Internal Energy, U

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Internal Energy, U

• Changes in U, dU or DU, are if energy of a
  system increases
• Energy here as heat ? heat added
• Work done on the system ?
• Sometimes it has been formulated that work done
  BY a system is in energy change ? not how we
  have it formulated above, difference in
  perspective..Be careful when reading other
  sources, this sign change confusion propogates
  through the rest of thermodynamics

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1st Law of Thermodynamics

• Aka the Law of conservation of energy, Gibbs in
  1873 stated energy cannot be created or
  destroyed, only transferred by any process
• The net change in energy is equal to the heat
  that flows across a boundary minus the work done
  BY the system
• DU q w
• Where q is heat and w is work
• Some heat flowing into a system is converted to
  work and therefore does not augment the internal
  energy

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Energy change with volume and heat

• Taking dU dq PdV from state 1 to state 2
• Yields U2-U1 (q2-q1) P(V2-V1)
• Make qpq2-q1, multiply PV terms and rearrange
• qp(E2PV2)-(E1-PV1)
• qp is MEASURABLE by measuring temperature
  changes resulting from energy changes (i.e. from
  a chemical reaction)

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Enthalpy (H)

• H U PV
• Total differential for Enthalpy is
• dH dU PdV VdP
• For our integrated change in state previous
• H1U1-PV1 and H2U2-PV2
• DH H2-H1 qp (AT constant P, V)
• Recall that energy is not known, only the change
  is meaningful
• Therefore change is measured from a reference
  state ? pure elements, 25ºC, 1 bar pressure have
  an enthalpy of zero ? H0f

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2nd Law of Thermodynamics

• 2nd Law introduces entropy, S
• Some of the enthalpy in a system is not
  convertible into work (PdV work for instance)
  because it is consumed by an increase in entropy
• Which could be restated that is requires some
  amount of work to increase entropy

(reversible)
(irreversible)
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• Combining the 1st and 2nd Laws of Thermodynamics
• dU dqrev dw TdS PdV
• If a process is at constant volume, V, and
  entropy, S ? dU 0 ? nothing happens, energy
  does not change in the system
• This is EQUILIBRIUM
• dUgt0 ? spontaneous rxn products to reactants
• dUlt0 ? spontaneous rxn reactants to products

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The Fundamental Equation

• Combining the first and second laws yields
• dU TdS PdV
• This is a key step, but the next one is the
  cornerstone of most thermodynamic calculations

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Free Energy

• Still need a function that describes reaction
  which occur at constant T, P
• G U PV TS H TS
• (dH dU PdV VdP)
• The total differential is
• dG dU PdV VdP TdS SdT
• G is therefore the energy that can run a process
  at constant P, T (though it can be affected by
  changing P and T)
• Reactions that have potential energy in a system
  independent of T, P ? aqueous species, minerals,
  gases that can react

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• Can start to evaluate G by defining total
  differential as a function of P and T
• dG dU PdV VdP TdS SdT
• Besides knowing volume changes, need to figure
  out how S changes with T
• For internal energy of a thing
• dU dqtot PdV determining this at constant
  volume ? dU CVdT
• where CV is the heat required to raise T by 1C

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Increasing energy with temp?

• The added energy in a substance that occurs as
  temperature increases is stored in modes of
  motion in the substance
• For any molecule modes are vibration,
  translation, and rotation
• Solid ? bond vibrations
• Gases ? translation
• Liquid water complex function

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Heat Capacity

• When heat is added to a phase its temperature
  increases (No, really)
• Not all materials behave the same though!
• dqCVdT ? where CV is a constant (heat capacity
  for a particular material)
• Or at constant P dqCpdT
• Recall that dqpdH then dHCpdT
• Relationship between CV and Cp

Where a and b are coefficients of isobaric
thermal expansion and isothermal compression,
respectively
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Enthalpy at different temps

• HOWEVER ? C isnt really constant.
• C also varies with temperature, so to really
  describe enthalpy of formation at any
  temperature, we need to define C as a function of
  temperature
• Another empirical determination
• Cpa(bx10-3)T(cx10-6)T2
• Where this is a fit to experimental data and a,
  b, and c are from the fit line (non-linear)

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Does water behave like this?

• Water exists as liquid, solids, gas, and
  supercritical fluid (boundary between gas and
  liquid disappears where this happens is the
  critical point)
• Cp is a complex function of
• T and P (H-bond affinities),
• does not ascribe to Maier-
• Kelley forms

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Heats of Formation, DHf

• Enthalpies, H, are found by calorimetry
• Enthalpies of formation are heats associated with
  formation of any molecule/mineral from its
  constituent elements

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Calorimetry

• Measurement of heat flow (through temperature)
  associated with a reaction
• Because dH q / dT, measuring Temperature change
  at constant P yields enthalpy

