04_ChemicalEquilibrium

7/28/2019 04_ChemicalEquilibrium

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III. CHEMICAL EQUILIBRIUM

VARIABLES

(a) Temperature (T) Commonly used scale is Celsius (oC)[ (but Kelvin (K) scale in thermodynamics]

(b) Pressure (P)Three units of pressure are in common use: bar, atmosphere (atm) and pascal. (Pa)

1 bar = 0.987 atm = 105 Pa = 0.1 MPa

The difference between the bar and the atmosphere is so small that for most purposes in geochemistry the

two may considered approximately equivalent.

(c) the concentration of chemical species (Xi) in the system under consideration

Concentrations ofsolids: - Usually expressed as ppm = g/tonne

Concentrations ofsolutes in an aqueous solution are expressed in various ways:

molalilty(m) = moles of solute/ kg of solvent [for pure water, 1 liter = 1 kg]

molarity(M) = moles of solute/ liter of solution

parts per million(ppm) = grams/106 g = tons/106 tons = mg/106 mg (=1 kg)

We will use molalilty for consistency (although in dilute solutions the difference between molalilty and

molarity is insignificant in most geologic calculations).

BASIC QUESTIONS TO BE ANSWERED

(a) In which direction should a chemical reaction go and why?

(b) When should it stop and why?

(c) Which mineral assemblage is stable and which is not at a given P-T condition, etc.

(d) What happens when we disturb the equilibrium?

Such questions may be answered by a variety of interrelated techniques: the Law of Mass Action (and the

Equilibrium Constant); Free Energy change for reactions; Chemical potentials, etc. -- and all of them would convergeon the same answer. We focus here on the simplest, and possibly somewhat familiar, technique involving equilibrium

constants.

THE LAW OF MASS ACTION

The Concept of Chemical Equilibrium

Many chemical reactions are incomplete because they are reversible - i.e., the same final product is obtained when

such a reaction is started from either end. For example, for a reversible reaction represented as

A + B Y + Z

the end product would contain the same mixture of four substances whether we mix A and B or Y and Z. In modern

language we would say that a reversible reaction stops when it has reached a state ofchemical equilibrium. Actually,

the reaction does not stop, but a dynamic equilibrium has been reached. No further change in the assemblage is

observed because of a balance between the two opposing processes going at equal rates. In fact, for a chemical

reaction, the terms reversibility and equilibrium are used interchangeably.

An irreversible process is unidirectional. No equilibrium can be achieved in the case of an irreversible process.


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From a state of equilibrium, the reaction can be made to go in a forward direction by adding more of A and/or B, and

in the reverse direction by adding more of Y and/or Z. The scientists who recognized this in the 1860s, it suggested

that the "driving force" of a reaction depended on the "masses" of the substances involved. This obscure formulation

has come down to us as theLaw of Mass Action.

Equilibrium Constant

For a generalized reversible chemical reaction represented by the equation

aA + bB yY + zZ (3.1a)

the state of equilibrium is represented by a quantity called theEquilibrium Constant(Keq), which is defined as

[Y]y [Z]z

Keq = ---------------- (3.1b)

[A]a [B]b

Note that the quantities [Y],.........in the expression for Keq represent concentrations (e.g., number of moles/g, ppm,

g/liter, etc), not masses (e.g., grams, liters, etc.).

Whether such a reaction is feasible or not has to be answered from other considerations, such as Free Energy changefor the reaction.

Actually, Keq as defined above is only an approximate constant; to obtain the correct value of Keq, we must use

activities (for solid and liquid phases)and fugacities(for gaseous phases). These are effective or thermodynamic

parameters related, respectively, to concentration and partial pressure.

