CHEMISTRY 161 Energy and Chemical Change Chapter 7 Energy an Chemical Change 1.Forms of Energy 2.SI Unit of Energy 3.Energy in Atoms and Molecules 4.Thermodynamics 5.Calculation of Heat and Energy Changes 6.Measuring Heat and Energy Changes 1. Forms of Energy 1. Kinetic energy energy of a moving microscopic or macroscopic object E = ½ m v2 2. Radiant energy energy in form of photons (‘light’) (solar energy) E = h (h = Planck’s constant) (Chapter 8) 3. Potential energy energy by changing object’s position in height E = m g h (h = height) 4. Thermal Energy energy associated with random motion of atoms and molecules Ekin = ½ M v2 = 3/2 R T (Chapter 7) M = m Na 5. Chemical Energy EXP1 energy stored in chemical bonds of substances (Chapter 7) LAW OF CONSERVATION OF ENERGY THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT 2. SI Unit of Energy 1 Joule = 1 J 1 cal = 4.184 J 1 J = 1 Nm = 1 kg m2 s-2 Ekin = ½ m v2 macroscopic versus microscopic 1 J vs. 1 kJ mol-1 3. Energy in Atoms and Molecules Atoms – Kinetic and Thermal Energy gases are constantly in motion and hold a kinetic energy Ekin = ½ M v2 = 3/2 R T EXP2 Molecules – Kinetic, Thermal, & Potential Energy (N2) molecules have different ‘internal’ (vibrational) energy when bond distances are changed EXP3/4 different bonds have different bond strength (stabilities) (H2 vs. N2) 4. Thermodynamics reactants products (different energies) THERMODYNAMICS HEAT CHANGE study of the energy associated with change THERMOCHEMISTRY study of the energy associated with chemical change 2 H2(g) + O2(g) → 2 H2O(l) + energy Hindenburg 1937 Challenger 1986 Surrounding Surrounding System System heat heat ENDOTHERMIC EXOTHERMIC EXP 2 HgO(s) → O2(g) + 2 Hg(l) 2 H2(g) + O2(g) → 2 H2O(l) Energy 2 H2(g) + O2(g) NH4NO3 (aq) Exothermic Endothermic (heat given off by system) (heat absorbed by system) 2 H2O(l) NH4NO3(s) + H2O (l) QUANTIFICATION Enthalpy of Reaction Enthalpy is the heat release at a constant pressure (mostly atmospheric pressure) DH = Hfinal - Hinitial DH = Hproducts - Hreactants board Hfinal > Hinitial : DH > 0 ENDOTHERMIC Hfinal < Hinitial : DH < 0 EXOTHERMIC Energy NH4NO3(aq) 2 H2(g) + O2(g) Hfinal < Hinitial Exothermic 2 H2O(l) Hfinal > Hinitial Endothermic NH4NO3(s) + H2O(l) Energy H2O(l) Hfinal > Hinitial DH = Hfinal – Hinitial Endothermic H2O(s) H2O(s) → H2O(l) ΔH = + 6.01 kJ mol-1 Energy H2O(l) Hfinal < Hinitial DH = Hfinal – Hinitial Exothermic H2O(s) H2O(l) → H2O(s) ΔH = - 6.01 kJ mol-1 THERMOCHEMICAL EQUATIONS H2O(l) → H2O(s) ΔH = - 6.01 kJ mol-1 CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g) ΔH=-890.4 kJ mol-1 Calculate the heat evolved when combusting 24.0 g of methane gas. 5. Calculation of Heat and Enthalpy Changes DHm = Hm,products – Hm,reactants molar REFERENCE SYSTEM e.g. oxidation numbers of elements are zero Standard Enthalpy of Formation DHfO heat change when 1 mole of a compound is formed from its elements at a pressure of 1 atm (T = 298 K) DHfO (element) = 0 kJ/mol DHfO (graphite) = 0 kJ/mol DHfO (diamond) = 1.9 kJ/mol ENTHALPY, H 0 C(s, graphite) + O2(g) Hreactants DHf0 = - 393.51 kJ mol-1 Hproducts -393.51 CO2(g) Standard Enthalpy of Formation C(s, graphite) + O2(g) CO2(g) DHf0 = - 393.51 kJ mol-1 C(s, graphite) + 2H2(g) CH4(g) DHf0 = - 74.81 kJ mol-1 ½ N2(g) + 3/2 H2(g) 0 -1 NH3(g) DHf = - 46.11 kJ mol (1/2) N2(g) + (1/2) O2(g) NO(g) DHf0 = + 33.18 kJ mol-1 Standard Enthalpy of Reaction aA+bB→cC+dD ENTHALPY, H aA+bB a × DHfO (A) + b × ΔHfO(B) Hreactants DHOrxn = ΣΔHf0(prod) – ΣΔHf0(react) Hproducts c × DHfO(C) + d × ΔHfO(D) cC+dD Standard Enthalpy