Chapter 10 Energy Changes in Chemical Reactions 1

Energy Changes in Chemical Rxns Most reactions give off or absorb energy Energy is the capacity to do work or supply heat. Heat: transfer of thermal (kinetic) energy between two systems at different temperatures (from hot to cold) Metal bar in water Metal bar drilled

Types of Energy Work (w): energy transfer when forces are applied to a system Heat (q): energy transferred from a hot object to a cold one Radiant energy heat from the sun Thermal energy  associated with motion of particles Potential energy  energy associated with object’s position or substance’s chemical bonds Kinetic energy  energy associated with object’s motion

Heat versus Temperature Describe the difference between the two. SI unit of energy: J 1 watt = 1 J/s, so a 100 Watt bulb uses 100 J each second We often use the unit of kJ to refer to chemical heat exchanges in a reaction. 1 kJ = 1000 J Energy is also reported in calories: Amount of energy needed to raise 1 gram of water by 1oC 1 cal = 4.184 J; 1 Cal = 4184 J Cal (or kcal) is used on food labels Molecular heat transfer

Energy and Energy Conservation Heat: form of energy transferred from object at higher temperature to one at lower temperature (from hot object to cold object) Thermochemistry: study of heat changes in chemical reactions, in part to predict whether or not a reaction will occur Thermodynamics: study of heat and its transformations First Law of Thermodynamics: Energy can be converted from one form to another but cannot be created or destroyed

System and Surroundings System loses heat (negative); gains heat (positive)

Endothermic vs Exothermic Endothermic reaction: q is positive (q > 0) Reaction (system) absorbs heat Surroundings feel cooler Exothermic reaction: q is negative (q < 0) Reaction releases heat Surroundings feel warmer

Enthalpies of Reaction Determine if the following processes are endothermic or exothermic… Combustion of methane Reacting Ba(OH)2 with NH4Cl Neutralization of HCl Melting CaCO3 (s)  CaO (s) + CO2 (g) 8

Answers to Enthalpies of Reaction Combustion of methane exothermic Reacting Ba(OH)2 with NH4Cl endothermic Neutralization of HCl exothermic Melting endothermic CaCO3 (s)  CaO (s) + CO2 (g) endothermic Combustion, neutralization, and combination reactions tend to be exothermic Decomposition reactions tend to be endothermic Melting, boiling, and sublimation are endothermic 9

Applications of heat emission/absorption 10

Specific Heat and Heat Capacity Specific heat (sp. ht.): amount of heat required to raise 1 gram of substance by 1oC Use mass, specific heat, and DT to calculate the amount of heat gained or lost: q = msDT  ms = C  q = CDT Heat capacity (C): amount of heat required to raise the temperature of a given quantity of a substance by 1oC; C = q / DT = J / oC Molar heat capacity (Cm): amount of heat that can be absorbed by 1 mole of material when temperature increases 1oC; q = (Cm) x (moles of substance) x (DT) = J / mol • oC

Specific Heat and Heat Capacity

Practice Problem Calculate the amount of heat transferred when 250 g of H2O (with a specific heat of 4.184 J/g·oC) is heated from 22oC to 98oC. q = msDT Is heat being put into the system or given off by the system? If a piece of hot metal is placed in cold water, what gains heat and what loses heat? Which one will have a positive q value and which will have a negative q value?

Practice Problem 34.8 g of an unknown metal at 25.2oC is mixed with 60.1 g of H2O at 96.2 oC (sp. ht. = 4.184 J/g·oC). The final temperature of the system comes to 88.4oC. Identify the unknown metal. Specific heats of metals: Al 0.897 J/g·oC Fe 0.449 J/g·oC Cu 0.386 J/g·oC Sn 0.228J/g·oC 14

Calorimetry and Heat Capacity Heat changes in a reaction can be determined by measuring the heat flow at constant pressure Apparatus to do this is called a calorimeter. Heat evolved by a reaction is absorbed by water; heat capacity of calorimeter is the heat capacity of water. 15

Example A 28.2 gram sample of nickel is heated to 99.8oC and placed in a coffee cup calorimeter containing 150.0 grams of water at 23.5oC. After the metal cools, the final temperature of the metal and water is 25.0oC. qabsorbed + qreleased = 0 Which substance absorbed heat? Which substance released heat? Calculate the heat absorbed by the substance you indicated above.

Group Quiz #25 A hot piece of copper (at 98.7oC, specific heat = 0.385 J/g•oC) weighs 34.6486 g. When placed in room temperature water, it is calculated that 915.1 J of heat are released by the metal. What gains heat? What loses heat? What is the final temperature of the metal? Watch signs!!!!

