General Chemistry II 2302102

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General Chemistry II2302102
Chemical Equilibrium for Gases and for
Sparingly-Soluble Ionic Solids
Part 1
i.fraser_at_rmit.edu.au Ian.Fraser_at_sci.monash.edu.a
u
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Chemical Equilibrium - 2 Lectures
Outline - 4 Subtopics

• Equilibrium and Le Chateliers Principle
• The Reaction Quotient and Equilibrium Constant
• Temperature and Pressure Effects
• Sparingly-Soluble Ionic Compounds in Aqueous
  Solution

3
Chemical Equilibrium
Objectives - Lecture 1

• By the end of this lecture AND completion of the
  set problems, you should be able to
• Understand the concepts of the chemical
  equilibrium condition, dynamic equilibrium as the
  balance of forward and reverse reaction rates.
• Know the definition of Le Chateliers Principle,
  and understand its application to the prediction
  of the direction of change in a chemical reaction
  at equilibrium, following changes in pressure,
  volume, temperature and amount of reactants and
  products.
• Understand the definition of the reaction
  quotient Q and the equilibrium constant K (or Kp
  Kc) for a chemical reaction.
• Calculate the final equilibrium conditions for
  gas-phase reactions, and for heterogeneous
  reactions involving gases, from given
  non-equilibrium initial conditions.

4
Equilibrium in Chemical Processes

• Equilibrium constants
• Take the simple process (at a given temperature)
• H2O (l) H2O (g)
• An equilibrium partial pressure of H2O known as
  the vapour pressure po(H2O) is reached at this
  point the rates of evaporation and condensation
  are equal.
• The equilibrium condition is very simply stated
  as
• pH2O po(H2O) Kp
• where Kp is a constant whose value characterises
  the equilibrium.

5

• A more complex equilibrium process is the
  chemical reaction
• 2NO2 (g) N2O4 (g)
• Here, colour changes
• (NO2 is brown, N2O4 is colorless)
• and total pressure changes
• (P pN2O4 pNO2 n RT/V )
• (the number of molecules changes)
• P can be monitored to measure progress and find
  the position of equilibrium.

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• For equilibrium in the reaction
• 2 NO2 (g) N2O4 (g) heat
• At any given temperature, one finds that
• pN2O4
• ------- has the same equilibrium value, Kp,
  whatever the pressures
• pNO22
• And, as the temperature rises, the value of Kp
  falls
• (this finding is what tells us that the
  reaction is exothermic).

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Le Chateliers Principle

• The Tendency Towards Equilibrium
• Any system which is not at equilibrium will tend
  to change spontaneously toward a state of
  equilibrium (i.e. without a need for us to adjust
  external variables such as temperature and
  pressure).
• It follows that if a chemical system at
  equilibrium is disturbed or stressed away from
  equilibrium then the system will tend to react so
  as to remove the stress and return to
  equilibrium.
• This response is known as Le Chateliers
  Principle
• (The approach to equilibrium may be fast,
  gradual or even undetectable).

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Ideal Gas Reaction Equilibrium Law
Law of Mass Action

• aA bB cC dD (all gaseous)
• pCcpDd
• at equilibrium constant, Kp
• pAapBb
• This ideal equilibrium relationship between
• partial pressures of products and reactants
• in any gas reaction
• is closely followed at normal temperatures
• and total gas pressures up to about 10
  atmospheres .

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Ideal Gas Reaction - Concentrations Kc
Law of Mass Action

• Remember pBV nBRT gt pB (nB/V) RT BRT
• aA bB cC dD (all gaseous)
• Cc Dd
• at equilibrium constant, Kc
• Aa Bb
• This ideal equilibrium relationship between
• concentrations of products and reactants
• is closely followed in any gas reaction at
  normal
• temperatures and total gas pressures up to about
  
• 10 atmospheres
• deviations at higher pressures arise from
• intermolecular forces

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Relationship between Kc and Kp
for gas phase reactions
Kp Kc(RT)??
?? ?gaseous products ?gaseous reactants
( c d ) - (a b .. ) gases only!
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Pressures, Concentrations Activities

• For any gas in a reaction, the numerical value of
  partial pressure in atmospheres is a good
  approximation to its activity (or tendency to
  react, taken by convention, relative to ideal gas
  behaviour at 1 atm)
• Any gas agas , pgas / atm
• For any solute in a reaction, the numerical value
  of concentration in moles per litre is a good
  approximation to its activity (or tendency to
  react, taken by convention, relative to ideal
  solute behaviour at 1 mol/L)
• Any solute, asolute , solute /
  M

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• For any pure solid or pure liquid in a reaction,
  its activity (or tendency to react) is taken
  (by convention) to be 1.000 areas affect rates
  but not activities
• Any pure solid or liquid apure solidor liquid
  1.000
• For any solid or liquid solvent (in dilute
  solution) the activity is taken as 1.00 (by
  convention, relative to pure solvent behaviour,
  assuming solvent mole fraction gt 98.5,
  otherwise, Raoults law gt ideal asolvent
  xsolvent/1.000
• Any solvent (dilute solution) asolvent ,
  1.00
• The above ideal approximations are easy to
  use.
• More accurate values can be obtained from
  measurements
• of vapour pressures, osmotic pressures, m.pts,
  b.pts, etc.

