chemical equilibrium chapter 15 chemistry: the molecular nature of matter, 6 th edition

Chemical Equilibrium

CHAPTER 15

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

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CHAPTER 15 Chemical Equilibrium

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

Learning Objectives:

Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium

Constant (K) Reaction Quotient (Q) Kc vs Kp

ICE Tables Quadratic Formula vs Simplifying Assumptions LeChatelier’s Principle van’t Hoff Equation

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Lecture Road Map:

① Dynamic Equilibrium

② Equilibrium Laws

③ Equilibrium Constant

④ Le Chatelier’s Principle

⑤ Calculating Equilibrium

4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Le Chatelier’s Principle


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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)


• 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

Le Chatelier Q & K Relationships


Le Chatelier Change in Concentration

Cu(H2O)42+(aq) + 4Cl–(aq) CuCl4

2–(aq) + 4H2O

blue yellow• Equilibrium mixture is blue-green

• Add excess Cl– (conc. HCl)– Equilibrium shifts to products

– Makes more yellow CuCl42–

– Solution becomes green



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

blue yellow

• 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?


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?

A.Towards the products

B.Towards the reactants

C.No change will occur



• 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


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)



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



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



• 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


Le Chatelier Change in Temperature

Ice water

Boiling water

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

blue yellow– Reaction endothermic– Adding heat shifts equilibrium toward products– Cooling shifts equilibrium toward reactants



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)



• 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



• 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


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


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


Le Chatelier How To Use Le Chatelier’s Principle

1. Write mass action expression for reaction

2. Examine relationship between affected concentration and Q (direct or indirect)

3. 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 22

GroupProblem

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


GroupProblem

The reaction

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

is exothermic.

What will happen if system is cooled?

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

heat


GroupProblem

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

pink blue

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

The reaction is:

A. endothermic

B. exothermic

C. cannot tell from the given information


GroupProblem

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

A. Ka = 2.2 × 10–3

B. Ka = 1.8 × 10–5

C. Ka = 4.0 × 10–10

D. Ka = 6.3 × 10–3

chemical equilibrium chapter 15 chemistry: the molecular nature of matter, 6 th edition

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