3. Equilibium Basics

Chemical Equilibrium Basics

Reading: Chapter 15 of Petrucci et al. Sections 15-1 through 15-6 Examples: 15-1 through 15-8 Assigned problems: Chapter 15 questions 1, 7, 12, 19 Suggested problems: Chapter 15 questions 26, 51, 52

Note: Section 15-7 and associated problems will be covered, but not as part of this set of notes.

York University CHEM 1001 3.0 Chemical Equilibrium 1 1

Chemical Equilibrium - Section 15.1 All chemical reactions are reversible; that is, if a reaction can occur in one direction:

CO(g) + 2H2(g) CH3OH(g)

then the reverse reaction can also occur: CH3OH(g) CO(g) + 2H2(g)

As a reaction proceeds, the forward reaction slows and the reverse reaction speeds up until they are equal. This balance is called equilibrium. The reverse reaction might be so slow that the reaction is effectively irreversible.


Approach to Equilibrium - Example Elementary reactions:

(1) N2O4(g) 2NO2(g) (forward)

(1) 2NO2(g) N2O4(g) (reverse)

rate(1) = k1[N2O4] rate(1) = k1[NO2]2

At t=0 let [N2O4] > 0 and [NO2] = 0, then rate(1) > rate(1). As time passes, [N2O4] decreases and [NO2] increases.

Rate(1) decreases and rate(1) increases until they are equal.

When rate(1) = rate(1), we have equilibrium.


Approach to Equilibrium - continued When rate(1) = rate(1), we have equilibrium. Therefore:

k1[N2O4] = k1[NO2]2

21 2

C-1 2 4

k [NO ]= = K = equilibrium constantk [N O ]

At equilibrium: Concentrations remain constant. Both reactions continue to occur. This is a dynamic equilibrium.


Constant Conditions In a static system, nothing happens. Example: Neon in the Earth's atmosphere is neither produced nor consumed.

At steady-state, the rate of production is balanced by the rate of consumption. Example: Oxygen in the atmosphere is produced by photosynthesis and consumed by respiration.

Dynamic equilibrium is a steady-state in which the consumption process is the exact opposite of the production process. Example: N2O4(g) 2NO2(g). People sometimes use equilibrium for any or all of these. This is unacceptable in chemistry.


Examples of Equilibrium In a closed container, H2O(l) will come into equilibrium with H2O(g).

(But H2O(g) and H2O(l) are not in equilibrium in the atmosphere.)

(a) I2 in H2O(l) (yellow).

(b) I2 equilibrium between H2O(l) (top) and CCl4(l) (bottom, purple). Most I2 is in the CCl . 4


Evidence of Dynamic Equilibrium AgI(s) Ag+(aq) + I(aq)

Make up a saturated solution with excess solid AgI.

Add a small amount of radioactive 131I(aq) to the solution.

The radioactivity will gradually appear in the solid AgI.


Approach to Equilibrium - Second Example Overall reaction: Cl2(aq) + 2Br(aq) 2Cl(aq) + Br2(aq)

Likely mechanism:

(1) Cl2 + Br Cl + ClBr (2) ClBr + Br Cl + Br2 Treat as reversible equilibria (1) Cl2 + Br Cl + ClBr (1) Cl + ClBr Cl2 + Br

(2) ClBr + Br Cl + Br2 (2) Cl + Br2 ClBr + Br


Approach to Equilibrium - Second Example Overall reaction: Cl2(aq) + 2Br(aq) 2Cl(aq) + Br2(aq)

Elementary reactions: (1) Cl2 + Br Cl + ClBr (1) Cl + ClBr Cl2 + Br (reverse of reaction 1)

At equilibrium: -11-

-1 2

k [Cl ][ClBr]= = Kk [Cl ][Br ]

(2) ClBr + Br Cl + Br2 (2) Cl + Br2 ClBr + Br (reverse of reaction 2)

At equilibrium: -2 22-

-2

k [Cl ][Br ]= = Kk [ClBr][Br ]


Approach to Equilibrium - Continued Overall reaction: Cl2(aq) + 2Br(aq) 2Cl(aq) + Br2(aq)

Multiply the two equilibrium constants:

This is the equilibrium constant for the overall reaction

22

1 2 22

[ ] [ ][ ][ ] CCl BrK K KCl Br

Equilibrium constants can be deduced from the chemical equation for elementary or overall reactions.

