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Thermochemistry: Chemical Energy

Chapter 8. Thermochemistry: Chemical Energy. Energy is the capacity to do work Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances

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Thermochemistry: Chemical Energy

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  1. Chapter 8 Thermochemistry: Chemical Energy

  2. Energyis the capacity to do work • Thermal energy is the energy associated with the random motion of atoms and molecules • Chemical energy is the energy stored within the bonds of chemical substances • Nuclear energy is the energy stored within the collection of neutrons and protons in the atom • Electrical energy is the energy associated with the flow of electrons • Potential energy is the energy available by virtue of an object’s position, or stored energy • Kinetic energy is moving energy. • E = qrxn + w

  3. Temperature = Thermal Energy 900C 400C Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. greater thermal energy

  4. SURROUNDINGS SYSTEM Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. closed isolated open energy nothing Exchange: mass & energy

  5. Calorimetry and Heat Capacity Measure the heat flow at constant pressure (DH).

  6. Calorimetry and Heat Capacity Measure the heat flow at constant volume (DE).

  7. First Law of Thermodynamics • Energy is neither created nor destroyed. • In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Use Fig. 5.5

  8. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

  9. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0

  10. Enthalpy Changes 01 • Enthalpies of Physical Change:

  11. Enthalpies of Physical and Chemical Change Enthalpy of Fusion (DHfusion): The amount of heat necessary to melt a substance without changing its temperature. Enthalpy of Vaporization (DHvap): The amount of heat required to vaporize a substance without changing its temperature. Enthalpy of Sublimation (DHsubl): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase.

  12. H2O (s) H2O (l) DH = 6.01 kJ Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

  13. DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

  14. 2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) DH = -6.01 kJ DH = 6.01 kJ DH = 2 x 6.01= 12.0 kJ Thermochemical Equations • The stoichiometric coefficients always refer to the number of moles of a substance • If you reverse a reaction, the sign of DH changes • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.

  15. How much heat is transfered when 266 g of white phosphorus (P4) burn in air? P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ x H2O (l) H2O (g) H2O (s) H2O (l) -3013 kJ 1 mol P4 x DH = 6.01 kJ DH = 44.0 kJ 1 mol P4 123.9 g P4 Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations. = -6470 kJ 266 g P4

  16. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = msDt q = CDt Dt = tfinal - tinitial

  17. J Specific heat (Water) = 4.18 g °C Calorimetry and Heat Capacity Assuming that a can of soda has the same specific heat as water, calculate the amount of heat (in kilojoules) transferred when one can (about 350 g) is cooled from 25 °C to 3 °C. q = (specific heat) x (mass of substance) x DT Mass = 350 g Temperature change = 3 C-25 °C = - 22 °C

  18. 1 kJ 4.18 J 1000 J g °C Calorimetry and Heat Capacity Calculate the amount of heat transferred. 350 g -22 °C x x Heat evolved = = -32 000 J -32 000 J = -32 kJ x

  19. Calorimetry and Heat Capacity

  20. Constant-Volume Calorimetry Bomb Calorimetry • Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. • E = qrxn + w • W = VΔP • For most reactions, the difference is very small. • ΔE almost = qrxn

  21. Constant-Volume Calorimetry qsys = qwater + qCal + qrxn qsys = 0 qrxn = - (qwater + qCal) qwater = msDt qCal = CCalDt No heat enters or leaves!

  22. Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves!

  23. N2H4(g) + H2(g) 2H2(g) + N2(g) 3H2(g) + N2(g) 3H2(g) + N2(g) N2H4(g) 2NH3(g) 2NH3(g) 2NH3(g) Hess’s Law Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Haber Process: DH° = -92.2 kJ Multiple-Step Process DH°1 = ? DH°2 = -187.6 kJ DH°1+2 = -92.2 kJ

  24. Hess’s Law DH°1 + DH°2 = DH°1+2 DH°1 = DH°1+2 - DH°2 = -92.2 kJ - (-187.6 kJ) = 95.4 kJ

  25. Hess’s Law 01 • Hess’s Law:The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. • 3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ

  26. Hess’s Law • The industrial degreasing solvent methylene chloride (CH2Cl2, dichloromethane) is prepared from methane by reaction with chlorine: • CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g) • Use the following data to calculate DH° (in kilojoules) for the above reaction: • CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) DH° = –98.3 kJ • CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) DH° = –104 kJ

  27. Standard Heats of Formation H2(g) + 1/2 O2(g)  H2O(l) DH°f = –286 kJ/mol 3/2 H2(g) + 1/2 N2(g)  NH3(g) DH°f = –46 kJ/mol 2 C(s) + H2(g)  C2H2(g) DH°f = +227 kJ/mol *Standard state (°): P = 1atm, t = 25°C, Concentration = 1 mol/Lit

  28. Standard Heats of Formation • Standard Heats of Formation (DH°f): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states. • The standard heat of formation for any element in its standard state is defined as being ZERO. DH°f = 0 for an element in its standard state

