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Text of Thermochemistry

PHYSICAL CHEMISTRYTemperature and Heat1. Thermochemistry is the study oI energy changes which take place during chemical reactions.2. Temperature is the degree oI hotness.3. Molar heat capacity: energy required to raise the temperature oI one mole oI the substance by one kelvin or (1oC)4. SpeciIic heat capacity (c): energy required to raise the temperature oI one g oI the substance by one kelvin or (1oC)5. The speciIic heat capacity Ior water 4.18 J g-1K-16. AH m c AT heat absorbed or released7. When zinc powder is added to plastic cup which contains 50 cm3oI Cu2(aq) solution, the temperature rises by 6.2oC. The speciIic heat capacity Ior Cu2solution is 4.2J g-1K-1Calculate the heat liberated during the reaction. |1.302 kJ|1. Enthalpy Change, , is given: The total enthalpy of products - The total enthalpy of reactants2. Exothermic process: heat given out to surrounding. The value has a negative sign as the heat content oI productis less than that oI the reactant. Define the endothermic process.Enthalpy ChangesEnergy Level Diagram1. When one mole oI methane is burnt in excess oxygen at a constant pressure, the heat liberated is 890 kJ. The thermochemical equation is:C4 (g) 2O2 (g)CO2 (g)22O (l) -890 kJ enthalpy change oI reaction under standard conditions.a) The total amount oI energy released or absorbed is directly proportional to the number oI moles oI reactants used.b) The enthalpy change Ior the reverse reaction is equal in magnitude but opposite in sign to the enthalpy change Ior the Iorward reaction.Thermochemical EquationsHess's Law iI a reaction is carried out in a series oI steps,Ior the reaction will be equal to the sum oI the enthalpy changes Ior the individual steps the overall enthalpy change Ior the process is independent oI the number oI steps or the particular nature oI the path by which the reaction is carried out. Consider the combustion reaction oI methane to Iorm CO2 and 6:/ H2O H4(g + 2O2(g -> O2(g + 2H2O( This reaction can be thought oI as occurring in two steps: In the Iirst step methane is combusted to produce water vapor: C4() 2O2() -~ CO2() 22O()-802 kJ In the second step water vapor condenses Irom the gas phase to the liquid phase: 22O() -~ 22O()-88 kJess`s LawCombining these equations yieIds the foIIowing:CH4()+2O2()+2H2O()CO2()+2H2O()+2H2O() AH = (-802) kJ + (-88) kJ= -890 kJ Determining AH for the reaction is difficult because some CO2 is also typically produced. However, complete oxidation of either C or CO to yield CO2 is experimentally pretty easy to do: We can invert reaction number 2 (making it endothermic) and have CO(g) as a product. (This describes the decomposition oI CO2 to produce CO and O2) AH = 171 iJ) + 2?10 iJ) = ll0 iJombining the Equations(graphite + O2(gO2(g AH -393.5 k1(diamond + O2(gO2(g AH -395.4 k1(graphite(diamond(graphite + O2(gO2(g AH -393.5 k1O2(g(diamond + O2(g AH +395.4 k1(graphite(diamond AH +1.9 k11. The standard heat oI Iormation, or standard enthalpy oI Iormation (represented asoI), Ior a substance is the enthalpy change in the Iormation oI 1 mole oI that substance Irom its individual elements at 1 atmosphere (1 atm) and 25o C (unless otherwise noted). 2. Elements Iound in many Iorms, or allotropes, which have diIIerent heats oI Iormation. To standardize this process, the allotrope with the most stability is used in determining heats oI Iormations. The most stable element will always have a heat oI Iormation 0. 3. The more negative the value, the more energetically stable the compound. $tandard Molar Enthalpy hange of Formation or example, carbon (C) is Iound as CO, CO2, C(graphite), etc. owever, C(graphite) is the most stable Iorm oI C. ThereIore, any heat oI Iormation equation using carbon will use C(graphite), which is represented Ior simplicity, as C(s): C(s) O2(g) CO2(g) C(s)C(diamond) C(s) 1/4 S8(rhombic)CS2(l) C(s) Ca(s) 3/2 O2(g) CaCO2(s) Example #1: Which oI the Iollowing equations represents a reaction that provides the heat oI Iormation oI ethanol (C3C2O)? A) C2 C2(g) 2O(l)C3C2O(l) B) 2C(s) 6 (g) O(g)C3C2O(l) C) 2 CO2(g) 6 2(g)C3C2O(l) 32O(l) D) 2 C(s) 32(g) 1/2 O2(g)C3C2O(l) E) 2 CO(g) 3 2(g) C3C2O(l) 1/2 O2(g) Equation A: CH2 = CH2(g) is not the most stable form of C, and H2O(l) is not the most stable form of H or O (C(s), H2(g) and O2(g) are). Equation B: H(g) is not the most stable form of H (H2(g) is). Equation C: CO2(g) is not the most stable form of C or O (C(s) and O2(g) are). Equation D: Correct. Equation E: CO is not the most stable form of C (C(s) is). Example #2: The heat oI Iormation oI CO2(g) is -394 kJ/mole and that oI 2O(l) is -286 kJ/mole. The heat oI combustion oI C512 is -3534 kJ/mole. What is the heat oI Iormation oI C512? C5