____________________________________________________________________________________________

Concentrations used for Equilibrium Constant

Pure solid phases (i.e., 1 component i) Activity (ai) = 1

Solid solution (e.g., component i in phase ) Activity (ai), depends on the solid solution

Solute (i) in a liquid solution Activity (ai ) = i mI [ m = molality ofi i = activity coefficient ofi]

lim ai/mI = i 1 as mI 0(i.e. for very dilute solutions, the activity is almost the same as

the molality, and for an ideal solution, ai = mI)

Ideal gas Partial pressure (Pi) = XiP [P = total pressure

Xi = mole fraction ofi]

Non-ideal gas Fugacity (f i) = iPi [ i= fugacity coefficient ofi]

lim fi/Pi = i 1 as P 0(i.e. all gases behave ideally at sufficiently low pressures)

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However, for the present, we will stick with the more familiar expressions of concentrations. In very dilute solutions,

as we are concerned with in this chapter, the concentration of a solute in a solution is numerically very close to its

activity and the partial pressure of a gas is very close to its fugacity.


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There are a few more points about Keq that need to be emphasized:

(a) Note that a chemical reaction is always written in terms of the number of moles of reactants and products

and is balanced.

(b) Keq for a reaction is a constant at a particular P-T condition. In fact, we will see later that the variation of

K with T and P forms the basis ofgeothermometry and geobarometry, respectively.

Generally, reactions are more sensitive to temperature than pressure (unless a fluid phase is involved in

the reaction). As a crude general guideline, the rate of most chemical reactions double every 10oC rise

in temperature.

(c) Some reactions in nature may fail to reach equilibrium (i.e., unidirectional reactions) because:

(i) some part of the reaction process is very slow (a kinetic barrier)

e.g., Feldspar weathers to kaolinite on the earths surface, but not the other way.

Anhydrite dissolves in water at 250C (to a very limited extent), but from a solution saturated

with anhydrite, what would precipitate is gypsum, not anhydrite.

(ii) the reactants and productions don't stay together long enough to achieve equilibrium (e.g., a reaction

such as

CaCO3 + 2HCl = 2Cl- + Ca2+ + H2O + CO2 (3.2)

produces CO2 that may escape into the atmosphere -- such a reaction may run to completion.)

(d) Commonly used notation:

= Balanced reaction

Reaction at equilibrium

Unidirectional reaction

How will you establish that a given reaction has achieved equilibrium?


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EXAMPLES OF CHEMICAL EQUILIBRIUM

(a) H2 (g) + Cl2 (g) 2HCl (g) [Homogeneous equilibria one gaseous phase] (3.3)

[HCl]2

KHCl = ----------------- = 2.6x108 = 108.4 at 1,000oC (measured) (3.4)

[H2] [Cl2]

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Equivalence of, for example, 2.5 x 108 and108.4 ||

||

log (2.6x108) = log 2.6 + log 108 = 0.4 + 8 = 8.4 || 108.4 = 100.4 x 108 = 2.5 x 108

antilog 8.4 = 108.4

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We can, of course, write the above reaction in several ways:

2HCl (g) H2 (g) + Cl2 (g) K'HCl = 1/K

1/2 H2 (g) + 1/2 Cl2 (g) HCl (g) K''HCl = (K)1/2

Thus, an Equilibrium Constant has no meaning except in reference to a specific reaction. Always balance the

reaction.---------------------------------------------------------------------------------------------------------------------------------------------

Assuming the gases to be ideal, their concentrations can be expressed aspartial pressures (Pi), so that we can write

KHCl = PHCl2/ PH2 PCl2

Dalton's Law of Partial Pressure: Pi = Xi P where Xi = mole fraction of i = ni / ni______________________________________________________________________________________________

Suppose PH2 = PCl2 = 1 bar; then, KHCl = PHCl2 and P

HCl=(108.4 bars) = 104.2 bars.

Now let us disturb the equilibrium by adding some H2 to the mixture so as to raise the partial pressure of H 2

momentarily to 10 bars. Some H2 must now react with an equivalent amount of Cl2 to produce HCl (forward

reaction) so that the value of KHCl is restored.

What will be the new values of the partial pressures when equilibrium is re-established?

Suppose the amount of H2 that reacts is such that the PH2 is reduced by the amount x.