The equilibrium constant for a sum of reactions is the product of the individual equilibrium constants.


Multiple Equilibria Example Consider the reactions:

(1) HF(aq) H+(aq) + F(aq) equil. const. = K1 (2) HCN(aq) H+(aq) + CN(aq) equil. const. = K2

For the reaction HF(aq) + CN(aq) F(aq) + HCN(aq)

The equilibrium constant for the reaction is K1/K2 You prove it


The Equilibrium Constant Reaction: aA + bB + ... gG + hH + ... At equilibrium the equilibrium constant is:

[ ] [ ][ ] [ ]

g h

C a bG HKA B

This applies to every step in a reaction mechanism. For the overall reaction, KC has the same form: products in the numerator, reactants in the denominator multiply concentrations raised to powers powers are the stoichiometric coefficients


Equilibrium Constant vs. Rate Laws These are similar, but different.

Both involve products of concentrations.

The rate law may or may not include reactants, products, and other species (catalysts)

The equilibrium constant includes all reactants and products, but only reactants and products.

In the equilibrium constant, the powers are the stoichiometric coefficients. Not so for rate laws (except elementary reactions).


Equilibrium Constant - Examples 2 4

2 2 2

[ ] [ ][ ] [ ]CCu SnKCu Sn

2Cu2+(aq) + Sn2+(aq) 2Cu+(aq) + Sn4+(aq)

4 1/ 2

2 2 1/ 2

[ ][ ][ ][ ]CCu SnKCu Sn

Cu

2+(aq) + Sn2+(aq) Cu+(aq) + Sn4+(aq)

2 2 2

2 4

[ ] [ ][ ] [ ]CCu SnKCu Sn

2Cu+(aq) + Sn4+(aq) 2Cu2+(aq) + Sn2+(aq)

The correct expression for the equilibrium constant depends on how the reaction is written.


Approach to Equilibrium 2Cu2+(aq) + Sn2+(aq) 2Cu+(aq) + Sn4+(aq)

Three different initial conditions.

Equilibrium reached in each case.

Different final conditions. All satisfy the

equilibrium condition.


Final State - Equilibrium


Beyond Concentrations In reality, rate laws do not always give the exact

dependence of reaction rate on concentration.

Also, equilibrium constants using concentrations, KC, are not constant (sometimes not nearly so).

The expressions we use here are approximations. The thermodynamic equilibrium constant, Keq, is

defined using activities instead of concentrations. We use concentrations as approximations to activities.


Activities Thermodynamic equilibrium constants use activities.

Activities are usually defined to be dimensionless.

Thermodynamic equilibrium constants are dimensionless.

Activities of pure solids and pure liquids are unity (1.0...). For dilute solutions:

solute activities approximately equal concentrations, the solvent activity is approximately unity.

For ideal gases, activities equal partial pressures (bar). We will treat solutions as dilute and gases as ideal.


Equilibrium Constant - More Examples

2H2(g) + O2(g) 2H2O(l) 2

2 2 2 2

2

2 2

1H Oeq

H O H

aK

a a P P

O

CaCO3(s) CaO(s) + CO2(g) 2

2

3

CaO COeq CO

CaCO

a aK P

a

3

2

32 [ ][ ]H O OH

eqH O

a aK H O OH

a 2H2O(l) H3O+(aq) + OH(aq)

3[ ][ ][ ]eq

H O ClKHCl

HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)


Equilibrium Constants Involving Gases CO(g) + 2H2(g) CH3OH(g)

For this reaction at 483 K, we find that:

3C 2

2

[CH OH]K = = 14.5[CO][H ]

The thermodynamic equilibrium constant is:

3

2

CH OHeq P2

CO H

PK = K

P P

These are not the same.


