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THERMOCHEMISTRY – 2 CALORIMETRY and HEATS of REACTION Dr

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THERMOCHEMISTRY – 2 CALORIMETRY AND HEATS OF REACTION Dr. Sapna Gupta HEAT CAPACITY • Heat capacity is the amount of heat needed to raise the temperature of the sample of substance by one degree Celsius or Kelvin. q = CDt • Molar heat capacity: heat capacity of one mole of substance. • Specific Heat Capacity: Quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one Kelvin) at constant pressure. q = m s Dt (final-initial) • Measured using a calorimeter – it absorbed heat evolved or absorbed. Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 2 EXAMPLES OF SP. HEAT CAPACITY The higher the number the higher the energy required to raise the temp. Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 3 CALORIMETRY: EXAMPLE - 1 Example: A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is 0.388 J/(g°C). Solution m = 35.8 g s = 0.388 J/(g°C) Dt = 28.00°C – 20.00°C = 8.00°C q = m s Dt 0.388 J q 35.8 g 8.00C = 111J gC Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 4 CALORIMETRY: EXAMPLE - 2 Example: Nitromethane, CH3NO2, an organic solvent burns in oxygen according to the following reaction: 3 3 1 CH3NO2(g) + /4O2(g) CO2(g) + /2H2O(l) + /2N2(g) You place 1.724 g of nitromethane in a calorimeter with oxygen and ignite it. The temperature of the calorimeter increases from 22.23°C to 28.81°C. The heat capacity of the calorimeter was determined to be 3.044 kJ/°C. Write the thermochemical equation for the reaction. Solution: q C Δt Heat evolved rxn cal 3.044 kJ qrxn 28.81C 22.23C 20.03 kJ C 20.03 kJ 61.04 g CH32 NO Convert the heat evolved to per mol. qrxn = -709 KJ 1.724 g CH3 NO 2 1mol CH 3 NO 2 Now the equation: 3 CH3NO2(l) + ¾O2(g) CO2(g) + /2H2O(l) + ½N2(g); DH = –709 kJ Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 5 CALORIMETRY: EXAMPLE - 3 Example: A metal pellet, 85.00 grams at an original temperature of 92.5C is dropped into a calorimeter with 150.00 grams of water at an original temperature of 23.1C. The final temperature of the water and the pellet is 26.8C. Calculate the heat capacity and the specific heat for the metal. Solution: Energy = qwater = msDT = (150.00 g) (4.184 J/gC) (3.7C) = 2300 J (water gained energy) = -2300 J (pellet released energy) Heat capacity of pellet: q = CDT C = q/DT = 2300 J/65.7C = 35 J/C Note: we could have calculated the sp. heat of the Specific heat of pellet: pellet in one calculation using 35 J/oC q = m s Dt, s = = 0.41J/goC 85.00g But since the question was also about Heat Capacity, we calculated that first and then just divide that by mass to get Dr. Sapna Gupta/Thermochemistry-2-Calorimetry the sp. heat of the metal. 6 CALORIMETRY: EXAMPLE - 4 Example: A snack chip with a mass of 2.36 g was burned in a bomb calorimeter. The heat capacity of the calorimeter 38.57 kJ/C. During the combustion the water temp rose by 2.70C. Calculate the energy in kJ/g for the chip. Solution: qrxn = − Ccal DT = −(38.57 kJ/C) (2.70C) = − 104 kJ Energy content is a positive quantity. = 104 kJ/2.36 g = 44.1 kJ/g Convert J to Cal. Food Calories: 10.5 Cal/g Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 7 HESS’S LAW • The change in enthalpy when reactants are converted to products is the same whether the reaction occurs in one step or a series of steps. • Usually used when heat of reaction cannot be determined directly. • Manipulation of equations is the same as before. • Add all the DH values after equations are manipulated. • Suppose you want the DH for the reaction: 2C(graphite) + O2(g) 2CO(g) However two other reactions are known: C(graphite) + O2(g) CO2(g); DH = -393.5 kJ x2 2CO2(g) 2CO(g) + O2(g); DH = – 566.0 kJ leave as is In order for these to add to give the reaction we want, we must multiply the first reaction by 2 (Note that we also multiply DH by 2.) Then add the two heat of reactions along with the DH. (-787+(-566.0))=1353 kJ Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 8 HESS’S LAW: EXAMPLE Example: Given the following equations: H3BO3(aq) HBO2(aq) + H2O(l) DHrxn = 0.02 kJ x 2 H2B4O7(aq) + H2O(l) 4 HBO2(aq) DHrxn = 11.3 kJ reverse, ÷2 H2B4O7(aq) 2 B2O3(s) + H2O(l) DHrxn = 17.5 kJ ÷ 2 Find the DH for this overall reaction. 2H3BO3(aq) B2O3(s) + 3H2O(l) Solution: 2H3BO3(aq) 2HBO2(aq) + 2H2O(l) DHrxn = 2(−0.02 kJ) = −0.04 kJ 2HBO2(aq) 1/2H2B4O7(aq) + 1/2H2O(l) DHrxn = +11.3 kJ/2 = 5.65 kJ 1/2H2B4O7(aq) B2O3(s)+ 1/2H2O(l) DHrxn = 17.5 kJ/2 = 8.75 kJ 2H3BO3(aq) B2O3(s) + 3H2O(l) DHrxn = 14.36 kJ Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 9 STANDARD ENTHALPY OF FORMATION • Standard state refers to the standard thermodynamic conditions for the substances when listing or comparing thermodynamic data: the atm pressure and temperature (25 oC). • Standard conditions are indicated with a degree sign “o”: eg DH and DHo – the second one is the standard enthalpy at 1 atm and 25 oC. • Elements can exist in more than one physical state e.g. carbon as graphite or diamond (allotropes). o • The standard enthalpy of formation, DHf , is the enthalpy change for formation of one mole of substance from its elements in their reference o forms and in their standard states. The DHf for an element in its reference state is zero. 1 • Example: H2(g) + /2O2(g) H2O(l) DHf° = –285.8 kJ • The equation below can be used to calculate the enthalpy. (n = number of moles) ΔHreaction nΔHf nΔHf products reactants Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 10 STANDARD ENTHALPY: EXAMPLE Calculate the DH° for the following reaction: CH3OH(l) CH3OH(g) Solution: Find the DHf° values for all the compounds and subtract product from reactant. kJ For liquid methanol : ΔH 238.7 f mol kJ For gaseous methanol : ΔH 200.7 f mol kJ kJ ΔHvap 1 mol 200.7 1 mol 238.7 = + 38.0 KJ mol mol Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 11 STANDARD ENTHALPY: EXAMPLE Methyl alcohol, CH3OH, is toxic because liver enzymes oxidize it to formaldehyde, HCHO, which can coagulate protein. Calculate DH° for the following reaction: 2CH3OH(aq) + O2(g) 2HCHO(aq) + 2H2O(l) Standard enthalpies of formation, DHf°: CH3OH(aq): −245.9 kJ/mol HCHO(aq): −150.2 kJ/mol ΔHreaction nΔHf nΔHf H2O(l): −285.8 kJ/mol products reactants o kJ kJ ΔHreacton 2 mol 150.2 2 mol 285.8 mol mol kJ kJ 2 mol 245.9 1 mol 0 mol mol o ΔHreaction 300.4 kJ 571.6 kJ 491.8 kJ o o ΔHreaction 872.0 kJ 491.8 kJ ΔHreaction 380.2 kJ Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 12 APPLICATIONS: FOOD Foods are fuel needed for the body. They undergo a combustion process in the body. For glucose, a carbohydrate: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l); DHf° = –2803 kJ For glycerol trimyristate, a fat: 127 C45H86O6(s) + /2O2(g) 45CO2(g) + 43H2O(l); DHf°= –27,820 kJ The average value for carbohydrates is 4.0 Cal/g and for fats is 9.0 Cal/g. Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 13 APPLICATIONS: FUEL • Fuels originate from organic substances: plants and animals, that have decayed over millions of years. This provides the coal, petroleum products and natural gas. • Coal is about 30% of US energy consumption. Variety of coal determines how much carbon is in it: anthracite – 80% carbon; bituminous coal – 45- 65% carbon. • Natural gas is about 23% of energy. It is a fluid and easy to transport. Pure natural gas is methane, CH4 with small amounts of ethane, propane and butane. • Petroleum is a mixture of hydrocarbon compounds. • Natural gas and petroleum products are in short supply while coal will last longer. • All these are used for energy sources. Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 14 KEY CONCEPTS • Measuring heats of reaction - calorimetry • Hess’s Law • Standard enthalpy of formation Dr. Sapna Gupta/Thermochemistry-2-Calorimetry 15.
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  • Isothermal Titration Calorimetry and Differential Scanning Calorimetry As Complementary Tools to Investigate the Energetics of Biomolecular Recognition