Since every mole of H2 would react with another mole of Cl2 to produce 2 moles of HCl, PCl2 must also be reduced

by x whereas PHCl will be increased by 2x. So, we can write

(104.2 + 2x)2

----------------- = KHCl = 108.4

(10-x) (1-x)

There should be no problem in solving this quadratic equation for x.

This is a standard technique for calculating new concentrations of reactant and product species when the equilibriumin a system is disturbed.

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For a quadratic equation of the form ax2 + bx + c = 0, the roots are given by:

-b + (b2 - 4ac)1/2

x = -----------------------------

2a

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But, to save time, we will use a shortcut to arrive at only a slightly less precise answer.

Note that x must be less than 1, so that 2x is negligible compared to 10 4.2. Neglecting 2x from the numerator

simplifies the equation to

(104.2)2 /(10-x) (1-x)=108.4

(10-x) (1-x) = 110x =11 (neglecting x2)

x = 0.91

We can even use another approximation that x is fairly small compared to 10 and therefore can be neglected, so that

10 (1-x) = 1

x = 0.90 (accurate enough !)

So, the new values of partial pressures when the equilibrium is re-established (K = 10 8.4):

PH2 = 10 - 0.9 = 9.1 bar

PCl2 = 1 - 0.9 = 0.1 bar

PHCl = 104.2

bars (neglecting x)

[You should substitute these values in our equation for KHCl and see what is the error in its value. The error will be

insignificant. You should always look for reasonable approximations while solving problems, but make sure to

explain the justification for your simplifying assumptions. Also, it is always a good idea to check the reasonableness

your assumptions by back-substitution.]

(b)CO2 (g) + H2O (liq) H2CO3 (liq) [Heterogeneous equilibria 2 phases] (3.5)

This is a very important reaction for the geochemical cycle of carbon, because it governs the distribution of CO 2

between the hydrosphere and the atmosphere.

The equilibrium constant for this reaction is:

[H2CO3] mH2CO3KCO2 = ------------- = ------------------ (3.6)

[CO2] [H2O] PCO2 mH2O

To evaluate KCO2, we use two standard conventions:

(1) Actually, mH2O = No. of moles of water per 1 kg (=1000 g) of water

= 1,000/18.016 (gram molecular wt. of H2O) / 1

= 55.5 / 1 = 55.5

But, the standard convention is to represent the concentration of H2O by its mole fraction in a solution, and inall dilute solutions this mole fraction is approximately 1

So, mH2O = 1

(2) It is convenient to adopt the convention that dissolved CO2 is all H2CO3 (although, as we will see later on, this

is not strictly correct), and to use equilibrium constants consistent with this convention.


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Thus, the equation for KCO2 can now be written as:

KCO2 = mH2CO3/PCO2 (3.7)

This relation establishes that, at given P (total) and T, for every PCO2 (or fCO2), there is a corresponding equilibrium

value ofmH2CO3 (or aH2CO3), and vice versa. Thus, in the literature, it is quite common to report mH2CO3 (or

aH2CO3) as the corresponding PCO2 (or fCO2), even when no gas phase is present.

Let us consider this reaction at T = 25oC and PCO2 = 1 bar

What is the value of mH2CO3?

Solubility of CO2 is: 0.76 liter of CO2/liter of water (= 1 kg of water) at 25oC and PCO2= 1 bar

So, how many moles of CO2 per 1 kg of water (i.e., molality of CO2)?

The volume of 1 mole of any gas at 25oC and 1 bar pressure = 24.5 liters

0.76 liters of CO2 under these conditions = 0.76/24.5 = 0.031 mole/kg of water

Since 1 mole of CO2 produces 1 mole of H2CO3, mH2CO3 = 0.031 mole/kg of water

Substituting the values of mH2CO3= 0.031 and PCO2 = 1 bar in the equation for KCO2

KCO2 = mH2CO3/1 = 0.031= 10-1.5

Thus, for the reaction CO2 (g) + H2O (liq) H2CO3 (liq) at 25oC and 1 bar, Keq = 10-1.5This is an important conclusion that will be used frequently in this course. Note that the higher the value of PCO2 ,

the higher the value of mH2CO3. Assumption: mH2CO3 (the molality of H2CO3 dissolved in water) is determined

solely by the equilibrium between water and atmospheric CO2.

