CHEM 163 Chapter 17

CHEM 163

Chapter 17

Spring 2009Instructor: Alissa Agnelloaagnello@sccd.ctc.edu

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What affects reaction rates?

Chemical EquilibriumMany reactions can go forward AND backwards

3

(g) 2NO(g)O(g)N 22

If the opposite reaction can occur,

(g)O (g) N(g) 2NO 22

then this reaction is a reversible reaction

What is equilibrium?

• …in terms of reaction rates?

• …in terms of reactant and product concentrations?

• Has the forward reaction stopped?• Has the reverse reaction stopped?

Chemical Equilibrium

• Reactions continue at equal (but opposite) rates• No further changes in concentrations of reactants

or products occurs5

Chemical Equilibrium

If fwd and rev reactions are both elementary steps, how would we write their rate laws?

6

(g) 2NO(g)ON 242

ratefwd = kfwd [N2O4]eq raterev = krev [NO2]2eq

kfwd [N2O4]eq krev [NO2]2eq=

42

22

rev

fwd

ON

NO

k

kK

Equilibrium Constants (K)

7

Krev

fwd

k

k

eq

eq

][reactants

[products]

Equilibrium Constants: Small K

Small K value:– Greater concentration of reactants or products?– Reaction favors reactants

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Keq

eq

][reactants

[products]

Equilibrium Constants: Large K

Large K value: – Greater concentration of reactants or products?– Reaction favors products

9

Keq

eq

][reactants

[products]

Equilibrium ConstantsWill each of the following favor reactants or products?• Reaction with Kc = 2.9 x 10-12

• Reaction with Kc = 0.001 x 105

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Reaction Quotient (Q)• K derived from rates• Q derived from concentrations

At a given temperature, a system will always return to the same [product] : [reactant] ratio

Q

What if K = Q?

Q reactants

products

Reaction Quotients

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D C B A dcba reactants products

coefficients

Q reactants

products ba

dc

BA

DC

(g)O(g)2SO 22 (g)2SO3

2-minute practice

Write a reaction quotient expression for the following:

Compare your answer with your neighbors!

(g)gg HI2)(I)(H 22

Q for Overall reaction

Many reactions take place in multiple steps.

• Add together the steps to get overall reaction• Multiply K for each step to get overall K • Multiply Q for each step to get overall Q

Other situations…In a reversible reaction:

revfwd Q

1Q

When the coefficients are multiplied by a common factor (n):

nQQ

(g)gg HI2)(I)(H 22

22

2

IH

HIQ fwd

2

22

HI

IHQ rev

(g)gg HI)(I)(H 221

221

2

121

22 IH

HI

21

22

2

IH

HI

fwdQ

If solids or liquids are present…

• heterogeneous equilibrium• Q and K only related to concentration that

change (gases)

(g)(s)(g)(s) H OFe OH Fe 2432

Q 42

42

OH

H

Rearrange the ideal gas law, so that concentration is isolated on one side.

What is the relationship between pressure and concentration?

2-minute exercise

Qp: using partial pressures(g)gg HI2)(I)(H 22

22

2

c IH

HIQ

22 IH

2HIQPP

Pp

concentration

pressure

2HI 2

2HI

V

n

2

2HI

RT

P

2H V

n2H

RT

P2H

2I V

n2I

RT

P2I

Shortcut for relating Qp to Qc

What is the ∆n for the reaction?

(g)gg HI2)(I)(H 22 nreactants = 2

nproducts = 2∆n = 0 Qp = Qc

If ∆n ≠ 0 Qp = Qc (RT)∆n

Kp = Kc

Kp = Kc (RT)∆n

Is a reaction at equilibrium?Compare Q and K!

• Q < K:– Too much reactant – Equilibrium “shifts” towards products

• Q > K: – Too much product– Equilibrium “shifts” towards reactants

• Q = K: – At equilibrium

Equilibrium Calculations

• If equilibrium concentrations and/or equilibrium constant is known:– Write Q expression, plug in concentrations, solve– At equilibrium Q = K

• Solving for K when concentrations not given: – Write Q expression– Set up table including:• Initial concentrations (or pressures)• Change• Equilibrium concentrations (or pressures)

Mix together graphite and carbon dioxide (P = 0.458 atm) to create carbon monoxide.

