Thermodynamics (That’s Hot!!! Or…. Not!)

The Nature of Energy Kinetic Energy Potential Energy Physics and Chemistry Agree Sometimes

First Law of Thermodynamics Energy in the universe is constant You’ve heard this stated another way Law of conservation of energy Energy can be neither created nor destroyed It is transferred.

What is Energy? When you think of energy, what comes to mind?

“Forms” of Energy We associate energy with “enthalpy” The “force” that holds things together Whenever there is a “change”, we can describe it as a change in Energy (enthalpy) How much of an energy change is there? How do we measure the change? Atoms →molecules molecules→ cells cells → plants / animals Plants / animals→ ecosystems Solid Liquid Gas Plasma Get students to mention the ways to measure. Include temperature and mass. Set up mentally for calorimetry

Foreshadowing Energy Entropy and Free energy will be discussed in the next chapter Enthalpy is what we are studying now Describe what bonds atoms and molecules to each other The production of heat (or absorbing)

Measuring Energy Temperature, Heat and Work Temperature - random motion of particles average kinetic energy Heat - transfer of energy due to a temperature difference Heat is not contained by an object If your hand is the same temperature as the radiator, would it feel hot? It is the temperature difference that feels hot The fire has a lot of heat. Work is force acting over a distance.

When Energy Transfers Occur, Does the Pathway Matter When Energy Transfers Occur, Does the Pathway Matter? Or Is It a State Function? Energy of a “system” is independent of where the energy came from or how it got there Energy of a system depends on what is in the system and their temperatures. Energy is a STATE FUNCTION

Internal Energy of a System Sum of the potential and kinetic energies A change in energy is done by work (motion) or heat E = q + w q = heat w = work

Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas). Work = Force x distance = Pressure x area x height = Pressure x Volume Pressure and Volumes are State Functions

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = -PV

Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy (H) of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV Since E, P, and V are State functions, then Enthalpy is a state function as well!

State Functions However, q and w are not state functions. Whether this battery is shorted out or is discharged by running the fan, its E is the same. But q and w are different in the two cases.

Heat and Work More proof that heat and work are not state functions Ex. 2: Imagine two magic boxes Each has 1g sugar or 1g of magnesium. There is a different amount of work (gases expanding against the atmosphere), light and heat produced for each gram of material burned.

Enthalpy When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) This can be written H = E + PV

Enthalpy Since E = q + w and w = -PV, we can substitute these into the enthalpy expression: H = E + PV H = (q+w) − w H = q So, at constant pressure, the change in enthalpy is the heat gained or lost.

Endothermicity and Exothermicity A process is endothermic when H is positive.

Endothermicity and Exothermicity A process is endothermic when H is positive. A process is exothermic when H is negative.

Signs and Thermodynamics Energy Change E < 0, or negative system loses energy E > 0, or positive system gains energy

Sign Conventions in Thermodynamics Heat q > 0 Increase in temp q < 0 Decrease in temp

Sign Conventions in Thermodynamics Work is more complicated W = - PV P = Pressure on the system V = Change in volume If a piston increases in size, it must move against the universe. So energy has left the system and gone into the universe. This is exothermic so E would be negative. Since the change in volume is positive, the sign is -PV. (Pg 247 in Book)

Sign Conventions So any work done by the system on the universe will be exothermic meaning w = - If work is done on the system by the universe, then it is endothermic meaning w = + . Pressure is measured in N/m2 which = 1Pa If you have 1kPa of Pressure and Volume is in L, then PV = J 1L•kPa = 1J 101.3kPa = 1 atm 1L•atm = 101.3 J (in case you were wondering)

Sample Problem A balloon is heated with 1.3 x 108 J and expands from 4.00 x 106 L to 4.5 x 106 L. Assuming constant outside pressure of 1.0 atm, calculate the change in energy E = q + w q = +1.3 x 108 J W = -P V = -1.0 atm x (4.50 x 106 L - 4.00 x 106 L) - 0.50 x 106 L atm x (101.3 J / L ATM ) = -5.1 x 107 J E = +1.3 x 108 J - 5.1 x 107J = 8.0 x 107 J

H= qp E = Eprod - Ereact CH4 + 2O2  CO2 + 2 H2O + energy qp = H = -890 Kj/mol notice negative value (energy left system) Stoichiometry type problems Energy is a reactant or product

Chemical Energy energy CH4 + O2 CO2 + 2 H2O  CH4 + O2  CO2 + 2 H2O + energy We must define “THE SYSTEM” The container with the methane and oxygen is a system The rest of the Universe is not the system. Heat flows out of the system – exothermic

Chemical Energy CH4 + 2O2  CO2 + 2 H2O + energy Chemical (potential) energy is converted to thermal (random kinetic energy) CH4 + 2O2 has more potential energy CO2 + 2 H2O has less potential energy The difference is the energy released

Chemical Energy N2O2 + energy 2 NO N2O2 has less potential energy than 2 NO The difference is the energy absorbed from universe This is endothermic

Enthalpy Problems CH4 + 2O2  CO2 + 2 H2O + energy qp = H = -890 Kj/mol A 5.8g sample of methane is burned in oxygen at a constant pressure. How much heat is given off? What is the enthalpy change? 5.8g x 1 mol x -890. kJ = -322 kJ 16.0g mol E = Eprod - Ereact = q = -322kJ q = 322 kJ

Ways to determine Enthalpies of Reactions Calorimetry – determine through experimentation and calculating heat flow. Hess’ Law – Calculate enthalpies of a series of reactions that when added together give the desired reaction Heats of Formations – Use determined energies of compounds to determine the overall enthalpy change. Bond Energies – Bonds broken – bonds formed (later)

Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity. We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K. Molar Heat Capacity is the amount of energy required to raise the temperature of 1 mole of a substance 1K. Remember a change of 1K and 1oC are the same!