Enthalpies of Physical/Chemical Changes Enthalpy (H) describes heat flow into and out of a system under constant pressure Enthalpy (a measure of energy) is heat transferred per mole of substance. At constant pressure, qp = DH = Hproducts – Hreactants DH > 0  endothermic (net absorption of energy from environment; products have more internal energy) DH < 0  exothermic (net loss of energy to environment; reactants have more internal energy)

Heating a Pure Substance (Water) Why does T become constant during melting and evaporating? Melting, vaporization, and sublimation are endothermic We can calculate total heat needed to convert a 15 gram piece of ice at -20oC to steam at 120oC. 2.09 J/goC 4.184 J/goC 2.080 J/goC 334 J/g 2250 J/g 19

Enthalpies of Phase Changes Heat of fusion (DHfus): Amount of heat required to melt (solid  liquid) Heat of vaporization (DHvap): Amount of heat required to evaporate (liquid  gas) Heat of sublimation (DHsub): Amount of heat required to sublime (solid  gas) Why are there no values for DHfreezing, DHcondendsation, or DHdeposition?

Thermochemical Equations Shows both mass and enthalpy relationships 2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s) DHo = -852 kJ Amount of heat given off depends on amount of material: 852 kJ of heat are released for every 2 mol Al, 1 mol Fe2O3, 2 mol Fe, and 1 mol Al2O3

Thermochemical Equations 2Al (s) + Fe2O3 (s)  2Fe (s) + Al2O3 (s) DHo = -852 kJ How much heat is released if 10.0 grams of Fe2O3 reacts with excess Al? What if we reversed the reaction? Heat would have to be put in to make the reaction proceed: 2Fe (s) + Al2O3 (s)  2Al (s) + Fe2O3 (s) DHo = +852 kJ

Hess’s Law If a compound cannot be directly synthesized from its elements, we can add the enthalpies of multiple reactions to calculate the enthalpy of reaction in question. Hess’s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps Look at direction of reaction and amount of reactants/products

Hess’s Law Value changes sign with direction Figure 8.5 24

Hess’s Law Values of enthalpy change DHT = DH1 + DH2 + DH3 + …. For a reaction in the reverse direction, enthalpy is numerically equal but opposite in sign Reverse direction, heat flow changes; endothermic becomes exothermic (and vice versa); sign of DH changes Proportional to the amount of reactant consumed Twice as many moles = twice as much heat; half as many moles = half as much heat DHT = DH1 + DH2 + DH3 + …. 25

Enthalpy of Chemical Reaction Thermochemical equation: H2(g) + I2(s)  2HI(g) DH = +53.00 kJ Two possible changes: Reverse the equation: 2HI(g)  H2(g) + I2(s) DH = -53.00 kJ Double the amount of material: 2H2(g) + 2I2(s)  4HI(g) DH = +106.00 kJ 26

Hess’s Law Calculate DHo for 2NO (g) + O2 (g)  N2O4 (g) DHo = ? N2O4 (g)  2NO2 (g) DHo = 57.2 kJ NO (g) + ½ O2 (g)  NO2 (g) DHo = -57.0 kJ

Hess’s Law We can use known values of DHo to calculate unknown values for other reactions P4 (s) + 3 O2 (g)  P4O6 (s) DH = -1640.1 kJ P4 (s) + 5 O2 (g)  P4O10 (s) DH = -2940.1 kJ What is DHo for the following reaction? P4O6 (s) + 2 O2 (g)  P4O10 (s) DH = ? 28

Hess’s Law

Hess’s Law Given: 2NH3(g)  N2H4(l) + H2(g) DH = 54 kJ N2(g) + H2(g)  NH3(g) DH = -69 kJ CH4O(l)  CH2O(g) + H2(g) DH = -195 kJ Find the enthalpy for the following reaction: N2H4(l) + CH4O(l)  CH2O (g) + N2(g) + 3H2(g) DH = ? kJ

Group Quiz #26 Given the following equations: 2CO2 (g)  O2 (g) + 2CO (g) DH = 566.0 kJ ½ N2 (g) + ½ O2 (g)  NO (g) DH = 90.3 kJ Calculate the enthalpy change for: 2CO (g) + 2NO (g)  2CO2 (g) + N2 (g) DH = ? 31