Let us calculate how much H2CO3 can be present in water exposed to ordinary atmosphere?

Qualitatively, it should be much less as the atmospheric PCO2 is much less than the 1 bar assumed for the earlier

calculation.

Atmosphere contains 0.03% CO2 by volume

So, volume fraction of CO2 in atmosphere = 0.03/100 = 0.0003

Assuming that mole fraction of CO2 volume fraction of CO2, XCO2 = 0.0003

PCO2 = XCO2 x P (atmospheric pressure) = 0.0003 x 1 = 0.0003

Substituting, KCO2 = mH2CO3/PCO2 = 0.031

So, mH2CO3= 0.031 x 0.0003 = 10-1.5 x 10-3.5 = 10-5m

Thus, water in equilibrium with the present atmosphere should contain 10-5m of CO2/kg of water. This is an

important conclusion. Note that the higher the value of PCO2, the higher the value of mH2CO3.

A molality of 10-5mseems to be a very small concentration of H2CO3acid in normal surface water, but it is

sufficient to make natural waters much better weathering agents than pure water.

A relevant question: what is the pH of water in equilibrium with atmospheric CO2?

The answer is complicated by the fact that the pH of the solution will be related to the dissociation of H 2CO3:

H2CO3 = HCO3- + H+.

We will take this up later.

But, for the time being, qualitatively, should you expect the Precambrian ocean water to have been more acidic or

less acidic compared to the present?


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SOLUBILITY SOLUBILITY PRODUCT

Solubility = the amount of a compound (or an ionic species) that dissolves in a saturatedsolution.

[Remember that in order to measure or calculate solubility of a species, the system must be at equilibrium.]

When a salt dissolves in water, it dissociates into the anions and cations of the acid and the base from which it was

derived. The solution may also contain neutral molecules of the salt, but their concentration is very low in most cases

and may be ignored.

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The Concept of Solubility

Let us conduct an experiment at by putting a lump of CaSO 4 (anhydrite) crystals in a beaker of pure water at room

temperature (25oC). Some amount of CaSO4 (actually, very little amount) will dissolve in the water a dissociation

reaction represented as:

CaSO4 = Ca2+ + SO42

So, how much CaSO4 has dissolved in the water after a given time can be determined by measuring the concentration

(e.g., molality) of Ca2+ in the water (or of SO42- in this case), because every mole of dissociated CaSO4 gives rise to 1mole of Ca2+ and 1 mole of SO4

2-).

If we monitor the concentration of Ca2+ as a function of time, we will get a graph as shown above: m Ca2+ increasing

with time up to a certain point and then staying at a constant value as equilibrium is established between the CaSO4and the water. At this point, the water is saturated with (dissolved) CaSO4 (i.e., the water cannot dissolve any more

CaSO4).

S

ConcentrationofCa2+inthewater(m

Ca2+)

Time

Equilibrium

Disequilibrium


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The solubility of CaSO4 is the number of moles of CaSO4 that will dissolve in pure water (at a given temperature

25oC in this experiment) to make the water (actually, now an aqueous solution) saturated with respect to CaSO 4

The solubility is determined by measuring the concentration of one of the dissociated species in the saturated

solution. For this experiment,

Solubility of CaSO4

= mCa2+

in the solution saturated with CaSO4

= mSO42- in the saturated solution of CaSO4

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CaSO4(s) Ca2+ (aq) + SO42- (aq) [heterogeneous equilibria 2 phases] (3.8)

Assuming equilibrium i.e., a saturated solution (how will you test the attainment of equilibrium?),

[Ca2+] [SO42-]

KCaSO4 = ------------------------- = 3.4 x 10-5 = 10-4.5 at 25oC and 1 bar (3.9)

[CaSO4]

In a saturated solution, we consider the concentration of the pure solid to be essentially constant (can you figure outthe reason?) and assign a value of 1 to this constant concentration (this convention does not apply to impure solids or

solid solutions). The concentrations of Ca2+ and SO42- in the aqueous solution can be represented by molalities.

