Thermochemistry

Thermochemistry • Internal Energy • Kinetic energy • Potential energy

Thermochemistry • InternalEnergy • 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 molH2O

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

Using Hess’s Law • Find ΔHofor 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

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)

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 • 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

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 • 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 1 g copper 100 J Increased 24 oC 25 oC 1 g water

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?)

Thermochemistry