Chem 1310: Introduction to physical chemistry Part 3: Equilibria

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Chem 1310 Introduction to physical chemistry
Part 3 Equilibria

• Peter H.M. Budzelaar

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Kinetics and Equilibria

• Kineticshow fast does a reaction go (initially)
• Equilibriumwhat will be the final
  compositionof the reaction mixture?

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What is "equilibrium" ?

• A system is at (dynamic) equilibrium when its
  composition doesn't change any more over time.
• There are equal numbers of molecules going to the
  left as to the right (and this number is not
  zero!).
• A mixture with a composition that does not appear
  to change over time is not necessarily at
  equilibrium. The reaction could just be very slow!

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How do you know that equilibrium has been
established?

• The reaction started, but then did not proceed
  any further.
• This is not an infallible criterion!
• The equilibrium was approached from both the
  product and the reactant side (in two separate
  experiments), and both resulted in the same
  composition.
• Molecules have no memory, so the equilibrium
  composition does not depend on how you got there.

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Approaching equilibrium from different sides
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How do you know that equilibrium has been
established?

• Heating accelerates approach to equilibrium, but
  also changes the equilibrium composition.
• On cooling, the reaction might slow down so much
  that the new equilibrium will not be established.
• To make sure you are in a dynamic equilibrium,
  you could also add one pure component and see how
  the system responds. If nothing happens, you are
  probably not at equilibrium.

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The equilibrium constant

• For a reaction A ? B at equilibrium
• For a reaction A B ? C 2 D at equilibrium
• KC is constant for a given temperature,
  independent of pressure, volume, etc.

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Units and equilibrium constants

• MSJ says (p678) that we should simply omit units
  from the equilibrium constant, because they are
  always mol/L.
• That is not the real reason. We do it because the
  equilibrium constant is actually something like
• where A etc are the concentrations at an
  agreed reference state (which is 1 mol/L for
  solutes). Because of this division, units cancel
  and KC is truly dimensionless. Because the
  reference state is 1 mol/L, we can ignore it when
  we use mol/L as units.

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Kinetics vs equilibria

• Rate laws can have a complicated dependence on
  concentrations, unrelated to the reaction
  stoichiometry. They might depend e.g. on
  concentrations of catalysts and poisons.
• Equilibrium constant expressions can be deduced
  directly from the reaction equation and do not
  depend on the reaction mechanism, nor on
  catalysts etc.

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Writing an equilibrium constant

• KC equilibrium constant expressed in
  concentrations. Preferably in a single phase,
  usually gas or solution.
• NH4 CH3COO- ? NH3 CH3COOH (in water)
• Adding NaOH or HCl may change the values of some
  of the concentrations involved, but KC will
  indeed remain constant.

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Writing an equilibrium constant

• NH3 ¾ O2 ? ½ N2 1½ H2O
• 4 NH3 3 O2 ? 2 N2 6 H2O
• KC(b) (KC(a))4
• An equilibrium constant belongs toa chemical
  equation as written.

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Reactions involvingsolids and liquids

• Pure solids and liquids are not included in KC.
• Nearly-pure solvents are also left out.
• If you have a vapour or solute in equilibrium
  with the pure solid or liquid, you also leave out
  that vapour or solute.
• So if you have ? S8(s) O2 ? SO2, it doesn't
  matter whether you also have some S8(g) present,
  or whether that is involved in the reaction. You
  don't need to include S8(g) as long as S8(s)
  remains present.

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Why drop pure solids and liquids?

• This is not simply because their concentrations
  are constant (MSJ 677). The correct KC would
  contain not S8(s) but S8(s)/S8(s), where
  S8(s) is the concentration in the reference
  state, which is the pure solid/liquid by
  convention.
• Since solids and liquids are not compressible,
  S8(s)/S8(s) will always be very close to 1,
  and there is no reason to include it. We are not
  just dropping a constant, we are dropping a
  factor 1.

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Reactions involvingsolids and liquids

• CH3COOH H2O ? CH3COO- H3O
• Hg(l) Cl2 ? HgCl2(s)
• Cl2(eq) 1/KC
• Hg(l) ? S8(s) ? HgS(s)
• No equilibrium constant! If product-favoured,
  reaction will proceed until one or both reactants
  consumed.

