chapter 15

21
Chapter 15 Chemical equilibria

Upload: plato-russo

Post on 03-Jan-2016

24 views

Category:

Documents


1 download

DESCRIPTION

Chapter 15. Chemical equilibria. Closed vial of NO 2 : NO 2 (g) + NO 2 (g)  N 2 O 4 (g) brown colorless Brown color becomes less intense, then, after some time period, color change stops. Concentrations of products and reactants remain constant with time (equilibrium) - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Chapter 15

Chapter 15

Chemical equilibria

Page 2: Chapter 15

Chemical Equilibrium Closed vial of NO2:

NO2(g) + NO2(g) N2O4(g)

brown colorless Brown color becomes less intense, then, after some time

period, color change stops.

Concentrations of products and reactants remain constant with time (equilibrium)

Applies to all reactions

Page 3: Chapter 15

Dynamic EquilibriumH2O(g) + CO(g) H2(g) + CO2(g)

Start off with closed flask of CO and H2O. Equilibrium sets up (chemical reaction occurring).

What happens to amounts of H2O(g) + CO(g) ?

What happens to amounts of H2(g) + CO2(g) ?

Reaction still occurring past dotted line (although concentrations don’t change): Dynamic Equilibrium

Page 4: Chapter 15

Dynamic Equilibrium

H2O(g) + CO(g) H2(g) + CO2(g)

Page 5: Chapter 15

Equilibrium Expression and K Law of mass action

jA + kB lC + mD

K = [C]l [D]m Equilibrium Expression [A]j [B]k

Concentrations of species at equilibrium K = equilibrium constant

H2O(g) + CO(g) H2(g) + CO2(g)

2NO2(g) N2O4(g)

Page 6: Chapter 15

Value of Equilibrium Constant3H2(g) + N2(g) 2NH3(g)

Value for K is always the same for a particular reaction,

Page 7: Chapter 15

Magnitude of K Equilibrium position (does equilibrium lie towards products or

reactants?); is this important (Synthesis of Aspirin) Large K

K1

Small K

Page 8: Chapter 15

Writing equilibrium expressionsAlways predicted from the balanced equation:

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

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

C4H10(g) + O2(g) CO2(g) + H2O(g)

Page 9: Chapter 15

Calculating values of K

Page 10: Chapter 15

Eq. laws for gaseous reactions3H2(g) + N2(g) 2NH3(g)

K = [NH3]2 Equilibrium constant found in terms of

[H2]3 [N2] concentrations of species (Kc)

K can also be found in terms of partial pressures (Kp)

PV = nRT

Kp = PNH32

(PH23)(PN2)

Relationship between Kp and Kc:

Kp = Kc(RT)ng

Page 11: Chapter 15

Heterogeneous Equilibria More than one phase exists in reaction mixture Thermal decomposition of CaCO3(s) CaO(s) + CO2(g)

Concentrations of pure solids and pure liquids are always constant (thus, can be removed from eq. expression)

Page 12: Chapter 15

Heterogeneous Equilibria Write equilibrium expressions (in terms of both K and Kp

for each of the following):

1. CaO(s) + SO2(g) CaSO3(s)

2. Decomposition of solid phosphorous pentachloride to liquid phosphorous trichloride and chlorine gas

3. Deep blue solid copper(II)sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II)sulfate

Page 13: Chapter 15

Reaction Quotient, Q In which direction will a particular reaction shift to reach

equilibrium?

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

If [NH3]0 = 0, shift to right to achieve equilibrium.

If [H2]0 or [N2]0 = 0, shift to left to achieve equilibrium.

If initial concentrations of all three species are nonzero, which way will shift occur to achieve equilibrium? More difficult to predict.

Use Reaction Quotient, Q Q is obtained by applying law of Mass Action to initial

concentrations of species involved.

Page 14: Chapter 15

Reaction Quotient, Q3H2(g) + N2(g) 2NH3(g) Q = [NH3]o

2

[H2]o3 [N2]o

Then, to determine in which direction a system will shift to reach equilibrium, compare values of Q and K:

1. Q = K System is at equilibrium

2. Q > K Ratio of initial conc. of products to initial conc. of reactants is too large.

System shifts to the left to reach equilibrium.

3. Q < K Ratio of initial conc. of products to initial conc. of reactants is too small.

System shifts to the right to reach equilibrium

Page 15: Chapter 15

‘ICE’ Tables

Calculating Equilibrium Concentrations

Page 16: Chapter 15

Le Chatelier’s Principle ‘If an outside influence upsets an equilibrium, the system

undergoes a change in the direction that counteracts the disturbing influence, and, if possible, returns the system to equilibrium.’

Outside influences?

Adding / removing a reactant / product Changing volume/pressure of gaseous reactions Changing T

Page 17: Chapter 15

Adding/removing a reactant or product Eq. shifts in direction that will partially consume a reactant or

product added Eq. shifts in direction that will partially replace a reactant or

product removed

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

Add some N2 to above equilibrium; which way will it shift to re-establish equilibrium?

Page 18: Chapter 15

Changing volume/pressure Reducing volume (what happens to the pressure?) – eq. shifts to

side with smaller # of gas molecules

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

Increasing volume? What would happen here?

Page 19: Chapter 15

Changing volume/pressure

2NO2(g) N2O4(g)

brown colorless

Page 20: Chapter 15

Changing T Increasing T shifts an equilibrium in direction that produces an

endothermic change (need to know energy involved in reaction to predict which direction this is)

3H2(g) + N2(g) 2NH3(g) H= -46.19 kJ/mol

Exothermic or endothermic reaction?

Increasing T: Decreasing T:

Page 21: Chapter 15

Changing T

Is this reaction exo- or endothermic?

2NO2(g) N2O4(g)

brown colorless