Equilibrium – Chapter 13In the 1st semester treated reactions as if they all went all the way.

But do all reactions go all the way?

If they do stop after only going part way, what stops them?

Think about dissolving table salt in water: NaCl(s) Na+(aq) + Cl¯(aq)⇄

How does the solid know when to stop dissociating in water?

2

Changes in Concentrations

CO + H2O ⇄ CO2 + H2

3

CO + H2O ⇄ CO2 + H2

Forward rate = kfwd[CO][H2O]Reverse rate = krev[CO2][H2]

The Changes with Time in the Rates of Forward and Reverse Reactions

At equilibrium, the forward rate equals the back rate. K is the ratio of the forward to back rate constants.

bak

fwd

k

kK

For a general reaction aA + bB cC + dD ⇄ba

dc

BA

DCK

][][

][][

For a general reaction aA + bB cC + dD ⇄ba

dc

BA

DCK

][][

][][

THE VALUE OF K IS A CONSTANT AT CONSTANT TEMPERATURE, NO

MATTER WHAT.

If you start with only A and B, the concentrations will change until the equilibrium condition is met.

If you start with only C and D, the concentrations will change until the same equilibrium condition is met.

The equilibrium constant expression for the reaction 4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g) is

Kc = [NO]4 [H

2O]6

[NH3]4 [O2]5.

b.

Kc = 4[NO] 6[H

2O]

4[NH3

] 5[O2

]

.

c.

Kc = [NH

3]4 [O

2]5

[NO]4 [H2O]6

Kc = 4[NH

3] 5[O

2]

4[NO] 6[H2

O]

Kc = [NO] [H

2O]

[NH3

] [O2

]

.

.

a.

.

e.

.

d.

Consider the equilibrium reaction2N2(g) + O2(g) 2N⇌ 2O(g)

In a particular experiment, the equilibrium concentration of N2 is 0.048 M, of O2, 0.093 M, and of N2O, 6.55 10-21 M. What is the value of the equilibrium constant K?

a. 1.5 10-18

b. 2.0 10-37

c. 2.2 10-36

d. 3.1 10-17

e. 5.0 1036

SUMMARY OF WHAT HAS BEEN SAID

1. When forward rate = back rate, you have an equilibrium

2. Equilibrium constant K = kfwd/kback

3. K is a constant (if temp constant)

4. K in terms of concentrations is ba

dc

BA

DCK

][][

][][

Products over reactants to the power of their stoichiometry.

Concentration units for K – somewhat different from the book

(most common units, called thermodynamic K)

Use Molarity for solutions

Use atmospheres for gases: from PV=nRT, n/V = P[1/RT]

Pure liquids or solids are defined to have a concentration of 1

K can be composed of mixed concentration units, but it has no units itself: 2H+(aq) + 2Na(s) 2Na+(aq) + H⇄ 2(g)

We will use units of atm and M. What do Kc and Kp mean?

Numerical value of K will depend on the units used.

Concentrations of pure liquids and solids are defined to be 1

Consider the reaction for dissolving AgCl in water:AgCl(s) ⇄ Ag+(aq) + Cl-(aq) K = 1.6x10-10

 

Since AgCl is a pure solid, in the equilibrium constant expression it is given a value of 1 Physically this makes sense - as the AgCl dissolves, the amount of AgCl will decrease, but its concentration will be constant -

]][[][

]][[)()(

)(

)()(aqaq

s

aqaqClAg

AgCl

ClAgK

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

[CaO(s)]PCO2K =

[CaCO3(s)]concentrations of pure solids and liquidsare constant are dropped from expression

K = PCO2

The equilibrium constant expression for the reaction shown below is Ag3PO4(s) ⇌ 3Ag+(aq) + PO4

3-(aq)

Kc =

[Ag3PO

4]

[Ag][PO43 ]

.

a.

Kc [Ag]3[PO43 ]

Kc [Ag][PO

43 ]

[Ag3

PO4

]

Kc [Ag]3[PO

43 ]

[Ag3

PO4

]

Kc [Ag

3PO

4]

[Ag]3[PO43 ]

.

.

b.

.

d.

.

c.

.

e.

