* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Thermodynamics
Survey
Document related concepts
Rutherford backscattering spectrometry wikipedia , lookup
Equilibrium chemistry wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Thermodynamics wikipedia , lookup
Eigenstate thermalization hypothesis wikipedia , lookup
Marcus theory wikipedia , lookup
Thermal conduction wikipedia , lookup
George S. Hammond wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Heat transfer physics wikipedia , lookup
Transition state theory wikipedia , lookup
Transcript
Thermochemistry • Internal Energy • Kinetic energy • Potential energy Thermochemistry • Internal Energy • Kinetic energy • Potential energy Chemical Energy Changes • System and Surroundings Exothermic Reaction Endothermic Reaction Thermochemistry • Thermochemisty is the study of the relationship between heat and chemical reactions. • 1. Kinetic energy is energy possessed by matter because it is in motion – Thermal energy-- random motion of the particles in any sample above 0 K – Heat -- causes a change in the thermal energy of a sample. Flows from hot to cold Heat Potential Energy • 2. Potential energy is energy possessed by matter because of its position or condition. – A brick on top of a building has potential energy that is converted to kinetic energy when it is dropped on your head – Chemical energy is energy possessed by atoms as a result of forces which hold the atoms together (Boxes!) Where is the Energy? • Definitions we will use: – System: Reaction (bonds) – Surrounding: solvent, reaction vessel, air, etc. • An everyday example: burning wood – Initially, much energy stored as potential in C-H bonds, little kinetic energy in the air – Finally, lower potential energy in the C=O bonds, higher kinetic energy in the air Total Energy • Total Energy = kinetic + potential • Law of Conservation of Energy - The total energy of universe is constant • Internal Energy - E - the sum of all the kinetic and potential energies of all the atoms and molecules in a sample. Change in Energy of System • Change in internal energy of system = heat + work • Convention: point of view of system Change in Internal Energy • DE = q + w • Work = Force x distance • What happens to your internal energy when you push a boulder? • What happens to your internal energy when you push a boulder on a rough surface? Chemical Work • • • • • ׀W׀ = ׀F x Dh׀ P = F/A ׀W׀ = ׀P x A x h׀ ׀W׀ = ׀PV׀ Sign Convention: W = -PDV Test Your Understanding • For the following three reactions: – Are they performed under constant pressure or not? – What is the sign of work in each case? State Function • State Function – Internal Energy – Pressure – Volume • Path Dependent – Work – heat Property depends only on present state Enthalpy • Most reactions are done in open containers, so P is constant • Need a term for constant pressure where only work is PV • At constant pressure, qp • ΔE = qp – PΔV • qp = ΔE + PV • ΔH = ΔE + PΔV (definition) • ΔH = qp Enthalpy • IF pressure is constant and only work is PV • Change in enthalpy is equal to flow of energy in form of heat – Measurable by Temperature – Change in enthalpy is “heat of reaction” If no net change in moles of gas, enthalpy ~ energy Enthalpy • Signs on ΔH – + heat is taken in by system – - heat is given off by system • Endothermic ΔH = + • Exothermic ΔH = - Exothermic 2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (s) + energy Which bonds have more potential energy? Which bonds are stronger? Endothermic Ba(OH)2. 8H2O (s) + 2 NH4SCN (s) + energy Ba(SCN)2 (aq) + 2 NH3 (g) + 10 H2O (l) Reaction Enthalpy • CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) + ENERGY • It is useful to know how much energy is released • CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) ΔH = -890. kJ • This is a stoichiometric amount • 890 kJ released = 1 mol CH4 = 2 mol O2 = 1 mol CO2 = 2 mol H2O Reaction Enthalpy • Depends on coefficients, direction, and phases – 2 CH4 (g) + 4 O2 (g) 2 CO2 (g) + 4 H2O (l) ΔH = -1780. kJ – CO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g) ΔH = +890. kJ – CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) ΔH = -802 kJ Reaction Enthalpy Standard Reaction Enthalpy • Standard State - a compound in its pure state at 1 atm pressure, all solutions are 1 M • Temperature can vary but usually 298.15 K • Standard Reaction Enthalpy (ΔHro)- reaction enthalpy when all products and reactants are in the standard state • CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) ΔHo = -890. kJ Enthalpy and Stiochiometry Application: Enthalpies of Combustion • Standard Enthalpy of Combustion (ΔHco) -change in enthalpy for the combustion of one mole substance at standard conditions • Combustion is combination with O2 to give CO2 and water • Specific Enthalpy -- the enthalpy of combustion per gram • enthalpy density -- enthalpy of combustion per liter Enthalpies of Combustion How Do We Determine Standard Reaction Enthalpies? • Tabular Data: Hess’s Law – Combining appropriate reactions – Heat of Formation – Bond enthalpies • Experimental – Calorimetry: constant pressure – Calorimetry: constant volume Hess’s Law Just a restatement of the first law How Do We Determine Standard Reaction Enthalpies? • Tabular