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Ideal Solution Reaction Equilibrium Law

• Many reactions occur between solutes in solutions
• aA bB cC dD (all in a
  liquid solvent)
• here we use concentrations/(mole per
  litre) to represent activities
• ( partial pressures are NOT relevant within
  solutions) - Thus
• Reaction Quotient Q
• (C/M)c (D/M)d
• --gt constant, K, at equilm
• (A/M)a (B/M)b
• This ideal relationship between the quotient of
• numerical values of concentrations/ (mol/L) at
  equilibrium,
• and the equilibrium constant, K,
• is closely followed in dilute solutions .

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10.3 The Equilibrium Constant

• Suppose we mix 0.6 atm of CO and 1.1 atm of Cl2.
  Say we can measure the equilibrium partial
  pressure of COCl2 and find that it is 0.1 atm.
  The temperature is 600oC.
• CO(g) Cl2(g) COCl2(g)
• 0.6-0.1 atm 1.1-0.1 atm 0.1 atm at
  equilm
• The equilibrium expression in terms of partial
  pressures is
• Q (pCOCl2/atm)/ (pCO/atm) x (pCl2 /atm) --gt K

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• Using the known stoichiometry we now insert the
  activities (as pressure/atm data), as follows
• (Take care if the stoichiometric coefficients
  are not all equal to one!!)
• For reasons to be covered in later chapters,
• K should be dimensionless. This is ensured by
  using activities (approximated here by dividing
  all partial pressures by the reference pressure
  of 1 atm, or the equivalent pressures 760 torr,
  101.3 kPa, etc, to obtain purely numerical
  values).
• A partial pressure of 1 atm is often referred to
  as the standard state for a gaseous substance.

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Favoured Reactions

• If K (for a gas reaction) is
• gtgt 1 then equilibrium tends to favour products
• ltlt 1 then equilibrium tends to favour reactants
• 1 then substantial amounts of both products
  and reactants will tend to be present at
  equilibrium
• this depends to some extent on the initial
  mixture that is used

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Relationships among Equilibrium Expressions

• (i) forward reaction K1 A ?
  B
• reversed equation K2 A ? B
• K2 1/K1
• (ii) Multiplying an equation by a factor n
• reaction (1) K1 A ? B
• reaction (x n) Kn K1n nA ? nB
• (iii) addition and subtraction of reactions
• reaction 1 K1 A ? B
• reaction 2 K2 C ? D
• 1 2 K12 K1K2 A C ?
  B D
• 1 - 2 K1-2 K1 / K2 A D ?
  B C

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Equilibrium Calculations for Gas-Phase Reactions

• Sometimes equilibrium problems lead to complex
  equations which can be solved via simple
  assumptions. For example
• CH4(g) H2O(g) CO(g) 3H2(g)
• for which K at 600 K is 1.8 x 10-7.
• Suppose our initial conditions are as follows
• Gas p/(atm)
• CH4 1.4
• H2O 2.3
• CO 1.6
• H2 0.0

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• Let's assume the partial pressure of CH4
  decreases by y atm to reach equilibrium. Again
  we write an equilibrium expression,
• At equilibrium the following pressures will
  apply
• Gas p/(atm)
• CH4 1.4-y
• H2O 2.3-y
• CO 1.6y
• H2 3y
• By substitution we obtain
• This looks like real trouble ( a quartic
  poynomial in y) !!
• DON'T TRY TO SOLVE THIS DIRECTLY.
• There is a much easier way.

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• We'll make the assumption that yltlt1.4 and check
  at the end that everything is OK.
• i.e.
• Which transposes to y3 1.34 x 10-8
• or y 2.38 x 10-3 (cube root)
• Clearly our assumption of yltlt1.4 is quite
  reasonable.
• At equilibrium, then, the partial pressure will
  be pH2/atm 3y 7.1 x 10-3 .

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• Example
• 2NO2 (g) N2O4 (g) heat
• A sample of atmospheric NO2 is concentrated by
  freezing as N2O4 and then vaporizing.
  Immediately upon vaporizing the vessel would
  contains 200 kPa of N2O4(g) before any reaction.
• What are the pressures of N2O4(g) and NO2(g) at
  equilibrium for T298 K ??
• Step 1 From tables of chemical data K298K
  11.3
• Step 2 Rough prediction.
• at time 0, we have only product (N2O4 and no
  reactant).
• so reaction must go backwards to reach
  equilibrium.
• Step 3 Write down equilibrium expression.