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Using equilibrium constants

• Calculate KC, given equilibrium concentrations.
• Just plug the concentrations in the formula.
• Calculate whether a reaction will go forward or
  backward, given KC and initial concentrations.
• Calculate QC (same formula as KC, but now for
  non-equilibrium concentrations).
• If QC lt KC reaction will go forward
• If QC gt KC reaction will go backward
• If QC KC reaction is at equilibrium

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Using equilibrium constants

• Calculate concentrations at equilibrium, given KC
  and initial conditions.
• This may involve serious calculations use the "x
  method" of the book (MSJ p682-684), e.g.

A B C
initial 1 1 0
change -x -x x
equilibrium 1-x 1-x x
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Dissociation of HI

• H2 I2 ? 2 HI
• We begin with 1 mol/L of HI, and let this reach
  equilibrium with H2 and I2 at 25C (KC 25, see
  MSJ p685). What will be the final concentrations?

H2 I2 HI
initial 0 0 1
change ½ x ½ x -x
equilibrium ½ x ½ x 1-x
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Dissociation of HI (2)

• Plug into KC equation
• Final concentrations HI 0.71, H2 I2
  0.14.
• Put into KC expression for final check.

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Dissociation of N2O4

• N2O4 ? 2 NO2
• We begin with 1 mol/L of N2O4, and let this reach
  equilibrium with NO2 at 500K (KC 46, see MSJ
  p702). What will be the final concentrations?

N2O4 NO2
initial 1 0
change -x 2 x
equilibrium 1-x 2 x
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Dissociation of N2O4 (2)
Quadratic equations

• Plug into KC equation
• Final concentrations N2O4 0.07, NO2
  1.85.
• Put into KC expression for final check.

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Concentration or pressure dependence

• Some equilibria are concentration- or
  pressure-dependent if all concentrations/pressure
  s change by the same factor (by compressing a gas
  mixture, or diluting a solution), they are no
  longer at equilibrium.
• This happens if the sums of exponents in
  numerator and in denominator of KC differ. For
  the case of N2O4 dissociation numerator 2,
  denominator 1. The HI dissociation equilibrium
  is not pressure dependent (both exponents 2).

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Dissociation of N2O4 (3)

• What will happen if we compress the previous
  equilibrium mixture (N2O4 0.07, NO2 1.85)
  to 1/10th of its original volume?
• Immediately after compression N2O4
  0.7,NO2 18.5, new QC 489 (should have been
  460, but there are rounding errors). So QC gt KC,
  the equilibrium will shift towards the reactant
  N2O4.

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Dissociation of N2O4 (4)

• Change table

N2O4 NO2
initial 0.7 18.5
change x - 2 x
equilibrium 0.7x 18.5-2 x
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Dissociation of N2O4 (5)

• Plug into KC equation
• (the other solution, x 27.1, would make NO2
  negative and is unphysical)
• Final concentrations N2O4 3.56, NO2
  12.8.
• Put into KC expression for final check.

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Le Chatelier's principle

• The system reacts to counteract an imposed
  change
• A B ? X Y
• Add A (reactant) drive reaction towards X, Y.
• Add X (product) drive reaction towards A, B.
• Add something non-reacting (within same
  volume)no change in concentrations, QC stays
  equal to KC, no change.
• Compress all concentrations change, but QC does
  not (no change in particles) no change.

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Le Chatelier's principle (2)

• A ? X Y
• Add A (reactant) drive reaction towards X, Y.
• Add X (product) drive reaction towards A.
• Add something non-reacting (within same
  volume)no change in concentrations, QC stays
  equal to KC, no change.
• Compress QC increases, so reaction is driven
  towards A (fewer particles), reducing pressure.

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Le Chatelier's principle (3)

• Effect of temperature
• heating drives an exothermic reaction (DH lt 0)
  towards reactants,an endothermic reaction (DH gt
  0) towards products.
• (but not always to completion the figure on
  p687 is incorrect!)
• cooling has the opposite effect.

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KP and KC

• For gases, instead of KC one usually uses KP
• where the pressures p are in bar (atm).
• For any component X, with the ideal-gas lawpV
  nRT, one has
• For the above example

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KP and KC

• Whenever the system is pressure-dependent(sums
  of exponents in numerator and denominator differ)
  you will also find RT terms in the conversion
  between KP and KC.
• If gases are involved, always make clear whether
  you are using KC/QC or KP/QP !!!!!

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Ammonia synthesis