LECTURE 21. THERMODYNAMICS—LET’S GET QUANTITATIVE Today we will examine the quantitative side of thermodynamics, while actually coming up with numerical values for ΔH, ΔS, etc. Up to this point we have been more concerned with the signs of thermo data and what they mean. These are summarized below and should be as much a part of you as your own name: (-) Exothermic, Heat Released, Hot ΔH(+) Endothermic, Heat Absorbed, Cold (-) Spontaneous ΔG(+) Non Spontaneous (+) Disordered ΔS (-) Ordered (-) A Bomb, Work Done on Surroundings w (+) “Arming a Bomb” Work Done on System But now we will be adding numbers: C3H8 + 5O2 3C02 + 4H2O Is NOT just ΔG = (-) It IS ΔG = -2070kJ which means it is really spontaneous. (Still, while we learn to do these calculations, don’t forget the importance of simply looking at a reaction and being able to assign the signs! It is knowing the signs that proves you understand thermo-- anyone can put numbers in a calculator but since half the questions I put on tests don’t require a calculator…..)How to complete the Thermodynamic Reaction Worksheet Predict the values for ΔH, Δngas, w, ΔS, and ΔG. At the least provide a sign, but if you can offer a guess of the magnitude without using a calculator, that is even better. First the signs. Can you explain how you know them? Reaction ΔH Δngas w ΔS ΔG CH4(g) + O2(g) CO2(g) + 2H2O(g) (-) 0 0 (-) (-) 2H2(g) + O2(g) 2H2O(g) (-) (-) + (-) (-) And now the calculations. Chemical Reaction ΔH (kJ) BE (kJ) Δngas w(kJ)= -ΔnRT Δnsys TΔS (kJ) ΔG (kJ) CH4(g) + O2(g) CO2(g) + 2H2O(g) Combustion reaction, would predict large heat and spontaneous reaction. Δn = 0 means no work, small entropy change -802 -802 0 0 0 -1 -801 2H2(g) + O2(g) 2H2O(g) Spontaneous combustion reaction, would predict large heat. Δn = negative which means entropy decreases and work on system -484 -482 -1 +2.5 -1 -27 -458 Two central concepts to remember as we do calculations: 1. Why signs are what they are. Remember, it is not just arbitrary to say ΔH is (-) for exothermic. It is a consequence of our having identified with energy flow to and from the system. 2. The fundamental concept behind calculations is Hess’ Law, which says the path does not matter, it is just Δ = (how it ended - how it started.) Which is why there will be more than one way to calculate Δ state functions for a reaction.First Thermodynamic calculation: Measuring Heat of the System, ΔH For example, in calculating enthalpy change, ΔH, Hess’ Law says that these 3 very different methods all yield the same answer for ΔHrxn and you will need to learn how to do each of them Three calculations of ΔH ΔH f° Heat of Formation BE Bond Energy ΔH calorimeter Bomb Calorimeter Let’s apply Hess’ Law here: Consider three ways to determine heat of reaction for a propane explosion Make into C H H O O Gas Atoms C H H O O Put Back Together C H H O O As Products Bond H H O O Energy O O C3H8 + 5O2 3CO2 + 4H20 Heat of Formation C H2 O2 Make into C H2 O2 Put Back Together Elements in C H2 O2 As Products Standard State H2 O2 O2The Bomb Calorimeter Calculation. Remember that exploding balloon? We want to know the energy released as heat by that balloon. Heat Change ΔH = mCΔT Temperature Change Of System Mass of Specific Heat System But this is hard to measure directly because the heat is given off in every direction, so the ΔT is not representative of the total energy. So why not isolate the system in a container Balloon in a Sealed Container But we still need to make sure ΔT is accurate and that we know C. So here is a trick that uses conservation laws. Since ΔEuniverse = 0 (and assuming the only E is heat) Then ΔHuniverse = 0 = ΔHsystem +ΔHsurrounding So ΔHsystem = mCΔT = ΔH surrounding = mCΔTsurroundingWhy is this good? Make the surroundings something you know, like H2O. for which you can measure ΔT. It is easy to find T1 + T2 of H2O and mass of water and CH20 = 1 cal/ 1°C H2O Calorimeter C3H8 System O2 H2O ΔHC3H8 = ΔHH2O = mCΔTH2O Surrounding This is how they measure caloric content in food. Put a marshmallow in a calorimeter, surround with water, and explode the marshmallow. All the heat goes to the water surrounding it. !! BOOM !! Marsh- Heat H20 Heat mallow + O2 CO2 H20 H20 T1 T2So we find ΔH marshmallow using the Calorimetry Equation ΔHmarshmallow = ΔHsystem = ΔHsurrounding = mCΔTwater On to to other methods to measure ΔH Method 2: B.E. (Bond Energy): In this method the molecules become gas atoms and you calculate ΔH from differences in energy to tear apart and reform molecules like a tinker toy set. Method 3: ΔHf (Heat of Formation): In this method the molecules become elements in standard states (298K, 1atm) and then reform products. A comparison: B.E. ΔHf 1. Easy conceptually 1. Tougher Conceptually 2. Uses tables of bond energy 2. Uses Appendixes of formation data 3 Only correct for gases 3. Applicable for all states of matter (g,l,s) 4. Can be used for ΔHf, ΔSf, ΔGf (See Appendix) Now let’s perform BE and ΔHf calculations on C3H8 + 5O2 3CO2 + 4H2O First find values for B.E. of various bonds as well as ΔHf, ΔSf, ΔGf (this information is in the text in appendices and on the web—you will always be given this info on an exam.)The propane combustion data from Tables and Appendices: B.E. values ΔHf, ΔSf, ΔGf C3H8 O2 CO2H2OC-C, C-H, O=O, 346kJ, 413kJ, 458kJ ΔH (kJ) -104 0 -374 -242S (J) 270 205 -214 -189C=O O-H 799kJ 463kJ ΔG (kJ) -23 0 -394 -229 And of course you have equations that are variations on Hess’s Law. For Bond Energies: B.E. = ΣBEreact –
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