KCaSO4 = mCa2+ mSO4

2- (3.10)

This kind of equilibrium constant is called a Solubility Product(Ksp)

For any slightly soluble salt (so that the solution is a very dilute one), a similar constant product of ionic

concentrations at equilibrium can be set up. For example,

AgCl Ag+ + Cl- Ksp (AgCl) = mAg+ mCl- (3.11)

CaF2 Ca2+ + 2F- Ksp (CaF2) = mCa2+ m2F- (3.12)As2S3 2As3+ + 3S2- Ksp (As2S3) = m

2As3+ m

3S2- (3.13)

(a) What is the solubility of CaSO4 (a salt) in pure water at 25oC and 1 bar?

Since every mole of CaSO4 on dissolution gives 1 mole each of Ca2+ and SO4

2-,

Solubility of CaSO4 = mCa2+ = mSO42-

Since mCa2+.mSO42-= Ksp (CaSO4)= 3.4 x 10

-5 = (solubility)2,

Solubility of CaSO4 in pure water = K1/2 = (3.4 x 10-5)1/2 = 5.8 x 10-3m

Note that this figure represents the solubility of CaSO4 when it is the sole source of Ca2+ and SO4

2- in the water. In

general, this is not true for geologic environments because most natural waters contain Ca2+ and SO42- from other

sources.


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(b) How much CaSO4 is dissociated

% dissociation = Amt of CaSO4 dissociated / Amt of CaSO4 originally present = [(5.8 x 10-3 / 1) x 100] = 0.58%

(c) How can we tell if a given solution is saturated in a particular solute?

By determining the Saturation Index (not applicable to hydrolysis reactions, which are controlled by pH):

Saturation Index (S.I.) = Ion Activity Product / Solubility Product

Supersaturated: S.I. > 1 IAP > Ksp the excess solute should precipitate out

Saturated: S.I. = 1 IAP = Ksp equilibrium condition

Undersaturated: S.I. < 1 IAP < Ksp the solution can dissolve more of the solute

Suppose a sample of water contains 5x10-2m of Ca2+ and 7x10-3m of SO42-. Is this solution saturated with

respect to Ca2+?

Ion Activity Product (IAP) = mCa2+ mSO42- = (5x10-2m) (7x10-3) = 35x10-5 = 10-3.45

[We are, of course, using concentrations, assuming that they are numerically equal to activities for very dilute

solutions.]

Compare IAP of CaSO4 in the given solution with Ksp (CaSO4) = 10-4.5

Since IAP > Ksp, the water sample is supersaturated with CaSO4. That is, some CaSO4 should precipitate out of this

solution.

(d) How much CaSO4 would precipitate from the solution for restoration of the equilibrium between CaSO4 and

solution?

[Actually, at room temperature, we will precipitate gypsum (CaSO4.2H2O) rather than anhydrite, but we will ignore

this fact as it does not make any difference in our calculations.]

Let x = moles of CaSO4 that will precipitate

Then, the concentrations of the ions at equilibrium will become:

mCa2+ = (5x10-2) - x

mSO42- = (7x10-3) x

Since, at equilibrium, Ksp (CaSO4) = 10-4.5 = mCa2+.mSO4

2-,

[(5x10-2) - x] [((7x10-3) - x] = 10-4.5

x2 - 5 x10-2 x - 7 x10-3 x + 35 x10-5 - 10-4.5 = 0

x2 - (5.7 x 10-2)x + (3.18 x10-4) = 0

The two roots of this quadratic equation are: x1 = 5.05 x 10-2

x2 = 6.45 x 10-3

x1 will make the values of mCa2+.and mSO42- negative. So, we accept x2.

The amount of CaSO4 that will precipitate = 6.45 x 10-3

moles/kg water

[After this amount of CaSO4 has precipitated, there should be equilibrium established between CaSO4 and water --

i.e., IAP of CaSO4 in this solution should be the same as Ksp (CaSO4) = 10-4.5. Let us check.

