IB CHEMISTRY Topic 5 Energetics/thermochemistry Higher level level Higher 5.1 Measuring energy changes OBJECTIVES Heat is a form of energy. Temperature is a measure of the average kinetic energy of the particles. Total energy is conserved in chemical reactions. Chemical reactions that involve transfer of heat between the system and the surroundings are described as endothermic or exothermic. The enthalpy change (H) for chemical reactions is indicated in kJ mol-1.
H values are usually expressed under standard conditions, given by H, including standard states. Calculation of the heat change when the temperature of a pure substance is changed using q=mcT. A calorimetry experiment for an enthalpy of reaction should be covered and the results evaluated. Temperature and Heat Temperature is the average kinetic energy of molecules (K). Heat is the amount of energy exchanged due to a temperature difference between two substances (J). Enthalpy (H) Enthalpy (H) the amount of energy or heat
content of a substance. The energy is stored in the chemical bonds. It includes kinetic and potential energy and is not measured directly but changes are measured (J). Standard enthalpy change of reaction (H) H) - the difference between the enthalpy of the products and the enthalpy of the reactants at 298K and 1.00 x 105Pa. Enthalpy Change in enthalpy (H) H) enthalpy of products less reactants. It is the heat energy change per mole. H = HH = HP HR For exothermic reactions the temperature of the surroundings increases and H = HH is negative, energy is given
out For endothermic reactions the temperature of the surroundings decreases and H = HH is positive, energy is absorbed The higher the enthalpy the less stable the substance Enthalpy level diagrams Enthalpy level diagrams The energy is measured by temperature. We can calculate the energy change only. Not the energy stored in the resultant products or reactants. Exothermic reactions
Examples include: 1. Combustion reactions 2. Neutralization reactions In an exothermic reaction products: have less stored energy are more stable have stronger bonds (harder to break) Heat energy change (Q) Q = mcH = HT m = mass in g c = specific heat capacity of the substance JK-1g-1 H = HT = change in temperature in K Q = Heat energy change H = HH = Heat energy change per mole
So, H = HH = Q /n Calorimetry A technique used to measure enthalpy change. Heat absorbed, Q = (m x c x T)liquid c water = 4.18 JK-1g-1 m water = volume in cm3 1 cm3 = 1g Styrofoam Calorimeter Bomb
Calorimeter Assumptions and sources of error That no heat is lost to or gained from the surroundings. Significant errors are associated with reactions: involving the evolution of gas involving combustion where hot gas is used to heat liquid in a calorimeter. Heat rises & T under these conditions are reported to be less than the literature stipulates. Thermometers often have precision uncertainty 0.1oC or greater Measuring Enthalpy changes graphically
Temperature is constantly changing to return to room temperature so lines are extrapolated to calculate the value of T. Problem 1: 1200 kJ of heat is evolved when 2 moles of magnesium react completely with 1 mole of oxygen. How much energy is released if 0.600g of Mg is burnt? 2Mg(s) + O2(g) 2MgO(s) So Hc(Mg) = -1200kJmol-1/2 = -600 kJmol-1 n(Mg) = m/M = 0.600g/24.3g/mol = 0.0247mol H = Q/n so Q = H x n = -600 kJmol-1 x 0.0247mol = 14.8 kJ Problem 2: 20.0cm3 of 2M NaOH is added to 30.0cm3 HCl of the same concentration. The
temperature increases by 12.0oC. Determine the heat released. Total volume is 50.0cm3, water density is 1.00 g cm-3 so mass of aqueous solution is 50.0g. Amount of heat required to heat the water, Q = mcT = 50.0 g x 4.18Jg-1C-1 x 12.0C = 2508 J or 2.51 kJ So, 2.51 kJ is the amount of heat energy evolved in the reaction: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) So sodium hydroxide is the limiting reagent. n(NaOH) = CV = 2M x 0.02dm3 = 0.0400 moles (so the enthalpy PER MOLE can be determined)