(pN2O4/atm)eq (pNO2/atm)eq2
K 11.3
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• Step 4 Work out expressions for the pressures.
• We start with 202 kPa of N2O4
• Say that a fraction ? reacts gt 1- ? unreacted
• pN2O4 (1- ?) 202 kPa (1- ?) x 2.00 atm.
• Next consider NO2
• pNO2 (2 ?) 202 kPa 2 ? x 2.00 at? 4.00
  ? atm
• Step 5 Substitute each p/atm in the
  equilibrium expression
• 2.00 (1 - ?) / (4.00 ?)2 11.3
• ?? 11.3 x (16.00 ? 2) 2.00 - 2.00 ?
• or 90.4 ? 2 1.00 ? - 1.00 0, a quadratic,
  in which
• ? 0.0998 (discarding the negative root)
• (i.e. about 10 of N2O4 has reacted)
• note a negative a is not physically sensible
  gt reject

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• So, at equilibrium
• pNO2(eq) 2 ? 202 kPa 39.9 kPa
  0.399 atm
• pN2O4(eq) (1- ?) 202 kPa 180.1 kPa
  1.801 atm
• ?? Ptotal 2.200 atm 222.9 kPa
  I.e. 223 kPa
• Notice that the total pressure has gone up during
  the reaction.
• Finally, it's a good idea to check our
  calculations by back-substituting each calculated
  value of pressure/atm (not /kPa) into the
  equilibrium constant expression
• All is well!!

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The Reaction Quotient

• The "reaction quotient, Q, contains
• activities of products / activities of
  reactants
• - activity of gas, a(gas) p(gas)/atm
• - activity of solute, a(solute) solute/M
• Pure liquid and solid substances need not appear
  in reaction quotients (their activities can be
  taken as 1.000 and normal pressures have
  negligible effect and areas ).
• Activity of a solvent (close to its mole
  fraction) is taken as 1.00.
• By definition, the value of Q at equilibrium is
  the "equilibrium constant", K.

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• Reactions always proceed in such a direction as
  to make Q  K. eg
• if Q lt K it follows that the reaction can only go
  forward.
• (eg. when we start with only reactants)
• if Q gt K it follows that the reaction can only go
  backward.
• (eg. When we start with only products)
• if Q K it follows that the reaction is at
  equilibrium
• Therefore by finding Q at any point in time
  compared with K we can predict the direction in
  which the reaction tends to proceed
• The value of Q compared with K tells us nothing
  about the rate of reaction, but the minimum work
  required to drive reaction is ?G RT ln Q/K
  (negative ?G gt spontaneous reaction).

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• Example
• The reaction
• SO2Cl2(g) SO2(g) Cl2(g)
• has an equilibrium constant K 2.4
• Suppose the initial values of partial
  pressure/atm are
• pSO2Cl2 /atm 1.0
• pSO2 /atm 0
• pCl2 /atm 0
• We would proceed by computing the reaction
  quotient
• Therefore since Q lt 2.4 the reaction tends to
  proceed spontaneously (unaided, without being
  driven) to the right.

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• Alternatively, if the conditions had been
• pSO2Cl2 0.01 atm
• pSO2 0.1 atm
• pCl2 0.4 atm
• Since Q gt 2.4 the reaction now tends to proceed
  spontaneously to the left (this reaction
  approaches equilibrium quite rapidly and without
  need of catalyst).

pSO2/atm x pCl2/atm 0.1 x 0.4 Q
----------------------- --------
4 pSO2Cl2/atm
0.01
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External Effects and Le Chatelier's Principle

• When a system at equilibrium is subject to
  change it will respond in such a way as to reduce
  the change.
• For example, in the reaction
• 2NO2 (g) N2O4 (g) initially at
  equilibrium
• Some possible changes we can explore are
• addition of NO2
• increase in total volume
• addition of an inert gas

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pN2O4/atm Q ----------------
(pNO2/atm)2

• First we need the reaction quotient
• If we increase pNO2 , Q will decrease to less
  than K and the reaction will adjust by forming
  more N2O4 (g).
• If we increase the volume of the system, then
  each partial pressure drops. Now Q increases,
  and the reaction must proceed in the reverse
  direction.
• What happens if we increase Ptotal (by adding an
  inert gas)?
• Nothing! The partial pressures do not change!!

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• Another example
• H2(g) I2(g) 2HI(g)
• Stage 1 reactants only gt Q 0 (Q lt K )
• ???? react. ?? right
• Equilibrium reached -gt Q K
• Stage 2 inject iodine -gt Q lt K ?? react. ??
  right
• System always adjusts to counteract the
  perturbation which takes it out of balance.