12(l) 8 O2(g) 5 CO2(g) 6 2O(l)Calculate the heats of formation and combustion for the individual elementsof the reaction: AHoc C5H12 = 1 x (-3534) = -3534 kJ/mole AHof O2 = 8 x 0 = 0 kJ/mole A Hof CO2 = 5 x (-394) = -1970 kJ/mole A Hof H2O = 6 x (-286) = -1716 kJ/mole (-1970) + (-1716) - Hof+ 0= -3534 Hof= -152 kJ/mole Example #3: alculate the heat evolved (k1 for the reaction of 2.40 g of H4.(Atomic weights: 12.01. H 1.008 H4 (g + 2 O2 (g O2 (g + 2 H2O (l Ho -890.3 k1/mole0.149 x (-890.3) = -133 kJ/moIe1. The heat oI combustion, or enthalpy oI combustion(represented asoc), is the amount oI energy released in the combustion oI one mole oI a substance. The standard enthalpy of formation of propyne. 3H4. is +185.4 k1/mole. alculate the heat of combustion of one mole of 3H4. The heats of formation of O2(g and H2O(l are -394 k1/mole and -285.8 k1/mole respectively. 1) Write out and balance the equation based on the inIormation given: C3

4 4 O23CO2 2 2OAHoc = -1939 kJ/mole$tandard Molar Enthalpy hange of ombustion!ractice:1. The standard enthalpy oI combustion (in kJ/mol) Ior ethanol and ethanal (C3CO) are -1371 and -1167 respectively. Calculate the standard enthalpy change Ior the oxidation oI ethanol to ethanal.xC3C2O O2 C3CO 2O-1371 -11672CO2 32O1. 2,000.0 g oI water in calorimeter has its temperature raised 3.0SC by an exothermic chemical reaction. ow much heat was transIerred? |or 25 kJ| " = mcAT note: since the temperature increased,AT will be a positive number = (2000.0 g) 4.184 J g SC 3.0SC =25,000 J or 25 kJ (to the correct number of significant digits)SimpIe CaIorimeteromb CaIorimeter1. A weighed sample in a platinum crucible is placed in the 'bomb.2. A volume oI water suIIicient to cover the bomb is added and the entire apparatus is closed.3. Pure oxygen is pumped and initial temperature recorded.4. The sample is ignited. The maximum temperature rise is recorded.5. The heat capacity oI calorimeter is the heat required to raise the temperature oI the whole apparatus by 1 K.!ractice:1. The combustion oI 0.625 g oI benzoic acid causes the temperature oI the water in the bomb calorimeter to increase by 1.572 K while the combustion oI 0.712 g oI an organic compound causes an increase oI 1.54 K. Calculate the standard enthalpy change oI combustion oI the organic compound.| C benzoic acid is -3230 kJmol-1. The RMM Ior benzoic acid and organic compound is 122 and 60.|$olution $olution:eat released oI burning oI acid: 0.625/122 x 3230 16.55 k11.572 K16.55 kJ1.54 K? kJ .. (16.2 kJ) released by burning oI organic compound.$tandard enthalpy change of combustion of the organic compound:16.2 kJ/ no. oI mol organic compound 16.2 /(0.712/60) 1361 k1mol-11. Enthalpy change when one mole oI acid reacts with one mole oI alkali to Iorm one mole oI water under standard condition.2. When strong acid is neutralized by strong base, heat oI neutralization is constant at -57.2 kJmol-1.3. neut Ior diprotic strong acid ~ -57.1 kJ/mol due to dilutionoI acid when alkali is added.

2SO4 NaONaSO4 2O neut -62.0 kJ/molNaSO4 NaONa2SO4 2Oneut -71.0 kJ/molAverage -66.5 k1/mol4. When weak acid or weak alkali is used, Iurther heat change is involved in ionizing the weak acid. Thus, heat oI neutralization is less exothermic (less negative) than -57 kJ/mol.$tandard Molar Enthalpy hange of Neutralization1. Enthalpy change when one mole oI gaseous atoms is Iormed Irom the element in its standard condition.This process requires energy (endothermic)Na (s) Na (g)at 107 kJmol-1 Cl2(g) Cl (g) at 242 kJmol-1$tandard Enthalpy hange of Atomization 1. The first electron affinity: enthalpy change when one electron is added to each gaseous atom in one moleto Iorm one mole oI x-1..2. The second electron affinity: enthalpy change when one electron is added to each gaseous anion in one moleto Iorm one mole oI x-2. O (g) eO-(g) EA1 -144 kJmol-1O-(g) eO2-(g) EA2 791 kJmol-1 1he secon/ eectron affntv s en/othermc beca:se energv s re6:re/ to overcome the rep:sve force between the eectron an/ the x-. Electron Affinity1. The lattice energy: energy given out when one mole oI an ionic crystalline solid Iormed Irom its ions.K(g) Cl- (g) KCl (s) lattice -701 kJmol-1KCl (s)K(g) Cl- (g) lattice dissociation 701 kJmol-12. Factors that affecting the magnitude of lattice energy:a) The attractive Iorces between the positive and negative ionsb) The distance between the ions/ ionic radius.3. a) The larger the charge oI the ions, the more exothermic is the lattice energy b) The smaller the ionic radius,the more exothermic is the lattice energy Energy changes in Forming Ionic $ubstances1. Consider the reaction between Li and Iluorine gas to Iorm lithium Iluoride, Li. The reaction can take place in several steps: Atomization enthalpy of metal (in this case lithium Ionization enthalpy of metal Atomization enthalpy of non-metal (in this case fluorine Electron affinity of non-metal Lattice enthalpy a Li (sLi (g ... atomizationAHat +107 k1/molb Li (gLi+(g .. first ionizationAHIE +496 k1/molc F2 (gF (g .. atomizationAHat +122 k1/mold F + e F- (g .... Electron affinityAHEA -349 k1/mole Li+(g + F- (gLiF(s .. Lattice energy AHLattice ??? k1/molfLi(s+