The value of K can be VERY big or VERY small - really any positive number -   for CH3O2 + NO2 ⇄ CH3O2NO2, K= 1.2 X 1033

 for O3 + NO ⇄ NO2 + O2, K = 5.8 10-34

 for I2 + H2 ⇄ 2HI Kc = 0.50

•When K is very large, the reaction goes nearly to completion

•When K is very small, the reaction goes very little in the direction written

•When K is moderate, the reaction stops somewhere near the middle.

New understanding

I2 + H2 ⇄ 2HI

An equilibrium expression defines a forward and back directions, but does not necessarily imply the direction

This reaction would equally well have been written as 2HI ⇄ I2 + H2  

New understanding 

direction the reaction will proceed in depends on two things: 1) starting conditions (concentrations of reactants and products initially)

2) value of K: large K suggests equilibrium position favors products, K<<1 suggests equilibrium position favors reactants (i.e., reaction, as written, does not go far towards products)

similarly, the terms products and reactants refer to the direction the reaction is written in, not necessarily the direction it will proceed in.

A B⇄

If K = 100, then at equilibrium [B] is 100x greater than [A].

Product is favored.kforward > kback

If K = 0.01, then at equilibrium [A] is 100x greater than [B]

Reactant is favored. kforward< kback

The value of K for the reaction A B ⇌is 1.4 1015. At equilibrium,

a. The amount of A is slightly less than the amount of B.

b. The amount of A is much larger than the amount of B.

c. The amount of A is much less than the amount of B.

d. The amount of A is very close to the amount of B.

e. More information is needed to make any statement about the relative amounts of A and B.

REMEMBER:

1.Reactions don’t all go to completion

2.K is the quantitative measure of how far the reaction goes

3.K = kfwd/kbak

4.Equilibrium is dynamic – even after it appears reaction has ended, microscopically both forward and back reactions continue to occur

The Value of K Depends On How You Write the Reaction

Products over reactants to the power of their stoichiometry.

The value K depends on how you write the reaction

Na(s) + H+(aq) ⇄ Na+

(aq) + ½H2(g)

2Na(s) + 2H+(aq) ⇄ 2Na+

(aq) + H2(g)

][

][ 2

1

2

H

PNaK H

2

2

][

][' 2

H

PNaK H

2' KK When you multiply a reaction by 2, you square the

value of K

The value K depends on how you write the reaction

Na(s) + H+(aq) ⇄ Na+

(aq) + ½H2(g)

Na+(aq) + ½H2(g) ⇄ Na(s) + H+

(aq)

][

][ 2

1

2

H

PNaK H

KK

1'

When you reverse a reaction, you take the reciprocal of K.

2

1

2][

]['

HPNa

HK

The equilibrium constant for the reaction

NO(g) + ½ O2(g) NO⇌ 2(g)

has a value of K = 1.23 at a certain temperature. What is the value of K for the reaction below?

2NO(g) + O2(g) 2NO⇌ 2(g)

a. 1.23b. 1.51c. 0.81d. I’m lost

The equilibrium constant for the reaction

NO(g) + ½ O2(g) NO⇌ 2(g)

has a value of K = 1.23 at a certain temperature. What is the value of K for the reaction below?

2 NO2(g) 2 NO(g) + O⇌ 2(g)

a. 2.46b. 1.51c. 0.66d. 0.41

e. I’m lost

2 HCl(g) H⇌ 2(g) + Cl2(g)

I2(g) + Cl2(g) 2 ICl(g)⇌

Sum the two reactions to get 2 HCl(g) + I2(g) H⇌ 2(g)

2122

HCl

ClH

P

PPK

22

2

2ClI

ICl

PP

PK

212

2

2

2 KKPP

PPK

IHCl

IClHsum

22

22

2

2ClI

ICl

HCl

ClH

PP

P

P

PP

+ 2 ICl(g)

When you add two reactions to get a new reaction, you multiply the Ks to get the new K.

What is K for the reaction:

HCl(g) + 1/2I2(g) ↔ ICl(g) + 1/2H2 (g)

Given the following K values:

2HCl(g) ↔ Cl2 (g) + H2 (g) K = 4.2x10-3

2ICl(g) ↔ Cl2 (g) + I2 (g) K=5.4x10-4