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Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson , Brady, & Hyslop. CHAPTER 15 Chemical Equilibrium. Learning Objectives: Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium Constant (K)

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slide1

Chemical Equilibrium

CHAPTER 15

Chemistry: The Molecular Nature of Matter, 6th edition

By Jesperson, Brady, & Hyslop

slide2

CHAPTER 15 Chemical Equilibrium

  • Learning Objectives:
  • Reversible Reactions and Equilibrium
  • Writing Equilibrium Expressions and the Equilibrium Constant (K)
  • Reaction Quotient (Q)
  • KcvsKp
  • ICE Tables
  • Quadratic Formula vs Simplifying Assumptions
  • LeChatelier’s Principle
  • van’t Hoff Equation

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide3

CHAPTER 15 Chemical Equilibrium

Lecture Road Map:

Dynamic Equilibrium

Equilibrium Laws

Equilibrium Constant

Le Chatelier’s Principle

Calculating Equilibrium

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide4

CHAPTER 15 Chemical Equilibrium

Le Chatelier’s Principle

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide5

Le Chatelier

Definition

  • Equilibrium positions
    • Combination of concentrations that allow Q = K
    • Infinite number of possible equilibrium positions
  • Le Châtelier’s principle
    • System at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress
      • System said to “shift to right” when forward reaction is dominant (Q < K)
      • System said to “shift to left” when reverse direction is dominant (Q > K)

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide6

Le Chatelier

Q & K Relationships

  • Q = K reaction at equilibrium
  • Q < K reactants go to products
    • Too many reactants
    • Must convert some reactant to product to move reaction toward equilibrium
  • Q > K products go to reactants
    • Too many products
    • Must convert some product to reactant to move reaction toward equilibrium

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide7

Le Chatelier

Change in Concentration

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O

blueyellow

  • Equilibrium mixture is blue-green
  • Add excess Cl– (conc. HCl)
    • Equilibrium shifts to products
    • Makes more yellow CuCl42–
    • Solution becomes green

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide8

Le Chatelier

Change in Concentration

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O

blueyellow

  • Add Ag+
    • Removes Cl–: Ag+(aq) + Cl–(aq) AgCl(s)
    • Equilibrium shifts to reactants
    • Makes more blue Cu(H2O)42+
    • Solution becomes increasingly more blue
  • Add H2O?

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide9

Le Chatelier

Change in Concentration: Example

For the reaction 2SO2(g) + O2(g) 2SO3(g)

Kc = 2.4 × 10–3 at 700 °C

Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture?

  • Towards the products
  • Towards the reactants
  • No change will occur

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide10

Le Chatelier

Change in Concentration

  • When changing concentrations of reactants or products
    • Equilibrium shifts to remove reactants or products that have been added
    • Equilibrium shifts to replace reactants or products that have been removed

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide11

Le Chatelier

Change in Pressure or Volume

  • Consider gaseous system at constant T and n

3H2(g) + N2(g) 2NH3(g)

  • If volume is reduced
    • Expect pressure to increase
    • To reduce pressure, look at each side of reaction
    • Which has less moles of gas
    • Reactants = 3 mol + 1 mol = 4 mol gas
    • Products = 2 mol gas
    • Reaction favors products (shifts to right)

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide12

Le Chatelier

Change in Pressure or Volume

Consider gaseous system at constant T and n

H2(g) + I2(g) 2HI(g)

  • If pressure is increased, what is the effect on equilibrium?
    • nreactant = 1 + 1 = 2
    • nproduct = 2
    • Predict no change or shift in equilibrium

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide13

Le Chatelier

Change in Pressure or Volume

2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g)

  • If you decrease volume of reaction, what is the effect on equilibrium?
    • Reactants: All solids, no moles gas
    • Products: 2 moles gas
    • Decrease in V, causes an increase in P
    • Reaction shifts to left (reactants), as this has fewer moles of gas

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide14

Le Chatelier

Change in Pressure or Volume

  • Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can
    • Increasing pressure
  • Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids
    • Substances are already almost incompressible
  • Changes in V, P and [X ] effect position of equilibrium (Q), but not K

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide15

Le Chatelier

Change in Temperature

Boiling water

Ice water

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O

blueyellow

  • Reaction endothermic
  • Adding heat shifts equilibrium toward products
  • Cooling shifts equilibrium toward reactants

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide16

Le Chatelier

Change in Temperature

Hf°=+6 kJ (at 0 °C)

  • Energy + H2O(s) H2O(l )
  • Energy is reactant
  • Add heat energy, shift reaction right

3H2(g) + N2(g) 2NH3(g) Hrxn= –47.19 kJ

  • 3 H2(g) + N2(g) 2 NH3(g) + energy
  • Energy is product
  • Add heat, shift reaction left

H2O(s) H2O(l)

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide17

Le Chatelier

Change in Temperature

  • Increase in temperature shifts reaction in direction that produces endothermic (heat absorbing) change
  • Decrease in temperature shifts reaction in direction that produces exothermic (heat releasing) change

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide18

Le Chatelier

Change in Temperature

  • Changes in T change value of mass action expression at equilibrium, so K changed
    • K depends on T
    • Increase in temperature of exothermic reaction makes K smaller
      • More heat (product) forces equilibrium to reactants
    • Increase in temperature of endothermic reaction makes K larger
      • More heat (reactant) forces equilibrium to products

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide19

Le Chatelier

Change with Catalyst

  • Catalyst lowers Ea for both forward and reverse reaction
  • Change in Ea affects rates k r and k f equally
  • Catalysts have no effect on equilibrium

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide20

Le Chatelier

Addition of Inert Gas at Constant Volume

Inert gas

    • One that does not react with components of reaction

e.g. argon, helium, neon, usually N2

  • Adding inert gas to reaction at fixed V (n and T), increase P of all reactants and products
  • Since it doesn’t react with anything
    • No change in concentrations of reactants or products
    • No net effect on reaction

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide21

Le Chatelier

How To Use Le Chatelier’s Principle

  • Write mass action expression for reaction
  • Examine relationship between affected concentration and Q (direct or indirect)
  • Compare Q to K
    • If change makes Q > K, shifts left
    • If change makes Q < K, shifts right
    • If change has no effect on Q, no shift expected

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide22

Group

Problem

Consider:

H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq)

What will happen if PO43– is removed?

  • Q is proportional to [PO43–]
  • Decrease [PO43–], decrease in Q
  • Q < K equilibrium shifts to right

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide23

Group

Problem

The reaction

H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq)

is exothermic.

What will happen if system is cooled?

heat

  • Since reaction is exothermic, heat is product
  • Heat is directly proportional to Q
  • Decrease in T, decrease inQ
  • Q < K equilibrium shifts to right

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide24

Group

Problem

The equilibrium between aqueous cobalt ion and the chlorine ion is shown:

[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O

pinkblue

It is noted that heating a pink sample causes it to turn violet.

The reaction is:

  • endothermic
  • exothermic
  • cannot tell from the given information

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

slide25

Group

Problem

The following are equilibrium constants for the reaction of acids in water, Ka. Which reaction proceeds the furthest to products?

  • Ka = 2.2 × 10–3
  • Ka = 1.8 × 10–5
  • Ka = 4.0 × 10–10
  • Ka = 6.3 × 10–3

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E