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Transcript
Chapter 8 Thermochemistry Thermodynamics Study of the changes in energy and transfers of energy that accompany chemical and physical processes. address 3 fundamental questions Will two (or more) substances react when they are mixed under specified conditions? If they do react, what energy changes and transfers are associated with their reaction? If a reaction occurs, to what extent does it occur? The First Law of Thermodynamics Exothermic reactions combustion of propane C 3 H 8 g + 5 O 2 g 3 CO 2 g + 4 H 2 O l + 2.22 10 kJ 3 combustion of n-butane 2 C4H10g + 13 O2g 8CO2g +10 H2Ol + 5.78 10 kJ 3 The First Law of Thermodynamics Exothermic reactions release specific amounts of heat as products Potential energies of products are lower than potential energies of reactants. The First Law of Thermodynamics 2 basic ideas of importance systems tend toward a state of minimum potential energy a. H 2 O flows downhill b. objects fall when dropped c. E potential mgh d. E potential mg h The First Law of Thermodynamics 2 basic ideas of importance systems tend toward a state of maximum disorder a. mirror shatters when dropped b. easy to scramble an egg c. food coloring disperses in water The First Law of Thermodynamics Also known as Law of Conservation of Energy The total amount of energy in the universe is constant. Energy can be converted from one form to another but cannot be created. Some Thermodynamic Terms System - substances involved in the chemical and physical changes under investigation for us this is what is happening inside the beaker Surroundings - rest of the universe outside the beaker Universe - system plus surroundings Some Thermodynamic Terms Thermodynamic State of a System - set of conditions that describe and define the system number of moles of each substance physical states of each substance temperature pressure State Functions - properties of a system that depend only on the state of the system capital letters Some Thermodynamic Terms state functions are independent of pathway climbing a mountain, taking two different paths E1 = energy at bottom of mountain E1 = mgh1 E2 = energy at top of mountain E2 = mgh2 E E2- E1 = mgh2 - mgh1 = mg(h) Some Thermodynamic Terms Properties that depend only on values of state functions are also state functions examples: • • • T P V Enthalpy Change, H Commonly, chemistry is done at constant pressure open beakers on a desk top are at atmospheric pressure H - enthalpy change change in heat content at constant pressure H = qp Hrxn - heat of reaction Hrxn = Hproducts - Hreactants Hrxn = Hsubstances produced - Hsubstances consumed Calorimetry coffee-cup calorimeter - used to measure the amount of heat produced (or absorbed) in a reaction at constant P measures qP Calorimetry exothermic reaction - heat evolved by reaction is determined from the temperature rise of the solution 2 part calculation Amount of heat Amount of heat Amount of heat released by reaction gained by calorimeter gained by solution Amount of heat gained by calorimeter is the heat capacity of the calorimeter or calorimeter constant value determined by adding a specific amount of heat to calorimeter and measuring T rise Calorimetry When 3.425 kJ of heat is added to a calorimeter containing 50.00 g of water the temperature rises from 24.000C to 36.540C. Calculate the heat capacity of the calorimeter in J/0C. The specific heat of water is 4.184 J/g 0C. Four part calculation Calorimetry Find the temperature change T = 36.54 - 24.00 C = 12.54 C 0 0 Find the heat absorbed by the water in going from 24.000C to 36.540C. q P mCT = 50.00 g 4184 . J 12.54 C 0 0 g C 262337 . J 2623 J Calorimetry Find the heat absorbed by the calorimeter. total amount of heat added to calorimeter - heat absorbed by water 3.425 kJ = 3425 J 3425 J - 2623 J = 802 J Find the heat capacity of the calorimeter (heat absorbed by the calorimeter)/(temperature change) 802 J J 0 64.00 J 0 63 . 