Heat Capacity and Specific Heat Specific heat (Cp), then, is used mathematically by the following equation:

Calorimetry Why does a piece of metal feel cold to your hand? Why does the sand heat up faster than the water at the beach? Why does the center of the United States have more temperature extremes than the coasts? Why do ceramic tiles prevent a spacecraft from burning up upon reentry into the atmosphere?

Calorimetry Constant Pressure Calorimetry (open) q = H q = mCp T. Constant Volume Calorimetry (Bomb) q = E (need to know heat capacity of calorimeter) qrxn = Ccal ΔT

Hess’ Law or State Functions NO2(g)  1NO(g) + 1/2O2(g) H = 56 kJ/mol Double the reaction = double the enthalphy 2NO2(g)  2NO(g) + O2(g)  H = 112 kJ/mol The reverse of the reaction, “negative” 2NO(g) + O2(g)  2NO2(g) H =- 112 kJ/mol

Hess’ Law N2(g) + 2O2(g)  2NO2(g) H = 68 kJ/mol Can be determined by: N2(g) + O2(g)  2NO(g)  H = 180 kJ/mol 2NO(g) + O2(g)  2NO2(g)  H =- 112 kJ/mol N2(g) + 2O2(g)  2NO2(g)  H = 68 kJ/mol

Hess’s Law Works because Enthalpy is a state function The pathway is not important How we get there is not important You can use a series of known equations to “add up” to equal the equation you desire. Arrange the steps to equal the overall equation Add up the enthalpies

2B(s) + 3H2(g)  B2H6 Pg 258 kJ 2B(s) + 3/2 O2(g)  B2O3(S) -1273 B2O3(s) + 3H2O(g)  B2H6(g) + 3O2 +2035 3H2(g) + 3/2O2(g)  3H2O(l) -858 3H2O(l)  3H2O(g) 132 2B(s) + H2(g)  B2H6 +36 Pg 258

2B(s) + 3H2(g)  B2H6 Opposite sign OK Backwards Not enough Not enough kJ 2B(s) + 3/2 O2(g)  B2O3(S) -1273 B2H6(g) + 3O2  B2O3(s) + 3H2O(g) -2035 H2(g) + 1/2O2(g)  H2O(l) -286 H2O(l)  H2O(g) 44 Backwards Not enough Not enough X 3 X 3

2B(s) + 3H2(g)  B2H6 +36 2B(s) + 3/2 O2(g)  B2O3(S) -1273 kJ 2B(s) + 3/2 O2(g)  B2O3(S) -1273 B2O3(s) + 3H2O(g)  B2H6(g) + 3O2 +2035 3H2(g) + 3/2O2(g)  3H2O(l) -858 3H2O(l)  3H2O(g) 132 +36 2B(s) + 3H2(g)  B2H6

Hess’s Law Calculate ΔH for the reaction 2 C(s) + H2(g)  C2H2(g) given the following chemical equations and their respective enthalpy changes: C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l) ΔH = - 1299.6 kJ C(s) + O2(g)  CO2(g) ΔH = -393.5 kJ H2(g) + ½ O2(g)  H2O(l) ΔH = -285.8 kJ

Enthalpies of Formation An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which one mole of a compound is made from its constituent elements in their elemental forms. Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). See Appendix of textbook for tables of Hf°.

Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Imagine this as occurring in three steps: C3H8 (g)  3 C (graphite) + 4 H2 (g) 3 C (graphite) + 3 O2 (g)  3 CO2 (g) 4 H2 (g) + 2 O2 (g)  4 H2O (l) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

Calculation of H We can use Hess’s law in this way: H = nHf°products – mHf° reactants where n and m are the stoichiometric coefficients. Take the sum of the products’ heats of formation and subtract the sum of the reactants’ heats of formation. The Standard Heat of Formation of any element in it’s standard state is 0!

Practice For which of the following reactions at 25 ºC would the enthalpy change represent a standard enthalpy of formation? 2 Na(s) + ½ O2(g)  Na2O (s) 2 K(l) + Cl2(g)  2 KCl(s) C6H12O6(s)  6 C(diamond) + 6 H2(g) + 3O2(g)

Practice Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to form CO2(g) and H2O(l). ΔHfº: CO2(g) = -393.5 kJ/mol H2O (l) = -285.8 kJ/mol C6H6(l) = 49.0 kJ/mol O2 (g) = 0 kJ/mol

Practice Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to form CO2(g) and H2O(l). ΔHfº: CO2(g) = -393.5 kJ/mol H2O (l) = -285.8 kJ/mol C6H6(l) = 49.0 kJ/mol O2 (g) = 0 kJ/mol -3267 kJ/mol

Practice The standard enthalpy change for the reaction CaCO3(s)  CaO(s) + CO2(g) is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) (-635.5 kJ) and CO2(g) (-393.5 kJ), calculate the standard enthalpy of formation of CaCO3(s).

Practice The standard enthalpy change for the reaction CaCO3(s)  CaO(s) + CO2(g) is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) (-635.5 kJ) and CO2(g) (-393.5 kJ), calculate the standard enthalpy of formation of CaCO3(s). -1207.1 kJ/mol