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Thermochemistry Definition: The study of the energy changes associated with a physical and/or chemical change of a system. Thermal Energy: arises from a particle’s position and motion. Systems: Chemical - _______________________________________________________ - _______________________________________________________ Surroundings - ___________________________________________________ Open -____________________________________________________________ Closed - __________________________________________________________ Isolated - _________________________________________________________ Others: Heat (q) - ________________________________________________________ Exothermic - ______________________________________________________ Endothermic - _____________________________________________________ Temperature - _____________________________________________________ Calculation: q = mc∆T where m = mass surroundings, g c = specific heat capacity, _____________________ ∆T = temperature change = ____________________ eg. Calculate the heat generated when 50.0 g of water (cH2O = 4.18 J/g•ºC) changes temperature from 12.5ºC to 47.0ºC? Solution: q = mc∆T = eg. Calculate the heat generated when 50.0 g of water and 125 g of the Al can (cAl = 0.900 J/g•ºC) changes temperature from 12.5ºC to 47.0ºC? Solution: q = mc∆T = mH2O x cH2O x ∆T + mAl x cAl x ∆T Heat Transfer and Energy Change Chemical system energies, PE & KE, are sources of Enthalpy, H - interactions between nuclei and electrons - electron motions - vibration of atoms within molecules - rotation and translation of molecules - nuclear potential energies - electronic PE of atoms connected by chemical bonds and molecules connected by forces of attraction Enthalpy Change, ∆H Results in the difference in enthalpies of the reactants and the products during a change, due to: - breakage of bonds or intermolecular forces ∆H = +ve (endo) - formation of bonds or intermolecular forces ∆H = -ve (exo) Determination of Enthalpy Change, ∆H - H and ∆H cannot be directly measured - measured relative to the energy transferred, heat (q), during the change in the system ∆Hsys = - qsurr = - mc∆Tsurr eg. Calculate the enthalpy change for a combustion reaction which results in the temperature of 500. mL of water in a 0.250 kg Al (cAl = 0.900 J/g•°C) tea kettle rising from 20.0 °C to 95.0 °C. ∆H = - (mc∆TH2O + mc∆TAl) ; = - (mcH2O + mcAl) ∆T = VH2O = mH2O as DH2O = 1.0 g/mL This is an exothermic reaction showing the change in PE of the SYSTEM is equal to the change in KE of the SURROUNDINGS PE is that of position or the ability to work. When objects that attract each other: Get closer together, PE __ Move further apart, PE __ KE is that of motion. average KE __ as T __ Chemical change involves 2 processes that alter the PE of atoms. Bond breaking ____________ energy and PE of the atoms ___. Bond forming ____________ energy and PE of the atoms ___. Depending on which effect is greater, the net change in PE energy may be an ___________ or a ________. The ___________________________________________________________, the __________________ the bond. Molar Enthalpies Molar Enthalpy, ∆Hrxn(X) the ∆H for 1 mole of any substance, (X), undergoing a reaction. standard unit is kJ/mol eg. rxn solution _______ combustion formation _______ neutralization _________ _________ Determination: H rxn (x) = - q - m surrcsurr T = nx nx = eg. - m surrcsurr T x Mx mx where X is the reactant where M x Molar Mass of X a) Calculate the ∆Hcomb (C8H18) if 6.593 g causes 1.000 L of H2O to increase its temperature by 75.2 °C. ( DH2O = 1 g/mL or 1kg/L) mH2O = Hcomb (C8H18 ) = -mH2Oc H2OT x M C 8 H18 mC8 H18 = = b) Express this enthalpy change 4 ways: i) ∆Hcomb(C8H18) = ii) 2 C8H18 + 25 O2 16 CO2 + 18 H2O + OR C8H18 + 25/2 O2 8 CO2 + 9 H2O + iii) C8H18 + 25/2 O2 8 CO2 + 9 H2O ; H = iv) PE diagram For comparison: c) Calculate the molar enthalpy of combustion of octane, C8H18, given that 6.593 g causes 150 mL of H2O in a 375 g Al can, to increase its temperature by 75.2 °C. H comb (C 8 H18 ) = - = (m H 2O c H 2O m Alc Al ) T m C8H1 8 x M C8H1 8 Hess’ Law and Heats of Formation A State Function: _____________________ Definition: When a reaction that can be expressed as the ____________ ____, , of two or more __________ reactions, the enthalpy of reaction, Hrxn, is the algebraic sum of the ___________________ rxn enthalpies, Hx. Standard Enthalpies of formation , Hfº Are often used to calculate _______ The enthalpy (_______ or _______ of heat energy) for the _____________ of 1 mole of the ______________ from its ____________ in their standard state. For an _____________ in its standard state, Hfº = 0 Standard state is SATP is 25ºC, 100. kPa These equations are created from its ________________ELEMENTS eg. Rules for applying Hess’ Law: 1) Use the ________ reaction steps or _______ the _____________ equations of the overall reaction and find the value of Hx or Hfº for each step. 2) ___________ intermediate steps (reactions) as needed and remember to __________ the _____ of the Ho for that reaction ( multiply by -1). eg. 3) ____________ intermediate reactions as necessary to _______ the coefficients in the overall equation. Remember to multiply the Ho values by the same multiplier. eg. 4) Determine the Horxn from the algebraic sum of Ho values for all the intermediate steps. 5) If applicable use Hrxnº = nHfº (products) - nHfº (reactants) eg.1 Determine the heat of reaction for the following: 4 NH3 (g) + 5 O2 (g) Using Formation equations: 4 NO (g) + 6 H2O (g) Hrxn = ? eg.2 Determine the heat of reaction for the following: C2H4 (g) + H2 (g) C2H6 (g) Hrxn = ? Using the following GIVEN reactions: 2 CO2 + 7/2 O2 2 CO2 + + 1/2 O2 H2O (l) (1) C2H4 + (2) C2H6 (3) H2 3 O2 + 2 H2O (l) 3 H2O (l) H1 = -1400.5 kJ H2 = -1550.0 kJ H3 = -258.8 kJ eg.3 Repeat question 2, but use the Summation Formula of Hess’ Law: Hrxnº = nHfº (products) - nHfº (reactants) Heat Capacities, c Substance aluminum calcium copper gold hydrogen iron lead lithium magnesium mercury Specific Heat Capacity at SATP (J/(g·°C)) 0.900 0.653 0.385 0.129 14.267 0.444 0.159 3.556 1.017 0.138 Substance nickel potassium silver sodium sulphur tin zinc ice water steam Specific Heat Capacity at SATP (J/(g·°C)) 0.444 0.753 0.237 1.226 0.732 0.213 0.388 2.01 4.18 2.01 Thermodynamic Properties of Organics At SATP ∆H °f Substance S° (kJ•mol-1) (J•K-1•mol-1) benzene, C6H6(l) bromoethane, CH3CH2Br(g) bromomethane, CH3Br (g) butanal, CH3CH2CH2CHO(l) butane, n-C4H10 (g) butan-1-ol, C4H9OH(l) but-1-ene, C4H8 (g) but-1-yne, C4H6 (g) carbon tetrachloride, CCl4 (l) CCl4 (g) chloroethene, CH3CH2Cl(g) chloromethane, CH3Cl (g) cyclopropane, (CH2)3 (g) 1,2-dichloroethane, (CH2Cl)2(g) ethanal, CH3CHO(g) ethane, C2H6 (g) ethane-1,2-diol, (CH2OH)2(l) ethanoic acid, CH3CO2H(l) (acetic) ethanol, C2H5OH(l) ethene, C2H4 (g) ethoxyethane, (CH3CH2)2O(g) ethyne, C2H2 (g) fluoromethane, CH3F (g) glucose, C6H12O6(s) +49.0 -90.5 -37.2 -241.2 -126.5 -327.4 -0.4 +165.2 -128.4 -96.0 -136.8 -82.0 +53.3 -165.0 -191.5 -83.8 -454.8 -484.5 172.8 ----246.3 ---310.1 228.0 305.6 ----216.4 309.9 263.9 234.5 --------160.2 229.5 166.9 159.8 -277.1 +52.5 -279.0 +228.2 -247.0 -1260.0 160.7 219.5 251.9 201.0 ----212.1 Substance ∆H °f S° (kJ•mol-1) (J•K-1•mol-1) hexane, n-C6H14 (l) -198.6 296.1 iodoethane, CH3CH2Cl(g) -40.7 ----iodomethane, CH3I (g) -15.5 163.2 methanal, HCHO(g) -108.7 218.7 (formaldahyde) methane, CH4 (g) -74.8 186.2 methanoic acid, CH3OH(l) -425.1 129.0 (formic) methanol, CH3OH(l) -239.1 239.7 methoxymethane, CH3OCH3(g) -184.0 266.7 methylpropane, C4H10 (g) -134.5 294.6 napthalene, C10H8 (s) +77.7 ----octane, n-C8H18 (l) -250.0 361.1 pentane, n-C5H12 (l) -173.1 262.7 phenylethene or styrene +103.8 345.1 phenol, C6H5OH(s) -165.0 ---propanal, CH3CH2CHO(l) -217.1 ---propane, C3H8 (g) -104.5 269.9 propanone(acetone) -248.0 198.8 propan-1-ol, C3H7OH(l) -302.7 196.6 propene, C3H6(g) +20.2 266.9 propyne, C3H4 (g) +186.6 248.1 sucrose, C12H22O11(s) -2221.0 360.2 2,2,4-trimethyl pentane -259.2 328.0 urea -335.5 104. Substance Al(s) Al2O3 (s) Al2(SO4)3 (s) BaCO3 (s) BaCl2 (s) BCl3 (g) B2O3 (s) Br2 (l) Ca(s) CaCO3 (s) CaBr2 (s) CaCl2 (s) CaO(s) Ca(OH)2 (s) Ca3(PO4)2 (s) CaSO4 (s) C(s) graphite C(s) diamond CO (g) CO2 (g) Cl2 (g) Cu (s) CuCl (s) CuCl2 (s) Cu2O (s) CuO (s) CuSO4 (s) CuSO4·5H2O s) F2 (g) H2 (g) H2O2 (l) HBr(g) HCl (g) HCl (aq) HCN(g) HF (g) HI (g) HNO3 (l) HNO3 (aq) H3PO4 (s) H2S (g) H2SO4 (l) H2SO4 (aq) I2 (s) I2 (g) Fe (s) FeO (s) Fe2O3 (s) FeCl2 (s) FeCl3 (s) Thermodynamic Properties of Inorganics At SATP) ° ° ° ° ∆H f S° ∆G f Substance ∆H f S° ∆G f (kJ•mol-1) (J•K-1•mol-1) (kJ•mol-1) (kJ•mol-1) (J•K-1•mol-1) (kJ•mol-1) 0 28.3 0 Pb (s) 0 64.8 0 -1675.7 50.9 -1582.3 PbCl2 (s) -359.4 136.0 -314.1 -3405.5 50.9 -1582.3 PbO (s) -219.0 66.5 -188.6 -1216.3 112.1 -1142.0 PbO2 (s) -277.4 68.6 -----860.2 124.1 -813.5 Mg (s) 0 32.7 0 -404 291.1 -390.1 MgCO3 (s) -1095.8 65.7 -----1273 54.18 -1144.1 MgCl2 (s) -641.3 89.63 -591.8 0 152 0 Mg(OH)2 (s) -924.5 63.24 ----0 41.4 0 MgO (s) -601.6 26.95 -569.4 -1207 92.9 -1128.8 N2 (g) 0 191.6 0 -682.8 130.2 -1352.4 NH3 (g) -45.9 192.78 -16.5 -795.8 104.6 -748.1 N2H4 (l) 50.6 121.2 149.3 -634.9 38.1 -566.5 N2H4 (g) +95.4 237.11 -----986.1 83.4 -901.7 NH4Cl (s) -314.4 94.6 -202.9 -4119 236.9 -3897.7 NH4NO3 (s) -365.6 151.08 -183.9 -1434.1 108.4 -1326.8 NO (g) 90.2 210.76 86.6 0 5.7 0 NO2 (g) 33.2 240.1 51.3 +1.88 2.38 +2.90 N2O (g) 82.1 219.9 104.2 -110.5 197.66 -137.2 N2O4 (g) 9.2 304.3 97.9 -393.5 213.78 -394.4 O2 (g) 0 205.1 0 0 223.1 0 O3 (g) 142.7 238.9 163.2 0 +33.2 0 PCl3 (g) -319.7 217.2 -----137.2 86.2 -119.9 PCl5 (g) -443.5 364.6 -----220.1 108.1 -175.7 K (s) 0 64.2 0 -168.6 93.1 -146.6 KCl (s) -436.7 82.55 -409.1 KClO3 (s) -397.7 143.1 -296.3 -157.3 42.6 -129.7 KOH (s) -424.8 78.9 -379.1 -771.4 109.2 -663.6 Ag 0 42.6 0 -2279 301.6 -1887.1 (s) AgBr (s) -100.4 107.11 -97.4 0 202.8 0 AgCl (s) -127.0 96.25 -109.8 0 130.7 0 AgNO3 (s) -124.4 140.9 -33.4 -187.8 109.6 -120.4 Ag2O (s) -31.1 121.8 -11.3 -36.3 198.7 -53.5 Na (s) 0 51.2 0 -92.3 186.9 -95.3 NaBr -361.1 86.82 -350.2 (s) -167.2 56.7 -131.8 Na2CO3 (s) -1130.7 135.0 -1044.0 +135.1 201.81 +125.2 NaCl (s) -411.2 115.5 ------271.1 +173.8 -273.2 NaF -571 51.7 -545.6 (s) +26.5 206.59 +1.75 NaOH -425.6 64.4 -379.5 (s) -174.1 155.6 -80.7 NaI (s) -287.8 98.50 -287.3 -207.0 --------S rhombic 0 31.8 0 8 (s) -1279.0 110.5 -1119.1 S (g) 278.8 167.8 283.3 -20.6 205.8 -33.6 SO -296.8 248.22 -300.2 2 (g) -813.8 156.9 -690.0 SO3 (g) -395.7 256.8 -371.1 -909.3 20.16 -743.4 SnO (s) -280.7 57.17 ----0 116.3 0 SnO -577.6 49.04 ----2 (s) +62.4 180.79 ----H O -285.8 69.95 -237.1 2 (l) 0 27.8 0 H -241.8 188.84 -228.6 2O (g) -272.0 57.6 +245.1 ZnO -350.5 43.65 ----(s) -824.2 87.4 -742.2 ZnS (s) -206.0 57.7 -----341.8 118.0 -302.8 -399.5 142.3 -344.0 Energy and Driving Forces: The Laws of Thermodynamics Spontaneous Reaction: given the required Ea, ________________ ___________the reaction continues to proceed by itself may be slow or fast Thermodynamics: Study of energy transformations. There are 3 Laws of Thermodynamics – used to predict reaction spontaneity 1st Law of Thermodynamics Law of Conservation of Energy Total energy of the universe is constant Energy cannot be created or destroyed, just transferred from one form to another For a chemical reaction ∆Hºuniverse = __________________________________________ In a chemical reaction, the PE of the reactants and products results in the transfer of energy from the: 1) surroundings to the chemical system (ENDO) 2) chemical system to the surroundings (EXO) Enthalpy changes and Spontaneity Bond energy (BE): the minimum energy required to break one mole of bonds between two atoms ______________ Equals the energy released when 1 mole of bonds are formed, _________ The greater the value, the more stable the bond SO, the Enthalpy of reaction can also be calculated by: ∆Hreaction = ___________________________________________ ∆Hreaction __________ indicates the formation of stronger bonds and more stable compounds is most likely to be spontaneous reaction Entropy Changes (∆S) and Spontaneity the greater the # of ways particles can arrange themselves, the less ordered they are the greater the # of ways a particular state can be achieved, the more likely that state is going to exist Entropy , Sº The measure of disorder or randomness increased disorder or entropy favours spontaneity, ____________ 2nd Law of Thermodynamics ∆Sºuniverse = disorder increases, since ____________ The following increases entropy: 1. The volume of a gas increases 2. The temperature of the system increases 3. Physical state solid to liquid to gas 4. Increase in the number of moles produced 5. Breaking complex molecules into smaller ones 3rd Law of Thermodynamics At 0K all motion ceases, the forces of attraction have reduced entropy to a minimum S = 0 at T = 0K As a result as T increases, S must increase S is a measure of the energy needed to achieve a level of disorder by overcoming the FA This is dependent on the substance and the temperature reached J/molK ∆Srxnº = ______________________________________ ∆Srxnº _________________________________ Entropy Calculations eg. Calculate ∆Sº for 2CO(g) + O2(g) 2CO2(g) ∆Srxnº = nSº products - nSº reactants = ___________________________________________________ This would mean it is nonspontaneous, but we need to consider the ∆Hrxnº ∆Hrxnº = n∆H º products - n∆H º reactants = Gibb’s Free Energy – Enthalpy and Entropy Unites the 2 reaction driving forces ∆G = ∆H – T∆S If: ∆G < 0 ∆G = 0 ∆G > 0 ___________________________ ___________________________ ____________________________________________ For Spontaneity: ∆H - ∆S + + - - - + + ∆G spontaneity Calculating ∆Gº from ∆Hº and ∆Sº eg. 2CO(g) + O2(g) ∆Hºrxn = -566.0 kJ ∆Sºrxn = -172.86 J/K 2CO2(g) = -0.17286 kJ/K ∆Gº = ∆Hº - T∆Sº = Calculating ∆Gº from ∆Gfº these values are not in the textbook, but are in the reference booklet Use: ∆Gº = ____________________________________ Predicting Change in Spontaneity Temperature and equilibrium As ∆G = 0 at equilibrium and ∆Gº = ∆H – T∆S then: 0 = ∆H – T∆S this gives the temperature at which the system changes spontaneity need to examine ∆H and ∆S to determine if the reaction is spontaneous above or below the temperature calculated eg. For previous example: 2CO(g) + O2(g) 2CO2(g) T = ∆H ∆S = as ∆H –ve and ∆S –ve , then spontaneous at low T T> T< Chemical Kinetics Definition: the study of reaction rates Rate of Reaction: The change in concentration, ∆[ ], of a _________ or a __________ per unit of time can be determined from a [ ] vs time graph Average Rate of Rxn: from the slope of the secant drawn between 2 points on the curve over a given time interval. Instantaneous Rate of Rxn: from the slope of the tangent to the curve at a specific moment in time. Reaction Rates are given as: as rates are __________ and as the rxn proceeds, ______________ the rates of the different reactants and products are ________ by the balanced chemical reaction. In General, for aA + bB cC + dD Rrxn req’d = rate of rxn given x mole ratio of req’d given eg. For 2 C2H6 + 7 O2 4 CO2 + 6 H2O The rate of reaction with respect to C2H6 is 4.0 x 10-4 mol/L•s. State the Rrxn with respect to each product and reactant. R rxn (C 2 H 6 ) - [C 2 H 6 ] 4.0 x 10-4 mol/L s t Measuring Raterxn • requires an _______________, _____________ change that doesn’t disturb the rxn: 1) ________________________________________________ 2) ________________________________________________ 3) ________________________________________________ Factors Affecting the Raterxn 1) Nature of the Reactant eg. aqueous, gas, liquid, reactivity • depends on the _______ of the ______________ particle forces. 2) Concentration ___________________________________ 3) Temperature ___________________________________ 4) Surface Area ___________________________________ 5) Catalyst The Rate Law and the Rate Law Equation As various factors (T, [ ], etc) affect the rate of the rxn, their impact MUST be determined empirically through experiment. aX + bY product For Rate Law: Raterxn _____________________ where m, n R determined experimentally i = initial concentration Rate Law Equation: Raterxn = ___________________ where k is the rate constant for a specific temp. Order of Rxn: m = ________________________ n = ________________________ m + n = ________________________ eg. Raterxn = k [A]1[B]2 [C]0 = k [A]1[B]2 order = 3 ∆[ ] 2 (doubles) 3 (triples) ∆Rate 0th order 1st order 2nd order _______ _______ _______ _______ _______ _______ ______________________________ For the following reaction, the data in the table was obtained: 2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g) Trial [NO] i (mol / L) 1 2 3 0.100 0.100 0.200 [H2] i (mol / L) Initial Rate (mol / Ls) 0.100 0.200 0.100 0.00123 0.00246 0.00492 Conclusion k can be determined from any of the expt trials: using Trial 3: [NO]i = 0.200 mol/L, [H2]i = 0.100 mol/L R = 0.00492 mol/L•s and R= k [NO]2[H2]1 Reaction Rate and Time • often rates of rxn are measured by the time that it takes for a certain point occurs in a rxn • usually measured by a colour change – called a clock rxn For A products Graphically: Reaction Mechanisms are the ______ of _______ that make up a ___________ . each one of these, called _______________ _______, involves the collision of __, __ or at MOST __ particles. each Elementary step has a _____ law that _________ the ______________ of the ___________. the Rate lawrxn ________ the Rate ____ for the ________ step 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) Occurs in 3 elementary steps: eg. 1) 2) 3) where N2O2 and N2O are __________ __________________, species that are ________________ and then ______________ Proposed Reaction Pathway Step 2 is the slow step, with the ___________ Ea this is called the _______ _______________________ ___________, _____ the overall Raterxn is proportional to the rate of this step only as IT _____________ THE ______ !!!!!! the Raterxn can be taken _________ from the ___________ in the ______ . as 2) N2O2 + H2 H2O + N2O slow but N2O2 is not a reactant in the overall equation (intermediate) so the rate law equation is rewritten to include reactants only. from 1) NO + NO N2O2 Developing the Reaction Mechanism from Experimental Data For the reaction NO2 + CO NO + CO2 [NO2]i Trial mol/L 1 0.0100 2 0.0200 3 0.0100 [CO]i mol/L 0.0100 0.0100 0.0200 Rate Conclusion 4.2 x 10-3 1.68 x 10-2 4.2 x10-3 Possible Pathway assume RDS step (1) - often true as reactants require bond breakage 1) 2) where NO3 is a reaction intermediate, but NO2 is a CATALYST – it is used up and then reformed. Chain Reactions are _____________ in which an _____________ formed in an ____________ step is _____________ and then acts as a ____________ . often involve ___________ conditions due to the vast amount of available ________, gases and the existence of _____ ________ which have an _________ ___ produced when ____________ molecules’ ______ are ______ and are ______________ reactive. Are _______ to be confused with other ______________ . eg. H2 + Cl2 UV 2 HCl Initiation Propagation Propagation Termination overall These free radicals also cause big problems in the ozone layer Normally: Problem: Initiation Propagation Propagation Termination overall Collision Theory of Rates of Reactions The 2 requirements that determine whether a potential collision between molecules result in a reaction are: 1) Sufficient energy 2) Correct Orientation If both of these conditions are meet then an unstable species, called the activated complex is formed. This can be: Raterxn Factors Affecting Raterxn and Collision Theory Raterxn = collision frequency I) Concentration x fraction effective II) Surface area III) Temperature i) collision frequency ii) fraction effective Maxwell – Boltzmann Distribution IV) Nature of reactant V) Catalyst with PE diagram