Data: Hess’s Law – Combining appropriate reactions – Heat of Formation – Bond enthalpies • Experimental – Calorimetry: constant pressure – Calorimetry: constant volume Using Hess’s Law • Find ΔHo for C (s) + ½ O2 (g) CO (g) • C (s) + O2 (g) CO2 (g) ΔHo = -393.5 kJ • 2 CO (g) + O2 (g) 2 CO2 (g) ΔHo = -566.0 kJ How Do We Determine Standard Reaction Enthalpies? • Tabular Data: Hess’s Law – Combining appropriate reactions – Heat of Formation – Bond enthalpies • Experimental – Calorimetry: constant pressure – Calorimetry: constant volume Standard Enthalpies of Formation • The standard enthalpy of formation is the enthalpy change when one mole of a substance in its standard state is formed from the elements in their standard states. Hof • Write an equation for the standard heat of formation of carbon dioxide – C(s) + O2 (g) CO2 (g) Hof (CO2) • Write an equation for DHof of CH3OH • Write an equation for DHof of N2 (g) and explain why its value is zero. Standard Enthalpies of Formation DHof can be compiled in table form Application of Heat of Formation • Hess’s Law • Without the limitations of combining limited number of reactions • Common starting point: elements Calculating Heat of Reaction CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) ΔHor = ΣHof (products) - ΣHof (reactants) How Do We Determine Standard Reaction Enthalpies? • Tabular Data: Hess’s Law – Combining appropriate reactions – Heat of Formation – Bond enthalpies • Experimental – Calorimetry: constant pressure – Calorimetry: constant volume Using Bond Enthalpies • Most versatile • Least exact • Must be able to draw Lewis Dot structures http://chemed.chem.wisc.edu/chempaths/GenChemTextbook/Bond-Enthalpies-718.html Bond Enthalpies • Bond Dissociation Energies – Positive values • Hess’s Law – In principle, “free atoms” formed • DHrxn = SBDE(broken) – SBDE(formed) Bond Enthalpies Calculate the heat of reaction for the combustion of formamide (CH3NO). 1. Equation 2. Lewis Dot 3. Calculate How Do We Determine Standard Reaction Enthalpies? • Tabular Data: Hess’s Law – Combining appropriate reactions – Heat of Formation – Bond enthalpies • Experimental – Calorimetry: constant pressure – Calorimetry: constant volume Heat Capacity • Heat Capacity (C) the heat required to raise the temperature of an object by 1 K • C = q/ΔT • extensive property Specific Heat Capacity • Specific heat capacity (Cs) - the heat required to raise 1 g of a substance by 1 K • Specific heat • Cs = C/m • intensive property • q = m Cs ΔT Each Substance “Stores” Heat Differently 25 oC 1 g copper 100 J 25 oC Increased 263 oC Increased 24 oC 1 g water • Putting the same amount of heat into two different substances will raise their temperature differently based on the specific heat of each substance • q=mCsDT Calorimetry • Calorimeter - an insulated container fitted with a thermometer • Open to atmosphere, so P is constant • qp = m Cs ΔT • qp = ΔH Calorimetry Problem In a coffee cup calorimeter, 50.0 mL of 0.100 M silver nitrate and 50.0 mL of 0.100 M HCl are mixed. The following reaction occurs: Ag+ (aq) + Cl- (aq) AgCl (s). If the two solutions are initially at 22.60 oC, and if the final temperature is 23.40 oC, calculate the change in enthalpy for the reaction. (What assumptions need to be made?) Bomb Calorimeter • • • • • • Volume is constant q = ΔE qsystem= -qsurroundings qrxn= -(qbomb+ qwater) qwater = mwaterCsΔT qbomb = mbombCsΔT or CDT Thermodynamics of Ideal Gas • • • • • Heat capacity of monoatomic gas From KMT, KE = 3/2RT KE = translational energy Heat required to raise temp 1 degree is 3/2R At constant volume, no work is done – Molar heat capacity Cv = 3/2R = 12.47 J/mol K Is this constant for all gases? Consider polyatomic gases • For polyatomic molecules, _____ energy has to be put into the same amount of gas to raise it by 1 degree • Where does this energy go? (Not translational!) • Trend?? Monoatomic Gas at Constant Pressure • Energy input does two things: increase translational energy (T) and expand gas (w) • w = PDV = nRDT • Molar heat capacity – Cp = 3/2R + R = 5/2R Polyatomic at Constant Pressure • Explain physical basis of this equation: • Cp =Cv + R • This is observed! • When would you see a deviation from Cp-Cv = R? Summary Conceptual Understanding of Gas Cycle Thermochemistry of Physical Change • Vaporization - endothermic process • Vapor has higher H that a liquid at the same temperature • Enthalpy of vaporization ΔHvap -• ΔHvap = Hvapor - Hliquid Freezing, Melting and Sublimation • • • • • Enthalpy of fusion ΔHfus (melting) ΔHfus = Hliquid - Hsolid Enthalpy of freezing = - ΔHfus Enthalpy of sublimation, ΔHsub ΔHsub = Hvapor - Hsolid Heating/Cooling Curve Enthalpy of Solution • Enthalpy (Heat) of Solution – ΔHsoln – heat change associated with the dissolution of a known amount of solute in a known amount of solvent. • ΔHsoln = Hsoln – Hcomponents Lattice Energy • STEP 1 • Lattice Energy (U) – the energy required to separate 1 mole of a solid ionic compound into gaseous ions. • NaCl (s) Na+(g) + Cl-(g) • U = 788 kJ/mol Heat of Hydration • Heat of hydration – ΔHhydr – Enthalpy change associated with the hydration process. • Na+(g) + Cl-(g) Na+(aq) + Cl-(aq) • ΔHhydr = -784 kJ ΔHsoln= U +ΔHhydr= 788 kJ/mol + (-784 kJ) = 4 kJ/mol