Reactions and thermodynamics OUTLINE

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Reactions and thermodynamics

Equilibrium?

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OUTLINE• Equilibrium and reactions• Chemical Thermodynamics

– Intro and The Phase Rule• Enthalpy• Entropy• The Gibbs Function• ΔG and K

START READING WHITE

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Mathematically describing chemical reactions

Reactions described by algebraic equations Material and charge "balanced”

Generic reaction: reactants A, B + products C, DBalance with stoichiometric coefficients a, b, c, d

both charge and atoms need balancing

aA +bB ↔ cC + dD

EquilibriumDescribing a chemical reactions: there’s an equation for that.

chemical equilibrium: if there is a unique mix of reactants and products in an assemblage of molecules that doesn't change with time because an energetically favorable balance is achieved:

Described withequilibrium constant (K): K= Cc Dd

Aa Bb

For: aA + bB ↔ cC + dD

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K= Cc Dd

Aa Bb

A, B, C and D are element activities (approximately = concentration (l) or pressure (g), but activity ≠ concentration;Activity adjusts for % component not available for rxn)

By definition: activities of solids and pure H2O(l) = 1 in the formulation of K.

K also depends on P & T

Non-Equilibrium

For non-equilibrium

reaction coefficient Q:

Q= Cc Dd

Aa Bb

Q < K: deficit of products. reaction proceeds in the direction written to correct this.

Q > K: excess of products. reaction will proceed in the reverse direction.

Q does not give reaction speed, that’s topic of kinetics

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Using K; what do the absolute values of K mean?

1) Qualitative solubility (Ksp ) NaCl(s) ↔ Na+(aq) + Cl-(aq) : large Ksp, so halite is very soluble in water

BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) : small Ksp,

so barite is sparingly soluble

2) Can add up equations and K’s(1) A + B ↔ AB K1= [AB]

[A] [B](2) AC ↔ A + C K2= [A] [C]

[AC]------------------------------------------------------------------------------(1 + 2) AC + B ↔ AB + C K3= [AB] [C] K3=K1•K2

[AC] [B]

K= Cc Dd

Aa Bb

Example: Combining Equilibria in the aqueous CO2 systemCO2 in natural waters (links to atmosphere, geosphere and biosphere)1.CO2(g) ↔ CO2(aq)

gas solubility – H in KH for “Henry’s law”2.CO2 (aq) + H2O ↔ H2CO3(aq)

bond reorganization – Keq

3.H2CO3 (aq) ↔ HCO3-(aq) + H+

1st acid dissociation – Ka1

4.HCO3- (aq) ↔ CO3

2- (aq) + H +

2nd acid dissociation – Ka2

5.CaCO3 (s) ↔ Ca2+ (aq) + CO32- (aq)

solubility/dissolution – Ksp

Combine 2 & 3 – H2CO3 = rare:CO2(aq) + H2O ↔ HCO3

-(aq) + H+ ( K'a1 )

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Combine for predictions rearrange K’s: [Ca2+] = Ksp/[CO3

2-][CO3

2-] = Ka2 [HCO3-]/[H]+

[HCO3-] = K'a1∙[CO2(aq)]/[H+]

Substituting:

For air vs water (P vs [C]) KH = [CO2(aq)]/PCO2[CO2(aq)] = KH∙PCO2

Substituting: [Ca2+(aq)] = Ksp ∙ [H+]2KH ∙ K'a1∙ Ka2 ∙ PCO2

So [Ca2+(aq)] depends on the amount of CO2 in the atmosphere and the pH (= -log[H+]) of the water. To be continued

Substituting: [Ca2+]= Ksp∙[H+]/Ka2∙[HCO3-]

[Ca2+]= Ksp∙[H+]∙[H+]/Ka2∙K'a1∙[CO2(aq)]

Chemical Thermodynamics

Thermodynamics is the study of the energetics of physical and chemical transformations of matter.

energetics of a "system " function of

physical quantities (e.g., T, P)

andchemical quantities (e.g., composition = X )

system can contain >1 chemical component and >1 physiochemical phase

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Definitions

• system - collection of matter in identifiable place & condition

• component

chemical entity (e.g., H2O) used to describe compositional variation in a system. Use minimum number of components needed to describe system: component need not physically exist (use end-members)

• phase

is a physical form of that chemical: for H2O: ice, water vapor, liquid water; SiO2: quartz, chert, cristobalite.

chemical entity (e.g., H2O) used to describe compositional variation in a system. Use minimum number of components needed to describe system: component need not physically exist (use end-members)

is a physical form of that chemical: for H2O: ice, water vapor, liquid water; SiO2: quartz, chert, cristobalite.