Converting between KC and KP The ideal gas law: PV = nRT So, for CO: PCO = (nCO/V)RT = [CO]RT

3

2

CH OH 3P 2 2 2

CO H 2

3 C2 2 2

2

P [CH OH]RTK = = P P [CO]RT[H ] (RT)

[CH OH] K = = [CO][H ] (RT) (RT)

KC = 14.5 at T = 483 K. R = 0.08206 L atm mol1 K1

So RT = 39.6 L atm mol1 and KP = 9.23103

( ) gasnP CK K RTIn General


Magnitudes of Equilibrium Constants -

Equilibrium constants have a very wide range of values.

Equilibrium constants can change enormously with changes in temperature.


Significance of Magnitude of K 2H2(g) + O2(g) 2H2O(l)

KP = 1.41083 at 298 K If pO2 = 0.21 bar, what is the equilibrium pH2?

H 22

2

2

P 2O

-42H 83

P O

1K = P P

1 1P = = =5.810 barK P (1.410 )(0.21)

This is less than one molecule in the atmosphere


Significance of K - continued A small K means that a reaction can not proceed

significantly in the direction written. It is unfavorable. A reasonably large K means that the reaction may proceed. It

is favorable. But it might be very slow. All reactions are reversible. So there are always some

reactants left at equilibrium, unless the phase changes. If K is so large that the equilibrium concentration of a

reactant is negligible, we say that the reaction "goes to completion".

If K is very small, the reverse reaction goes to completion. 'Large' and 'small' depend on how much material is available.


The Equilibrium Constant - Summary All reactions are reversible. When the rates of the forward

and reverse reactions are equal, we have equilibrium. Equilibrium is a dynamic state, both forward and reverse

reactions continue to occur.

The equilibrium constant, K, gives the condition that must obtain at equilibrium.

K can be used to determine the direction in which a reaction is allowed to proceed. But the reaction might be very slow.

K is expressed in terms of activities. Using concentrations or partial pressures is an approximation.


Predicting the Direction of Change CO(g) + 2H2(g) CH3OH(g)

If we start with only CO and H2, the reaction will initially proceed to the right.

If we start with only CH3OH, the reaction will initially proceed to the left.

What if we start with a mixture of reactants and products? In what direction will the reaction proceed?

The answer depends on where the reaction stops - so it depends on the equilibrium constant. But how?


The Reaction Quotient The reaction quotient, Q, is defined to be the same as

the equilibrium constant, but using initial conditions instead of equilibrium conditions.

Q is used to decide the direction of change from a specific set of initial conditions.

If Q < K, the reaction will proceed to the right. If Q > K, the reaction will proceed to the left. If Q = K, the reaction is at equilibrium.

QC is defined using concentrations; QP is defined using partial pressures. Must be consistent with KC or KP.


Using the Reaction Quotient CO(g) + 2H2(g) CH3OH(g) KC = 14.5 at 483 K Initial conditions: [CO] = 1.0 M,

[H2] = 0.30 M, [CH3OH] = 2.5 M

We calculate QC

3C C2 2

2

[CH OH] 2.5Q = = = 27.8 > K[CO][H ] 1.00.30

The reaction will proceed to decrease QC


Using the Reaction Quotient - Continued What happens as the reaction proceeds?

CO(g) + 2H2(g) CH3OH(g) We have QC > KC, so the reaction proceeds to the left. In other words, the rate of the reverse reaction is greater than the rate of the forward reaction.

Thus: [CH3OH] decreases, [CO] increases, [H2] increases QC decreases until QC = KC When QC = KC, equilibrium is reached. The forward and reverse rates are equal.


More on Using the Reaction Quotient What if conditions change?

CO(g) + 2H2(g) CH3OH(g) At equilibrium, QC = KC. What if we then add more CO?

Adding CO makes QC smaller. Then QC < KC.

