Transcript Thermodynamics

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
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‫׀‬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 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/GenChemTextbook/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).
1. Equation
2. Lewis Dot
3. 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
25 oC
1 g copper
100 J
25 oC
Increased
263 oC
Increased
24 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
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Volume is constant
q = ΔE
qsystem= -qsurroundings
qrxn= -(qbomb+ qwater)
qwater = mwaterCsΔT
qbomb = mbombCsΔT or CDT
Thermodynamics of Ideal Gas
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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
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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