F = C - f + 2F = # degrees of freedom

The number of intensive parameters that must be specified in order to completely determine the system think P,T(X)

Can also think of this as “how many parameters can be varied independently from each other”

The Phase Rule – describes a system

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The number of intensive parameters that must be specified in order to completely determine the system

f = # of phases, sometimes written as Pphases are mechanically separable constituents Examples: minerals, liquids, …

The Phase Rule


The number of intensive parameters that must be specified in order to completely determine the system

f = # of phasesphases are mechanically separable constituents

C = minimum # of components (chemical constituents that must be specified in order to define all phases)Not necessarily atoms, SiO2, FeO, mineral endmembers,…

The Phase Rule

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The Phase RuleF = C - f + 2

F = # degrees of freedomThe number of intensive parameters that must be specified in

order to completely determine the system f = # of phases

phases are mechanically separable constituentsC = minimum # of components (chemical constituents that must

be specified in order to define all phases)2 = 2 intensive parameters -- balances the equation:

Usually temperature (T), pressure (P), composition (X) for us geologists. Variables #: C + pressure + temperature (C+2), while # equations or # phases = f. Fix P, -> C+1

Examples:

b) dissolved silicaSiO2 (s) + H2O (l) ↔ H4SiO4 (aq)

P=2 (l+s)C=2 (SiO2 and water)F=2 (e.g., T & P)

c)”salt” water NaCl (s) + H2O (l) ↔ Na+ (aq) + Cl- (aq)

P=2 (l+s)C=3 (water, sodium ions, chloride ions)F=3 (e.g., T & P and either Na+ or Cl-)

d) forsterite-fayalite-melt system...Mg2SiO4 + Fe22SiO4 form the solid solution mineral olivine: count as 1 phase

P = 2 (melt + olivine)C = 2 (forsterite + fayalite)F = 2 (e.g., T & P)

a)

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Minimize components 2In a complex natural system (sea water, magma) there are significant number of components (think trace elements)

• USUALLY approximate overall system with major components

• Add trace constituents for refinement, though large differences are needed to affect whole system

Chemical Thermodynamics – G, H, S

► Please refresh your memory on:enthalpy (H)entropy (S)Gibbs function (G)and their relationships to measures of chemical equilibrium (and chemical equilibrium constants, K).

° next to the variable is for STP (298.15°K = 25°C, 1 atm)

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Energy and Chemical Equilibrium

Gibbs Free Energy, ΔG, total system energy available for chemical work

► ΔGreaction = 0 at chemical equilibrium (no more change Δ)

ΔG°reaction related to equilibrium constant for a reaction:

ΔGo = -RT lnK (° = STP) Derivations in book, if you’re interested

For aA +bB ↔ cC + dD, substitute for K:

ΔGo = -RT ln Cc Dd

Aa Bb

Add reactants: positive ΔGo, back to 0 making more products

►If not at chemical equilibrium: ΔG = ΔGo + RT lnQ

Thermodynamic Quantities and RelationshipsBook derives the equations for the relevant variables

• Not going to derive thermodynamic expressions from first principles and partial differential equations. • We’ll use "finite" changes (e.g., ΔH, ΔG) for our applications.

G, H and S refer to reference materials, Δ defines change from reference state, but values also change in reactions

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Enthalpy – H

Enthalpy is heat energy (measure of molecular motion)Aka: the energy consumed or released in changes of state of a system

From H = U + PV and “taking care of” W=PΔV (ΔU=ΔQ-ΔW):ΔH = ΔQ + VΔPconstant pressure:ΔH = ΔQ (change is the heat gained or lost)ΔH = Cp ΔT (T change * Cp, heat capacity)constant temperature:ΔH = TΔS + VΔP(relation to changes in entropy and pressure)

More Enthalpy – H

enthalpy change of a chemical reaction at STP:For A + B ↔ C + D

ΔH°reaction = [ ΔHf°(C) + ΔHf° (D) ] - [ ΔHf° (A) + ΔHf°(B) ]

or simply the difference between both sides

The enthalpy of a chemical reaction reflects changes in 3 basic types of system heat; heat associated with

1. chemical bonds between different atom pairs2. molecular ordering in solids versus liquids versus gasses.3. molecular motion (kinetic energy)

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Enthalpy – H 1. "Bond Heat"

Heat of Formation Hf is the heat input/output when making a molecule from its constituent elements.