3C 2

2

[CH OH]Q =[CO][H ]

The reaction proceeds to the right. As the reaction proceeds [CH3OH] , [CO] , [H2] , QC . Eventually QC = KC and we have a new equilibrium. All concentrations are different from our first equilibrium.


Le Chteliers Principle When an equilibrium system is subjected to a change, it responds by attaining a new equilibrium that partially offsets the impact of the change.

Input (Change) Response increase inside pressure increase volume

increase outside pressure reduce volume

increase in temperature absorb heat

increase in concentration shift away from that species (as directed by Q)


Le Chteliers Principle - Example Closed, flexible container with liquid water and water vapor in equilibrium:

H2O(l) H2O(g) Increase external pressure: Volume decreases,

some H2O(g) changes to H2O(l) (shift to left).

Add He(g) to increase internal pressure: Volume increases (flexible container), some H2O(l) changes to H2O(g) (shift to right).

Increase temperature: Some H2O(l) changes to H2O(g) (shift to right absorbs heat).


Adding/Removing Reactants or Products The equilibrium will shift to partially offset the change.

CO(g) + 2H2(g) CH3OH(g) Change: adding CH3OH

Equilibrium shifts to: a) left b) right c) no change

Change: adding CO Equilibrium shifts to: a) left b) right c) no change

Change: removing CH3OH Equilibrium shifts to: a) left b) right c) no change


Effect of Changing the Volume The equilibrium will shift to partially offset the change.

CO(g) + 2H2(g) CH3OH(g) A shift to the right reduces the pressure (replaces 3 moles

of gas with 1 mole of gas).

A shift to the left increases the pressure.

Change: decreased volume (increased pressure) Response: equilibrium shifts to right

Change: increased volume (decreased pressure) Response: equilibrium shifts to left


Effect of Inert Gas CO(g) + 2H2(g) CH3OH(g)

(1) Add He(g) while keeping the volume fixed. No effect on the reactants or products.

No effect on the equilibrium.

(2) Add He(g) while keeping the total pressure fixed. Volume increases.

Partial pressures of reactants and products decrease.

Equilibrium shifts to left.


Exothermic and Endothermic Reactions Breaking chemical bonds requires energy. Forming

bonds releases energy.

Chemical reactions may absorb or release energy. This energy is often transferred as heat.

Exothermic reactions release energy as heat (H < 0): CH4 + 2O2 CO2 + 2H2O H2SO4(l) + 2H2O(l) 2H3O+(aq) + SO42(aq)

Endothermic reactions absorb energy as heat (H > 0): H2O(l) H2O(g) NaOH(s) Na+(aq) + OH(aq)


Effect of a Change in Temperature The equilibrium will shift to partially offset the change.

For endothermic reactions, the equilibrium constant increases when the temperature goes up.

For exothermic reactions, the equilibrium constant decreases when the temperature goes up. Example:

CaCO3(s) CaO(s) + CO2(g) At 298 K, KP = 1.01023. At 1200 K, KP = 1.0.

This reaction is endothermic.


Effect of a Change in Temperature CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

When the temperature increases, the equilibrium constant will become

a) smaller

b) larger

c) there is no way to tell

Think!


Effect of a Catalyst A catalyst is a substance that increases the rate of a chemical reaction without being consumed by the reaction.

A catalyst: Does not change the position of equilibrium.

Can decrease the time it takes to reach equilibrium.

Will not make an unfavorable reaction occur.

Can help a favorable reaction to proceed faster.



The Direction of Change - Summary Equilibrium is not achieved for just one set of

concentrations. It is achieved for any concentrations that satisfy the equilibrium constant.

The reaction quotient, Q, gives the direction of change from a specific set of initial conditions.

Le Chteliers Principle: An equilibrium system responds to a change in conditions by partially offsetting the change.

The effect of changes in amount, pressure, or volume can also be determined by using Q.

The effect of temperature on K depends on whether the reaction is endothermic or exothermic.

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