Chapter 6: Thermochemistry

Transcription

1 Chapter 6: Thermochemistry Thermochemistry: energy considerations associated with chemical and physical change Energy: the capacity of a system to do work or produce heat potential energy energy of position; stored energy kinetic energy energy of motion kinetic energy =! mυ 2 kinetic energy = 3 2 RT In a system, energy can be transferred in the form of heat or work: An Example of a System Configured for Energy Transferred as Work: heat, q energy transfer results in temperature change (ΔT); ΔT = Tfinal Tinitial work, w energy transfer during the act of moving an object against an opposing force overall: ΔU = q + w

2 The System, The Surroundings, The Universe & The First Law of Thermodynamics Units of Energy SI unit of energy: Joule, J 1 Joule is the amount of energy required to raise the temperature of g H2O by 1ºC. 1 kj = 103 J 1 MJ = 106 J thermochemical calorie, cal 1 calorie is the amount of energy required to raise the temperature of 1 g of H2O by 1ºC. 1 cal = J 1 kcal = 103 cal 1 nutritional calorie, 1 Cal = 1000 cal Endothermic Change: energy is absorbed by the system energy flow is from the surroundings into the system Esys increases Esurr decreases for the system: Efinal > Einitial if energy transferred in the form of heat, Tsys increases and Tsurr decreases E lost by surroundings = E gained by the system The System: what we re interested in The Surroundings: everything else The Universe: System + Surroundings = the Universe The 1st Law of Thermodynamics: The energy of the Universe is constant.

3 energy is released by the system energy flow is from the system into the surroundings Esys decreases Esurr increases Exothermic Change for the system: Efinal < Einitial if energy transferred in the form of heat, Tsys decreases and Tsurr increases E lost by system = E gained by the surroundings Changes in Thermchemical Quantities: importance of sign and magnitude if E is + Efinal > Einitial energy is absorbed by the system endothermic change ΔX = Xfinal Xinitial if ΔX is + : Xfinal > Xinitial if ΔX is! : Xfinal < Xinitial if E is! Efinal < Einitial energy is released by the system exothermic change if q is + qfinal > qinitial heat is absorbed by the system endothermic if q is! qfinal < qinitial heat is released by the system exothermic Calculate the change in internal energy of a system that releases 26.8 kj of heat as it does 68.7 kj of work on the surroundings. if w is + surroundings do work on the system endothermic if w is! system does work on the surroundings exothermic

4 PV Work expansion or compression of an ideal gas w =!P V notes: w and V are always opposite in sign; why? Calculate the quantity of work done (in kj) by an ideal gas as it expands from an initial volume of 2.50 L to a final volume of L against a constant external pressure of atm. 1 L atm = J Enthalpy, H H = E + PV We will be focussing on changes in enthalpy that accompany processes that occur at constant P. at constant P: H = qp H = Hfinal! Hinitial if H is + : heat is absorbed by the system endothermic process if H is! : heat is released by the system exothermic process Enthalpy Changes for Chemical Reactions: Thermochemical Equations Hrxn = Hfinal! Hinitial OR Hrxn = Hproducts! Hreactants Thermochemical Equation: balanced chemical equation + thermochemical data 2 Na (s) + 2 H2O (l)! 2 NaOH (aq) + H2 (g); H = kj notes: reaction is exothermic as written specifically, kj of energy is released when: 2 mol Na & 2 mol H2O are consumed 2 mol NaOH & 1 mol H2 are formed

5 An Enthalpy Diagram for This Reaction: Enthalpy Changes for Chemical Reactions: Thermochemical Equations interpretation of energy changes associated with a reaction must take into consideration the following: phases of reactants and products ex. 2 H2 (g) + O2 (g)! 2 H2O (g); H = kj vs. 2 H2 (g) + O2 (g)! 2 H2O (l); H = kj direction of the reaction: a reaction is endothermic in one direction and exothermic in the reverse direction ex. CH4 (g) + 4 Cl2 (g)! CCl4 (l) + 4 HCl (g); H = 433 kj vs. CCl4 (l) + 4 HCl (g)! CH4 (g) + 4 Cl2 (g); H = +433 kj Enthalpy Changes for Chemical Reactions: Thermochemical Equations stoichiometry of balanced equation energy released or absorbed is an extensive property (depends on size of sample) ex. combustion of propane C3H8 (g) + 5 O2 (g)! 3 CO2 (g) + 4 H2O (g); Hrxn = 2200 kj combustion of 100 g (2.27 mol) C3H8 generates ~5000 kj vs. combustion of 2.5 kg (56.7 mol) C3H8 generates ~125,000 kj N2 (g) + 3 H2 (g)! 2 NH3 (g); Determine the quantity of heat released (in kj) in the synthesis of 907 kg NH3. Stoichiometric Calculations: Using Hrxn as a Conversion Factor H = 91.8 kj 12.0 L N2 (g) at STP and 5.00 L H2 (g) at 30ºC and 3.02 atm are combined and allowed to react. What is the maximum amount of heat (in kj) that can be generated?