Hf°=0 for form that is stable at STPEx. C (graphite) ↔ C (diamond) involves transfer of heatGet ΔHf° (standard heat of formation) of diamond from a table:

ΔHf° of C (diamond) = -453 cal/mol (= -1.897 kJ/mol)

453 calories or -1.897 kJ of heat must be added to a mole of graphite to transform it into diamond at STP.

Enthalpy – H 1. "Bond Heat"

If two reactions can be mathematically combined to make a third, we can also combine their ΔHf° to calculate ΔHf° of the new reaction.

Example:1. C (graphite) + O2 (g) ↔ CO2 ΔHf°= 94.051 kcal/mol2. CO2 ↔ C (diamond) + O2 (g) ΔHf°= -94.504 kcal/mol

add the 2 to getC (graphite) ↔ C (diamond)

What do we add to what?

ΔHf° = (94.051 - 94.504) kcal/molΔHf° = -0.453 kcal/mol = -1.897 kJ/mol

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Enthalpy – H 2. "phase change heat"

Heat of Fusion Δhfus heat absorbed to melt, -Δhfus heat released to crystallize

Heat of Vaporization Δhvap heat absorbed to vaporize-Δhvap heat released to condense

Snowflake photos by Wilson Bentley circa 1902http://en.wikipedia.org/wiki/Wilson_Bentley

+ heat

- heat

3. other "kinetic energy heat"If no phase changes and chemical reactions, heat is gained or lost by temperature changes: ΔH = Cp ΔT

Entropy - S

Entropy = system disorder (chaos)

Naturals systems prefer minimum energy, maximum entropy

Qr= TΔS, at constant P

ΔS = Cp/ΔT (where Cp = heat capacity)

The standard state for any compound S° is determined relative to S° of H+ = 0 at STP.

Like ΔH, ΔS°reaction:

ΔS°reaction = ΣΔSf° (products)- ΣΔSf° (reactants)

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"Free Energy", or the Gibbs Function – G

Total system energy available for chemical work is known as Gibbs Free Energy, ΔG

► ΔGreaction = 0 at chemical equilibriumLike other variables of state, ΔG is additive:

ΔG °reaction = ΣΔGf° (products) - ΣΔGf° (reactants)

Since ΔG = ΔG° + RT lnQ (where Q is the reaction coefficient, not heat)

and since AT EQUILIBRIUM Q=K and ΔG = 0, we derive:ΔG° = -RT lnK


In closed systems, G is related to enthalpy H and entropy S by:

G = H-TS

For isothermal conditions:

ΔG = ΔH - TΔS and ΔGf° = ΔHf° - TΔSf°

In other words, the free energy change in a closed system reflects the heat and the degree of disorder.

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To adjust for different P-T:

ΔG = VΔP - SΔT

Consider a P-T phase diagram like this one

Now change P and T (dP, dT):

dΔG = ΔVdP – ΔSdT = 0

Rearranging, we arrive at the famous

Clapeyron Equation (slope in P-T plot):

dT = ΔV and because DS = DH/T, dT = TΔVdP ΔS dP ΔH

"Free Energy", or the Gibbs Function – Gsolve a problem using relationship of G and K

What is Keq for the precipitation reaction of themineral malachite from water?

2Cu2+(aq) + CO32-(aq) + 2 OH-(aq) ↔ Cu2(OH)2CO3 (s)

"copper 2+ "carbonate "hydroxyl "malachite"cation" anion" anion"

From the equation:reactants products

material: Cu 2 2CO3 1 1 OH 2 2

charge: Cu 4 (2x2) --CO3 -2 --OH -2 (-1x2) --Cu2(OH)2CO3 -- 0

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Next, ΔG°reaction = ΣΔGf°(products) - ΣΔGf°(reactants)

Make sure your units match (R has units too)!

2Cu2+(aq) + CO32-(aq) + 2 OH-(aq) ↔ Cu2(OH)2CO3 (s)

ΔG°reaction=-46.02kcal/mol = -46,020 cal/mol

From ΔGo = -RT lnK, we get ΔGo/(RT) = -lnK

-46,020/(298.15∙1.987) = -lnK K = e46,020/(298.15∙1.987)

K= 5.45 x 1033 or about 1033.7

What if you measured the values for K, you can solve for…?

ΔG°reaction=-216.44-(2∙15.50+ 2∙-37.60 -126.22) kcal/mol

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Reactions and thermodynamics OUTLINE