Lecture 3

Report
The Third E ….
Lecture 3
Enthalpy
But first…
Uniting First & Second
Laws
• First Law:
dU = dQ + dW
• With substitution:
dU ≤ TdS – PdV
• For a reversible change:
dUrev = TdSrev – PdV
Defining Enthalpy
• Enthalpy is the heat gained or lost by the system if
the change takes place at constant pressure and
P–V work is the only work done by the system
dH = ∆QP
• More formally:
H = U + PV
• Consequently:
dH = dU +PdV +VdP
• But recalling that
dU ≤ TdS – PdV
• What do we have left?
Enthalpy and Heat
dH ≤ TdS + VdP
• For a reversible reaction:
dHrev = TdSrev + VdP
• And, if the (reversible) change takes place at
constant pressure:
dHP = TdSP = dQP
Enthalpies of Reaction,
Formation, Fusion, …
• For an isobaric experiment in which:
2H2 + O2 = 2H2O
o the enthalpy is simply the heat given off by the reaction. We can measure
this in a calorimeter.
o This is called a heat of reaction, ∆Hr
o If the reactants are elements, it is the heat of formation: , ∆Hf
• Another experiment we might conduct is a phase
change, e.g., melting ice to form (liquid) water.
o Enthalpy, in this case, is just the energy we would have to add to the ice
to melt it.
o Formally, this is called the latent heat of fusion; to avoid confusion, we will
call it the heat of melting, ∆Hm (heat of crystallization is just the negative of
this).
o We can similarly define latent heat of evaporation (or condensation).
Value of Enthalpy
• It is usually easier to conduct experiments at
constant pressure rather than constant volume.
• These kinds of experiments (e.g., combusting
hydrogen, melting ice) provide the basic data of
thermodynamics that allow us ultimately to predict
the outcome of natural reactions (like weathering
of rock). We can calculate other thermodynamic
properties from the enthalpy.
Deriving Entropies
• Since:
• dHrev = TdSrev + VdP
• If we conduct an experiment at constant pressure,
then:
dHrev = TdSrev
• And if we also did the experiment at constant
temperature, e.g., melting ice:
∆S = ∆H/T
• If the reaction is not at constant temperature
(melting of rock would not be), we would have to
integrate:
S=ò
dH (T )
T
Calorimetry
• A calorimeter in simply
a device in which we
conduct a reaction,
then measure the
change in temperature
in the vessel.
• But, we need to know
something additional:
How much energy change does
that temperature change
correspond to?
In other words, how much energy
would we have to put into the
system to change its temperature
by 1 K?
Heat Capacity
• The thing we need to know the Heat Capacity.
• We define it as:
C = dQ/dT
• At constant pressure:
CP = (∂H/∂T)P
• At constant volume since dU = dQ –PdV
CV = (∂U/∂T)V
• However, as we noted, experiments are usually not
conducted at constant volume (except for gases), so
mostly we will work with CP.
Heat Capacities of Solids
• Heat capacities of solids are complex functions of
temperature as vibrational motions of atoms in solids is
quantized. Theoretical calculation requires knowing all
possible vibrational motions and energy levels.
• In practice, heat capacities are empirically determined.
In geochemistry, the 5-term Haas-Fisher formulation is
generally used:
c
2
-1/2
CP = a+bT - 2 + fT + gT
T
• Where a, b, c, ƒ, g are empirically determined constants.
• We will use the simpler Maier-Kelley formulation:
CP = a+bT - c2
T
Boltzmann Distribution
Law
• While energy is on average shared equally among atoms or
molecules of a substance, this does not mean that in detail
they all have exactly the same energy.
• Theoretical calculation of heat capacity requires knowing
how energy is shared between atoms or molecules.
• This is given by the Boltzmann Distribution Law, which states
that the probability of an atom being in energy state εi is:
e- ei /kT
Pi =
- e n /kT
e
å
n
• k is Boltzmann’s constant, the microscopic equivalent of the
gas constant k = R/Av (where Av is Avagadro’s number)
and the sum is over all possible energy states.
Boltzmann’s Distribution
Law
In English:
• Energy among atoms is
like money among
men: the poor are
many and the rich few.
Occupation of vibrational energy levels. The
probability of an energy level associated with the
vibrational quantum number n calculated from
the Boltzmann distribution function of n for a
hypothetical diatomic molecule at 273 K and
673 K.
The Partition Function
• The denominator of the Boltzmann Distribution Law
is a key quantity in statistical mechanics called the
Partition Function:
Q = å e-en /kT
n
• Text explains how it is related to energy and
entropy.
• We will use it later in the course to calculate
equilibrium constant and isotope fractionation
factors.
Relationship of Entropy to
other State Variables
æ
ç
ç
è
¶S ö÷ = CV
¶T ÷øV T
æ
ç
ç
è
æ
ç
ç
è
¶S ö÷ = CP
¶T ÷ø P T
¶S ö÷ = -aV
¶P ÷ø T
æ
ç
ç
è
¶S ö÷ = a
¶V ÷ø T b
Calculating Entropy and
Enthalpy Changes
• Since:
o then
æ ¶H ö
çè
÷ø = CP
¶T V
dH = CP dT
• The enthalpy change of a substance heated from
T1 to T2 is then:
T2
∆ H = ò CP dT
T1
• For an entropy change it is:
T2 C
P dT
T1 T
∆S=ò
What if both Temperature
and Pressure Change?
• Entropy dependence
on pressure is:
æ
ç
ç
è
¶S ö÷ = -aV
¶P ÷ø T
P2
• So: ∆ S = òP -aVdP
1
o (for gases, V will be a function
of P; we can sometimes
assume solids are
incompressible).
• We need to do both
the T and P integrals!
• But, in a natural
process, T and P will
likely change
simultaneously!
o For example, a rising parcel of
air or sinking parcel of
seawater will experience
simultaneous changes of P and
T.
• Does it matter?
No!
• For state functions, the result
does not depend on the
path taken.
• Entropy and enthalpy are
state functions.
• The entropy change for an
isothermal pressure change
followed by an isobaric
temperature change will be
the same as a simultaneous
change of both.
• (Caveat: we must use the
appropriate values of
constants or functions (e.g.,
α at the relevant T; CP at the
relevant P).
Hess’s Law
• Once we have experimentally measured enthalpies of
formation for substances, we can calculate enthalpies
of reaction using Hess’s Law:
∆ H r = ån i H f ,i
i
• Where ν is the stoichiometric coefficient (by convention
negative for reactants, positive for products) and the
sum is over all compounds in the reaction.
• So, for the reaction
2MgO + SiO2 = Mg2SiO4
∆Hr = Hf,Mg2SiO4 - 2Hf,MgO- Hf,SiO2
Standard State Enthalpies
• We define the enthalpies of elements (or elemental
compounds such as O2) in the standard state of
298.16K and 0.1 MPa (25˚C, 1 atm) as 0.
o Units of enthalpies are what?
• We can then determine standard states of
formation (from the elements) of compounds such
as SiO2 as the enthalpy of reaction to form the
compounds under standard state conditions:
• ∆Hof,SiO2 = ∆Hr where the reaction is:
Si + O2 = SiO2

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