H(NaOH) = Q/n = 2.51kJ/0.04moles = 62.7 kJ/mol The energy is evolved, exothermic, so H(NaOH) = -62.7 kJ/mol Problem 3: 300 mL of 0.2 M aqueous KOH neutralizes 150 mL of aqueous 0.2 M H 2SO4. We go from an average initial temperature of 22.3 oC to a maximum of 29.2 oC. Calculate the molar heat enthalpy of neutralisation of KOH and the molar heat of enthalpy of the reaction. 2KOH + H2SO4 K2SO4 + 2H2O Q = mcH = HT = (300 + 150 g)(4.19)(29.2-22.3) =13009.95 J n(KOH) = CV = 0.2mol/L(0.300 L) = 0.06 mol n(H2SO4) = CV = 0.2mol/L(0.150 L) = 0.03 mol For molar enthalpy of neutralisation of KOH:
H = HH = Q/n = -13.10 kJ/0.06 moles = -217 kJ/mol of KOH For molar enthalpy of the reaction you need to watch for stoichiometric ratios. There are 2 moles of KOH so divide by 2: H = HH = Q/n = -13.10 kJ/0.03 moles = -109 kJ/mol 5.2 Hess's Law OBJECTIVES The enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps. Application of Hesss Law to calculate enthalpy changes. Calculation of reactions using H reactions using Hf data. Determination of the enthalpy change of a reaction that is the sum of multiple reactions with known enthalpy changes.
Hesss Law In a chemical reaction the total change in chemical potential energy (enthalpy change) must be equal to the energy lost or gained by the reaction system. Enthalpy level cycles Therefore according to Hesss Law: H = HH = H = HH1 + H = HH2 + H = HH3 Problem 1: C + O2 CO H = 110.5 kJ
CO + O2 CO2 H = 283.0 kJ C + CO + O2 CO + CO2 H = 393.5 kJ C + O2 CO2 H = 393.5 kJ Standard enthalpy change of formation (H = H H ) f Standard enthalpy change of formation (HHH ) is the f
enthalpy change that results when one mole of a compound is formed from its elements at 298K and 1.00 x 105Pa. All reactants and products are in their standard states. Eg. C(s) + 2H2(g) CH4(g) 6C(s) + 3H2(g) + O2(g) C6H5OH(s) Using Hesss Law to determine enthalpy of formation Hf = Hf of products Hf of reactants Problem 1: Calcium carbonate reacts with hydrochloric acid according to the
following equation: CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + H2O (l) + CO2 (g) Calculate the enthalpy change for this reaction Horeaction = Hoproducts Horeactants Solution HoCaCO3 -1207 Ho HCl (aq) -167 -796 -286
-394 HoCaCl2 Ho H2O (l) Ho CO2 (g) Hoproducts =(-796)+(-286)+(-394) = -1476 kJ/mol Horeactants =(-1207)+(2)(-167) = -1541 kJ/mol Horeaction = -1476-(H-1541) = +75 kJ/mol Problem 2: Calculate the enthalpy change for the burning of 11 grams of propane
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Horeaction = Hoproducts Horeactants Ho C3H8 -104 H O2 (Hg) 0 Ho H2O (Hg) -242 -394
o Ho CO2 (g) Solution Hoproducts =(3)(-394)+(4)(-242) = -2150 kJ/mol Horeactants =(-104)+(5)(0) = -104 kJ/mol Horeaction = -2150-(H-104) = -2046 kJmol-1 Now 11 grams = 0.25 mole of propane (H11 g/44 g mol -1) (H0.25 mol )(H-2046 kJ mol-1) = 511.5 kJ Standard Enthalpy of Combustion (Hc)
Standard enthalpy change of combustion (HHHC) is the enthalpy change that results when one mole of a compound reacts with oxygen at 298K and 1.00 x 105Pa. All reactants and products are in their standard states. Eg. C(graphite) + O2(g) CO2(g) H2(g) + O2(g) H2O(l) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) Using Hesss Law to determine enthalpy of combustion Hc = Hc of reactants Hc of products
Problem 1: Calculate the standard enthalpy of formation of methane; the standard enthalpies of combustion of carbon, hydrogen and methane are -394, -286 and -890 kJmol-1 . C(graphite) + 2H2(g) CH4(g) Hc = Hc of reactants Hc of products = [ (1 x Hc of C) + (2 x Hc of H2) ] - [ 1 x Hc of CH4] = 1 x (-394) + 2 x (-286) - 1 x (-890) = -76 kJmol-1 SUMMARY Horeaction = Hoproducts Horeactants Horeaction = Hof products Hof reactants Horeaction = Hoc reactants Hoc products
Horeactant = Horeaction / mole ratio 5.3 Bond enthalpies OBJECTIVES Bond-forming releases energy and bond-breaking requires energy. Average bond enthalpy is the energy needed to break one mol of a bond in a gaseous molecule averaged over similar compounds. Calculation of the enthalpy changes from known bond enthalpy values and comparison of these to experimentally measured values. Sketching and evaluation of potential energy profiles in determining whether reactants or products are more stable and if the reaction is exothermic or endothermic. Discussion of the bond strength in ozone relative to oxygen in its importance to the atmosphere.