955 0 C C 12.54 C Calorimetry A coffee-cup calorimeter is used to determine the heat of reaction for the acid-base neutralization CH3COOHaq + NaOHaq NaCH3COOaq + H2Ol When we add 25.00 mL of 0.500 M NaOH at 23.0000C to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.9470C. Calorimetry The heat capacity of the calorimeter has previously been determined to be 27.8 J/0C. Assume that the specific heat of the mixture is the same as that of water, 4.18 J/g0C and that the density of the mixture is 1.02 g/mL. Two part calculation: a Calculate the amount of heat given off in the reaction. Calorimetry temperature change T = 25.947 - 23.000 C = 2.947 C 0 0 heat absorbed by calorimeter q = 2.947 C 27.8 0 J0 C 819. J mass of solution in calorimeter 102 . g 510 . g 25.00 mL + 25.00 mL mL Calorimetry heat absorbed by solution q = mCT q = 51.0 g 418 . J 2.947 C 628 J 0 g0 C total amount of heat produced by reaction q = 81.9 J + 628 J = 709.9 J Calorimetry b Determine H for the reaction under the conditions of the experiment. • • must determine the number of moles of reactants consumed limiting reactant calculation Calorimetry C H 3C O O H aq + N aO H aq N a C H 3C O O aq + H 2O 0 .5 0 0 m m o l N a O H 2 5 . 0 0 m L N a O H 1 m L N aO H 1 m m o l N a C H 3 C O O 1 2 .5 m m o l N a C H C O O 3 1 m m ol N aO H 2 5 .0 0 0 .6 0 0 m m o l C H 3 C O O H m L C H 3C O O H 1 m L C H 3C O O H 1 m m o l N a C H 3C O O 1 5 .0 m m o l N a C H 3 C O O 1 m m o l C H 3C O O H l Calorimetry finally, calculate H based on the limiting reactant calculation H rxn 12.5 mmol = 0.0125 mol 709.9 J 56792 J / mol 56.8 kJ / mol 0.0125 mol Thermochemical Equations Thermochemical equations are a balanced chemical reaction plus the H value for the reaction. for example: C 5 H12 l 8 O 2 g 5 CO 2 g + 6 H 2 O l 3523 kJ 1 mol 8 mol 5 mol 6 mol coefficients in thermochemical equations must be interpreted as numbers of moles 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions Thermochemical Equations Equivalent method of writing thermochemical equations C5 H12l 8 O2g 5 CO2g + 6 H2Ol H = -3523 kJ H < 0 designates an exothermic reaction H > 0 designates an endothermic reaction Thermochemical Equations Write the thermochemical equation for the previous reaction. CH 3COOH aq + NaOH aq NaCH 3COO aq + H 2 O l H = -56.8 kJ / mol Standard States & Standard Enthalpy Changes Thermochemical standard state conditions T = 298.15 K P = 1.0000 atm Thermochemical standard states pure substances in their liquid or solid phase - standard state is the pure liquid or solid gases - standard state is the gas at 1.00 atm of pressure • gaseous mixtures - partial pressure must be 1.00 atm aqueous solutions - 1.00 M concentration Standard Molar Enthalpies of Formation, Hfo Standard molar enthalpy of formation symbol is Hfo defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements for example: Mg s Cl 2 g MgCl 2s 6418 . kJ o H f MgCl 2 s 6418 . kJ / mol Standard Molar Enthalpies of Formation, Hfo Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 8.3 and Appendix 1 in the text. Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero. Example: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for which Horxn = -1281 kJ. P in standard state is P4 phosphoric acid in standard state is H3PO4(s) Standard Molar Enthalpies of Formation, Hfo Hess’s Law Hess’s Law of Heat Summation - enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps true because H is a state function we know the following Ho’s 1 4 FeO s O2 g 2 Fe2 O3 2 2 Fe s O2 g 2 FeO s 3 4 Fe s 3 O2 g 2 Fe2 O3 s Ho 560 kJ Ho 544 kJ Ho 1648 kJ Hess’s Law We could calculate the Ho for [1] by properly using the 0 Ho’s for [2] and [3] H 2 x [2] 2(2 FeO s 2 Fe s O 2 g ) 2( 544 ) kJ 3 4 Fes 3 O 2 g 2 Fe 2 O 3s 1 4 FeO s O 2 g 2 Fe 2 O 3 1648 kJ 560 kJ Hess’s Law Example : Given the following equations and Ho values H o kJ [1] 2 N 2 g O 2 g 2 N 2 O g 164.1 [2] N 2 g + O 2 g 2 NO g 180.5 [3] N 2 g + 2 O 2 g 2 NO 2 g 66.4 calculate Ho for the reaction below. N 2O g + NO 2g 3 NO g H o ? Hess’s Law Use a little algebra and Hess’s Law to get the appropriate Ho values o H (kJ) 1 2 1 N 2 O g N 2 g + 3 2 2 1 2 3 NO 2 g 3 2 3 N 2 g + 1 2 2 1 2 O 2 g - 82.05 O 2 g 3 NO g 270.75 N 2 g + O 2 g N 2 O g NO 2 g 3 NO g - 33.2 155.5 Hess’s Law The + sign of the Ho value tells us that the reaction is endothermic. The reverse reaction is exothermic, i.e., 3 NO g N 2 O g + NO 2 g H o = - 155.5 kJ Hess’s Law Hess’s Law in a more useful form any chemical reaction at standard conditions, the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants o H 298 n n o H f products n n o H f reactants Hess’s Law Example: Calculate Ho298 for the following reaction from data in Appendix 1. C 3 H 8 g + 5 O 2 g 3 CO 2 g + 4 H 2 O l Hess’s Law Example: Calculate Ho298 for the following reaction from data in Appendix 1. C3H8g + 5 O2g 3 CO2g + 4 H2Ol o H298 o 3Hf CO2 g o 4Hf H2Ol o Hf C3H8 g o 5Hf O2 g 3(3935 . ) 4(2858 . ) (1038 . ) 5(0) kJ = -2211.9 kJ Ho298 22119 . kJ, and so the reaction is exothermic. Hess’s Law Application of Hess’s Law and more algebra allows us to calculate the Hfo for a substance participating in a reaction for which we know Hrxno , if we also know Hfo for all other substances in the reaction. Hess’s Law Example: Given the following information, calculate Hfo for H2S(g) 2 H2Sg + 3 O2g 2 SO2g +2 H2Ol Hof ? (kJ / mol) 0 -296.8 -285.8 Ho298 = -1124 kJ Hess’s Law 1124 kJ 2(2968 . ) 2(2858 . 2H o H298 o 2Hf SO2g o 2Hf H2Ol o 2Hf H2Sg o f H2S g now we solve for Hof H2Sg 2Hof H2Sg 412 . kJ Hof H2Sg 206 . kJ o 3Hf O2g 3(0) kJ Bond Energies Bond energy - amount of energy required to break the bond and separate the atoms in the gas phase A - B g bond energy A g + B g H - Cl g 432 kJ mol H g + Cl g Bond Energies Table of average bond energies Molecule Bond Energy (kJ/mol) F2 159 Cl2 243 Br2 192 O2 (double bond) 498 N2 (triple bond) 946 Bond Energies Bond energies can be calculated from other Ho298 values Bond Energies Example: Calculate the bond energy for hydrogen fluoride, HF. Bond Energies Example: Calculate the bond energy for hydrogen fluoride, HF. H - F g BE HF H g F g atoms NOT ions or H - F g H g F g H o298 BE HF H o298 H of H g H of F g H of HF g o H 298 218.0 kJ + 78.99 kJ 271 kJ H o298 568.0 kJ BE for HF Bond Energies Example: Calculate the average N-H bond energy in ammonia, NH3. Bond Energies NH 3g N g + 3 H g H o 298 3 BE N -H H H o 298 H H o 298 (472.7) 3(218) 46.11 kJ H o 298 1173 kJ o f Ng average BE N -H 3 H o f Hg o f NH 3 g 1173 kJ 391 kJ mol N -H bonds 3 Bond Energies In gas phase reactions Ho values may be related to bond energies of all species in the reaction. o H 298 BE reactants BE products Bond Energies Example: Use the bond energies listed in Table 8.4 to estimate the heat of reaction at 25oC for the reaction below. CH4g 2 O2g CO2g 2 H2Og o H298 4 BEC-H 2 BEO=O 2 BEC=O 4 BEO-H Bond Energies Example: Use the bond energies listed in Table 8.4 to estimate the heat of reaction at 25oC for the reaction below. CH 4g 2 O2g CO2g 2 H 2 Og o H 298 4 BE C-H 2 BE O=O 2 BE C=O 4 BE O-H H o298 4(414) 2(498) 2(741) 4(464) kJ o H 298 686 kJ