6 Calorimetry A laboratory technique in which a reaction proceeds in an insulated container (i.e. calorimeter) at constant P or constant V. adiabatic conditions for the system; no heat is lost to or gained from the surroundings record an observed T calculate H or E at constant P: H = qp at constant V: E = qv Heat Capacity (C) and Specific Heat (s) How does a substance respond to heating? Compare the observed temperature changes when 50.0 J energy are added to 10.0 g samples of diamond and tungsten: diamond m = 10.0 g q = 50.0 J T = 9.8ºC tungsten m = 10.0 g q = 50.0 J T = 37.3ºC Heat Capacity of a substance, C heat required to raise the temperature of a sample by 1º (C or K). heat supplied Heat Capacity = ; units J ºC or K T produced OR C = q T Specific Heat, s heat required to raise the temperature of 1g of substance by 1º (C or K) heat supplied Specific Heat = ; units J/g ºC or K ( T)(mass) OR s = q/(m T) note: use molar mass to convert between specific heat and molar heat capacity of a substance

7 Calculate the amount of heat required (in J) to raise the temperature of a 25.0 g block of nickel from 22ºC to 104ºC. For Ni, s = 0.44 J/g ºC g H2O at 280 K is mixed with 50.0 g H2O at 330 K. What will be the final temperature of the mixture? For H2O, s = J/g ºC. Constant Pressure Calorimetry to Determine H adiabatic conditions: no heat exchanged between system and surroundings Hsys = 0 if: Hsys = H1 + H2 then: H1 = H2 for an exothermic reaction: H1 = heat released by chemical reaction occurring in calorimeter H2 = heat absorbed by solution & calorimeter resulting in increase in T What do you measure in lab? amounts of reactants (mass, volume, mol) observed T order of determination (for the following examples): H2 H1 molar enthalpy of reaction +, units kj!, units kj!, units kj/mol Consider the neutralization reaction of hydrochloric acid and sodium hydroxide: HCl (aq) + NaOH (aq)! NaCl (aq) + H2O (l) Determine the molar enthalpy change (in kj/mol) for this reaction if, in a constant pressure calorimetry experiment ml of 1.20 M HCl (aq) is combined with 42.0 ml of 1.20 M NaOH (aq) resulting in an increase in temperature from an initial 25.00ºC to a final 31.80ºC. You can assume that, for the solutions, volume is additive; d = 1.00 g/ml; s = J/gºC.

8 When 23.6 g CaCl2 is dissolved in water in a constant pressure calorimeter, the temperature rose from 25.0ºC to 38.7ºC. If the heat capacity of the calorimeter and solution is 1258 J/ºC (i.e. Ccal = 1258 J/ºC), determine the heat released by the dissolution of 1.20 mol CaCl2. CaCl2 (s)! Ca 2+ (aq) + 2 Cl (aq) Constant Volume Calorimetry to Determine E adiabatic conditions: no heat exchanged between system and surroundings Esys = 0 if: Esys = E1 + E2 then: E1 = E2 for an exothermic reaction: E1 = heat released by chemical reaction occurring in calorimeter E2 = heat absorbed by solution & calorimeter resulting in increase in T The combustion of ethanol is studied in a constant volume (bomb) calorimeter with a calorimeter constant of 9.63 kj/ºc (i.e Ccal = 9.63 kj/ºc). The combustion of 2.84 g C2H5OH results in an increase in temperature from Ti = 25.00ºC to Tf = 33.73ºC. Determine the energy released in the combustion of 1 mol of C2H5OH. Determine the heat of combustion per gram of ethanol (in kj/g) Hess s Law of Heat Summation We can determine of H for one chemical reaction based on the H values for related reactions using Hess s Law: For a chemical reaction that can be written as the sum of 2 or more steps, H for the overall reaction is equal to the sum of H s for the individual steps. so: for a reaction that is a result of the sum of 3 individual steps: Hrxn = H1 + H2 + H3