Bond enthalpy Average bond enthalpy is the amount of energy required to break one mole of bonds in the gaseous state averaged across a range of compounds containing that bond. Breaking bonds is endothermic Making bonds in exothermic If the overall reaction is exothermic, the bonds have lost energy, are more stable, and have higher bond enthalpies If the overall reaction is endothermic, the bonds have gained energy, are less stable, and have lower bond enthalpies Using Hesss Law and bond enthalpies to determine enthalpy of reaction
H = bond enthalpies bond enthalpies of reactants of products ATOMS SUM OFTHE BOND ENTHALPIES OF THE REACTANTS SUM OFTHE BOND ENTHALPIES OF THE PRODUCTS REACTANTS
H PRODUCTS Bond Enthalpy Table The average bond enthalpies are in your data booklet: Problem 1: Calculate the enthalpy change for the reaction N2 + 3H2 2NH3 Bonds broken 1 N=N: 3 H-H: = 945 3(435) = 1305
Total = 2250 kJ/mol Bonds formed 2x3 = 6 N-H: 6 (390) = - 2340 kJ/mol Net enthalpy change H = reactants products = + 2250 - 2340 = - 90 kJ/mol Bond enthalpies of ozone Ozone is decomposed more easily than oxygen. O2 requires 498kJ/mol to break O3 requires 364kJ/mol to break
O2(g) O(g) + O(g) O3(g) O2(g) + O(g) The energy profile shows that ozone molecule is more stable than the oxygen molecule with an oxygen radical OBJECTIVES Representative equations (eg M+(g) M+(aq)) can be used for enthalpy/energy of hydration, ionization, atomization, electron affinity, lattice, covalent bond and solution. Enthalpy of solution, hydration enthalpy and lattice enthalpy are related in an energy cycle.