of Reaction DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(react) CaO(s) + CO2(g) → CaCO3(s) -635.6 -393.5 -1206.9 DHOrxn = -177.8 kJ/mol [kJ/mol] Standard Enthalpy of Reaction CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(react) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) 2 H2O(g) → 2 H2O(l) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Hess’s Law The overall reaction enthalpy is the sum of the reaction enthalpies of the steps in which the reaction can be divided CH4(g) + 2O2(g) ENTHALPY, H - 802 kJ Reactants - 890 kJ CO2(g) + 2H2O(g) - 88 kJ CO2(g) + 2H2O(l) Products DHrxn for S(s) + 3/2 O2(g) SO3(g) S(s) + O2(g) SO2(g) DH1 = -320.5 kJ SO2(g) + 1/2 O2(g) SO3(g) DH2 = -75.2 kJ S solid direct path + 3/2 O2 DH3 = -395.7 kJ SO3 gas +O2 Indirect Path DH1 = -320.5 kJ SO2 gas + 1/2 O 2 DH2 = -75.2 kJ 6. State Functions THERMODYNAMICS quantitative study of heat and energy changes of a system CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) the state (condition) of a system is defined by T, p, n, V, E the state (condition) of a system is defined by T, p, n, V, E STATE FUNCTIONS properties which depend only on the initial and final state, but not on the way how this condition was achieved Hess Law ΔV = Vfinal – Vinitial Δp = pfinal – pinitial ΔT = Tfinal – Tinitial ΔE = Efinal – Einitial Energy is a STATE FUNCTION ΔE = m g Δh IT DOES NOT MATTER WHICH PATH YOU TAKE Hess Law ENTHALPY, H CH4(g) + 2O2(g) - 802 kJ Reactants - 890 kJ CO2(g) + 2H2O(g) - 88 kJ CO2(g) + 2H2O(l) Products Applications Zeroth Law of Thermodynamics a system at thermodynamical equilibrium has a constant temperature heat is spontaneous transfer of thermal energy two bodies at different temperatures T1 > T2 spontaneous T2 T1 EXP LN2/Metal/H2O First Law of Thermodynamics energy can be converted from one form to another, but cannot be created or destroyed CONSERVATION OF ENERGY SURROUNDINGS + - SYSTEM THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT First Law of Thermodynamics ΔEsystem = ΔQ + ΔW ΔQ heat change ΔW work done DQ > 0 ENDOTHERMIC DQ < 0 EXOTHERMIC ? mechanical work ΔW = - p ΔV M ΔV < 0 the energy of gas goes up M ΔV > 0 the energy of gas goes down 6. Measurement of Heat Changes Surrounding System heat temperature increase DH = ΔQ ∞ ΔT (pressure is constant) Where does the ‘heat’ go? DH = ΔQ ∞ ΔT DH = ΔQ = const × ΔT DH =ΔQ = C ΔT temperature change enthalpy change heat capacity C=ms s = specific heat capacity DH = ΔQ = m s ΔT EXP specific heat capacity capability of substances to store heat and energy s = J g-1 K-1 the J necessary to increase the temperature of 1 g of a compound by 1 K DH = ΔQ = m s ΔT 1.prepare two styrofoam cups 2. carry out chemical reaction in a compound with known s s (H2O) = 4.184 J g-1 K-1 3. measure temperature change 4. determine ΔH calorimeter 100 ml of 0.5 M HCl is mixed with 100 ml 0.5 M NaOH in a constant pressure calorimeter (scup = 335 J K-1). The initial temperature of the HCl and NaOH solutions are 22.5C, and the final temperature of the solution is 24.9C. Calculate the molar heat of neutralization assuming the specific heat of the solution is the same as for water. DH = ΔQ = C ΔT DH = ΔQ = (c1 + c2) ΔT 1.Neutralization reactions 2.Redox reactions 3. Precipitation reactions Constant Volume Calorimeter ΔQ = (m s(H2O) + cbomb) ΔT Energy an Chemical Change 1.Forms of Energy 2.SI Unit of Energy 3.Energy in Atoms and Molecules 4.Thermodynamics 5.Calculation of Heat and Energy Changes 6.Measuring Heat and Energy Changes
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