9 Why does Hess s Law of Heat Summation work? Enthalpy is a state property: Hess s Law of Heat Summation dependent only on initial and final states independent of path Hess s Law of Heat Summation Determine H for the following reaction using equations 1 & 2 given below: 2 C (s) + O2 (g)! 2 CO (g) H =???? use: (1) 2 C (s) + 2 O2 (g)! 2 CO2 (g) H1 = 787 kj (2) 2 CO2 (g)! 2 CO (g) + O2 (g) H2 = +566 kj overall: 2 C (s) + O2 (g)! 2 CO (g) H = 221 kj An Enthalpy Diagram for This Process: To solve a Hess s Law of Heat Summation problem: consider the equations you have to work with identify how to manipulate them so that when you add them together, you end up with your target, overall equation your options are: multiply all stoichiometric coefficients in the equation by some factor, n " multiply H by the same factor, n write an equation in reverse (i.e. change direction) " change the sign on H

10 Determine H for: W (s) + C (s)! WC (s); H =???? using: (1) 2 W (s) + 3 O2 (g)! 2 WO3 (s) H = kj (2) C (s) + O2 (g)! CO2 (g) H = kj (3) 2 WC (s) + 5 O2 (g)! 2 WO3 (s) + 2 CO2 (g) H = kj solution: Determine H for: W (s) + C (s)! WC (s); H =???? using: (1) 2 W (s) + 3 O2 (g)! 2 WO3 (s) H = kj (2) C (s) + O2 (g)! CO2 (g) H = kj (3) 2 WC (s) + 5 O2 (g)! 2 WO3 (s) + 2 CO2 (g) H = kj actions to take: reverse equation (3) multiply equation (1) by factor of " multiply equation (3) by factor of " do nothing to equation (2) solution: Determine H for: W (s) + C (s)! WC (s); H =???? using: (1) W (s) + 3/2 O2 (g)! WO3 (s) H = kj (2) C (s) + O2 (g)! CO2 (g) H = kj (3) WO3 (s) + CO2 (g)! WC (s) + 5/2 O2 (g) H = kj overall: W (s) + C (s)! WC (s) H = 38.0 kj Standard Reaction Enthalpies Standard Reaction Enthalpies, Hº: reaction enthalpy corresponding to reactants in their standard states forming products in their standard states standard state: the physical state of a pure substance at P = 1 atm and a defined temperature

11 Standard Enthalpies of Formation, Hºf Hºf corresponds to the formation of 1 mol of a substance in its standard state from its elements in their standard states. ex: Na (s) + " Cl2 (g)! NaCl (s) 6 C (s) + 6 H2 (g) + 3 O2 (g)! C6H12O6 (s) Hºf of an element in its most stable form = 0 units of Hºf are kj/mol tabulated in Table 6.2 and Appendix C of text use Hºf s to calculate Hºrxn Using Standard Enthalpies of Formation, Hºf Hºrxn = n Hºf products n Hºf reactants Calcualte Hº (in kj) for the following reaction using the given Hºf values: 3 NO2 (g) + H2O (l)! 2 HNO3 (aq) + NO (g) Hºf: (in kj/mol) Determine Hºf (in kj/mol) for HCl (g) using the following data: CH4 (g) + 4 Cl2 (g)! CCl4 (l) + 4 HCl (g); Hº = kj Hºf : ??? (in kj/mol)

12 Determine Hºf (in kj/mol) for HCl (g) using the following data: CH4 (g) + 4 Cl2 (g)! CCl4 (l) + 4 HCl (g); Hº = kj Hºf : ??? (in kj/mol) solution: let x = Hºf of HCl (g) kj = [(1 mol CCl4)( 139 kj/mol) + (4 mol HCl)(x)]! [(1 mol CH4)( 74.9 kj/mol)] kj = 139 kj + (4 mol)(x) kj = 4 mol(x) so x = Hºf for HCl (g) = 92.3 kj/mol