Construction of Born-Haber cycles for group 1 and 2 oxides and chlorides. Construction of energy cycles from hydration, lattice and solution enthalpy. For example dissolution of solid NaOH or NH4Cl in water. Calculation of enthalpy changes from Born-Haber or dissolution energy cycles. Relate size and charge of ions to lattice and hydration enthalpies. Perform lab experiments which could include single replacement reactions in aqueous solutions. Higher Higher level level 15.1 Energy cycles Standard Enthalpy of formation Hf
Standard enthalpy of formation Hof for an ionic compound is the energy required to produce one mole of formula units from atoms in their natural state at 298K and 1.00 x 105Pa. M(s) + X MX(s) eg. Na(s) + Cl2(g) NaCl(s) Hof = -411kJ/mol Higher Higher level level KEY TERMS:
Enthalpy of atomization Hat Enthalpy of atomization Hoat is the energy change required to change one mole of atoms from their standard state to a gaseous state. M(s) M(g) Hoat > 0 X2(g) X(g) Hoat > 0 Higher Higher level level KEY TERMS:
Ionization energy HIE Ionization energy HoIE is the minimum energy required to remove of one mole of electrons from one mole of atoms or positive ions in the gaseous phase. IE1: M(g) M+(g) + e- >0 IE2: M+(g) M2+(g) + e- > 0 Higher Higher level
level KEY TERMS: Electron affinity HEA Electron affinity HoEA is the energy change that occurs when one mole of electrons joins to one mole of atoms in the gaseous phase. X(g) + e- X-(g) HoEA < 0 Higher Higher level level
KEY TERMS: Lattice enthalpy Hlat Lattice enthalpy Holat The energy required to convert one mole of the solid compound into gaseous ions. MX(s) M+(g) + X- (g) Eg. NaCl(s) Na+(g) + Cl-(g) Holat > 0 Holat = +771 kJ/mol Higher Higher level
level KEY TERMS: Depends on: Size of the charge on the ions Greater the charge the greater the attraction thus larger the lattice energy Size of the ions Larger the ions the further apart they are and the more separate the charges the smaller the lattice energy
Higher Higher level level Magnitude of lattice enthalpy Cl Br F O2- Na+ -780 -742 -918 -2478
K+ -711 -679 -817 -2232 Rb + -685 -656 -783 Mg2+ -2256 Units: kJ mol-1 -3791 Ca2+ -2259 Smaller ions will have a greater attraction for each other because of their higher charge density. They
will have larger lattice enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions. Higher Higher level level Magnitude of lattice enthalpy Na+ Cl K+
Cl The sodium ion has the same charge as a potassium ion but is smaller. It has a higher charge density so will have a more effective attraction for the chloride ion. More energy will be released when they come together. Higher Higher level level Magnitude of lattice enthalpy
The Born-Haber cycle is a diagrammatic representation of the formation of ionic compounds, starting from atoms, moving to gaseous atoms to gaseous ions to a solid compound. We use it to determine lattice enthalpy. Higher Higher level level Born-Haber cycle +800 +700 +600
Born-Haber Cycle for Sodium Chloride 1 Atomization of Sodium +500 +400 +300 +200 +100 0 -100 -200 -300 -400
Lattice Enthalpy for Sodium Chloride + - -100 H = 786 kJmol-1 lat -200 -300 -
+ - + - + - -
+ -400 NaCl(Hs) Higher Higher level level kJmol-1 HatmNa + HatmCl + HIE + HEA - Hlat = Hf Rearrange to find the lattice energy: - Hlat = Hf - (HatmNa + HatmCl + HIE + HEA)
Higher Higher level level Applying Hesss Law to Born-Haber cycles Covalent character increases down a group and across a period Covalent character increase as the difference between experimental and theoretical values of lattice enthalpy increases. Higher Higher level
level Difference between experimental and theoretical lattice enthalpies Solvation is when intermolecular forces form between a solvent and solute. The type of ion formed affects the strength of the bond. Higher Higher level level Solvation
Enthalpy of solution Hsol Enthalpy of solution Hosol is the energy change that occurs when one mole of substance dissolves in excess solvent. The value may be positive or negative. MX(s) M+(aq) + X-(aq) Higher Higher level level KEY TERMS:
Enthalpy of hydration Hhyd Enthalpy of hydration Hohyd is the energy change that occurs when one mole of gaseous ions are added to water to form a dilute solution. The value is always negative. If another solvent other than water is used it is enthalpy of solvation. M+(g) M+(aq) X-(g) X-(aq) Higher Higher level level KEY TERMS:
level Problem 1: Calculate the enthalpy of solution for sodium hydroxide using your data booklet and given the lattice enthalpy for NaOH is 900kJ/mol. OBJECTIVES Entropy (S) refers to the distribution of available energy among the particles. The more ways the energy can be distributed the higher the entropy. Gibbs free energy (G) relates the energy that can be obtained from a chemical reaction to the change in enthalpy (H = HH), change in entropy (H = HS), and absolute temperature (T). Entropy of gas>liquid>solid under same conditions.
