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Chapter 13 Chemical Equilibrium

REVERSE REACTION  reciprocal K

1 ADD REACTIONS ADD REACTIONS  Multiply Ks  Multiply Ks

-8.4

-8.4

LE CHATELIER’S PRINCIPLE LE CHATELIER’S PRINCIPLE

CO2 + H2  H2O(g) + CO a drying agent is added to absorb H2O

Shift to the right. Continuous removal of a will force any reaction to the right

H2(g) + I2(g)  2HI(g) Some nitrogen gas is added

No change; N2 is not a component of this reaction system.

2 LE CHATELIER’S PRINCIPLE LE CHATELIER’S PRINCIPLE

+ – NaCl(s) + H2SO4(l)  Na2SO4(s) + HCl(g) AgCl(s)  Ag (aq) + Cl (aq) reaction is carried out in an open container some NaCl is added to the

Because HCl is a gas that can escape from the system, Shift to left due to increase in Cl– . the reaction is forced to the right. This is known as the common ion effect on solubility. This is the basis for the commercial production of hydrochloric .

H2O(l)  H2O(g) N2 + 3 H2  2 NH3 evaporates from an open container a catalyst is added to speed up this reaction

No change. Continuous removal of water vapor forces the reaction to the right, Catalysts affect only the rate of a reaction; so equilibrium is never achieved they have no effect at all on the composition of the equilibrium state

LE CHATELIER’S PRINCIPLE , Q

+ O2  oxyhemoglobin K is thus the special value that Q has when the reaction is at equilibrium

Take up in lungs at high O2 pressure Release in cells at low O2 concentration

Br2(g)  2 Br (g) Pressure increased

shift to left To reduce number of molecules or atoms

3 REACTION QUOTIENT, Q REACTION QUOTIENT, Q

K is thus the special value that Q has when the reaction is at equilibrium

X X THERMODYNAMICS and equilibrium THERMODYNAMICS and equilibrium

1. The of an endothermic reaction

(ΔH° = +) increases if the is raised.

2. The equilibri um constant of an exothihermic reacti on

(ΔH° = −) decreases if the temperature is raised.

NB: understand this from Le Chatelier’s principle!

4 K IS DIMENSIONLESS! HABER-BOSCH:

N +3H+ 3 H  2NH2 NH + E 2 2 3 • in mol/liter (M) •pressures in atmospheres (atm) •ignore solids •ignore solvents

Equilibrium calculation EXAMPLE Equilibrium calculation EXAMPLE X

0.001 mol Br2 0.001 mol Br2 - - 0.005 mol IO3 0.005 mol IO3 0.02 mol Br- 0.02 mol Br- First calculate Q to know the direction 1.00 mol H+ 1.00 mol H+

Solid I2 Solid I2

What will be the concentrations at equilibrium?

So which way does it go?

5 Equilibrium calculation EXAMPLE

SOLUBILITY PRODUCT Ksp

0.001 mol Br2 - 0.005 mol IO3 0.02 mol Br- Make an ICE table 1.00 mol H+ Ksp = equilibrium constant Solid I Solve for x 2 of a reaction that forms a precipitate

SOLUBILITY PRODUCT Ksp SOLUBILITY PRODUCT Ksp

COMMON ION EFFECT

C

C + S

6 SOLUBILITY PRODUCT Ksp SOLUBILITY PRODUCT Ksp

COMMON ION EFFECT

2+ 3 3- 2 -26 Ksp = [Ca ] [PO4 ] = 1.0 x 10 2+ 3 3- 2 -26 Ksp = [Ca ] [PO4 ] = 1.0 x 10 3 2 -26 = (3x)3(2x)2 = 1.0 x 10-26 = (3x) (0.10 + 2x) = 1.0 x 10

X X

7 I¯ I¯ X SOLUBILITY PRODUCT Ksp SOLUBILITY PRODUCT Ksp

SEPARATION BY PRECIPITATION SEPARATION BY PRECIPITATION

Which will form a precipitate first? Starting with 0.01 M of each, 2+ can you precipitate 99.99% of Hg2 Higher or lower Ksp? without losing any Pb2+?

BrO3¯ X SOLUBILITY PRODUCT Ksp Gas – solution eaquilibrium KH

SEPARATION BY PRECIPITATION Henry’s Law

When (BrO —) is added to a solution containing 3 CO2 dissolves in water: + + equal concentrations of Ag and Pb2 , which will precipitate first and why? -2 CO2(g) + H2O <==> H2CO3 (aq) KH = 3.4 x 10

Ksp = 5.49 x 10-5 for AgBrO -4 3 at a CO2 pressure of 3 x 10 atmospheres, -5 Ksp = 3 .23 x 10 for Pb(BrO 3)2 whihhat is the concentrati on of fh the carb bionic acid idih in the water?

10-5 M

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