Chem 1211 1st Edition Lecture 14Outline of Last Lecture I. TitrationOutline of Current Lecture I. Specific HeatII. State Functions and Changes of StateIII. Thermochemical EquationsCurrent LectureI. Specific Heata. ConcepTest- Calculate the amount of heat required to raise the temperature of 100 g of water from 0 degrees C to 100 degrees Celcius.a. ANSWER- 49 kJ (multiply mass x specific heat x change in temperature and you get the energy in joules. The specific heat of water is 4.184)b. ConcepTest: At room temperature, a typical dorm room contains 30 kg of air and 300 g of water vapor (at 50% relative humidity). How much energy is required to warm the room 10 degrees Celsius if the specific heat for water vapor is 1.86 J/gCand for air is 1.0 J/gC?a. ANSWER- 305 kjb. REASONING: Heat (air) = 30 kg x (1000g/1kg) x (1.0 J/gC) x 10 degrees C= 300,000 J1. Heat (H2O) = 300g x (1.86 J/gC) x 10 degrees C= 5,580 J2. Answer= 305,580 J = 305 kJII. State Functions and Changes of Statea. State functions: set of conditions and properties that only depend on the state of the system1. Do not depend on the pathway by which systems arrived at the state2. Indicated by capital letters3. Includes temperatures likeThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. T (temperature)b. P (pressure)c. V (volume)d. E (energy)e. H(enthalpy) heatf. S (enthropy) disorder4. Often interested in changes in state functions, of change in x = xfinal– xinitial5. Heat vs. Temperature Plota. Temperature on x axis (degrees Celcius), energy on y axis (in kJ)b. Warm icec. Melt iced. Heat liquid watere. Evaporate waterf. Heat steamb. Enthalpy Changes and Reactions1. Reactions typically generate heat2. Amounts of heat are exact and related to potential energy of reactants and products3. Exothermic reactions have products of lower potential energy than reactants 4. The slope on the plot defines how much energy is in the moleculec. Exothermic: a process where heat is evolved1. On a graph, you end at a lower potential energyd. The barrier to the reaction happing spontaneously is the high point of thegraph1. Heat needed to make the reaction happen is independent from the energy given ofIII. Thermochemical Equationsa. Balanced chemical reaction plus the change in Ha. Example: C5H12 + 8 O2 5 CO2 + 6 H2O + 3523 J1. New idea: there is an amount of energy associated with each reactionb. Heat released during reaction (3523 kJ) assuming stoichiometric units are moles1. 1 mol of C5H12 reacts with 8 mole of 02 to produce 5 mol of CO2, 6 mole of H20b. Change in H < 0 (negative number) designates an exothermic reactionc. Change in H > 0 (positive number) designates an endothermic reactiond. Hess’s Law: the enthalpy change for a reaction is the same whether it occurs by one step of by any (hypothetical ) series of stepsa. Hess’s Law is true because H is a state functionb. The change in HO values for reactions (2) and (3) can determine the change in H0 for any reaction (1)c. Example:1. 4 FeO + O2 2Fe2O3 change in H= ??2. 2 FeO 2 Fe + O2 change = +544 kJ3. 4Fe +3O2 2 Fe2O3 change= -1648kJa. Manipulate the equations to find the answer you are interested in d. Hess’s Law- you can add equations…1. A+B C delta H= 100kg2. 2A+2B 2C delta H=200kga. C A+B delta H…e. Hess’s Law using Delta Hf1. For chemical reaction at standard conditions, standard enthalpy change is sum of standard molar enthalpies of formation of products (each multiplied by its coefficient in balanced chemical equation) minus corresponding sum for reactants2. If you know the heat of formationa. Heat of formation= delta Hf is making a compound from anelementb. Add up the heats of the starting materials (reactants)c. Add up the heats of productsd. Subtract the reactants from the producte. That gives you the heat of reaction f. Standard enthalpies of formation, delta Hf1. Standard molar enthalpy of formation of elements in their most stable forms at 298.15K and 1.000 atm are 0 a. Delta Hf (elements) = 02. Others are
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