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Heterogeneous Equilibrium

• So far we have concentrated on homogeneous
  equilibria involving gases. We now consider
  heterogeneous reactions in which at least two
  different physical states of matter are present.
• Pure solids and liquids appear in Q as 1.000
• H2(g) I2(s) 2HI(g)
• Q (pHI/atm)2 / (pH2/atm) x 1.000
• C(s) H2O(l) Cl2(g) COCl2(g)
  H2(g)
• Q (pCOCl2/atm) x (pH2/atm) / 1.000 x 1.00 x
  (pCl2/atm)

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• For a liquid solvent in a reaction equation
• its mole fraction should be used in Q,
• this can normally be taken as 1.00
• I2(s) I2(aq)
• Q I2(aq) / 1.00 ---gt
  solubility/(mol/L) at equilm
• CaCO3(s) CaO(s) CO2(g)
• Q 1.000 x pCO2/atm / 1.000 ---gt equilm
  pressure/atm
• Q lt K ? K ? gt K ?

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The Lime Kiln

• CaCO3(s) CaO(s) CO2(g)

An industrially important example of
heterogeneous equilibria involves the production
of calcium oxide (CaO - builders lime,
quicklime) from limestone (CaCO3 - garden lime,
agricultural lime).
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The Lime Kiln

• CaCO3(s) CaO(s) CO2(g)

At any given temperature the equilibrium pressure
of CO2 (/atm) is equal to the equilibrium
constant K and is independent of the relative
amounts of CaO and CaCO3 present. K increases
with T (gt endothermic reaction) and exceeds 1 at
above 1000oC (--gt equilm pCO2 gt 1 atm). It is
then simply necessary to allow the CO2 to escape
to obtain more CaO from the CaCO3.
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• In all equilibria involving pure solids with
  gases, the partial pressures of gases are
  independent of the amounts of solid present.
  (n.b. as long as we have some solid present)
• Similar behaviour is found for solid-liquid
  equilibria. For example, a sparingly soluble
  salt like PbSO4 dissolves in water to form ions
• PbSO4(s) Pb2(aq) SO42-(aq)
• Q (Pb2/M) x (SO42-/M) / 1.000 ---gt K
• (at equilm)
• The concentrations of Pb2(aq) and SO42-(aq) are
  independent of the amount of PbSO4(s) that is
  present.

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• Some more general rules for the Law of Mass
  Action follow from these observations.
• 1. Each gas enters a reaction quotient as a
  numerical activity (partial pressure in
  atmospheres, p(gas)/atm)
• 2. Each dissolved species enters as a numerical
  activity (concentration in moles per litre,
  solute/M)
• 3. Activity of a pure solid or pure liquid need
  not appear in a reaction quotient (except as a
  1.000), neither need a solvent taking part in the
  reaction, provided the solution is dilute (take
  a(solvent) as 1.00).
• 4. The reaction quotient Q contains the
  numerical activity of each product in the
  numerator and of each reactant in the
  denominator (each raised to the power of its
  coefficient in the balanced chemical equation).

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• Example 1
• Consider the equilibrium
• H2(g) I2(s) 2HI(g)
• Given K 0.345 at 25C
• If pH2 1.00 atm and I2(s) is present, determine
  pHI
• Solution
• For the reaction given we have Q
• so that pHI/atm 0.5874 x 1.00
• thus pHI 0.587 atm

2
p
HI

K

p
x 1.000
H

2
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• Example 2
• Find the equilibrium expression for the
  extraction of gold from ore with cyanide via the
  reaction
• 4 Au(s) 8 CN(aq) O2(g) 2H2O(l)
• 4 Au(CN)2(aq) 4 OH(aq)

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• 4 Au(s) 8 CN(aq) O2(g) 2 H2O(l)
  
• 4 Au(CN)2(aq) 4 OH(aq)
• Answer
• K

1.0004
1.002
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Chemical Equilibrium - End of Lecture 1
Objectives Covered in Lecture 1

• After studying this lecture you should be able
  to
• Understand the concepts of the chemical
  equilibrium condition, dynamic equilibrium as the
  balance of forward and reverse reaction rates.
• Know the definition of Le Chateliers Principle,
  and understand its application to the prediction
  of the direction of change in a chemical reaction
  at equilibrium, following changes in pressure,
  volume, temperature and amount of reactants and
  products.
• Understand the definition of the reaction
  quotient Q and the equilibrium constant K (or Kp
  and Kc) for a chemical reaction.
• Calculate the final equilibrium conditions for
  gas-phase reactions, and for heterogeneous
  reactions involving gases, from given
  non-equilibrium initial conditions.