Energy2 Outline

Bomb Calorimetry • • • • • constant volume often used for combustion reactions heat released by reaction is absorbed by calorimeter contents need heat capacity of calorimeter qcal = –q rxn = qbomb + qwater Example 4 When 0.187 g of benzene, C6H6 , is burned in a bomb calorimeter, the surrounding water bath rises in temperature by 7.48ºC. Assuming that the bath contains 250.0 g of water and that the calorimeter has a heat capacity of 4.90 kJ/ºC, calculate the energy change for the combustion of benzene in kJ/g. C6H6 (l) + 15/2 O2 (g) → 6 CO2 (g) + 3 H2O (l) qcal = (4.90 kJ/ºC)(7.48ºC) = 36.7 kJ qrxn = -qcal = -36.7 kJ / 0.187 g C6H6 = -196 kJ/g qV = ∆E = -196 kJ/g Energy and Enthalpy Most physical and chemical changes take place at constant pressure • Heat transferred at constant P - enthalpy (H) – Can only measure ∆H – ∆H = H final - H initial = q P – sign of ∆H indicates direction of heat transfer heat system heat system ∆H > 0 endothermic ∆H < 0 exothermic 1 Energy and Enthalpy Enthalpy and Phase Changes Enthalpy and Phase Changes • Melting and freezing – Quantity of thermal energy that must be transferred to a solid to cause melting - heat of fusion (q fusion) – Quantity of thermal energy that must be transferred from a solid to cause freezing - heat of freezing (q freezing) – qfusion = - qfreezing – heat of fusion of ice = 333 J/g at 0°C 2 Enthalpy and Phase Changes • Vaporization and condensation – Similarly: q vaporization = - q condensation – heat of vaporization of water = 2260 J/g at 100°C – 333 J of heat can melt 1.00 g ice at 0°C but it will boil only: 333 J x (1.00 g / 2260 J) = 0.147 g water Enthalpy and Phase Changes H2 O H2 O (l) H2 O H2 O (g) (s) (l) → H2O (g) → H2 O (l) endothermic → H2O (l) → H2O (s) exothermic State Functions • Value of a state function is independent of path taken to get to state - depends only on present state of system • Internal energy is state function 3 State Functions • q and w not state functions Enthalpy and PV Work H - state function so how can ∆H = q?? q - not a state function How do internal energy and enthalpy differ? ΔE = q + w ΔH = q P answer: work Enthalpy and PV Work 4 Example 5 A gas is confined to a cylinder under constant atmospheric pressure. When the gas undergoes a particular chemical reaction, it releases 89 kJ of heat to its surroundings and does 36 kJ of PV work on its surroundings. What are the values of ∆H and ∆E for this process? q = -89 kJ w = -36 kJ @ constant pressure: ∆H = q P = -89 kJ ∆E = ∆H + w = -89 kJ - 36 kJ = -125 kJ Thermochemical Equations ∆H = Hfinal - Hinitial = H(products) - H(reactants) ∆H rxn - enthalpy or heat of reaction 2 H2 (g) + O2 (g) → 2 H2 O (l) ∆H° = -571.66 kJ coefficients of equation represent # of moles of reactants and products producing this energy change Thermochemical Equations ∆H°   standard enthalpy change defined as enthalpy change at 1 bar pressure and 25°C 5 “Rules” of Enthalpy • Enthalpy is an extensive property - magnitude of ∆H depends on amounts of reactants consumed CH4 (g) + 2 O2 (g) → CO2 (g) 2 CH4 (g) + 4 O2 (g) → 2 CO2 + 2 H2 O (g) ∆H = -890 kJ (l) + 4 H2O (l) ∆H = -1780 kJ • Enthalpy change of a reaction is equal in magnitude but opposite in sign to ∆H for the reverse reaction CO2 (g) + 2 H2 O (l) → CH4 (g) + 2 O2 ∆H = +890 kJ (g) • Enthalpy change for a reaction depends on the states of the reactants and products H 2 O (l) → H2 O ∆H = +44 kJ (g) H 2 O (s) → H2O (g) ∆H = +50 kJ Example 6 Hydrogen peroxide can decompose to water and oxygen by the reaction: 2 H2 O2 (l) → 2 H2O (l) + O2 (g) ∆H = -196 kJ Calculate the value of q when 5.00 g of H2 O2 (l) decomposes. Example 7 Consider the following reaction, which occurs at room temperature and pressure: 2 Cl (g) → Cl2 (g) ∆H = -243.4 kJ Which has the higher enthalpy under these conditions, 2 Cl or Cl2 ? 2 Cl (g) 6 Example 8 When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: Ag+ (aq) + Cl(a) (aq) → AgCl (s) ∆H = -65.5 kJ Calculate ∆H for the formation of 0.200 mol of AgCl by this reaction. Example 8 (cont’d) Ag+ (aq) + Cl- (aq) → AgCl (s) ∆H = -65.5 kJ (b) Calculate ∆H when 0.350 mmol AgCl dissolves in water. AgCl (s) → Ag+ (aq) + Cl- (aq) ∆H = +65.5 kJ Bond Enthalpies • • • • during chemical reaction bonds are broken and made breaking bonds requires energy input (endothermic) formation of bonds releases energy (exothermic) weaker bonds broken and stronger bonds formed 7 Hess’s Law • • • we can calculate ∆H for a reaction using ∆Hs for other known reactions ∆H is a state function - result is same no matter how we arrive at the final state Hess’s Law - if a reaction is carried out in a series of steps, ∆H for overall reaction is equal to sum of ∆Hs for steps Hess’s Law C (s) + O2 (g) → CO2 (g) CO (g) + 1/2 O 2 (g) → CO2 (g) ∆H = -393.5 kJ ∆H = -283.0 kJ What is ∆H for C (s) + 1/2 O2 (g) → CO (g) ??? C (s) + O2 (g) → CO2 (g) CO2 (g) → CO (g) + 1/2 O2 (g) ∆H = -393.5 kJ ∆H = +283.0 kJ C (s) + 1/2 O2 (g) → CO (g) ∆H = -110.5 kJ Example 9 Calculate ∆H for the conversion of S to SO3 given the following equations: S (s) + O2 (g) → SO2 (g) ∆H = -296.8 kJ SO2 (g) + 1/2 O2 (g) → SO 3 (g) ∆H = -98.9 kJ want S (s) → SO3 (g) S (s) + O2 (g) → SO2 (g) SO2 (g) + 1/2 O2 (g) → SO3 (g) ∆H = -296.8 kJ ∆H = -98.9 kJ S (s) + 3/2 O2 (g) → SO3 (g) ∆H = -395.7 kJ 8 Enthalpies of Formation • tables of enthalpies (∆Hvap, ∆Hfus, etc.) • ∆H f - enthalpy of formation of a compound from its constituent elements. • magnitude of ∆H - condition dependent • standard state - state of substance in pure form at 1 bar and 25°C • ∆H f ° - change in enthalpy for reaction that forms 1 mol of compound from its elements (all in standard state) • ∆H f ° of most stable form of any element is 0 CO2 : C (graphite) + O2 (g) → CO2 (g) ∆Hf ° = -393.5 kJ/mol Calculating ∆Hrxn° from ∆Hf° • we can use ∆Hf° values to calculate ∆H rxn° for any reaction • ∆H rxn° = ∑ [n ∆H f° (products)] - ∑ [n ∆Hf° (reactants)] C6 H6 (l) + 15/2 O 2 (g) → 6 CO2 (g) + 3 H2O (l) ∆H rxn° = [(6 mol)(-393.5 kJ/mol) + (3 mol)(-285.83 kJ/mol)] - [(1 mol)(49.0 kJ /mol) + (15/2 mol)(0 kJ/mol)] ∆H rxn° = -3267 kJ/mol Example 10 Styrene (C8 H8), the precursor of polystyrene polymers, has a standard heat of combustion of -4395.2 kJ/mol. Write a balanced equation for the combustion reaction and calculate ∆H f ° for styrene (in kJ/mol). ∆Hf ° (CO2) = -393.5 kJ/mol ∆Hf ° (H2O) = -285.8 kJ/mol C8H8 (l) + 10 O2 (g) → 8 CO2 (g) + 4 H2O (l) ∆H rxn ° = -4395.2 kJ/mol = [(8 mol)(-393.5 kJ/mol) + (4)(-285.8)] - [(1) ∆Hf ° (C8H8 ) + (10)(0)] ∆H f ° (C8 H8) = 104.0 kJ/mol 9