Pressure (atm) CO2 (g) + C (graphite) ↔ 2CO (g)

CO2 (g) + C (graphite) ↔ 2CO (g)

Once equilibrium is reached, the pressure in the vessel (from CO2 and CO) is 0.757 atm.

Initial

Change

Equilibrium

0.458 0

- x + 2x

0.458 - x + 2x

0.458 – x + 2x = 0.757Ptotal =

Solving for “x”To solve for x, you may need to use the quadratic formula.

Set up your equation: a*x2 + b*x + c = 0 a

acbbx

2

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You will end up with two values. Which value is right?Remember: • K can’t be negative• We can’t have a higher concentration of reactant

than our initial concentration.

Steps for Solving1. Write balanced equation2. Set up table 3. Solve for x by…

1. Setting up Q expression 2. Setting total pressures equal to final pressure

4. Solve for equilibrium concentrations or pressures(using x)

5. To check: plug in solved concentration or pressures into Q expression and compare to known K value

3-minute PracticeConsider the reaction:

0.45 mole H2S is placed in a 3.0 L container. Make a table for this situation.

Kc = 9.30 x 10-8

at 700 °C H2S (g) ↔ H2 (g) + S2 (g)2 2

Calculate the equilibrium concentration of H2 (g) at 700 °C

Le Châtelier’s Principle• At equilibrium, concentrations of substances

do not change. • If a stress is put on the reaction at equilibrium,

the equilibrium will shift to relieve the stress.

What changes count as stress?• Concentration – Adding or removing reactant or product

• Volume (Pressure)• Temperature

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Effect of Concentration Changes2NO2 (g) NO (g) + O2 (g)

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What is the effect on the concentration of each substance?

• Add NO2?

• Add NO?

• Add O2?

• Remove NO2?

• Remove NO?

• Remove O2?

2

Calculations: adding/removing substances

• Make a table!

Pressure (atm) CO2 (g) + C (graphite) ↔ 2CO (g)

Original Equil.

Disturbance

New Initial

0.159 0.598

+ 0.1

0.698

Change

New Equil.

0.159

+ x - 2x

0.159 + x 0.698 - 2x

Predict direction of shift

Effect of Volume Changes• Each mole of gas exerts a certain pressure• Decrease the volume…– Increase the pressure…– Shifts to side of reaction with FEWER moles of gas

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2NO2 (g) 2NO (g) + O2 (g)

• What happens if we increase the volume?• Shifts to side with MORE moles

• What happens if we increase the pressure?• Shifts to side with FEWER moles

Effect of Volume Changes

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With volume changes and concentration changes, Equilibrium shifts to relieve new stress…in order to return to equilibrium

Effect of Temperature Changes• If T is increased, equilibrium shifts to remove the heat• If T is decreased, equilibrium shifts to create heat• Kc changes!– Unlike for concentration and volume changes

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A (g) B (g) + C (g) Endothermic:

Heat +

Increase the temperature?• Shifts to the products (to use up heat)

• Kc increases

Effect of T on K

van’t Hoff Equation:

• Exothermic (∆H° < 0): increasing T decreases Kc

• Endothermic (∆H° > 0): increasing T increases Kc

121

2 11 ln

TTR

H

K

K orxn R = 8.314 J/mol∙K

3-minute Practice

Consider the exothermic reaction between nitrogen gas and hydrogen gas, creating ammonia gas (NH3).

Write a balanced equation for this reaction.

Which direction will the equilibrium shift if: • T is increased?• Ammonia is removed?• Volume of the container is decreased?• A catalyst is used?• Hydrogen is added?

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Chapter 17 Homework

Due: Tuesday, 4/14

#18, 23, 29, 34, 45, 55, 63, 76, 77, 91, 114