Prediction of whether a change will result in an increase or decrease in entropy by considering the states of the reactants and products. Calculation of entropy changes (H = HS) from given standard entropy values (S). Application of H = HG = H = HH - TH = HS in predicting spontaneity and calculation of various conditions of enthalpy and temperature that will affect this. Relation of H = HG to position of equilibrium. Higher Higher level level 15.2 Entropy and spontaneity Higher Higher level
level Is this your room? Then you already know about entropy Entropy (S) is the measure of the distribution of total available energy between the particles. The greater the disorder of the particles, the greater the entropy. Higher Higher level level
Entropy (S) For better or for worse, nature 'likes' chaos, disorder, high entropy... In fact, much of our life consists in fighting this disorder! Higher Higher level level A system (such as a room) is in a state of high entropy
when its degree of disorder is high. Higher Higher level level 220 ways to arrange? This can be explained in terms of probabilities. Disordered states are simply more likely to exist (or emerge) than ordered states. The spontaneous direction of change is from a less probable to a more probable state, as illustrated above. S(HBr2 liquid) < S(HBr2 gas)
S(HH2O solid) < S(HH2O liquid) Higher Higher level level Entropy and States of Matter S increases slightly with T S increases a large amount with phase changes
Higher Higher level level Entropy, Phase & Temperature The Entropy of a substance increases with temperature. Molecular motions of heptane, C7H16 Molecular motions of heptane at different temps.
Higher Higher level level Entropy and Temperature The change in the number of gaseous particles is usually the largest factor to be considered, even if a comparable reaction has a greater increase in total particles but these particles are still in the liquid or solid state. Higher
Higher level level Predicting entropy changes Units J K-1 mol-1. An increase in disorder produces a +ve H = HS: 1. Decomposition reactions 2. Changes in state 3. Dissolution reactions 4. Increases in temperature Higher Higher level level
Entropy calculations aA + bB cC +dD Standard entropy change of reaction H = HSrxn: H = HS = H = HS (products) - H = HS (reactants) = [c H = HS (C) + d H = HS (D)] - [a H = HS (A) + b H = HS (B)] Higher Higher level level Standard entropy calculations H = HS = H = HS (products) - H = HS (reactants)
= H = HS(H2O) [H = HS(O2) + H = HS(H2)] = 70 (130.7 + x 205.1) = -163.3J/K Higher Higher level level Problem 1: Calculate the standard entropy change for the following reaction: H2(g) + O2(g) H2O(l) H = HS(H2) = 130.7J/Kmol H = HS(O2) = 205.1J/Kmol Gibbs free energy (G) is the energy available to do
work. H = HG =H = HH - TH = HS (Temp. in K) If H = HG is negative the reaction will be spontaneous If H = HG is positive the reaction will not be spontaneous. For any element in its standard state H = HG = 0. Higher Higher level level Gibbs free energy
Higher Higher level level Gibbs free energy useful equations for calculations Higher Higher level level Spontaneity H = HH Enthalpy Energy
-ve H = HS Randomness +ve Spontaneity -ve Entropy H = HG Free
Energy Higher Higher level level Summary of natural trends C2H5OH(l) 3O2(g) 2CO2(g) + 3H2O(g) H = HG(O2) = 0 as in its standard state H = HG = H = HG (products) - H = HG (reactants) = [2H = HG(CO2) + 3H = HG(H2O) ] [H = HG(C2H5OH)] = [2 x -394.4 + 3 x -228.6] [-175] = -1299.6kJ/mol
Higher Higher level level Problem 1: Calculate Gibbs free energy change of reaction for the combustion of ethanol (C2H5OH). H = HG = H = HH - TH = HS = 137 (298 x 120/1000) = 101kJ As this value is positive, the reaction will not be spontaneous at 298K. Higher Higher level
level Problem 2: Determine if the following reaction will be spontaneous at 298K. C2H6(g) C2H4(g) + H2(g) H = HH = 137kJ H = HS = 120J/K The point at which Gibbs free energy is at a minimum determines if a reaction goes to completion or reaches a balance
of reactants and products (equilibrium). Higher Higher level level Gibbs free energy and equilibrium
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