The new IAP = [(5x10-2) - x] [((7x10-3) - x]

= [(5 x 10-2) - (6.45 x 10 -3)] - (7 x 10-3 - (6.45 x 10-3)]

= [(5 - 0.645) 10-2] [(7 - 6.45) 10-3]

= [4.355 x 10-2] [0.55 x 10-3] = 2.39 x 10-5 = 10-4.6


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This turns out to be 10-4.6, instead of 10-4.5, because I have taken the solution of the quadratic equation from Faure,

1991, who uses Ksp (CaSO4) = 10-4.6]

(e) What is the solubility of CaF2 in pure water (25oC, 1 bar)?

CaF2 Ca2+ + 2F- (3.14)

Ksp (CaF2)= mCa2+.m2F-

= 10-10.4 at 25oC and 1 bar

Let x be the solubility of CaF2 (i.e., x moles of CaF2 can dissolve in 1 mole of water to make a saturated solution)

Since every mole of CaF2 that dissolves gives 1 mole of Ca2+ and 2 moles of F-,

Ksp (CaF2) = mCa2+.m2F = (x) (2x)

2 = 4x3 = 10-10.4

x = mCa2+ = (0.25 x 10-10.4)1/3 = 10-3.7m = solubility of CaF2

Thus, Solubility of CaF2 = mCa2+ = 1/2 mF

Note that this figure is valid only for CaF2 in contact with pure water containing no Ca2+ or F- from other sources.

(d) What is the solubility of CaSO4 in a solution of 0.1m CaCl2?

The task here is to find out how is the solubility affected by the presence of other ions in the solution, particularly by

an excess of Ca2+ or SO4-2 from other sources?

Let x be the solubility of CaSO4 (i.e., x moles ofCaSO4 will dissolve in 1 kg of water)

The concentration of Ca 2+ in 1 kg of water = x moles (from CaSO4

) + 0.1 moles (contribution from CaCl2

)

The concentration of SO42- in 1 kg of water = x moles/1 kg of water

Ksp (CaSO4) = mCa2+ mSO42-= (0.1+x) (x) = 3.4 x 10-5

0.1x + x2 = 3.4 x 10-5

We can solve this quadratic equation for x, but again we will take a short cut for an approximate solution.

Experience tells us that x will not be larger than the solubility we calculated for pure water; hence the term x 2 should

be small compared to 0.1x.

Neglecting x2, 0.1x = 3.4 x 10-5 i.e., x = 3.4 x 10-4m

[We can and should check the validity of our assumption that x2 is small by back-substitution.]

The decrease in solubility (compared to pure water) is a result we could have guessed qualitatively by noting that the

excess Ca 2+ would speed up the reverse reaction (formation of CaSO4).

The decrease in the solubility of a salt due to the presence of one of its own ions in solution is called the common-

ion effect. The presence of ions different from those furnished by the salt itself generally increases the solubility, but

this is not something we could predict from simple equilibrium reasoning.


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(e) Effect of the solubility of one salt on the solubility of another in the same solution

As the last example, let us calculate the value K eq for the reaction at 25oC and 1 bar

Ba2+ + CaSO4 Ca2+ + BaSO4 (3.15)

Keq = mCa2+/ mBa2+

Keq for the above reaction can be calculated from the solubility products of the two component reactions:

CaSO4(s) Ca2+ (aq) + SO42-(aq) Ksp(CaSO4) = mCa2+ mSO42- = 3.4 x 10

-5 = 10-4.5

BaSO4(s) Ba2+ (aq) + SO42-(aq) Ksp(BaSO4) = mBa2+ mSO4

2- =1.0 x 10-10= 10-10

Dividing to eliminate mSO42-, Keq = mCa2+ / mBa2+=10

5.5 = 3.4 x 105

Evidently, the concentration of Ca2+ is 340,000 times more than that of Ba2+ in the solution. That is, CaSO4 is much

more soluble than BaSO4 when the solution is saturated in both.