Standard Heats of Formation Standard heat of formation (DHof): heat needed to make 1 mole of a substance from its stable elements in their standard states DHof = 0 for a stable (naturally occurring) element Which of these have DHof = 0? CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s), P4(s) Do the following equations represent standard enthalpies of formation? Why or why not? 2Ag (l) + Cl2 (g)  2AgCl (s) Ca (s) + F2 (g)  CaF2 (s) 32

Standard Enthalpies of Formation Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book) DHorxn = SnDHof (products) – SnDHof (reactants) S = sum; n = number of moles (coefficients) Direct calculation of enthalpy of reaction if the reactants are all in elemental form Sr (s) + Cl2 (g)  SrCl2 (g) DHorxn = [DHof (SrCl2)] – [DHof (Sr) + DHof (Cl2)] = -828.4 kJ/mol

Standard Enthalpies of Formation Some Common Substances (25oC)

Heats of Formation DHorxn = S DHof,products - S DHof,reactants Calculate values of DHo for the following rxns: 1) CaCO3 (s)  CaO (s) + CO2 (g) 2) 2C6H6 (l) + 15O2 (g)  12CO2 (g) + 6H2O (l) DHof values: CaCO3: -1207.1 kJ/mol; CaO: -635.5 kJ/mol; CO2: -393.5 kJ/mol; C6H6: 49.0 kJ/mol; H2O(l): -285.8 kJ/mol 35

Group Quiz #27 Use Standard Heat of Formation values to calculate the enthalpy of reaction for: C6H12O6(s)  C2H5OH(l) + CO2(g) Hint: Is the equation balanced? DHof (C6H12O6(s)) = -1260.0 kJ/mol DHof (C2H5OH(l)) = -277.7 kJ/mol DHof (CO2(g)) = -393.5 kJ/mol

Bond Dissociation Energies Bond Dissociation Energy (or Bond Energy, BE): energy required to break a bond in 1 mole of a gaseous molecule Reactions generally proceed to form compounds with more stable (stronger) bonds (greater bond energy) H2 Bond Energy 37

Bond Dissociation Energies Bond energies vary somewhat from one mole- cule to another so we use average bond dissociation energy (D) H-OH 502 kJ/mol Avg O-H = 453 H-O 427 kJ/mol kJ/mol H-OOH 431 kJ/mol 38

Bond Dissociation Energies 39

Bond Dissociation Energies DHorxn = SBE (reactants) + - SBE (products) endothermic exothermic energy input energy released SBE(react) > SBE(prod)  endothermic SBE(react) < SBE(prod)  exothermic Use only when heats of formation are not available, since bond energies are average values for gaseous molecules 40

Heats of Reaction Use bond energies to calculate the enthalpy change for the following reaction: N2(g) + 3H2(g)  2NH3(g) DHrxn = [BEN  N + 3BEH-H] + [-6BEN-H] DHrxn = [945 + 3(436)] – [6(390)] = -87 kJ measured value = -92.2 kJ Why are the calculated and measured values different? 41

Heats of Reaction Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl3 (g)  N2 (g) + Cl2 (g) How many distinct bond types are there in each molecule? How many of each bond type do we need to calculate DHrxn? BE(N-Cl) = 200 kJ/mol BE(N≡N) = 945 kJ/mol BE(Cl-Cl) = 243 kJ/mol 42

Answer 6(N-Cl) + -1(N N) + -3(Cl-Cl) 6(200) + -(945) + -3(243) = -474 kJ ≡

Thermochemistry Calculation Summary Use q = msDT (s = J/g·oC) If given mass of reactant, convert to moles and multiply by enthalpy to find total heat transferred If given multiple equations with enthalpies, use Hess’s Law If given DHof values: products – reactants If given bond energy (BE) values: +reactants + -products

Practice Problems Identify how to set up the following problems: Calculate the DHo of reaction for: C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) DHof C3H8(g): -103.95 kJ/mol; DHof CO2(g): -393.5 kJ/mol; DHof H2O(l): -285.8 kJ/mol 8750 J of heat are applied to a 170 g sample of metal, causing a 56oC increase in its temperature.  What is the specific heat of the metal? Which metal is it?

Practice Problems C2H4(g ) + 6F2(g)  2CF4(g) + 4HF(g) DHo = ? H2 (g) + F2 (g)  2HF (g) DHo = -537 kJ C (s) + 2F2 (g)  CF4 (g) DHo = -680 kJ 2C (s) + 2H2 (g)  C2H4 (g) DHo = 52.3 kJ Use average bond energies to determine the enthalpy of the following reaction CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g) (BEC-Cl = 328 kJ/mol)

The End