CHEM 101 1st Edition Lecture 15Outline of Last Lecture I. Energy and Changes of StateII. The First Law of ThermodynamicsOutline of Current Lecture I. EnthalpyII. Internal EnergyIII. Hess’s LawCurrent Lecture - Enthalpyo We can only measure the change of enthalpy in a system, which is independent from the path between Hinitial (A) and Hfinal (B):ΔH = Hfinal – HinitialTherefore, enthalpy is a state functiono Since enthalpies are state functions, we must specify the conditions at which they are measured.o H(T,P): Enthalpy is a function of temperature and pressureo The standard state of the element or compound is the most stable for of the substance in the physical state that exists under standard conditionso Standard state conditions are: 100kPa = 1.00 bar Usually 298.15K or 25°C “H°” indicates that the enthalpy is taken at standard state conditionso Reactants ProductsΔH = Hfinal – HinitialEnthalpy of reaction = ΔrH = Hproducts – Hreactantso CH4(g) + 2 O2(g) CO2(g) + 2H2O(g) ΔrHO = -802kJCO2(g) + 2 H2O(g) CH4(g) + 2 O2(g) ΔrHO = +802kJ o Energy can be considered as a product just like CO2 or H2Oo From the chemical equation:802kJ of Energy released 802kJ of energy released1 mol CH4(g) consumed 2 mol H2O (g) producedo CH4(g) + 2 O2(g) CO2(g) + 2H2O(g) ΔrHO = -802kJIt’s also common to write: ΔrH° = -802 kJ/molo Ex: what is the enthalpy change when 128.5g of methane, CH4(g) is combusted (ΔrHO = -802 kJ)? molar mass(g) mols energy (enthalpy) in J 128.5g CH4 x 1 mol CH 4 x “1 mol r x n” x -802 kJ = -6.43 x 10^(3) kJ 16.04 g 1 mol CH4 “1 mol r x n”o CH4(g) + 2 O2(g) CO2(g) + 2H2O(g) ΔrHO = -802kJ½ x [CH4(g) + 2 O2(g) CO2(g) + 2H2O(g)] [ΔrHO = -802kJ] x ½ ½ CH4(g) + O2(g) ½ CO2(g) + H2O(g) ΔrHO = -401kJo A constant pressure calorimeter “coffee-cup calorimeter” can be used to measure the amount of energy transferred as heat under constant pressure conditions, (= the enthalpy change ΔH for a chemical reaction)o Ex: 5.44 g of NH4NO3(s) was added to 150.0 mL of water in a coffee-cup calorimeter. A decrease in temperature from 18.6°C to 16.2°C was observed. Calculate the enthalpy change for dissolving NH4NO3(s) in water in kJStep 1: calculation of moles of NH4NO35.44 g NH4NO3 x 1 mol NH 4 NO 3 = 0.0680 mol NH4NO3 80.04 gStep 2: determine qsolution m x C x ΔTqsolution = 154.4 g x 4.18 J x (16.2°C -18.6°C) = -1.55 x 10^(3) g °CStep 3: qsolution = qreaction = 0qrxn = -qsolution = 1.55x10^(3)Step 4: calculation of enthalpy per moleΔrH = q rxn = 1.55x10^(3) J x 1 kJ = 22.8 kJ/mol mol reaction 0.0680 mol NH4NO3 10^(3) J- Internal Energyo Under conditions of constant volume, any heat transferred is equal to a change of internal energy ΔrU:qv =ΔrUo The hear of reaction is found by:qrxn + qbomb + qwater = 0o Ex: octane, a primary component of gasoline, combusts by the following reaction: C9H18(l) + 25/2 O2(g) 8CO2(g) + 9 H2O(l)A 1.00 g sample of octane is burned in a bomb calorimeter that contains 1.20 kg of water surrounding the bomb. The temperature of the water rises to 33.20 °C from 25.00°C when the octane is reacted. The heat capacity of the bomb is 837 J/°C. What is the heat of the combustion per mole of octane?qrxn + qbomb + qwater = 0qrxn = -qbomb - qwater qrxn = mwater x Cwater x ΔTwater x qbomb x ΔTwater- Hess’s Law- if a reaction is the sum of two or more reactions, ΔrH° for the overall processis the sum of the ΔrH° values of those reactions C(s) + ½ O2(g) CO2(g) ΔrH1° = ? ΔrH3° = ΔrH1° + ΔrH2°+ CO(g) = ½ O 2 (g) CO 2 (g) Δ r H 2 ° = -283.0 kJ -393.5 kJ = ? + -283.0 kJ C(s) + O2(g) CO2(g) ΔrH3° = -393.5 kJ ΔrH1° = -110.5- Standard Molar enthalpy of formationC(s) + O2(g) CO2(g) ΔfH° = -393.5 kJWhere ΔH is enthalpy change, f is the formation of 1 mol of a compound directly from the elements in their standard states, and ° is the standard state; most stable form of thesubstance in the physical state that exists at 1 bar and 298.15K (25°C)- Clicker question: Which equation below defines the standard molar enthalpy of formation of gaseous methanol, CH3OH?A. CH4(g) = ½ O2(g) CH3OH(g)B. C(s) + 2 H2(g) + ½ O2(g) CH3OH(g) correctC. CO(g) +2 H2(g) CH3OH(g)D. H2O(g) + C(s) + H2(g) CH3OH(g)- Enthalpy Change for a ReactionfH° = Σn fH°(products) - Σn fH°(reaction)- Clicker Question: CaCO3(s) CaO(s) + CO2(g) rH° = ? fH° [CaO(s)] = -635.5 kJ/mol fH° [CO2(g)] = -393.5 kJ/mol fH° [CaCO3(s)] = -1207 kJ/molA. 178.0 kJ/mol correctB. -178.0 kJ/molC. 2236 kJ/molD. -965.0 kJ/molE. 965.0
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