UNIT IX – EQUILIBRIUM

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I. Equilibrium – a dynamic condition in which a forward reaction proceeds at the same rate as the reverse reaction; no changes will be observed II. Chemical Equilibrium * physical equilibrium – equilibrium in physical processes (example: vapor-liquid equilibrium) * chemical equilibrium – chemical changes

H2(g) + I2(g)

2 HI(g)

K eq

2 HI ] [ = [ H2 ][ I 2 ]

Understanding Equilibrium The concept of equilibrium is based on the understanding that a reaction will always proceed in a direction that gets it to the “desired” equilibrium state, regardless of the starting amounts of reactant or product. Assuming the conditions don’t change, the equilibrium state (represented by the numerical value of Keq) will be the same for the same reaction.

2A+B

A2B

2 A2 B ] [ K eq = [A]2 [B]

* in this example, Keq = 1

A

B

A2B final conditions

starting conditions

( A2 B ) ( A) 2 ( B ) ( 2) = = 0.0185 (6) 2 (3) Q < K , so rxns goes

Q=

( A2 B ) ( A) 2 ( B ) ( 0) = =0 (10) 2 (5) Q < K , so rxns goes

Q=

Q=

right (more product)

right (more product)

at equilibrium

( A2 B ) ( A) 2 ( B ) ( 4) = =1 (2) 2 (1) Q = K , so rxn is

UNIT IX – EQUILIBRIUM

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starting conditions ( 4) =1 (2) 2 (1) Q = K , so rxn is at equilibrium

(5) =∞ (0) 2 (0) Q > K , so rxns goes

Q=

Q=

left (more reactant)

* mass-action statement * example:

c d C ] [ D] [ Keq = [ A]a [ B]b

cC + dD

aA + bB

S. Give the mass-action statement for: 1)

2A+3B

C+4D

2)

2 A2B

4 A + B2

3)

N2O4(g)

[A] [B2 ] K eq = [A2 B]2

[C ][D]4 K eq = [A]2 [B]3 For the following equilibrium:

2 SO2(g) + O2(g)

o

2 NO2(g)

2 [ NO2 ] K eq = [N 2 O4 ]

4

2 SO3(g)

If at 250 C, the equilibrium concentrations are: [SO2] = 0.75 M, [O2] = 0.30 M, [SO3] = 0.15 M, what is the numerical value of K? 2 2 [ SO3 ] [ 0.15] K eq = = = 0.13 → K < 1, so reactants are favored [SO2 ]2 [O2 ] [0.75]2 [0.30]

* K shows relationship between amount of products and reactants at equilibrium * If K>1, equilibrium favors products. If K K , so equilibrium shifts left (to reactants) since reactant is the precipitate, the precipitate will form

UNIT IX – EQUILIBRIUM

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VIII. The Common Ion Effect and Solubility 1) Calculate the solubility of silver chloride (in g/L) in pure water.

Ksp = 1.6 x 10-10

AgCl ⇔ Ag + + Cl − x x K sp = ( Ag + )(Cl − ) 1.6 x10 −10 = x 2 x=

1.3 x10 −5 mol 144 g ∗ = 1.8 x10 −3 g / L L 1mol

2) Calculate the solubility of silver chloride (in g/L) in a 0.0065M silver nitrate solution. −

0.0065M = ( Ag + ) i = ( NO3 ) i → only Ag is in equilibrium, so ignore NO 3 Ag + + Cl − 0.0065+ x x = ( Ag + )(Cl − )

AgCl ⇔ K sp

1.6 x10 −10 = (0.0065 + x)( x) 1.6 x10 −10 = (0.0065)( x) ← 5%rule x=

2.5 x10 −8 mol 144 g ∗ = 3.5 x10 − 6 g / L L 1mol

* Note that the solubility of AgCl in a silver nitrate solution is 500 times less than that of AgCl in pure water. This is the essence of the common ion effect. Having ions already present in solution reduces the solubility of a salt containing that ion. 3) What is the molar solubility of lead(II) iodide in a 0.050M solution of sodium iodide? Ksp = 1.4 x 10-8

0.050 M = ( Na + ) i = ( I − ) i → only I is in equilibrium, so ignore Na PbI 2 ⇔ Pb + 2 + 2 I − x 0.050+2 x K sp = ( Pb + 2 )( I − ) 2 1.4 x10 −8 = ( x)(0.050 + 2 x) 2 1.4 x10 −8 = ( x)(0.050) 2 ← 5%rule x = 5.6 x10 −6 mol / L

UNIT IX – EQUILIBRIUM

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4) Calculate the final concentrations of K+(aq), C2O4-2(aq), Ba+2(aq), and Br-1(aq) in a solution prepared by adding 0.100L of 0.200M K2C2O4 to 0.150L of 0.250M BaBr2. For BaC2O4, Ksp = 2.3 x 10-8

BaC 2 O4 ⇔ Ba +2 + C 2 O4

−2

−2

→ K sp = ( Ba +2 )(C 2 O4 )

total volume of solution = 0.100L +0.150L = 0.250L -2

−2

(0.100 L)(0.200 M ) = 0.0200mol C 2 O4 = 0.0400mol K + → 2x as many K' s (K 2 C 2 O4 → 2 K + + C 2 O4 ) (0.150 L)(0.250 M ) = 0.0375mol Ba + 2 = 0.0750mol Br -1 → ( BaBr2 → Ba + 2 + 2 Br − ) K + and Br - don' t form the ppt, so they are spectators, ∴ you can calculate their final concentrations now : 0.0400mol 0.0750mol = 0.160 M → ( Br − ) = = 0.300 M 0.250 L 0.250 L LR −2 + ⇔ BaC 2 O4 Ba + 2 C 2 O4 0mol 0.0375mol 0.0175mol 0.0200mol → ( Ba + 2 ) = = 0.0700 M 0.250 L − 0.0200mol − 0.0200mol + 0.0200mol NA( solid ) 0.0175mol ~ 0mol

(K + ) =

BaC 2 O4 ⇔ Ba + 2 + C 2 O4 0.0700 + x

+2

−2

x −2

K sp = ( Ba )(C 2 O4 ) 2.3 x10 −8 = (0.0700 + x)( x) 2.3 x10 −8 = (0.0700)( x) ← 5%rule −2

x = 3.3 x10 − 7 M = (C 2 O4 )