CHEM 102 1nd Edition Lecture 5 Outline of Previous Lecture I. Concentration and Freezing PointII. Boiling Point and Freezing PointIII. Osmotic PressureOutline of Current LectureI. Heat Units and EquationsII. Calculating ΔHCurrent LectureI. Heat Units and Equations- Joule (J) = kg m2s2 = 107 ergs- Calorie: the amount of energy needed to raise 1 gram of water by 1°Co 1 calorie = 4.184 Jo Calorie = I kcal = 1,000 calories Food calorie, if used in this course, it will be called a Food Calorie How many Joules are 250 Food Calories?- 250 FC = 1,000 cal1 FC×4.184 J1 cal=1.046 ×106J- Q = mcΔTo Q is the amount of heato M is the masso C is the specific heat of the elemento ΔT is the change in temperature- ΔH = Hf - Hi, usually given as just ΔHo A positive H is an endothermic equationo A negative H is an exothermic equation- Anything that combusts has had oxygen burned and is exothermic- Evolved = exothermicII. Calculating ΔH- There are various ways to determine H depending on what you are given- Heating Curveo Q=mcΔTThese 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.o Q=mΔH(fus or vap)o Calculate the heat required to transform 15 g ice at -10°C to water at 30°C Break up the transformation into three parts- Ice -10 to 0o Q=mcΔTo Q = 15 x 2.04 x 10o 306J- Solid to liquid (no temperature)o Q=mΔH( fus)o Q = 15 x 335o 5025 Jo Fus is used when transforming between solid and liquido Vap is used when transforming between liquid and gas- Liquid 0 to 30o Q=mcΔTo Q = 15 x 4.184 x 20o 1882.8J Add all together to get: 7213.8J- Calorimetero |Qrxn|= |QH2o + Qcal|o |Qrxn| = |mcΔT| + |KΔT| (broken down version of previous equation) M is the mass of the water C is the specific heat of water ΔT is the change in temperature of the water K is the calorimeter constant ΔT is the change in temperature of the system- Proportiono If you are given a chemical equation with ΔH, then just as moles of elements are proportional to other elements, moles of an element is proportional to the ΔH.o Make a proportion to determineo Ex. C2H12O11 + 12O2 12CO2 + 11H2O where ΔH is -5645 kJ If there is only 52 g of Oxygen, what will H be? 52 g of Oxygen ×1 mol Oxygen32 g Oxygen×−5645 kJ12molesO2=−9173.125 kJ- Equationo Given different equations with ΔH, you can combine the equations to findthe ΔH. Ex. H2 + F2 2 HF where ΔH is -537 kJ C + 2F2 CF4 where ΔH is -680 2 C + 2 H2 C2H4 where ΔH is 52.3 Find ΔH of 6 F2 + C2H4 4 HF + 2 CF42 H2 +2 F2 4 HF 2 x -537 J2 C + 4 F2 2 CF42 x -680 JC2H4 2 H2 +2 C -1 x 52.3 J6 F2 +C2H4 4 HF + 2 CF4-2486.3 J So, after you have manipulated the equations to where they have the right coefficients, you can also switch the equations around to where the ΔH switches from + to - or – to +. Once you have all the right coefficients, if the elements or compounds are on opposite sides of an arrow, they can be canceled out- Such as the hydrogen in the first equation and the hydrogen in the 3 equation cancel each other out after youflip the equation around- ΔHrxn = ΔH°fo Each element or compound has a heat of formation, or ΔH°f.o ΔH = ∑ΔH°f(products) -
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