Equilibrium Law
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Transcript of Equilibrium Law
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Equilibrium Law
1) Equilibrium Constant, Keq:
If, aA + bB cC + dD
then beq[B] a
eq[A]
deq[D] c
eq[C] eqK unitless
temperature dependent!
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eg. For A2 + B2 2AB
then]2[B ]2[A
2[AB] eqK
Note: all “eq”
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2) Significance of Keq magnitude:
• If Keq is very large then ][reactants
[products]
is large and the reaction nears completion.
If Keq is very small then there is no reaction.
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3) Keq and the Balanced Chemical Eq’n any action performed on the chemical rxn,
the Keq expression is raised to that action.
eg. H2 + I2 2HI ]
2[I ]
2[H
2[HI] eqK
Flip the rxn or x by -1:eg. 2HI H2 + I2
2[HI]
]2
][I2
[H '
eqK
eqK1
1- eqK '
eqK
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]2
[I ]2
[H
2[HI] eqK
by 2 or x 1/2:
eg. 1/2 H2 + 1/2 I2 HI
1/2]2
[I 2/1]2
[H
[HI] 'eqK eqK
1/2eqK '
eqK
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4) Keq and Reaction Kinetics:
If A + B C + D
and both the forward and reverse reactions are elementary steps, then
Ratef = kf [A]1[B]1 and Rater = kr [C]1[D]1
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At equilibrium:
kf [A]1[B]1 = kr [C]1[D]1
Ratef = Rater
and
then
eqK [B] [A][D] [C]
rkf
k
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5) Keq and the Effect of Temperature:
usually Eaf ≠ Ear.
an in T won’t affect the forward and reverse reactions equally, so,
Ratef ≠ Rater
and the equilibrium will change
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][reactants
[products] '
eqK
If exothermic:At higher temperature, Rater creates more reactants than at lower T, and
If endothermic then the opposite occurs:
][reactants
[products] '
eqK
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6) Heterogeneous Equilibria:
reactants and products in different physical states (s, l ,g, aq)
Pure liquids (not aq) and solids have constant densities
as a result, pure solids and liquids are not written in the Keq expression.
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Vm D
M x Vn D
constant a is C
thenVn C and
Mn x m
M x C D MD Cor
VMn x D
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eg. 2H2O(l) 2H2 (g) + O2 (g)
then ]2[O2]2[H eqK
eg. Zn(s) + Cu2+(aq) Cu (s) + Zn2+
(aq)
]2[Cu
]2[Zn eqK
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Equilibrium and Spontaneity
∆Gº = -RT ln Keq R = 8.314 J/mol•K
Calculate ∆Gº from Keq
eg. PCl3 + Cl2 PCl5 ; Keq = 0.18
∆Gº = -RT ln Keq
= (-8.314 J/ mol•K x 298K) x ln (0.18)= 4.2 kJ/mol
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Calculate Keq from ∆Gº
∆Gº = - RT ln Keq
ln Keq = - (∆Gº) / (RT)
Keq = e-(∆Gº)/(RT) * Note : ∆Gº in J/mol
eq. 2CO(g) + O2(g) 2CO2(g)
and ∆Gº = -514.5 kJKeq =
e-(∆Gº)/(RT)= e-(-514.5 x 103 J)/(8.314 J/mol•K x 298 K)
= 1.54 x 1090
the products are highly favoured!