Thermodynamics. Energy is neither created or destroyed during chemical or physical changes, but it...
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Transcript of Thermodynamics. Energy is neither created or destroyed during chemical or physical changes, but it...
Thermodynamics
Energy is neither created or destroyed during chemical or
physical changes, but it is transformed from one form
to another.
Euniverse = 0
TYPES of ENERGY
Kinetic Potential
Mechanical Gravitational
Thermal Electrostatic
Electrical Chemical
Radiant
Energy Conversion Examples:1. dropping a rock2. using a flashlight3. driving a car
Endo: heat added to system Exo: heat released by system
SYSTEMS and SURROUNDINGSSystem: The thing under studySurroundings: Everything else in the universe
Energy transfer between system and surroundings:
HEAT: What happens to thermal (heat) energy?
Three possibilities:
• Warms another object
• Causes a change of state
• Is used in an endothermic reaction
Example 1: 5 g wood at 0 oC + 5 g wood at 100 oCExample 2: 10 g wood at 0 oC + 5 g wood at 100 oCExample 3: 5 g copper at 0 oC + 5 g copper at 100 oCExample 4: 5 g wood at 0 oC + 5 g copper at 100 oC
Choices: 1: 0 oC 2: 33 oC 3: 50 oC 4. 67 oC 5: 100 oC 6: other
Temperature Changes from Heat Exchange
What happens to thermal (heat) energy?
When objects of different temperature meet:
• Warmer object cools
• Cooler object warms
• Thermal energy is transferred
• qwarmer = -qcooler
Quantitative: Calculating Heat Exchange: Specific Heat Capacity
Specific Heat Capacity
The energy required to heat one gram of a substance by 1 oC.
Usefulness: #J transferred = S.H. x #g x T
How much energy is used to heat 250 g water from 17 oC to 100 oC?
What happens to thermal (heat) energy?
When objects of different temperature meet:
• Warmer object cools• Cooler object warms• Thermal energy is transferred
• qwarmer = -qcooler
specific heat x mass x T = specific heat x mass x T
warmer object cooler object
Heat transfer between substances:
owood o
Jq = 1.8 5 g (-18 C)
g C
= -160 J
oCu o
Jq = 0.385 5 g (+82 C)
g C
= +160 J
Conceptually Easy Example with Annoying Algebra:
If we mix 250 g H2O at 95 oC with 50 g H2O at 5 oC,
what will the final temperature be?
Thermal Energy and Phase Changes
First: What happens?
Thermal Energy and Phase Changes
First: What happens?
Thermal Energy and Phase Changes
First: What happens?
Warming:
• Molecules move more rapidly
• Kinetic Energy increases
• Temperature increases
Melting/Boiling:
• Molecules do NOT move more rapidly
• Temperature remains constant
• Intermolecular bonds are broken
• Chemical potential energy (enthalpy) increases
But what’s really happening?
Energy and Phase Changes: Quantitative TreatmentMelting:
Heat of Fusion (Hfus) for Water: 333 J/g
Boiling:
Heat of Vaporization (Hvap) for Water: 2256 J/g
Total Quantitative AnalysisConvert 40.0 g of ice at –30 oC to steam at 125 oC
Warm ice: (Specific heat = 2.06 J/g-oC)
oo
J2.06 40.0 g (30.0 C) = 2470 J
g C
oo
J4.18 40.0 g (100.0 C) = 16720 J
g C
333 J40.0 g = 13320 J
g
Melt ice:
Warm water (s.h. = 4.18 J/g-oC)
Total Quantitative AnalysisConvert 40.0 g of ice at –30 oC to steam at 125 oC
Boil water:
oo
J1.92 40.0 g (25.0 C) = 1920 J
g C
2256 J40.0 g = 90240 J
g
Warm steam (s.h. = 1.92 J/g-oC)
Lots of different types of energy.
We use Enthalpy:
Heat exchanged under constant pressure.
Energy and Chemical Reactions
Energy/Enthalpy Diagrams
Some Examples of Enthalpy Change
2 C(s) + 2 H2(g) C2H4(g) H = +52 kJ
Enthalpy Change and Chemical Reactions
H is usually more complicated, due to solvent and solid interactions.
So, we measure H experimentally.
Calorimetry
Run reaction in a way that the heat exchangedcan be measured. Use a “calorimeter.”
Bomb Calorimetry Experiment
N2H4 + 3 O2 2 NO2 + 2 H2O
Energy released = E absorbed by water +E absorbed by calorimeter
Ewater =
Ecalorimeter =
Total E =
H = energy/moles =
0.500 g N2H4
600 g water
420 J/oC
Enthalpy Change and Bond Energies
H = energy needed to break bonds – energy released forming bonds
Example: formation of water:
H = [498 + (2 x 436)] – [4 x 436] kJ = -482 kJ
General Rule:
SO2 + ½ O2 SO3 dH = -98.9 kJ
2 SO3 2 SO2 + O2 dH = ?
Hess’s Law
If reactions can be “added”so can their H values.
Thermochemistry Lab Calculations
Goal: What is H for the formation of MgO from Mg(s) and O2(g)?
Mg(s) + ½ O2(g) MgO(s) H = ? kJ/molData:From lab measurements:
Mg(s) + 2 H+(aq) Mg2+(aq) + H2(g) H1 = ___________ kJ/mol
MgO(s) + 2 H+(aq) Mg2+(aq) + H2O(l) H2 = ___________ kJ/mol
From a table: H2(g) + ½ O2(g) H2O(l) H3 = -285.8 kJ/mol
Task: Find a way to add these three reactions to get the desired reaction. Manipulate the H values as needed, and add them.
Calculating Heat Production
Heat of Formation
Heat of Formation: The general idea
Find the enthalpy change for burning ethyl alcohol