  29. Standard Heats of Formation • Calculating DH° for a reaction: DH° = DH°f (Products) – DH°f (Reactants) • For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. aA + bB cC + dD DH° = [cDH°f(C) + dDH°f(D)] – [aDH°f(A) + bDH°f(B)]

  30. CO(g) -111 C2H2(g) 227 Ag+(aq) 106 CO2(g) -394 C2H4(g) 52 Na+(aq) -240 H2O(l) -286 C2H6(g) -85 NO3-(aq) -207 NH3(g) -46 CH3OH(g) -201 Cl-(aq) -167 N2H4(g) 95.4 C2H5OH(g) -235 AgCl(s) -127 HCl(g) -92 C6H6(l) 49 Na2CO3(s) -1131 Standard Heats of Formation Some Heats of Formation, Hf° (kJ/mol)

  31. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DH0 DH0 - [ ] [ + ] = rxn rxn [ 12x(–393.5) + 6x( -286 ) – [ 2x(49.04) ] = -6542.0 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3271.0 kJ/mol C6H6 6DH0 (H2O) -6542.0 f 2 mol Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. See Page 292 for Standard Heat of Formation

  32. 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) Standard Heats of Formation Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O. DH° = ? DH° = [DH°f (C6H12O6(s))] - [6 DH°f (CO2(g)) + 6 DH°f (H2O(l))] DH° = [(1 mol)(-1260 kJ/mol)] - [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] = 2816 kJ

  33. Bond Dissociation Energy • Bond Dissociation Energy:Can be used to determine an approximate value for DH°f. DH°f = D (Bonds Broken) – D (Bonds Formed) ΔHorxn = D (Reactant bonds) ­ - D (Product bonds) • For the reaction: H-H (g) + Cl-Cl (g)2H-Cl (g) DH°rxn = (D(H–H) + D(Cl-Cl))-(2 X D(H–Cl) ) DH°rxn= -185 kJ DH°f = -92.5 KJ/mol

  34. Bond Dissociation Energy 02

  35. H H H + H-O-H H C C C H H H O-H H C H Calculate an approximate DH° (in kilojoules) for the synthesis of ethyl alcohol from ethylene:C2H4(g) + H2O(g)  C2H5OH(g) ΔHorxn = D (Reactant bonds) ­- D (Product bonds) ΔHorxn = (DC=C + 4 DC­H + 2 DO­H) - ­ (DC­C + DC­O + 5 DC­H + DO­H)

  36. ΔHorxn = (DC=C + 4 DC­H + 2 DO­H) ­- (DC­C + DC­O + 5 DC­H + DO­H) ΔHorxn = [(1 mol)(611 kJ/mol) + (4 mol)(410 kJ/mol) +(2 mol)(460 kJ/mol)]­ - [(1 mol)(350 kJ/mol) + (1 mol)( 350 kJ/mol) + (5 mol)(410 kJ/mol) + (1 mol)(460 kJ/mol)] ΔHorxn = ­39 kJ

  37. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

  38. Predicting spontaneity of a chemical Reaction • Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. • A spontaneous process is one that proceeds on its own without any continuous external influence. • A nonspontaneous processtakes place only in the presence of a continuous external influence.

  39. Introduction to Entropy 02 • The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. • Entropy has units of J/K (Joules per Kelvin). DS = Sfinal – Sinitial • Positive value of DS indicates increased disorder. • Negative value of DS indicates decreased disorder.

  40. Introduction to Entropy 03

  41. Introduction to Entropy 05 • Predict whether DS° is likely to be positive or negative for each of the following reactions. • a. 2 CO(g) + O2(g)  2 CO2(g)b. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)c. C2H4(g) + Br2(g)  CH2BrCH2Br(l)d. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)

  42. Introduction to Entropy 04 • To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: • Spontaneous process: Decrease in enthalpy (–DH), Increase in entropy (+DS). • Nonspontaneous process: Increase in enthalpy(+DH),Decrease in entropy (–DS).

  43. Introduction to Free Energy 01 • Gibbs Free Energy Change (DG): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process. DG = DH – TDS DG < 0 Process is spontaneous DG = 0 Process is at equilibrium DG > 0 Process is nonspontaneous

  44. Which of the following reactions are spontaneous under standard conditions at 25°C? • AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) DG° = –55.7 kJ • 2 C(s) + 2 H2(g)  C2H4(g)DG° = 68.1 kJ • N2(g) + 3 H2(g)  2 NH3(g) DH° =-92.2 kJ DS° = -199 J/K ΔGo = ΔHo ­-TΔSo = (­-92.2 kJ) ­- (298 K)(-­0.199 kJ/K) = ­-32.9 kJ Because ΔGo is negative, the reaction is spontaneous.

  45. Introduction to Free Energy 04 • Equilibrium (DG° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? • N2(g) + 3 H2(g)  2 NH3(g) DH° = –92.0 kJ DS° = –199 J/K • Equilibrium is the point where DG° = DH° – TDS° = 0 • T = 462 K

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