HOW ARE EQUILIBRIUM EXPRESSIONS WRITTEN?
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HOW ARE HOW ARE EQUILIBRIUMEQUILIBRIUM
EXPRESSIONS WRITTEN?EXPRESSIONS WRITTEN?
Q VS. K
“Systems”: two reactions that differ only in direction
• Any reversible reaction
H2
+ I2
↔ 2HI
noted by the double arrow; ↔
Reversible Reactions
H2 + I2 ↔ 2HI • the products may react back to original
reactants.
• “closed system”: ONLY if all reactant are present
• If one piece is completely gone it has ”gone to competition” and no longer reversible
Equilibrium
•The state in which a chemical reaction and its reverse reaction occur at the same rate.
Equilibrium = No change in amount over time
6Properties of an Properties of an Equilibrium Equilibrium
Equilibrium systems are
• DYNAMIC (in constant motion)
• REVERSIBLE
• can be approached from either direction
Equilibrium systems are
• DYNAMIC (in constant motion)
• REVERSIBLE
• can be approached from either direction
E + Co(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2 H22OO
Le Chatelier’s Principle
•If a system at equilibrium is stressed, it will react to undo the
stress.
Can we explain WHY?
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The Reacton The Reacton Quotient, QQuotient, Q
In general, all reacting chemical systems are
characterized by their REACTION QUOTIENT
at equilibrium, Q = K
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Reaction QuotientReaction QuotientAt any point in the reaction
H2 + I2 --->---> 2 HI
At any point in the reaction
H2 + I2 --->---> 2 HI
Q reaction quotient = [HI]2
[H2 ][I2 ]
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Equilibrium ConstantEquilibrium Constant
[HI]2
[H2 ][I2 ] = 55.3 = K
K = equilibrium constant
Equilibrium achieved
In the equilibrium region
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THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTFor any type of chemical equilibrium of the
type
a A + b B ---> c C + d Dthe following is a CONSTANT (at a given T)
K =[C]c [D]d
[A]a [B]b
conc. of products
conc. of reactantsequilibrium constant
If K is known, then we can predict concs. of products or
reactants..
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Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Solids NEVER appear in equilibrium expressions.
S(s) + O2(g) ---> SO2(g)
K [SO2 ][O2 ]
K [SO2 ][O2 ]
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Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Writing and Manipulating Writing and Manipulating K ExpressionsK Expressions
Liquids NEVER appear in equilibrium expressions.
NH3(aq) + H2O(liq) ---> NH4+(aq) + OH-(aq)
K [NH4
+][OH- ][NH3 ]
K [NH4
+][OH- ][NH3 ]
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Writing Equilibrium Expressions
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Product- or Reactant Product- or Reactant FavoredFavored
Product-favored Reactant-favored
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For: N2(g) + 3 H2(g) ---> 2 NH3(g)Using K: Is the reaction product-favored
or reactant-favored?
Kc = [NH3 ]2
[N2 ][H2]3 = 3.5 x 108Kc =
[NH3 ]2
[N2 ][H2]3 = 3.5 x 108
When K is much greater than 1the reaction is strongly
product-favored.
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For
AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 1.8 x 10-5
If K is If K is much lessmuch less than 1 than 1
The reaction is strongly The reaction is strongly
reactant-reactant-favoredfavored..
AgAg++(aq) + Cl(aq) + Cl--(aq)(aq) AgCl(s) AgCl(s)is product-favored.is product-favored.
AgAg++(aq) + Cl(aq) + Cl--(aq)(aq) AgCl(s) AgCl(s)is product-favored.is product-favored.
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calculating Equilibrium constants:
product or reactant favored?
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Using K: Using K: Can determine if the reaction is
at equilibrium.
H
H
H
H
H
H
H
H
H H H
CH
HHH H
H—C—C—C—C—H H—C—C—C—H
K =
n-butane iso-butane
[iso]
[n] = 2.5
H
H
H
H
H
H
H
H
H H H
CH
HHH H
H—C—C—C—C—H H—C—C—C—H
K =
n-butane iso-butane
[iso]
[n] = 2.5
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If [iso] = 0.35 M and [n] = 0.15 M,
are you at equilibrium?
H
H
H
H
H
H
H
H
H H H
CH
HHH H
H—C—C—C—C—H H—C—C—C—H
K =
n-butane iso-butane
[iso]
[n] = 2.5
If not, which way does the reaction “shift” to approach equilibrium?
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REACTION QUOTIENT, Q
Characterize all chemical systems
Q = product concentrationsreactant concentrations
Q = product concentrationsreactant concentrations
Q = conc. of isoconc. of n
= 0.350.15
= 2.3Q = conc. of isoconc. of n
= 0.350.15
= 2.3
Q (2.33) < K (2.5) Reaction is NOT at equilibrium, [iso] must ________ and [n] must ____________._.
If Q = K, then system is at equilibrium..
25Experimental Experimental Determination of Determination of
Equilibrium Constant, K Equilibrium Constant, K
Experimental Experimental Determination of Determination of
Equilibrium Constant, K Equilibrium Constant, K 2 NOCl(g) 2 NOCl(g) --->---> 2 NO(g) + Cl2 NO(g) + Cl22(g)(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO.
Calculate K.
Set of an “ICE” table of concentrations
[NOCl] [NO] [Cl2]
Initial 2.00 0 0
Change
Equilibrium 0.66
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Determining KDetermining KDetermining KDetermining K2 NOCl(g) ---> 2 NO(g) + Cl2(g)
[NOCl] [NO] [Cl2]
Initial 2.00 0 0
Change -0.66 +0.66 +0.33
Equilibrium 1.34 0.66 0.33
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2 NOCl(g) ---> 2 NO(g) + Cl2(g)
[NOCl] [NO] [Cl2]
Initial 2.00 0 0
Change -0.66 +0.66 +0.33
Equilibrium 1.34 0.66 0.33
K [NO]2[Cl2 ]
[NOCl]2 =
(0.66)2(0.33)
(1.34)2 = 0.080K
[NO]2[Cl2 ]
[NOCl]2 =
(0.66)2(0.33)
(1.34)2 = 0.080