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Problem When 50.mL of 1.0M HCl and 50.mL of 1.0M
NaOH are mixed in a calorimeter, the temperature
of the resultant solution increases from 21.0oC
to 27.5oC. Calculate the enthalpy change per
mole of HCl for the reaction carried out at
constant pressure, assuming that the calorimeter
absorbs only a negligible quantity of heat, the
total volume of the solution is 100. mL, the
density of the solution is 1.0g/mL and its
specific heat is 4.18 J/g-K.
qrxn - (cs solution J/g-K) (mass of solution g)
(DT K) - (4.18 J/g-K) (1.0g/mL)(100 mL) (6.5
K) - 2700 J or 2.7 kJ DH 2.7 kJ Enthalpy
change per mole of HCl (-2.7 kJ)/(0.050 mol)
- 54 kJ/mol
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• Hesss Law
• Known values of DH for reactions can be used to
  determine DHs for other reactions.
• DH is a state function, and hence depends only on
  the amount of matter undergoing a change and on
  the initial state of the reactants and final
  state of the products.
• If a reaction can be carried out in a single step
  or multiple steps, the DH of the reaction will be
  the same regardless of the details of the process
  (single vs multi- step).

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• CH4(g) O2(g) --gt CO2(g) 2H2O(l) DH -890
  kJ
• If the same reaction was carried out in two
  steps
• CH4(g) O2(g) --gt CO2(g) 2H2O(g) DH -802
  kJ
• 2H2O(g) --gt 2H2O(l) DH -88 kJ

Hesss law if a reaction is carried out in a
series of steps, DH for the reaction will be
equal to the sum of the enthalpy change for the
individual steps.
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Volume Changes (Equation of State)
For Minerals
Volume is related to energy changes
Mineral volume changes as a function of T a,
coefficient of thermal expansion Mineral
volume changes as a function of P b, coefficient
of isothermal expansion
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Volume Changes (Equation of State)

• Gases and liquids undergo significant volume
  changes with T and P changes
• Number of empirically based EOS solns..
• For metamorphic environments
• Redlich and Kwong equation
• V-bar denotes a molar quatity, aRw and bRK are
  constants

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3rd Law of Thermodynamics

• The heat capacities of pure crystalline
  substances become zero at absolute zero
• Because dq CdT and dS dq / T
• We can therefore determine entropies of formation
  from the heat capacities (which are measureable)
  at very low temps

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Reference States

• We recall that we do not know absolute
  energies!!!
• We can describe any reaction or description of
  reaction relative to another ? this is all we
  need to describe equilibrium and predict reaction
  direction, just need an anchor
• Reference States
• Standard state 1 atm pressure, 25C
• Absolute states where can a value be defined? ?
  entropy at 0 Kelvin
• Most directly attributable to defined values for
  entropy, but uncommon in most applications

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Heat of Reaction

• Heat of reaction DH0R
• DH0R is positive ? exothermic
• DH0R is negative ? endothermic
• Example 2A 3B ? A2B3
• DH0R H0f(A2B3)-2H0f(A) 3H0f(B)

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Entropy of reaction

• Just as was done with enthalpies
• Entropy of reaction S0R
• When DS0R is positive ? entropy increases as a
  result of a change in state
• When DS0R is negative ? entropy decreases as a
  result of a change in state

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J. Willard Gibbs

• Gibbs realized that for a reaction, a certain
  amount of energy goes to an increase in entropy
  of a system.
• G H TS or DG0R DH0R TDS0R
• Gibbs Free Energy (G) is a state variable,
  measured in KJ/mol
• Tabulated values of DG0R are in Appendix B

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G is a measure of driving force

• DG0R DH0R TDS0R
• When DG0R is negative ? forward reaction has
  excess energy and will occur spontaneously
• When DG0R is positive ? there is not enough
  energy in the forward direction, and the BACKWARD
  reaction will occur
• When DG0R is ZERO ? reaction is AT equilibrium

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Free Energy Examples

• DG0R DH0R TDS0R

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Phase Relations

• Rule At equilibrium, reactants and products have
  the same Gibbs Energy
• For 2 things at equilibrium, can investigate the
  P-T relationships ? different minerals change
  with T-P differently
• For DGR DSRdT DVRdP, at equilibrium, DG0,
  rearranging

Clausius-Clapeyron equation
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• DV for solids stays nearly constant as P, T
  change, DV for liquids and gases DOES NOT
• Solid-solid reactions linear ? S and V nearly
  constant, DS/DV constant
• For metamorphic reactions involving liquids or
  gases, volume changes are significant, DV terms
  large and a function of T and P (and often
  complex functions) slope is not linear and can
  change sign!

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Phase diagram

• Need to represent how mineral reactions at
  equilibrium vary with P and T

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Gibbs Phase Rule

• The number of variables which are required to
  describe the state of a system
• pfc2 fc-p2
• Where p of phases, c of components,
• f degrees of freedom
• The degrees of freedom correspond to the number
  of intensive variables that can be changed
  without changing the number of phases in the
  system

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Variance and f

• fc-p2
• Consider a one component (unary) diagram
• If considering presence of 1 phase (the liquid,
  solid, OR gas) it is divariant
• 2 phases univariant
• 3 phases invariant