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Dynamic Equilibria-Two opposing prosesses that
occur simultaneous at equal rates |
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aA
+ bB = cC |
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Rf = k1 [A]a
[B]b |
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Rr = k-1[C]c |
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At equilibrium: |
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Rf = Rr |
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k1[A]a [B]b = k-1[C]c |
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K1 /k-1 = [C]c /
[A]a [B]b |
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Kc =K-1 /k1 = [C]c
/ [A]a [B]b |
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aA(g)
+ bB(g) =
pP(g) + qQ(g) |
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Based on Molar Concentration |
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Kc = [P]p [Q]q /
[A]a [B]b |
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Based on Partial Pressures |
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Kp = Pp Pq /
Pa Pb |
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Kc and Kp remains constant
at a fixed temperature |
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3H2(g) + N2(g) =
2NH3(g) |
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Kc = [NH3]2 / [N2]
[H2]3 |
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Kp = (PNH3)2 /
(PH2)3 PN2 |
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K increases then components on the right side
predominate in the equilibrium mixture |
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K decreases then the components on the left side
predominate in the equilibrium mixture |
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Example |
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aA(g) = bB(g)
K1 = [B]b / [A]a |
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bB(g) = aA(g)
K2 = [A]a / [B]b |
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Therefore: K1 = 1 / K2 |
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Heterogeneous Equilibrium –equilibria with two
or more physical states represented. |
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Liquid and solid states do not affect equilibria
because of incompressibility |
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Examples |
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CaCO3(s) = CaO(s) +
CO2(g) |
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Kc = [CO2] |
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Kp = PCO2 |
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FeO(s)
+ H2(g) =
Fe(s) + H2O(g) |
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Kc = [H2O] / [H2] |
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Kp = PH2O / PH2 |
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Given the equilibrium concentrations(or Partial
Pressures) of all components |
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Given the initial concentration (or partial
pressure) of components |
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Kp = Kc (RT) Dn |
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where Dn = mols of gaseous component on right – mols
of gaseous components on left |
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Example: |
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For H2(g) + I2(g) =
2HI(g) |
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Q = quotient = [HI]2 / [H2]
[I2] |
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If Kc = Q Rx in Equilibrium |
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If Kc < Q then Rx must proceed Left
(favor reverse reaction) to reach equilibrium (Kc = Q) |
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If Kc > Q then Rx must proceed to Right
(favor forward reaction) to reach equilibrium |
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Example |
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Given the equilibrium concentrations(or partial
pressures) of all components but one and the Kc(or Kp) |
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Given the initial concentration (or partial
pressures) of the reactants and the Kc(or Kp) |
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When an external stress is applied to a system in equilibrium,
the equilibrium is upset and will respond by shifting in such a way as to
undo the stress applied. |
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Change in the concentration of the components
(adding or removing) |
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Change in the volume of the container (alters
pressures of reacting gases) |
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Change in the temperature |
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Increasing the concentration of a component
(adding component) of a component drives the reaction to the opposite side
(uses up the component and relieves the stress) |
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Decreasing the Concentration of a
component(removing component) drives the reaction to the same side (produces
more component thereby relieving the stress by replacing what was removed) |
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Examples |
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Increasing the volume of a container decreases
the pressures of the gaseous components(Boyles Law) which drives the
reaction toward the side with the most total mols of gas(brings pressure
back up) |
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Decreasing the volume of the container increases
the pressure of the gaseous components (Boyles Law) which drives the
reaction toward the side with the least total mols of gas (brings pressure
back down) |
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Examples |
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Increasing the temperature will always favor the
endothermic process (uses up energy added by increased temperature) |
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Decreasing the temperature will always favor the
exothermic process(replaces the energy removed by stress) |
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Examples |
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Catalyst alters the Energy of Activation of both
processes |
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Therefore equilibrium is not changed |
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