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 Path Dependent 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 Endothermic ΔH = + Exothermic 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 It is useful to know how much energy is released 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/GenChem-Textbook/Bond-Enthalpies-718.html

Bond Enthalpies Bond Dissociation Energies Hess’s Law 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). Equation Lewis Dot 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 Increased 263 oC 25 oC 1 g copper 100 J Increased 24 oC 25 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