    Isothermal Titration Calorimetry and Differential Scanning Calorimetry As Complementary Tools to Investigate the Energetics of Biomolecular Recognition

    JOURNAL OF MOLECULAR RECOGNITION J. Mol. Recognit. 1999;12:3–18 Review Isothermal titration calorimetry and differential scanning calorimetry as complementary tools to investigate the energetics of biomolecular recognition Ilian Jelesarov* and Hans Rudolf Bosshard Department of Biochemistry, University of Zurich, CH-8057 Zurich, Switzerland The principles of isothermal titration calorimetry (ITC) and differential scanning calorimetry (DSC) are reviewed together with the basic thermodynamic formalism on which the two techniques are based. Although ITC is particularly suitable to follow the energetics of an association reaction between biomolecules, the combination of ITC and DSC provides a more comprehensive description of the thermodynamics of an associating system. The reason is that the parameters DG, DH, DS, and DCp obtained from ITC are global properties of the system under study. They may be composed to varying degrees of contributions from the binding reaction proper, from conformational changes of the component molecules during association, and from changes in molecule/solvent interactions and in the state of protonation. Copyright # 1999 John Wiley & Sons, Ltd. Keywords: isothermal titration calorimetry; differential scanning calorimetry Received 1 June 1998; accepted 15 June 1998 Introduction ways to rationalize structure in terms of energetics, a task that still is enormously difficult inspite of some very Specific binding is fundamental to the molecular organiza- promising theoretical developments and of the steady tion of living matter. Virtually all biological phenomena accumulation of experimental results. Theoretical concepts depend in one way or another on molecular recognition, have developed in the tradition of physical-organic which either is intermolecular as in ligand binding to a chemistry.