It means that when a saturated solution of barite comes into contact with anhydrite, barite will be precipitated, as it is

much less soluble than anhydrite in the mixture, and anhydrite will be dissolved in the solution. This replacement

anhydrite by barite will go on till the mCa2+ / mBa2+ ratio in the solution approaches the value of Keq, at which point

equilibrium will be restored.

Note: In such situations, which are common in geology, the more soluble mineral is replaced by the less soluble one.

How can we calculate solubility of Ca2+ (= mCa2+) and solubility of Ba2+ (= mBa2

+)?

The obvious technique is to first calculate mSO42-

We can determine the SO4-2 concentration in the solution by some algebraic manipulation.

To satisfy the requirement of electrical neutrality: mCa2+ + mBa2+ = (mSO42-)

K sp(CaSO4) Ksp (BaSO4)

mSO42- = ------------------- + ---------------------

mSO42- mSO4

2-

(mSO42-)2 = K sp(CaSO4) + Ksp (BaSO4) = 10

-4.5 + 10-10

Since 10-4.5 >> 10-10, barite obviously contributes little to the concentration of SO42-(aq) in the solution.

Therefore, we can neglect the term 10-10, giving

(mSO42-)2 = 10-4.5

log(mSO42-)2= log 10

-4.5

2 log mSO42- = -(4.5)

log mSO42- = -(4.5) /2 = -2.25

mSO42- = 10-2.25 mole/kg water

We can now calculate mCa2+ and mBa2+ using equations forK sp (CaSO4) and Ksp (BaSO4), respectively.


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In general, solubility increases with increase in temperature (although at different rates for different salts), but this

cannot be predicted from the elementary rules presented here.

LE CHATELIER'S PRINCIPLE

A system in equilibrium responds to any disturbance imposed on the system by trying to counteract the effects of this

disturbance.

Le Chatelier's Principle does not help us with any quantitative prediction, but is useful for predicting qualitatively

which way an equilibrium will shift under different circumstances.

How does a system respond to a change in temperature?

Through some process that will absorb heat (endothermic reaction) or give off heat (exothermic reaction)

How does a system respond to a change in total pressure?

Through some process that would produce denser minerals (increase in pressure) or lighter minerals (decrease in

pressure).

THE CONCEPT OF STABILITY

The stability of a substance always implies a given set of conditions: P, T, associated substances.

A substance may actually be unstable in terms of affinity for reaction (A r), but still continue to exist because the rate

of the relevant reaction (Vr) is extremely sluggish -- this is metastability.

Stable: Vr = 0; Ar = 0 (Stable w.r.t. to equilibrium)* (3.16)

Metastable: Vr = 0; Ar +ve

Unstable Vr +ve; Ar +ve

* Unless mentioned otherwise, we will use stability in the sense of equilibrium.

TIPS FOR PROBLEM SOLVING

(1) Make sure that any reaction you are working with is balanced.

(2) All the components in each reaction must be expressed in equivalent concentration units.

(3) Where appropriate, make reasonable assumptions for approximate answers (often adequate for your purpose), but

check the reasonableness of your assumptions.

(4) Do your work neatly without omitting steps in the long run, it pays to be patient and neat.

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PROBLEM SET 1 Chemical Equilibrium

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RECAPITULATION

Terms [should be able to define or explain these, with sketches as necessary]

Molality Molarity Mole Fraction

Partial Pressure Daltons Law of Partial Pressure

Activity Fugacity

Equilibrium (chemical)

Law of Mass Action

Equilibrium Constant

Solubility Solubility Product

Saturation Index

Concepts / Applications [should be able to explain these, providing justification as necessary]

The Solubility Product of a salt can be used to determine if a solution with known dissolved concentration of the salt

is undersaturated, saturated, or oversaturated with respect to the salt.

Common-ion effect Solubility of a salt decreases in the presence of one of its own ions in the solution.

Le Chateliers Principle.

Stability vs. Metastability

Computational Techniques

Reasonable approximations for quick approximate solutions of problems.

Calculation of mH2CO3 of water in equilibrium with the atmosphere.

Calculation of solubility under specified conditions.

04_ChemicalEquilibrium

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