WELCOME TO CHEM 45.5 Enthalpy is just another word for heat (q) given off or absorbed by a system at constant pressure.

ΔE = heat + work = q - PΔV

CASE 1 CASE 2

At constant pressure: At constant volume: ΔV=0

ΔE = qP - PΔV ΔE = qV - PΔV

qP = ΔE + PΔV ΔE = qV

Enthalpy = ΔH = qP = ΔE + PΔV An energy level diagram summarizes the enthalpy changes from an initial state to a final state in a chemical reaction. It’s a state function!

H2O(l) H2O(s) + heat H2O(l) + heat H2O(g)

H2O(l) H2O(g) Hfinal Hinitial , H , H ΔH < 0 heat out ΔH > 0 heat in Enthalpy Enthalpy

H2O(s) H2O(l) Hfinal Hinitial

A Exothermic process B Endothermic process Why Do We Care About State Functions? • A change in a state function tells us that the function has one unique value between any two arbitrary states. This fact is why thermodynamics is so powerful and useful.

Final State mgh2

Arbitray State Δh

Initial State h = 1 Chemists measure the enthalpies many chemical reactions, give them names and tabulate these values in Handbooks. They are useful in the real- world. The heat of fusion and the heat of vaporization are intensive properties that tell us something about the IMF between these molecules (tabulated in Handbook of Physics and Chemistry). ΔH˚vap ΔH˚fus • Molar Heat of Fusion (∆Hfus): The energy required to melt one mole of solid (in kJ).

• Molar Heat of Vaporization

(∆Hvap): The energy (in kJ) required to vaporize one mole of liquid.

• Molar Heat of Sublimation

(∆Hsub): The differences in energy required to The energy (in kJ) required to vaporize or melt a pure substance tells vaporize one mole of solid to us something about the IMF’s that try to gas. keep these molecules together in the condensed phase. A Small Table of Heats of Fusion and Heats of Vaporization

Notice that ΔH˚vap is greater than ΔH˚fus suggesting that IMF are more stronger in the liquid phase. We can graphically show the quantitative aspects of a phase change in a “heating curve” or a plot of temperature vs heat added to a substance.

Heating steam past 100˚C

Boiling all water to steam at 100˚C

Heating water to boiling 100˚C

Melting solid Heating solid ice to 0˚C ice to 0˚C water

Total heat = qA-B + qB-C + qC-D + qD-E + qE-F Enthalpies (ΔH) of chemical reactions can be determined using a lab method called calorimetry. We need to understand the physics of heat transfer (not really chemistry) and apply to simple concepts.

1. Law of Conservation of Energy n

qsystem + qsurroundings = 0 qi = 0 i=1 ! 2. Specific Heat Capacity of Substances q = s m ΔT = C ΔT Heat (q) absorbed depends on specific heat capacity, mass and temperature change. Some Physics: The magnitude of heat, q, transferred from an object of higher temperature to an object at lower temperature is proportional to the difference in temperature of the two objects!

Heat transferred = q ∝ ΔT q = C ΔT

C = heat capacity

q q = C ΔT material dependent constant of proportionality we call “heat capacity” Solving for C we get: C = q / ΔT

C has units of Energy per degree T (J/°C, J/K, cal/°C, cal/K)

Heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity of the substance by one degree Celsius or Kelvin (units of J/˚C or J/K or cal/˚C). Experiments also show that the heat capacity, C, of a material is proportional to mass (m) of a substance under question. We can write:

C ∝ m

C = m s

where m is mass of the substance and s is a proportionality constant called specific heat capacity of a substance. Units of s: C = m s = q / ΔT J/g °C, J/g K

(S is a measure of how thermally sensitive a substance is to the addition of energy) The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

q = C ΔT C = m s

(1) (2) Substituting specific heat (1) for heat capacity C in equation 2 gives:

Heat (q) absorbed or released: q = s m ΔT The molar heat capacity of a substance is the amount of heat (q) required to raise the temperature of one mole of the substance by one degree Celsius.

Heat (q) absorbed or released: q = s m ΔT

Units: ˚C or K Units: Units: moles J/mol °C or J/mol K How much heat, q, is required to raise the temperature of 1000. kg of iron from 25 to 75˚C?

We look up the Specific heat capacity in a table and find that s of Fe = 0.444 J/g ˚C How much heat, q, is required to raise the temperature of 1000. kg of iron from 25 to 75˚C?

We look up the Specific heat capacity, s, of Fe = 0.444 J/g ˚C

Heat (q) absorbed or released:

q = s m ΔT Now let’s apply the heat stuff to real-world stuff like phase changes and the energy or cost it takes to carry it out. A “heating curve”...... a plot of temperature of a substance vs heat added to a substance.

emperature ˚C

T

Heat Added (Joules) Here’s the same curve now applying the conservation of energy (sum of the heats)

q = mice ∆Hfus q = mH2O ∆Hvap q = sstm mstm ∆T q = sice mice ∆T q = sH2O mH2O ∆T Phase Phase F 220 transition transition Temperature does Temperature does not change during not change during a phase transition. a phase transition. Steam Heating Ice + Water 100 solid ice Water D Water + E to 0˚C mix steam mix 0 C Ice B

emperature ˚C Heating Boiling all

T Heating water to water to Melting steam past boiling steam -100 solid ice to 100˚C A 0˚C water 100˚C 100˚C 4.12 38 79.3 305 309 Heat Added (kJ/mole) Calculate the amount of heat required to convert 500 grams of ice at -20.0˚C to steam at 120.˚C. The specific heat capacities of water, ice and water vapor are 4.18, 2.06 and 1.84 J/g ˚C respectively, and the latent heat of fusion and vaporization, ΔHf and ΔHv, are 6.02 and 40.7 kJ/mol respectively. n

qi = 0 sum the q’s baby i=1 ! for heating non-phase transitions q = si mi ΔT

qsolid=>liquid = # moles ΔH˚fusion for phase transitions qliquid=>gas = # moles ΔH˚vaporization Calculate the amount of heat required to convert 500 grams of ice at -20˚C to steam at 120˚C. The specific heat capacities of water, ice and water vapor are 4.18, 2.06 and 1.84 J/g ˚C respectively, and the latent heat of fusion and vaporization, ΔHf and ΔHv, are 6.02 and 40.7 kJ/mol respectively. 1. Heat ice from -20˚C to ice at 0˚C = 500. g x 2.06 J/g ˚C x 20˚C 2. Melt ice at 0˚C to water at 0˚C = 500. g/(18 g/mol) x 6.02 kJ/mol 3. Heat water from 0˚C to water at 100˚C= 500. g x 4.18 J/g ˚C x 100˚C 4. Evap water at 100˚C to vap at 100˚C = 500. g/(18 g/mol) x 40.7 kJ/mol 5. Heat vap from 100˚C to vap at 120˚C = 500. g x 1.84 J/g ˚C x 20˚C 1. = 20.6 kJ 2. = 167.2 kJ 3. = 209.0 kJ 4. = 1130.5 kJ 5. = 18.4 kJ

Total = 1545.6 kJ