side of eqn (1), so multiply eqn (1) by 2: 2C(graphite) + 2O 2 (g) 2CO 2 (g) H = 787 kJ mol 1 (4) (2) (3) (1) First product is C 2 H 2 (g). One of them on the product side of the target reaction. Two of them on the reactant side of eqn (3), so multiply eqn (3) by 1 / 2 : 2CO 2 (g) + H 2 O(l) C 2 H 2 (g) + 5 / 2 O 2 (g) H = +1299.4 kJ mol 1 (5) Notice how I skipped the second reactant (H 2...
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This note was uploaded on 08/29/2010 for the course CH 52375 taught by Professor Ruth during the Spring '09 term at University of Texas at Austin.
 Spring '09
 RUTH

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